[{"video_title": "Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3", "Sentence": "All right, let's see if we can find the indefinite integral of one over five x squared minus 30x plus 65 dx. Pause this video and see if you can figure it out. All right, so this is going to be an interesting one. It'll be a little bit hairy, but we're gonna work through it together. So immediately, you might try multiple integration techniques and be hitting some walls. And what we're going to do here is actually try to complete the square in this denominator right over here. And then by completing the square, we're gonna get it in the form that it looks like the derivative of arctan."}, {"video_title": "Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3", "Sentence": "It'll be a little bit hairy, but we're gonna work through it together. So immediately, you might try multiple integration techniques and be hitting some walls. And what we're going to do here is actually try to complete the square in this denominator right over here. And then by completing the square, we're gonna get it in the form that it looks like the derivative of arctan. And if that's a big hint to you, once again, pause the video and try to move forward. All right, now let's do this together. So I'm just gonna try to simplify this denominator so that my coefficient on my x squared term is a one."}, {"video_title": "Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3", "Sentence": "And then by completing the square, we're gonna get it in the form that it looks like the derivative of arctan. And if that's a big hint to you, once again, pause the video and try to move forward. All right, now let's do this together. So I'm just gonna try to simplify this denominator so that my coefficient on my x squared term is a one. And so I can just factor a five out of the denominator. And if I did that, then this integral will become 1 5th times the integral of one over, so I factored a five out of the denominator. So it is x squared minus six x plus 13 dx."}, {"video_title": "Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3", "Sentence": "So I'm just gonna try to simplify this denominator so that my coefficient on my x squared term is a one. And so I can just factor a five out of the denominator. And if I did that, then this integral will become 1 5th times the integral of one over, so I factored a five out of the denominator. So it is x squared minus six x plus 13 dx. And then as I mentioned, I'm gonna complete the square down here. So let me rewrite it. So this is equal to 1 5th times the integral of one over, and so x squared minus six x."}, {"video_title": "Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3", "Sentence": "So it is x squared minus six x plus 13 dx. And then as I mentioned, I'm gonna complete the square down here. So let me rewrite it. So this is equal to 1 5th times the integral of one over, and so x squared minus six x. It's clearly not a perfect square the way it's written. Let me write this plus 13 out here. Now what could I add, and then I'm gonna have to subtract if I don't wanna change the value of the denominator, in order to make this part right over here a perfect square?"}, {"video_title": "Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3", "Sentence": "So this is equal to 1 5th times the integral of one over, and so x squared minus six x. It's clearly not a perfect square the way it's written. Let me write this plus 13 out here. Now what could I add, and then I'm gonna have to subtract if I don't wanna change the value of the denominator, in order to make this part right over here a perfect square? Well, we've done this before. You take half of your coefficient here, which is negative three, and you square that. So you wanna add a nine here."}, {"video_title": "Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3", "Sentence": "Now what could I add, and then I'm gonna have to subtract if I don't wanna change the value of the denominator, in order to make this part right over here a perfect square? Well, we've done this before. You take half of your coefficient here, which is negative three, and you square that. So you wanna add a nine here. But if you add a nine, then you have to subtract a nine as well. And so this part is going to be x minus three squared. And then this part right over here is going to be equal to a positive four."}, {"video_title": "Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3", "Sentence": "So you wanna add a nine here. But if you add a nine, then you have to subtract a nine as well. And so this part is going to be x minus three squared. And then this part right over here is going to be equal to a positive four. And we of course don't wanna forget our dx out here. And so let me write it in this form. So this is going to be equal to 1 5th times the integral of one over, get myself some space, x minus three squared plus four, which could also write as plus two squared."}, {"video_title": "Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3", "Sentence": "And then this part right over here is going to be equal to a positive four. And we of course don't wanna forget our dx out here. And so let me write it in this form. So this is going to be equal to 1 5th times the integral of one over, get myself some space, x minus three squared plus four, which could also write as plus two squared. Actually, let me do it that way. Plus two squared dx. Now many of y'all might already be saying, hey, this looks a lot like arctangent, but I'm gonna try to simplify it even more so it becomes very clear that it looks like arctangent is going to be involved."}, {"video_title": "Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3", "Sentence": "So this is going to be equal to 1 5th times the integral of one over, get myself some space, x minus three squared plus four, which could also write as plus two squared. Actually, let me do it that way. Plus two squared dx. Now many of y'all might already be saying, hey, this looks a lot like arctangent, but I'm gonna try to simplify it even more so it becomes very clear that it looks like arctangent is going to be involved. I'm actually gonna do some u substitution in order to do it. So the first thing I'm gonna do is, let's factor a four out of the denominator here. So if we do that, then this is going to become 1 5th times 1 4th, which is going to be 1 20th times the integral of one over x minus three squared over two squared."}, {"video_title": "Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3", "Sentence": "Now many of y'all might already be saying, hey, this looks a lot like arctangent, but I'm gonna try to simplify it even more so it becomes very clear that it looks like arctangent is going to be involved. I'm actually gonna do some u substitution in order to do it. So the first thing I'm gonna do is, let's factor a four out of the denominator here. So if we do that, then this is going to become 1 5th times 1 4th, which is going to be 1 20th times the integral of one over x minus three squared over two squared. And then this is going to be a plus one. And of course, we have our dx. And then we could write this as, and I'm trying to just do every step here."}, {"video_title": "Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3", "Sentence": "So if we do that, then this is going to become 1 5th times 1 4th, which is going to be 1 20th times the integral of one over x minus three squared over two squared. And then this is going to be a plus one. And of course, we have our dx. And then we could write this as, and I'm trying to just do every step here. A lot of these you might have been able to do in your head. One over, and I'll just write this as x minus three over two squared plus one. Plus one, and then dx."}, {"video_title": "Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3", "Sentence": "And then we could write this as, and I'm trying to just do every step here. A lot of these you might have been able to do in your head. One over, and I'll just write this as x minus three over two squared plus one. Plus one, and then dx. And now the u substitution is pretty clear. I am just going to make the substitution that u is equal to x minus three over two. Or we could even say that's u is equal to 1 1 2 x minus 3 1 2, that's just x minus three over two."}, {"video_title": "Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3", "Sentence": "Plus one, and then dx. And now the u substitution is pretty clear. I am just going to make the substitution that u is equal to x minus three over two. Or we could even say that's u is equal to 1 1 2 x minus 3 1 2, that's just x minus three over two. And du is going to be equal to 1 1 2 dx. And so what I can do here is, actually let me start to re-engineer this integral a little bit, so that we see a 1 1 2 here. So if I make this a 1 1 2, and then I multiply the outside by two, so I divide by two, multiply by two is one way to think about it."}, {"video_title": "Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3", "Sentence": "Or we could even say that's u is equal to 1 1 2 x minus 3 1 2, that's just x minus three over two. And du is going to be equal to 1 1 2 dx. And so what I can do here is, actually let me start to re-engineer this integral a little bit, so that we see a 1 1 2 here. So if I make this a 1 1 2, and then I multiply the outside by two, so I divide by two, multiply by two is one way to think about it. This becomes 1 1. And so doing my u substitution, I get 1 1, that's that 1 1 there, times the integral of, well, I have 1 1 2 dx right over here, which is the same thing as du. So I could put the du either in the numerator, I could put it out here."}, {"video_title": "Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3", "Sentence": "So if I make this a 1 1 2, and then I multiply the outside by two, so I divide by two, multiply by two is one way to think about it. This becomes 1 1. And so doing my u substitution, I get 1 1, that's that 1 1 there, times the integral of, well, I have 1 1 2 dx right over here, which is the same thing as du. So I could put the du either in the numerator, I could put it out here. And then I have one over, this is u squared, u squared plus one. Now you might immediately recognize, what's the derivative of arc tan of u? Well, that would be one over u squared plus one."}, {"video_title": "Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3", "Sentence": "So I could put the du either in the numerator, I could put it out here. And then I have one over, this is u squared, u squared plus one. Now you might immediately recognize, what's the derivative of arc tan of u? Well, that would be one over u squared plus one. So this is going to be equal to 1 1 0 times the arc tangent of u. And of course, we can't forget our big constant c, because we're taking an indefinite integral. And now we just wanna do the reverse substitution."}, {"video_title": "Integration using completing the square and the derivative of arctan(x) Khan Academy.mp3", "Sentence": "Well, that would be one over u squared plus one. So this is going to be equal to 1 1 0 times the arc tangent of u. And of course, we can't forget our big constant c, because we're taking an indefinite integral. And now we just wanna do the reverse substitution. We know that u is equal to this business right over here, so we deserve a little bit of a drum roll. This is going to be equal to 1 1 0 times the arc tangent of u. Well, u is just x minus three over two, which could also be written like this."}, {"video_title": "Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "It has a vertical tangent at the point three comma zero. So three comma zero has a vertical tangent. Let me draw that. So it has a vertical tangent right over there. And a horizontal tangent at the point zero comma negative three, zero comma negative three, so it has a horizontal tangent right over there. And also has a horizontal tangent at six comma three. Six comma three, let me draw the horizontal tangent."}, {"video_title": "Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So it has a vertical tangent right over there. And a horizontal tangent at the point zero comma negative three, zero comma negative three, so it has a horizontal tangent right over there. And also has a horizontal tangent at six comma three. Six comma three, let me draw the horizontal tangent. Just like that. Select all the x values for which f is not differentiable. Select all that apply."}, {"video_title": "Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Six comma three, let me draw the horizontal tangent. Just like that. Select all the x values for which f is not differentiable. Select all that apply. So f prime, f prime, I'll write it in shorthand. So we say no f prime under, it's going to happen under three conditions. The first condition, you could say, well we have a vertical tangent."}, {"video_title": "Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Select all that apply. So f prime, f prime, I'll write it in shorthand. So we say no f prime under, it's going to happen under three conditions. The first condition, you could say, well we have a vertical tangent. Vertical tangent. Why is a vertical tangent a place where it's hard to define our derivative? Well remember, our derivative, we're really trying to find our rate of change of y with respect to x."}, {"video_title": "Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "The first condition, you could say, well we have a vertical tangent. Vertical tangent. Why is a vertical tangent a place where it's hard to define our derivative? Well remember, our derivative, we're really trying to find our rate of change of y with respect to x. When you have a vertical tangent, you change your x a very small amount, you have an infinite change in y, either in the positive or the negative direction. So that's one situation where you have no derivative. And they tell us where we have a vertical tangent in here, where x is equal to three."}, {"video_title": "Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Well remember, our derivative, we're really trying to find our rate of change of y with respect to x. When you have a vertical tangent, you change your x a very small amount, you have an infinite change in y, either in the positive or the negative direction. So that's one situation where you have no derivative. And they tell us where we have a vertical tangent in here, where x is equal to three. So we have no, f is not differentiable at x equals three because of the vertical tangent. You might say, what about horizontal tangents? No, horizontal tangents are completely fine."}, {"video_title": "Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And they tell us where we have a vertical tangent in here, where x is equal to three. So we have no, f is not differentiable at x equals three because of the vertical tangent. You might say, what about horizontal tangents? No, horizontal tangents are completely fine. Horizontal tangents are places where the derivative is equal to zero. So f prime of six is equal to zero. F prime of zero is equal to zero."}, {"video_title": "Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "No, horizontal tangents are completely fine. Horizontal tangents are places where the derivative is equal to zero. So f prime of six is equal to zero. F prime of zero is equal to zero. What are other scenarios? Well another scenario where you're not going to have a defined derivative is where the graph is not continuous. It's not continuous."}, {"video_title": "Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "F prime of zero is equal to zero. What are other scenarios? Well another scenario where you're not going to have a defined derivative is where the graph is not continuous. It's not continuous. And we see right over here at x equals negative three, our graph is not continuous. So x equals negative three, it's not continuous. And those are the only places where f is not differentiable that they're giving us options on."}, {"video_title": "Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "It's not continuous. And we see right over here at x equals negative three, our graph is not continuous. So x equals negative three, it's not continuous. And those are the only places where f is not differentiable that they're giving us options on. We don't know what the graph is doing to the left or the right. These I guess would be interesting cases, but they haven't given us those choices here. And we already said, at x equals zero, the derivative is zero, it's defined, it's differentiable there."}, {"video_title": "Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And those are the only places where f is not differentiable that they're giving us options on. We don't know what the graph is doing to the left or the right. These I guess would be interesting cases, but they haven't given us those choices here. And we already said, at x equals zero, the derivative is zero, it's defined, it's differentiable there. And at x equals six, the derivative is zero. We have a flat, flat tangent. So once again, it's defined there as well."}, {"video_title": "Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And we already said, at x equals zero, the derivative is zero, it's defined, it's differentiable there. And at x equals six, the derivative is zero. We have a flat, flat tangent. So once again, it's defined there as well. Let's do another one of these. Oh, and actually I didn't include, I think that this takes care of this problem, but there's a third scenario in which we have, I'll call it a sharp turn. A sharp turn."}, {"video_title": "Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So once again, it's defined there as well. Let's do another one of these. Oh, and actually I didn't include, I think that this takes care of this problem, but there's a third scenario in which we have, I'll call it a sharp turn. A sharp turn. And this isn't the most mathy definition right over here, but it's easy to recognize. A sharp turn is something like that. Or like, or like, well no, that doesn't look too sharp."}, {"video_title": "Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "A sharp turn. And this isn't the most mathy definition right over here, but it's easy to recognize. A sharp turn is something like that. Or like, or like, well no, that doesn't look too sharp. Or like this. And the reason why I think where you have these sharp bends or sharp turns as opposed to something that looks more smooth like that, the reason why we're not differentiable there is as we approach this point, as we approach this point from either side, we have different slopes. Notice our slope is positive right over here."}, {"video_title": "Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Or like, or like, well no, that doesn't look too sharp. Or like this. And the reason why I think where you have these sharp bends or sharp turns as opposed to something that looks more smooth like that, the reason why we're not differentiable there is as we approach this point, as we approach this point from either side, we have different slopes. Notice our slope is positive right over here. As x increases, y is increasing, while our slope is negative here. So as you try to find the limit of our slope as we approach this point, it's not going to exist because it's different on the left-hand side and the right-hand side. So that's why the sharp turns, I don't see any sharp turns here, so it doesn't apply to this example."}, {"video_title": "Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Notice our slope is positive right over here. As x increases, y is increasing, while our slope is negative here. So as you try to find the limit of our slope as we approach this point, it's not going to exist because it's different on the left-hand side and the right-hand side. So that's why the sharp turns, I don't see any sharp turns here, so it doesn't apply to this example. Let's do one more examples. And actually this one does have some sharp turns, so this could be interesting. The graph of function f is given to the left right here."}, {"video_title": "Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So that's why the sharp turns, I don't see any sharp turns here, so it doesn't apply to this example. Let's do one more examples. And actually this one does have some sharp turns, so this could be interesting. The graph of function f is given to the left right here. It has a vertical asymptote at x equals negative three. We see that. And horizontal asymptotes at y equals zero."}, {"video_title": "Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "The graph of function f is given to the left right here. It has a vertical asymptote at x equals negative three. We see that. And horizontal asymptotes at y equals zero. Yep, this end of the curve, as x approaches negative infinity, it looks like y is approaching zero. And it has another horizontal asymptote at y equals four. As x approaches infinity, it looks like our graph is trending down to y is equal to four."}, {"video_title": "Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And horizontal asymptotes at y equals zero. Yep, this end of the curve, as x approaches negative infinity, it looks like y is approaching zero. And it has another horizontal asymptote at y equals four. As x approaches infinity, it looks like our graph is trending down to y is equal to four. Select the x values for which f is not differentiable. So first of all, we could think about vertical tangents. Doesn't seem to have any vertical tangents."}, {"video_title": "Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "As x approaches infinity, it looks like our graph is trending down to y is equal to four. Select the x values for which f is not differentiable. So first of all, we could think about vertical tangents. Doesn't seem to have any vertical tangents. Then we could think about where we are not continuous. Well, we're definitely not continuous where we have this vertical asymptote right over here. So we're not continuous at x equals negative three."}, {"video_title": "Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Doesn't seem to have any vertical tangents. Then we could think about where we are not continuous. Well, we're definitely not continuous where we have this vertical asymptote right over here. So we're not continuous at x equals negative three. We're also not continuous at x is equal to one. And then the last situation where we are not going to be differentiable is where we have a sharp turn, or you could kind of view it as a sharp point on our graph. And I see a sharp point right over there."}, {"video_title": "Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So we're not continuous at x equals negative three. We're also not continuous at x is equal to one. And then the last situation where we are not going to be differentiable is where we have a sharp turn, or you could kind of view it as a sharp point on our graph. And I see a sharp point right over there. Notice, as we approach from the left-hand side, the slope looks like a constant, I don't know, it looks like a positive 3 1\u20442. Well, as we go to the right side of that, it looks like our slope turns negative. And so if you were to try to find the limit of the slope as we approach from either side, which is essentially what you're trying to do when you try to find the derivative, well, it's not going to be defined because it's different on either side."}, {"video_title": "Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And I see a sharp point right over there. Notice, as we approach from the left-hand side, the slope looks like a constant, I don't know, it looks like a positive 3 1\u20442. Well, as we go to the right side of that, it looks like our slope turns negative. And so if you were to try to find the limit of the slope as we approach from either side, which is essentially what you're trying to do when you try to find the derivative, well, it's not going to be defined because it's different on either side. So f is also not differentiable at the x value that gives us that little sharp point right over there. And if you were to graph the derivative, which we will do in future videos, you will see that the derivative is not continuous at that point. So let me mark that off."}, {"video_title": "Differentiability at a point graphical Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And so if you were to try to find the limit of the slope as we approach from either side, which is essentially what you're trying to do when you try to find the derivative, well, it's not going to be defined because it's different on either side. So f is also not differentiable at the x value that gives us that little sharp point right over there. And if you were to graph the derivative, which we will do in future videos, you will see that the derivative is not continuous at that point. So let me mark that off. And then we could check x equals zero. X equals zero is completely cool. We're at a point that our tangent line is definitely not vertical."}, {"video_title": "Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "What I want to do in this video is a few examples that test our intuition of the derivative as a rate of change or the steepness of a curve or the slope of a curve or the slope of a tangent line of a curve, depending on how you actually want to think about it. So here it says f prime of five. So this notation, prime, this is another way of saying, well, what's the derivative? Let's estimate the derivative of our function at five. And when we say f prime of five, this is the slope, slope of tangent line, tangent line at five. Or you could view it as the, you could view it as the rate of change of y with respect to x, which is really how we define slope, with respect to x of our function f. So let's think about that a little bit. We see they put the point, the point five comma f of five right over here."}, {"video_title": "Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Let's estimate the derivative of our function at five. And when we say f prime of five, this is the slope, slope of tangent line, tangent line at five. Or you could view it as the, you could view it as the rate of change of y with respect to x, which is really how we define slope, with respect to x of our function f. So let's think about that a little bit. We see they put the point, the point five comma f of five right over here. And so if we want to estimate the slope of the tangent line, if we want to estimate the steepness of this curve, we could try to draw a line that is tangent right at that point. And so let me see if I can do that. So if I were to draw a line starting there, if I just wanted to make it tangent, it looks like it would do something like that, that right at that point, that looks to be about how steep that curve is."}, {"video_title": "Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "We see they put the point, the point five comma f of five right over here. And so if we want to estimate the slope of the tangent line, if we want to estimate the steepness of this curve, we could try to draw a line that is tangent right at that point. And so let me see if I can do that. So if I were to draw a line starting there, if I just wanted to make it tangent, it looks like it would do something like that, that right at that point, that looks to be about how steep that curve is. Now what makes this an interesting thing in nonlinear is that it's constantly changing. The steepness, it's very low here, and it gets steeper and steeper and steeper as we move to the right for larger and larger x values. But if we look at the point in question, when x is equal to five, remember, f prime of five would be, if we were estimating it, this would be the slope of this line here."}, {"video_title": "Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So if I were to draw a line starting there, if I just wanted to make it tangent, it looks like it would do something like that, that right at that point, that looks to be about how steep that curve is. Now what makes this an interesting thing in nonlinear is that it's constantly changing. The steepness, it's very low here, and it gets steeper and steeper and steeper as we move to the right for larger and larger x values. But if we look at the point in question, when x is equal to five, remember, f prime of five would be, if we were estimating it, this would be the slope of this line here. And the slope of this line, it looks like for every time we move one in the x direction, we're moving two in the y direction. Delta y is equal to two when delta x is equal to one. So our change in y with respect to x, at least for this tangent line here, which would represent our change in y with respect to x right at that point, is going to be equal to two over one, or two."}, {"video_title": "Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "But if we look at the point in question, when x is equal to five, remember, f prime of five would be, if we were estimating it, this would be the slope of this line here. And the slope of this line, it looks like for every time we move one in the x direction, we're moving two in the y direction. Delta y is equal to two when delta x is equal to one. So our change in y with respect to x, at least for this tangent line here, which would represent our change in y with respect to x right at that point, is going to be equal to two over one, or two. And they told us to estimate it, but all of these are way off. Having a negative two derivative would mean that as we increase our x, our y is decreasing. So if our curve looks something like this, we would have a slope of negative two."}, {"video_title": "Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So our change in y with respect to x, at least for this tangent line here, which would represent our change in y with respect to x right at that point, is going to be equal to two over one, or two. And they told us to estimate it, but all of these are way off. Having a negative two derivative would mean that as we increase our x, our y is decreasing. So if our curve looks something like this, we would have a slope of negative two. If having slopes in this, a positive of.1, that would be very flat. Down here we might have a slope closer to.1. Negative.1, that might be closer on this side, where it's downward sloping, but very close to flat."}, {"video_title": "Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So if our curve looks something like this, we would have a slope of negative two. If having slopes in this, a positive of.1, that would be very flat. Down here we might have a slope closer to.1. Negative.1, that might be closer on this side, where it's downward sloping, but very close to flat. A slope of zero, that would be right over here at the bottom where right at that moment, as we change x, y is not increasing or decreasing. The slope of the tangent line right at that bottom point would have a slope of zero. So I feel really good about that response."}, {"video_title": "Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Negative.1, that might be closer on this side, where it's downward sloping, but very close to flat. A slope of zero, that would be right over here at the bottom where right at that moment, as we change x, y is not increasing or decreasing. The slope of the tangent line right at that bottom point would have a slope of zero. So I feel really good about that response. Let's do one more of these. All right, so they're telling us to compare the derivative of g at four to the derivative of g at six. And which one of these is greater?"}, {"video_title": "Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So I feel really good about that response. Let's do one more of these. All right, so they're telling us to compare the derivative of g at four to the derivative of g at six. And which one of these is greater? And like always, pause the video and see if you could figure this out. Well, this is just an exercise. Let's see if we were to make a line that indicates the slope there."}, {"video_title": "Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And which one of these is greater? And like always, pause the video and see if you could figure this out. Well, this is just an exercise. Let's see if we were to make a line that indicates the slope there. And you could view this as a tangent line. So let me try to do that. So, no, that doesn't do a good job."}, {"video_title": "Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Let's see if we were to make a line that indicates the slope there. And you could view this as a tangent line. So let me try to do that. So, no, that doesn't do a good job. So right over here at, so that looks like a pretty, I think I can do a better job than that. No, that's too shallow. Let's see, not shallow, that's too flat."}, {"video_title": "Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So, no, that doesn't do a good job. So right over here at, so that looks like a pretty, I think I can do a better job than that. No, that's too shallow. Let's see, not shallow, that's too flat. So let me try to really, okay, that looks pretty good. So that line that I just drew seems to be indicative of the rate of change of y with respect to x or the slope of that curve or that line, you could view it as a tangent line so that we can think about what its slope is going to be. And then if we go further down over here, this one is, it looks like it is steeper, but in the negative direction."}, {"video_title": "Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Let's see, not shallow, that's too flat. So let me try to really, okay, that looks pretty good. So that line that I just drew seems to be indicative of the rate of change of y with respect to x or the slope of that curve or that line, you could view it as a tangent line so that we can think about what its slope is going to be. And then if we go further down over here, this one is, it looks like it is steeper, but in the negative direction. So it looks like it is steeper for sure, but it's in the negative direction. As we increase, think of it this way, as we increase x one here, it looks like we are decreasing y by about one. So it looks like g prime of four, g prime of four, the derivative when x is equal to four, is approximately, I'm estimating it, negative one."}, {"video_title": "Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And then if we go further down over here, this one is, it looks like it is steeper, but in the negative direction. So it looks like it is steeper for sure, but it's in the negative direction. As we increase, think of it this way, as we increase x one here, it looks like we are decreasing y by about one. So it looks like g prime of four, g prime of four, the derivative when x is equal to four, is approximately, I'm estimating it, negative one. While the derivative here, when we increase x, if we increase x by, if we increase x by one, it looks like we're decreasing y by close to three. So g prime of six looks like it's closer to negative three. So which one of these is larger?"}, {"video_title": "Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So it looks like g prime of four, g prime of four, the derivative when x is equal to four, is approximately, I'm estimating it, negative one. While the derivative here, when we increase x, if we increase x by, if we increase x by one, it looks like we're decreasing y by close to three. So g prime of six looks like it's closer to negative three. So which one of these is larger? Well, this one is less negative, so it's gonna be greater than the other one. And you could have done this intuitively. If you just look at the curve, this is some type of a sinusoid here."}, {"video_title": "Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So which one of these is larger? Well, this one is less negative, so it's gonna be greater than the other one. And you could have done this intuitively. If you just look at the curve, this is some type of a sinusoid here. You have right over here, the curve is flat. It's, you have right at that moment, you have no change in y with respect to x. Then it starts to decrease at a, then it decreases at an even faster rate, then it decreases at a faster rate."}, {"video_title": "Derivative as slope of curve Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "If you just look at the curve, this is some type of a sinusoid here. You have right over here, the curve is flat. It's, you have right at that moment, you have no change in y with respect to x. Then it starts to decrease at a, then it decreases at an even faster rate, then it decreases at a faster rate. Then it starts, it's still decreasing, but it's decreasing at slower and slower rates, decreasing at slower rates. And right at that moment, it's not, you have your slope of your tangent line is zero. Then it starts to increase, increase, so on and so forth."}, {"video_title": "Interpreting the meaning of the derivative in context AP Calculus AB Khan Academy.mp3", "Sentence": "We're told that Eddie drove from New York City to Philadelphia. The function d gives the total distance Eddie has driven in kilometers t hours after he left. What is the best interpretation for the following statement? D prime of two is equal to 100. So pause this video and I encourage you to write it out. What do you think this means? And be sure to include the appropriate units."}, {"video_title": "Interpreting the meaning of the derivative in context AP Calculus AB Khan Academy.mp3", "Sentence": "D prime of two is equal to 100. So pause this video and I encourage you to write it out. What do you think this means? And be sure to include the appropriate units. All right, now let's do this together. If d is equal to the distance driven, then to get d prime, you're taking the derivative with respect to time. So one way to think about it is, it is the rate of change of d. So we could view this as d prime is going to give you the instantaneous rate."}, {"video_title": "Interpreting the meaning of the derivative in context AP Calculus AB Khan Academy.mp3", "Sentence": "And be sure to include the appropriate units. All right, now let's do this together. If d is equal to the distance driven, then to get d prime, you're taking the derivative with respect to time. So one way to think about it is, it is the rate of change of d. So we could view this as d prime is going to give you the instantaneous rate. And they are both functions of t. So one way to view d prime of two is equal to 100, that would mean, well, what is our time now? Well, that is our t, and that's in hours. So two hours, actually, let me color code it."}, {"video_title": "Interpreting the meaning of the derivative in context AP Calculus AB Khan Academy.mp3", "Sentence": "So one way to think about it is, it is the rate of change of d. So we could view this as d prime is going to give you the instantaneous rate. And they are both functions of t. So one way to view d prime of two is equal to 100, that would mean, well, what is our time now? Well, that is our t, and that's in hours. So two hours, actually, let me color code it. So two hours after leaving, after leaving, Eddie drove, and this means, let me be grammatically correct, drove at an instantaneous, instantaneous, instantaneous rate of, and let me use a different color now for this part, of 100, and what are the units? Well, the distance was given in kilometers, and now we're gonna be thinking about kilometers per unit time, kilometers per hour. So this is 100 kilometers, kilometers per hour."}, {"video_title": "Interpreting the meaning of the derivative in context AP Calculus AB Khan Academy.mp3", "Sentence": "So two hours, actually, let me color code it. So two hours after leaving, after leaving, Eddie drove, and this means, let me be grammatically correct, drove at an instantaneous, instantaneous, instantaneous rate of, and let me use a different color now for this part, of 100, and what are the units? Well, the distance was given in kilometers, and now we're gonna be thinking about kilometers per unit time, kilometers per hour. So this is 100 kilometers, kilometers per hour. So that's the interpretation there. Let's do another example. Here we are told a tank is being drained of water."}, {"video_title": "Interpreting the meaning of the derivative in context AP Calculus AB Khan Academy.mp3", "Sentence": "So this is 100 kilometers, kilometers per hour. So that's the interpretation there. Let's do another example. Here we are told a tank is being drained of water. The function v gives the volume of liquid in the tank in liters after t minutes. What is the best interpretation for the following statement? The slope of the line tangent to the graph of v at t equals seven is equal to negative three."}, {"video_title": "Interpreting the meaning of the derivative in context AP Calculus AB Khan Academy.mp3", "Sentence": "Here we are told a tank is being drained of water. The function v gives the volume of liquid in the tank in liters after t minutes. What is the best interpretation for the following statement? The slope of the line tangent to the graph of v at t equals seven is equal to negative three. So pause this video again and try to do what we just did with the previous example. Write out that interpretation, and make sure to get the units right. All right, so let's just remind ourselves what's going on."}, {"video_title": "Interpreting the meaning of the derivative in context AP Calculus AB Khan Academy.mp3", "Sentence": "The slope of the line tangent to the graph of v at t equals seven is equal to negative three. So pause this video again and try to do what we just did with the previous example. Write out that interpretation, and make sure to get the units right. All right, so let's just remind ourselves what's going on. V is going to give us the volume as a function of time. Volume is in liters, and time is in minutes. And so if they're talking about the slope of the tangent line to the graph, the slope of the tangent line to the graph of v, that's just v prime."}, {"video_title": "Interpreting the meaning of the derivative in context AP Calculus AB Khan Academy.mp3", "Sentence": "All right, so let's just remind ourselves what's going on. V is going to give us the volume as a function of time. Volume is in liters, and time is in minutes. And so if they're talking about the slope of the tangent line to the graph, the slope of the tangent line to the graph of v, that's just v prime. So if you take the derivative with respect to time, that's going to give you v prime, and these are all functions of t. These are all functions of t. And they say at t equals seven, it's equal to negative three. So this, which is the same thing as the slope of tangent line slope of tangent, tangent line, and they tell us that v prime of, at time equals seven minutes, our rate of change of volume with respect to time is equal to negative three. And so you could say, if we were to write it out, this means that after, after seven minutes, seven minutes, the tank is being drained at an instantaneous, instantaneous."}, {"video_title": "Interpreting the meaning of the derivative in context AP Calculus AB Khan Academy.mp3", "Sentence": "And so if they're talking about the slope of the tangent line to the graph, the slope of the tangent line to the graph of v, that's just v prime. So if you take the derivative with respect to time, that's going to give you v prime, and these are all functions of t. These are all functions of t. And they say at t equals seven, it's equal to negative three. So this, which is the same thing as the slope of tangent line slope of tangent, tangent line, and they tell us that v prime of, at time equals seven minutes, our rate of change of volume with respect to time is equal to negative three. And so you could say, if we were to write it out, this means that after, after seven minutes, seven minutes, the tank is being drained at an instantaneous, instantaneous. That's why we need that calculus for that instantaneous rate. At an instantaneous rate of, now, you might be tempted to say it's being drained at an instantaneous rate of negative three liters per minute. But remember, the negative three just shows that the volume is decreasing."}, {"video_title": "Interpreting the meaning of the derivative in context AP Calculus AB Khan Academy.mp3", "Sentence": "And so you could say, if we were to write it out, this means that after, after seven minutes, seven minutes, the tank is being drained at an instantaneous, instantaneous. That's why we need that calculus for that instantaneous rate. At an instantaneous rate of, now, you might be tempted to say it's being drained at an instantaneous rate of negative three liters per minute. But remember, the negative three just shows that the volume is decreasing. So one way to think about it is, this negative is already being accounted for when you're saying it's being drained. If this was positive, that means it is being filled. So it is being drained at an instantaneous rate of three liters per minute."}, {"video_title": "Interpreting the meaning of the derivative in context AP Calculus AB Khan Academy.mp3", "Sentence": "But remember, the negative three just shows that the volume is decreasing. So one way to think about it is, this negative is already being accounted for when you're saying it's being drained. If this was positive, that means it is being filled. So it is being drained at an instantaneous rate of three liters per minute. Three liters per minute. And how did I know the units were liters per minute? Well, the volume function is in terms of liters, and the time is in terms of minutes, and then I'm taking the derivative with respect to time, so now it's going to be liters per minute, and we are done."}, {"video_title": "Derivative of inverse sine Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And like always, I encourage you to pause this video and try to figure this out on your own. And I will give you two hints. First hint is, well, we don't know what the derivative of sine inverse of x is, but we do know what the derivative of the sine of something is. And so maybe if you rearrange this and use some implicit differentiation, maybe you can figure out what dy dx is. Remember, this right over here, this right over here is our goal. We essentially want to figure out the derivative of this with respect to x. So I'm assuming you've had a go at it, so let's work through this together."}, {"video_title": "Derivative of inverse sine Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And so maybe if you rearrange this and use some implicit differentiation, maybe you can figure out what dy dx is. Remember, this right over here, this right over here is our goal. We essentially want to figure out the derivative of this with respect to x. So I'm assuming you've had a go at it, so let's work through this together. So if y is the inverse sine of x, that's just like saying that, that's equivalent to saying that sine of y is equal to x. Sine of y is equal to x. So now we have things that we're a little bit more familiar with, and now we can do a little bit of implicit differentiation."}, {"video_title": "Derivative of inverse sine Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So I'm assuming you've had a go at it, so let's work through this together. So if y is the inverse sine of x, that's just like saying that, that's equivalent to saying that sine of y is equal to x. Sine of y is equal to x. So now we have things that we're a little bit more familiar with, and now we can do a little bit of implicit differentiation. We could take the derivative of both sides with respect to x. So derivative of the left-hand side with respect to x and the derivative of the right-hand side with respect to x. Well, what's the derivative of the left-hand side with respect to x going to be?"}, {"video_title": "Derivative of inverse sine Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So now we have things that we're a little bit more familiar with, and now we can do a little bit of implicit differentiation. We could take the derivative of both sides with respect to x. So derivative of the left-hand side with respect to x and the derivative of the right-hand side with respect to x. Well, what's the derivative of the left-hand side with respect to x going to be? And here we just apply the chain rule. It's going to be the derivative of sine of y with respect to y, which is going to be cosine of y times the derivative of y with respect to x. So times dy dx."}, {"video_title": "Derivative of inverse sine Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Well, what's the derivative of the left-hand side with respect to x going to be? And here we just apply the chain rule. It's going to be the derivative of sine of y with respect to y, which is going to be cosine of y times the derivative of y with respect to x. So times dy dx. Times dy dx. And the right-hand side, what's the derivative of x with respect to x? Well, that's obviously just going to be equal to 1."}, {"video_title": "Derivative of inverse sine Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So times dy dx. Times dy dx. And the right-hand side, what's the derivative of x with respect to x? Well, that's obviously just going to be equal to 1. And so we could solve for dy dx, divide both sides by cosine of y, and we get the derivative of y with respect to x is equal to 1 over cosine of y. Now, this still isn't that satisfying because I have the derivative in terms of y. So let's see if we can re-express it in terms of x."}, {"video_title": "Derivative of inverse sine Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Well, that's obviously just going to be equal to 1. And so we could solve for dy dx, divide both sides by cosine of y, and we get the derivative of y with respect to x is equal to 1 over cosine of y. Now, this still isn't that satisfying because I have the derivative in terms of y. So let's see if we can re-express it in terms of x. So how could we do that? Well, we already know that x is equal to sine of y. Let me rewrite it."}, {"video_title": "Derivative of inverse sine Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So let's see if we can re-express it in terms of x. So how could we do that? Well, we already know that x is equal to sine of y. Let me rewrite it. We already know that x is equal to sine of y. So if we could rewrite this bottom expression in terms, instead of cosine of y, if we could use our trigonometric identities to rewrite it in terms of sine of y, then we'll be in good shape because x is equal to sine of y. Well, how can we do that?"}, {"video_title": "Derivative of inverse sine Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Let me rewrite it. We already know that x is equal to sine of y. So if we could rewrite this bottom expression in terms, instead of cosine of y, if we could use our trigonometric identities to rewrite it in terms of sine of y, then we'll be in good shape because x is equal to sine of y. Well, how can we do that? Well, we know from our trigonometric identities, we know that sine squared of y plus cosine squared of y is equal to 1. Or if we want to solve for cosine of y, subtract sine squared of y from both sides, we know that cosine squared of y is equal to 1 minus sine squared of y, or that cosine of y, just take the principal root of both sides, is equal to the principal root of 1 minus sine squared of y. So we can rewrite this as being equal to 1 over, 1 over, instead of cosine of y, we could rewrite it as 1 minus sine squared of y."}, {"video_title": "Derivative of inverse sine Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Well, how can we do that? Well, we know from our trigonometric identities, we know that sine squared of y plus cosine squared of y is equal to 1. Or if we want to solve for cosine of y, subtract sine squared of y from both sides, we know that cosine squared of y is equal to 1 minus sine squared of y, or that cosine of y, just take the principal root of both sides, is equal to the principal root of 1 minus sine squared of y. So we can rewrite this as being equal to 1 over, 1 over, instead of cosine of y, we could rewrite it as 1 minus sine squared of y. Now why is this useful? Well, sine of y is just x. So this is the same, if we just substitute back in, and let me just write it that way so it's a little bit clearer."}, {"video_title": "Derivative of inverse sine Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So we can rewrite this as being equal to 1 over, 1 over, instead of cosine of y, we could rewrite it as 1 minus sine squared of y. Now why is this useful? Well, sine of y is just x. So this is the same, if we just substitute back in, and let me just write it that way so it's a little bit clearer. I could write it as sine y squared. We know that this thing right over here is x. So this is going to be equal to, and we deserve a little bit of a drum roll, 1 over the square root of 1 minus, instead of sine of y, we know that x is equal to sine of y."}, {"video_title": "Derivative of inverse sine Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So this is the same, if we just substitute back in, and let me just write it that way so it's a little bit clearer. I could write it as sine y squared. We know that this thing right over here is x. So this is going to be equal to, and we deserve a little bit of a drum roll, 1 over the square root of 1 minus, instead of sine of y, we know that x is equal to sine of y. So 1 minus x squared. So there you have it. The derivative with respect to x of the inverse sine of x is equal to 1 over the square root of 1 minus x squared."}, {"video_title": "Derivative of inverse sine Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So this is going to be equal to, and we deserve a little bit of a drum roll, 1 over the square root of 1 minus, instead of sine of y, we know that x is equal to sine of y. So 1 minus x squared. So there you have it. The derivative with respect to x of the inverse sine of x is equal to 1 over the square root of 1 minus x squared. So let me just make that very clear. If you were to take the derivative with respect to x of both sides of this, you would get dy dx is equal to this on the right-hand side. Or we could say the derivative with respect to x of the inverse sine of x is equal to 1 over the square root of 1 minus x squared."}, {"video_title": "Derivative of inverse sine Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "The derivative with respect to x of the inverse sine of x is equal to 1 over the square root of 1 minus x squared. So let me just make that very clear. If you were to take the derivative with respect to x of both sides of this, you would get dy dx is equal to this on the right-hand side. Or we could say the derivative with respect to x of the inverse sine of x is equal to 1 over the square root of 1 minus x squared. Now you could always reprove this if your memory starts to fail you, and actually that is the best way to really internalize this. But this is also just a good thing to know, especially as we go into more and more calculus and you might see this in expression. You might say, oh, okay, that's the derivative of the inverse sine of x, and that might prove to be useful."}, {"video_title": "_-substitution intro AP Calculus AB Khan Academy.mp3", "Sentence": "So first when you look at it, it seems like a really complicated integral. We have this polynomial right over here being multiplied by this exponential expression and over here in the exponent, we essentially have another polynomial. It seems kind of crazy. And the key intuition here, the key insight, is that you might want to use a technique here called u-substitution. Substitution. And I'll tell you in a second how I would recognize that we have to use u-substitution. And then over time, you might even be able to do this type of thing in your head."}, {"video_title": "_-substitution intro AP Calculus AB Khan Academy.mp3", "Sentence": "And the key intuition here, the key insight, is that you might want to use a technique here called u-substitution. Substitution. And I'll tell you in a second how I would recognize that we have to use u-substitution. And then over time, you might even be able to do this type of thing in your head. U-substitution is essentially unwinding the chain rule. In the chain, well, I'll go in more depth in another video where I really talk about that intuition. But the way I would think about it is, well, I have this crazy exponent right over here."}, {"video_title": "_-substitution intro AP Calculus AB Khan Academy.mp3", "Sentence": "And then over time, you might even be able to do this type of thing in your head. U-substitution is essentially unwinding the chain rule. In the chain, well, I'll go in more depth in another video where I really talk about that intuition. But the way I would think about it is, well, I have this crazy exponent right over here. I have the x to the third plus x squared. And this thing right over here happens to be the derivative of x to the third plus x squared. The derivative of x to the third is three x squared."}, {"video_title": "_-substitution intro AP Calculus AB Khan Academy.mp3", "Sentence": "But the way I would think about it is, well, I have this crazy exponent right over here. I have the x to the third plus x squared. And this thing right over here happens to be the derivative of x to the third plus x squared. The derivative of x to the third is three x squared. Derivative of x squared is two x, which is a huge clue to me that I could use u-substitution. So what I do here is this thing, or this little expression here, where I also see its derivative being multiplied, I can set that equal to u. So I can say u is equal to x to the third plus x squared."}, {"video_title": "_-substitution intro AP Calculus AB Khan Academy.mp3", "Sentence": "The derivative of x to the third is three x squared. Derivative of x squared is two x, which is a huge clue to me that I could use u-substitution. So what I do here is this thing, or this little expression here, where I also see its derivative being multiplied, I can set that equal to u. So I can say u is equal to x to the third plus x squared. Now what is going to be the derivative of u with respect to x? du dx, well, we've done this multiple times. It's going to be three x squared plus two x."}, {"video_title": "_-substitution intro AP Calculus AB Khan Academy.mp3", "Sentence": "So I can say u is equal to x to the third plus x squared. Now what is going to be the derivative of u with respect to x? du dx, well, we've done this multiple times. It's going to be three x squared plus two x. And now we can write this in differential form. And du dx, this isn't really a fraction of a differential of du divided by a differential of dx. It really is a form of notation."}, {"video_title": "_-substitution intro AP Calculus AB Khan Academy.mp3", "Sentence": "It's going to be three x squared plus two x. And now we can write this in differential form. And du dx, this isn't really a fraction of a differential of du divided by a differential of dx. It really is a form of notation. But it is often useful to kind of pretend that it is a fraction. And you could kind of view this if you wanted to just get a du, if you just wanted to get a differential form over here. How much does u change for a given change in x?"}, {"video_title": "_-substitution intro AP Calculus AB Khan Academy.mp3", "Sentence": "It really is a form of notation. But it is often useful to kind of pretend that it is a fraction. And you could kind of view this if you wanted to just get a du, if you just wanted to get a differential form over here. How much does u change for a given change in x? You could multiply both sides times dx. So both sides times dx. And so if we were to pretend that they were fractions, and it will give you the correct differential form, you're going to be left with du is equal to, du is equal to three x squared plus two x dx."}, {"video_title": "_-substitution intro AP Calculus AB Khan Academy.mp3", "Sentence": "How much does u change for a given change in x? You could multiply both sides times dx. So both sides times dx. And so if we were to pretend that they were fractions, and it will give you the correct differential form, you're going to be left with du is equal to, du is equal to three x squared plus two x dx. Now why is this over here? Why did I go through the trouble of doing that? Well, we see we have a three x squared plus two x, and then it's being multiplied by a dx right over here."}, {"video_title": "_-substitution intro AP Calculus AB Khan Academy.mp3", "Sentence": "And so if we were to pretend that they were fractions, and it will give you the correct differential form, you're going to be left with du is equal to, du is equal to three x squared plus two x dx. Now why is this over here? Why did I go through the trouble of doing that? Well, we see we have a three x squared plus two x, and then it's being multiplied by a dx right over here. I could rewrite this original integral. I could rewrite this as the integral of, let me do it in that color, of three x squared plus two x times dx times e, let me do that in that other color, times e to the x to the third plus x squared. Now, what's interesting about this?"}, {"video_title": "_-substitution intro AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we see we have a three x squared plus two x, and then it's being multiplied by a dx right over here. I could rewrite this original integral. I could rewrite this as the integral of, let me do it in that color, of three x squared plus two x times dx times e, let me do that in that other color, times e to the x to the third plus x squared. Now, what's interesting about this? Well, the stuff that I have in magenta here is exactly equal to du. This is exactly equal to du. And then this stuff I have up here, x to the third plus x squared, that is what I set u equal to."}, {"video_title": "_-substitution intro AP Calculus AB Khan Academy.mp3", "Sentence": "Now, what's interesting about this? Well, the stuff that I have in magenta here is exactly equal to du. This is exactly equal to du. And then this stuff I have up here, x to the third plus x squared, that is what I set u equal to. That is going to be equal to u. So I can rewrite my entire integral, and now you might recognize why this is going to simplify things a good bit. It's going to be equal to, and what I'm gonna do is I'm gonna change the order."}, {"video_title": "_-substitution intro AP Calculus AB Khan Academy.mp3", "Sentence": "And then this stuff I have up here, x to the third plus x squared, that is what I set u equal to. That is going to be equal to u. So I can rewrite my entire integral, and now you might recognize why this is going to simplify things a good bit. It's going to be equal to, and what I'm gonna do is I'm gonna change the order. I'm gonna put the du, this entire du, I'm gonna stick it on the other side here so it looks like more of the standard form that we're used to seeing our indefinite integrals in. So it's going to be, we're gonna have our du, our du, at times e, times e to the u, times e to the u. And so what would the antiderivative of this be in terms of u?"}, {"video_title": "_-substitution intro AP Calculus AB Khan Academy.mp3", "Sentence": "It's going to be equal to, and what I'm gonna do is I'm gonna change the order. I'm gonna put the du, this entire du, I'm gonna stick it on the other side here so it looks like more of the standard form that we're used to seeing our indefinite integrals in. So it's going to be, we're gonna have our du, our du, at times e, times e to the u, times e to the u. And so what would the antiderivative of this be in terms of u? Well, the derivative of e to the u is e to the u. The antiderivative of e to the u is e to the u. So it's going to be equal to e to the u, e to the u."}, {"video_title": "_-substitution intro AP Calculus AB Khan Academy.mp3", "Sentence": "And so what would the antiderivative of this be in terms of u? Well, the derivative of e to the u is e to the u. The antiderivative of e to the u is e to the u. So it's going to be equal to e to the u, e to the u. Now, there's a possibility that there was some type of a constant factor here, so let me write that. So plus c. And now, to get it in terms of x, we just have to unsubstitute the u. We know what u is equal to, so we could say that this is going to be equal to e. Instead of writing u, we could say u is x to the third plus x squared."}, {"video_title": "_-substitution intro AP Calculus AB Khan Academy.mp3", "Sentence": "So it's going to be equal to e to the u, e to the u. Now, there's a possibility that there was some type of a constant factor here, so let me write that. So plus c. And now, to get it in terms of x, we just have to unsubstitute the u. We know what u is equal to, so we could say that this is going to be equal to e. Instead of writing u, we could say u is x to the third plus x squared. U is x to the third plus x squared, x to the third plus x squared. And then we have our plus c. And we are done. We have found the antiderivative."}, {"video_title": "Theorem for limits of composite functions when conditions aren't met AP Calculus Khan Academy.mp3", "Sentence": "In this video, we'll do a few more examples that get a little bit more involved. So let's say we wanted to figure out the limit as x approaches zero of f of g of x, f of g of x. First of all, pause this video and think about whether this theorem even applies. Well, the first thing to think about is what is the limit as x approaches zero of g of x to see if we meet this first condition. So if we look at g of x right over here, as x approaches zero from the left, it looks like g is approaching two. As x approaches zero from the right, it looks like g is approaching two. And so it looks like this is going to be equal to two."}, {"video_title": "Theorem for limits of composite functions when conditions aren't met AP Calculus Khan Academy.mp3", "Sentence": "Well, the first thing to think about is what is the limit as x approaches zero of g of x to see if we meet this first condition. So if we look at g of x right over here, as x approaches zero from the left, it looks like g is approaching two. As x approaches zero from the right, it looks like g is approaching two. And so it looks like this is going to be equal to two. So that's a check. Now let's see the second condition. Is f continuous at that limit, at two?"}, {"video_title": "Theorem for limits of composite functions when conditions aren't met AP Calculus Khan Academy.mp3", "Sentence": "And so it looks like this is going to be equal to two. So that's a check. Now let's see the second condition. Is f continuous at that limit, at two? So when x is equal to two, it does not look like f is continuous. So we do not meet this second condition right over here. So we can't just directly apply this theorem."}, {"video_title": "Theorem for limits of composite functions when conditions aren't met AP Calculus Khan Academy.mp3", "Sentence": "Is f continuous at that limit, at two? So when x is equal to two, it does not look like f is continuous. So we do not meet this second condition right over here. So we can't just directly apply this theorem. But just because you can't apply the theorem does not mean that the limit doesn't necessarily exist. For example, in this situation, the limit actually does exist. One way to think about it, when x approaches zero from the left, it looks like g is approaching two from above."}, {"video_title": "Theorem for limits of composite functions when conditions aren't met AP Calculus Khan Academy.mp3", "Sentence": "So we can't just directly apply this theorem. But just because you can't apply the theorem does not mean that the limit doesn't necessarily exist. For example, in this situation, the limit actually does exist. One way to think about it, when x approaches zero from the left, it looks like g is approaching two from above. And so that's going to be the input into f. And so if we are now approaching two from above, here's the input into f, it looks like our function is approaching zero. And then we can go the other way. If we are approaching zero from the right, right over here, it looks like the value of our function is approaching two from below."}, {"video_title": "Theorem for limits of composite functions when conditions aren't met AP Calculus Khan Academy.mp3", "Sentence": "One way to think about it, when x approaches zero from the left, it looks like g is approaching two from above. And so that's going to be the input into f. And so if we are now approaching two from above, here's the input into f, it looks like our function is approaching zero. And then we can go the other way. If we are approaching zero from the right, right over here, it looks like the value of our function is approaching two from below. Now, if we approach two from below, it looks like the value of f is approaching zero. So in both of these scenarios, our value of our function f is approaching zero. So I wasn't able to use this theorem, but I am able to figure out that this is going to be equal to zero."}, {"video_title": "Theorem for limits of composite functions when conditions aren't met AP Calculus Khan Academy.mp3", "Sentence": "If we are approaching zero from the right, right over here, it looks like the value of our function is approaching two from below. Now, if we approach two from below, it looks like the value of f is approaching zero. So in both of these scenarios, our value of our function f is approaching zero. So I wasn't able to use this theorem, but I am able to figure out that this is going to be equal to zero. Now let me give you another example. Let's say we wanted to figure out the limit as x approaches two of f of g of x. Pause this video, and we'll first see if this theorem even applies."}, {"video_title": "Theorem for limits of composite functions when conditions aren't met AP Calculus Khan Academy.mp3", "Sentence": "So I wasn't able to use this theorem, but I am able to figure out that this is going to be equal to zero. Now let me give you another example. Let's say we wanted to figure out the limit as x approaches two of f of g of x. Pause this video, and we'll first see if this theorem even applies. Well, we first wanna see what is the limit as x approaches two of g of x. When we look at approaching two from the left, it looks like g is approaching negative two. When we approach x equals two from the right, it looks like g is approaching zero."}, {"video_title": "Theorem for limits of composite functions when conditions aren't met AP Calculus Khan Academy.mp3", "Sentence": "Pause this video, and we'll first see if this theorem even applies. Well, we first wanna see what is the limit as x approaches two of g of x. When we look at approaching two from the left, it looks like g is approaching negative two. When we approach x equals two from the right, it looks like g is approaching zero. So our right and left-hand limits are not the same here, so this thing does not exist. Does not exist. And so we don't meet this condition right over here, so we can't apply the theorem."}, {"video_title": "Geometric series interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "So as we talked about in the last video, we've seen many examples of starting with a geometric series, expand it out, and then assuming that its common ratio, that the absolute value of the common ratio is less than 1, finding what the sum of that might be. And we've proven with this formula in previous videos. But now let's go the other way around. Let's try to take some function. Let's say h of x being equal to 1 over 3 plus x squared. And let's try to put it in this form. And then once we put it in that form, we can think about what a and our common ratio is and then try to represent it as an actual geometric series."}, {"video_title": "Geometric series interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let's try to take some function. Let's say h of x being equal to 1 over 3 plus x squared. And let's try to put it in this form. And then once we put it in that form, we can think about what a and our common ratio is and then try to represent it as an actual geometric series. So I encourage you to pause the video and try to do that right now. So let's see. The first thing that you might notice is we have a 1 here instead of a 3."}, {"video_title": "Geometric series interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "And then once we put it in that form, we can think about what a and our common ratio is and then try to represent it as an actual geometric series. So I encourage you to pause the video and try to do that right now. So let's see. The first thing that you might notice is we have a 1 here instead of a 3. So let's try to factor out a 3. So this is equal to 1 over 3 times 1 plus x squared over 3. And now, since we don't want that 3 in the denominator, we can think about this as 1 over 3."}, {"video_title": "Geometric series interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "The first thing that you might notice is we have a 1 here instead of a 3. So let's try to factor out a 3. So this is equal to 1 over 3 times 1 plus x squared over 3. And now, since we don't want that 3 in the denominator, we can think about this as 1 over 3. So we could say this is 1 third over 1. And we don't want to just add something. We want to subtract our common ratio."}, {"video_title": "Geometric series interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "And now, since we don't want that 3 in the denominator, we can think about this as 1 over 3. So we could say this is 1 third over 1. And we don't want to just add something. We want to subtract our common ratio. So 1 minus, and let me write our common ratio here in yellow, 1 minus negative x squared over 3. So now we've written this in that form. And so now we could say that the sum, let me write it here in, let me do it in a new color."}, {"video_title": "Geometric series interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "We want to subtract our common ratio. So 1 minus, and let me write our common ratio here in yellow, 1 minus negative x squared over 3. So now we've written this in that form. And so now we could say that the sum, let me write it here in, let me do it in a new color. So let me do it in blue. So now we could say that the sum from n equals 0 to infinity of, see our first term is 1 third, 1 third times our common ratio to the nth power. Common ratio is negative x squared over 3."}, {"video_title": "Geometric series interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so now we could say that the sum, let me write it here in, let me do it in a new color. So let me do it in blue. So now we could say that the sum from n equals 0 to infinity of, see our first term is 1 third, 1 third times our common ratio to the nth power. Common ratio is negative x squared over 3. And if we wanted to expand this out, this would be equal to. So the first term is 1 third times all of this to the 0th power. So it's just going to be 1 third."}, {"video_title": "Geometric series interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "Common ratio is negative x squared over 3. And if we wanted to expand this out, this would be equal to. So the first term is 1 third times all of this to the 0th power. So it's just going to be 1 third. And so each successive term is just going to be the previous term times our common ratio. So 1 third times negative x squared over 3 is going to be negative 1 ninth x squared. To go from that to that, you have to multiply by, let's see, 1 third to negative 1 third."}, {"video_title": "Geometric series interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "So it's just going to be 1 third. And so each successive term is just going to be the previous term times our common ratio. So 1 third times negative x squared over 3 is going to be negative 1 ninth x squared. To go from that to that, you have to multiply by, let's see, 1 third to negative 1 third. You have to multiply by negative 1 third. And we multiply by x squared as well. And now our next term, we're going to multiply by negative x squared over 3 again."}, {"video_title": "Geometric series interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "To go from that to that, you have to multiply by, let's see, 1 third to negative 1 third. You have to multiply by negative 1 third. And we multiply by x squared as well. And now our next term, we're going to multiply by negative x squared over 3 again. So it's going to be plus, the negative times the negative is a positive, plus 1 over 27 x to the 4th. x squared times x squared, x to the 4th power. And we just keep going on and on and on."}, {"video_title": "Geometric series interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "And now our next term, we're going to multiply by negative x squared over 3 again. So it's going to be plus, the negative times the negative is a positive, plus 1 over 27 x to the 4th. x squared times x squared, x to the 4th power. And we just keep going on and on and on. And when this converges, so over the interval of convergence, this is going to converge to h of x. Now what is the interval of convergence here? And I encourage you to pause the video and think about it."}, {"video_title": "Geometric series interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "And we just keep going on and on and on. And when this converges, so over the interval of convergence, this is going to converge to h of x. Now what is the interval of convergence here? And I encourage you to pause the video and think about it. Well, the interval of convergence is the interval over which your common ratio, the absolute value of your common ratio, is less than 1. So let me write this right over here. So our absolute value of negative x squared over 3 has to be less than 1."}, {"video_title": "Geometric series interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "And I encourage you to pause the video and think about it. Well, the interval of convergence is the interval over which your common ratio, the absolute value of your common ratio, is less than 1. So let me write this right over here. So our absolute value of negative x squared over 3 has to be less than 1. Well, the absolute value, this is going to be a negative number. This is the same thing as saying, let me scroll down a little bit, this is the same thing as saying that the absolute value of x squared over 3 has to be less than 1. And this is another way of saying, well, one thing that might jump out at you is that x squared, this is going to be positive no matter what."}, {"video_title": "Geometric series interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "So our absolute value of negative x squared over 3 has to be less than 1. Well, the absolute value, this is going to be a negative number. This is the same thing as saying, let me scroll down a little bit, this is the same thing as saying that the absolute value of x squared over 3 has to be less than 1. And this is another way of saying, well, one thing that might jump out at you is that x squared, this is going to be positive no matter what. Or I guess I say this is going to be non-negative no matter what. So this is another way of saying that x squared over 3 has to be less than 1. I don't want to confuse you in this step right over here, but the absolute value of x squared over 3 is just going to be x squared over 3, because this is never going to take on a negative value."}, {"video_title": "Geometric series interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "And this is another way of saying, well, one thing that might jump out at you is that x squared, this is going to be positive no matter what. Or I guess I say this is going to be non-negative no matter what. So this is another way of saying that x squared over 3 has to be less than 1. I don't want to confuse you in this step right over here, but the absolute value of x squared over 3 is just going to be x squared over 3, because this is never going to take on a negative value. And so we can multiply both sides by 3. I'll go up here now to do it. Multiply both sides by 3 to say that x squared needs to be less than 3."}, {"video_title": "Geometric series interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "I don't want to confuse you in this step right over here, but the absolute value of x squared over 3 is just going to be x squared over 3, because this is never going to take on a negative value. And so we can multiply both sides by 3. I'll go up here now to do it. Multiply both sides by 3 to say that x squared needs to be less than 3. And so that means that the absolute value of x needs to be less than the square root of 3. Or we could say that x is greater than the negative square root of 3, and it is less than the square root of 3. So this is the interval of convergence for this series, for this power series."}, {"video_title": "Geometric series interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "Multiply both sides by 3 to say that x squared needs to be less than 3. And so that means that the absolute value of x needs to be less than the square root of 3. Or we could say that x is greater than the negative square root of 3, and it is less than the square root of 3. So this is the interval of convergence for this series, for this power series. It's a geometric series, which is a special case of a power series. And over the interval of convergence, that is going to be equal to 1 over 3 plus x squared. So as long as x is in this interval, it's going to take on the same values as our original function, which is a pretty neat idea."}, {"video_title": "Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3", "Sentence": "At the point x equals a. And we've already seen this with the definition of the derivative. We could try to find a general function that gives us the slope of the tangent line at any point. So let's say we have some arbitrary point. Let me define some arbitrary point x right over here. Then this would be the point x comma f of x. And then we could take some x plus h. So let's say that this right over here is the point x plus h. And so this point would be x plus h f of x plus h. We can find the slope of the secant line that goes between these two points."}, {"video_title": "Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3", "Sentence": "So let's say we have some arbitrary point. Let me define some arbitrary point x right over here. Then this would be the point x comma f of x. And then we could take some x plus h. So let's say that this right over here is the point x plus h. And so this point would be x plus h f of x plus h. We can find the slope of the secant line that goes between these two points. So that would be your change in your vertical. Which would be f of x plus h minus f of x over the change in the horizontal. Which would be x plus h minus x."}, {"video_title": "Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3", "Sentence": "And then we could take some x plus h. So let's say that this right over here is the point x plus h. And so this point would be x plus h f of x plus h. We can find the slope of the secant line that goes between these two points. So that would be your change in your vertical. Which would be f of x plus h minus f of x over the change in the horizontal. Which would be x plus h minus x. And these two x's cancel. So this would be the slope of this secant line. And then if we want to find the slope of the tangent line at x, we would just take the limit of this expression as h approaches 0."}, {"video_title": "Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3", "Sentence": "Which would be x plus h minus x. And these two x's cancel. So this would be the slope of this secant line. And then if we want to find the slope of the tangent line at x, we would just take the limit of this expression as h approaches 0. As h approaches 0, this point moves towards x. And that slope of the secant line between these two is going to approximate the slope of the tangent line at x. And so this right over here, this we would say is equal to f prime of x."}, {"video_title": "Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3", "Sentence": "And then if we want to find the slope of the tangent line at x, we would just take the limit of this expression as h approaches 0. As h approaches 0, this point moves towards x. And that slope of the secant line between these two is going to approximate the slope of the tangent line at x. And so this right over here, this we would say is equal to f prime of x. This is a function of x. You give me an x, you give me an arbitrary x where the derivative is defined. I'm going to plug it into this."}, {"video_title": "Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3", "Sentence": "And so this right over here, this we would say is equal to f prime of x. This is a function of x. You give me an x, you give me an arbitrary x where the derivative is defined. I'm going to plug it into this. Whatever this ends up being, it might be some nice clean algebraic expression. And then I'm going to give you a number. So for example, if you wanted to find, you could calculate this somehow."}, {"video_title": "Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3", "Sentence": "I'm going to plug it into this. Whatever this ends up being, it might be some nice clean algebraic expression. And then I'm going to give you a number. So for example, if you wanted to find, you could calculate this somehow. Or you could even leave it in this form. And then if you wanted f prime of a, you would just substitute a into your function definition. And you would say, well, that's going to be the limit as h approaches 0 of every place you see an x, replace it with an a. I'll stay in this color for now."}, {"video_title": "Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3", "Sentence": "So for example, if you wanted to find, you could calculate this somehow. Or you could even leave it in this form. And then if you wanted f prime of a, you would just substitute a into your function definition. And you would say, well, that's going to be the limit as h approaches 0 of every place you see an x, replace it with an a. I'll stay in this color for now. Blank plus h minus f of blank. All of that over h. I left those blanks so I could write the a in red. Notice, every place where I had an x before, it's now an a."}, {"video_title": "Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3", "Sentence": "And you would say, well, that's going to be the limit as h approaches 0 of every place you see an x, replace it with an a. I'll stay in this color for now. Blank plus h minus f of blank. All of that over h. I left those blanks so I could write the a in red. Notice, every place where I had an x before, it's now an a. So this is the derivative evaluated at a. So this is one way to find the slope of the tangent line when x equals a. Another way, and this is often used as the alternate form of the derivative, would be to do it directly."}, {"video_title": "Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3", "Sentence": "Notice, every place where I had an x before, it's now an a. So this is the derivative evaluated at a. So this is one way to find the slope of the tangent line when x equals a. Another way, and this is often used as the alternate form of the derivative, would be to do it directly. So this is the point a comma f of a. Let's just take another arbitrary point, some here, some place. So let's say this is the value x."}, {"video_title": "Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3", "Sentence": "Another way, and this is often used as the alternate form of the derivative, would be to do it directly. So this is the point a comma f of a. Let's just take another arbitrary point, some here, some place. So let's say this is the value x. This point right over here on the function would be x comma f of x. And so what's the slope of the secant line between these two points? It would be change in the vertical, which would be f of x minus f of a, over change in the horizontal, over x minus a."}, {"video_title": "Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3", "Sentence": "So let's say this is the value x. This point right over here on the function would be x comma f of x. And so what's the slope of the secant line between these two points? It would be change in the vertical, which would be f of x minus f of a, over change in the horizontal, over x minus a. So let me do that in a purple color, over x minus a. Now, how can we get a better and better approximation for the slope of the tangent line here? Well, we could take the limit as x approaches a."}, {"video_title": "Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3", "Sentence": "It would be change in the vertical, which would be f of x minus f of a, over change in the horizontal, over x minus a. So let me do that in a purple color, over x minus a. Now, how can we get a better and better approximation for the slope of the tangent line here? Well, we could take the limit as x approaches a. As x gets closer and closer and closer to a, the secant line slope is going to better and better and better approximate the slope of the tangent line, this tangent line that I have in red here. So we would want to take the limit as x approaches a here. Either way, we're doing a very similar, we're doing the exact same thing."}, {"video_title": "Formal and alternate form of the derivative Differential Calculus Khan Academy.mp3", "Sentence": "Well, we could take the limit as x approaches a. As x gets closer and closer and closer to a, the secant line slope is going to better and better and better approximate the slope of the tangent line, this tangent line that I have in red here. So we would want to take the limit as x approaches a here. Either way, we're doing a very similar, we're doing the exact same thing. We're finding, we're taking, we have an expression for the slope of a secant line, and then we're bringing those x values of those points closer and closer together, closer and closer together, so the slopes of those secant lines better and better and better approximate that slope of the tangent line, and at the limit, it does become the slope of the tangent line. That is the derivative of, or that's the definition of the derivative. So this is the kind of the more standard definition of a derivative."}, {"video_title": "Approximating limits using tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "In this video, we're going to try to get a sense of what the limit as x approaches three of x to the third minus three x squared over five x minus 15 is. And when I say get a sense, we're gonna do that by seeing what values for this expression we get as x gets closer and closer to three. Now one thing that you might wanna try out is well what happens to this expression when x is equal to three? Well then it's going to be three to the third power minus three times three squared over five times three minus 15. So at x equals three, this expression's gonna be and you see the numerator, you have 27 minus 27, zero over 15 minus 15 over zero. So this expression is actually not defined at x equals three we get this indeterminate form, we get zero over zero. But let's see, even though the function, even though the expression is not defined, let's see if we can get a sense of what the limit might be."}, {"video_title": "Approximating limits using tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Well then it's going to be three to the third power minus three times three squared over five times three minus 15. So at x equals three, this expression's gonna be and you see the numerator, you have 27 minus 27, zero over 15 minus 15 over zero. So this expression is actually not defined at x equals three we get this indeterminate form, we get zero over zero. But let's see, even though the function, even though the expression is not defined, let's see if we can get a sense of what the limit might be. And to do that, I'm gonna set up a table. So let me set up a table here. And actually I'm gonna set up two tables."}, {"video_title": "Approximating limits using tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "But let's see, even though the function, even though the expression is not defined, let's see if we can get a sense of what the limit might be. And to do that, I'm gonna set up a table. So let me set up a table here. And actually I'm gonna set up two tables. So this is x and this is x to the third minus three x squared over five x minus 15. And actually I'm gonna do that again and I'll tell you why in a second. So this is gonna be x and this is x to the third minus three x squared over five x minus 15."}, {"video_title": "Approximating limits using tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And actually I'm gonna set up two tables. So this is x and this is x to the third minus three x squared over five x minus 15. And actually I'm gonna do that again and I'll tell you why in a second. So this is gonna be x and this is x to the third minus three x squared over five x minus 15. The reason why I set up two tables, I didn't have to do two tables, I could have done it all in one table but hopefully this will make it a little bit more intuitive what I'm trying to do is on this left table, I'm gonna, let's try out x values that get closer and closer to three from the left, from values that are less than three. So for example, we could go to 2.9 and figure out what the expression equals when x is 2.9. But then we could try to get even a little bit closer than that."}, {"video_title": "Approximating limits using tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So this is gonna be x and this is x to the third minus three x squared over five x minus 15. The reason why I set up two tables, I didn't have to do two tables, I could have done it all in one table but hopefully this will make it a little bit more intuitive what I'm trying to do is on this left table, I'm gonna, let's try out x values that get closer and closer to three from the left, from values that are less than three. So for example, we could go to 2.9 and figure out what the expression equals when x is 2.9. But then we could try to get even a little bit closer than that. We could go to 2.99. And then we could go even closer than that. We could go to 2.999."}, {"video_title": "Approximating limits using tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "But then we could try to get even a little bit closer than that. We could go to 2.99. And then we could go even closer than that. We could go to 2.999. And so one way to think about it here is as we try to figure out what this expression equals as we get closer and closer to three, we're trying to approximate the limit from the left. So limit from the left. And why do I say the left?"}, {"video_title": "Approximating limits using tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "We could go to 2.999. And so one way to think about it here is as we try to figure out what this expression equals as we get closer and closer to three, we're trying to approximate the limit from the left. So limit from the left. And why do I say the left? Well, if you think about this on a coordinate plane, these are the x values that are to the left of three but we're getting closer and closer and closer. We're moving to the right but these are the x values that are on the left side of three, they're less than three. But we also, in order for the limit to exist, we have to be approaching the same thing from both sides, from both the left and the right."}, {"video_title": "Approximating limits using tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And why do I say the left? Well, if you think about this on a coordinate plane, these are the x values that are to the left of three but we're getting closer and closer and closer. We're moving to the right but these are the x values that are on the left side of three, they're less than three. But we also, in order for the limit to exist, we have to be approaching the same thing from both sides, from both the left and the right. So we could also try to approximate the limit from the right. And so what values would those be? Well, those would be x values larger than three."}, {"video_title": "Approximating limits using tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "But we also, in order for the limit to exist, we have to be approaching the same thing from both sides, from both the left and the right. So we could also try to approximate the limit from the right. And so what values would those be? Well, those would be x values larger than three. So we could say 3.1 but then we might wanna get a little bit closer. We could go 3.01 but then we might wanna get even closer to three, 3.001. And every time we get closer and closer to three, we're gonna get a better approximation for, or we're gonna get a better sense of what we are actually approaching."}, {"video_title": "Approximating limits using tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Well, those would be x values larger than three. So we could say 3.1 but then we might wanna get a little bit closer. We could go 3.01 but then we might wanna get even closer to three, 3.001. And every time we get closer and closer to three, we're gonna get a better approximation for, or we're gonna get a better sense of what we are actually approaching. So let's get a calculator out and do this. And you could keep going, 2.99999, 3.00001. Now one key idea here to point out before I even calculate what these are going to be, sometimes when people say the limit from both sides or the limit from the left or the limit from the right, they imagine that the limit from the left is negative values and the limit from the right are positive values."}, {"video_title": "Approximating limits using tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And every time we get closer and closer to three, we're gonna get a better approximation for, or we're gonna get a better sense of what we are actually approaching. So let's get a calculator out and do this. And you could keep going, 2.99999, 3.00001. Now one key idea here to point out before I even calculate what these are going to be, sometimes when people say the limit from both sides or the limit from the left or the limit from the right, they imagine that the limit from the left is negative values and the limit from the right are positive values. But as you can see here, the limit from the left are to the left of the x value that you're trying to find the limit at. So these aren't negative values, these are just approaching the three right over here from values less than three. This is approaching the three from values larger than three."}, {"video_title": "Approximating limits using tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Now one key idea here to point out before I even calculate what these are going to be, sometimes when people say the limit from both sides or the limit from the left or the limit from the right, they imagine that the limit from the left is negative values and the limit from the right are positive values. But as you can see here, the limit from the left are to the left of the x value that you're trying to find the limit at. So these aren't negative values, these are just approaching the three right over here from values less than three. This is approaching the three from values larger than three. So now let's fill out this table. And I'm speeding up my work so that you don't have to sit through me typing everything into a calculator. So based on what we're seeing here, I would make the estimate that this looks like it's approaching 1.8."}, {"video_title": "Approximating limits using tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "This is approaching the three from values larger than three. So now let's fill out this table. And I'm speeding up my work so that you don't have to sit through me typing everything into a calculator. So based on what we're seeing here, I would make the estimate that this looks like it's approaching 1.8. So is this equal to 1.8? As I said, in the future we're gonna be able to find this out exactly. But if you're not sure about this, you could try even closer and closer and closer values."}, {"video_title": "If function u is continuous at x, then _u_0 as _x_0 AP Calculus AB Khan Academy.mp3", "Sentence": "But the result we're gonna look at is if we have some function u, which is a function of x, and we know that it is continuous, it is continuous at x equals c. So if we know this, then that's going to imply that the change in u goes to zero as our change in x in this region around c goes to zero. This is what I want to get an intuition for, that if u is continuous at c, then as our change in x around c gets smaller, or it gets smaller and smaller and smaller as it approaches zero, then our change in u approaches zero as well. So let's just, to think about this, or even kind of prove it to ourselves a little bit more rigorously, let's think about what it means to be continuous at x equals c. Well, the definition of continuity is, so this literally is the same thing as saying that the limit as x approaches c of u of x is equal to u of c. That the limit that our function approaches as x approaches c is equal to the value of the function at c. We don't have a point discontinuity or a jump discontinuity. If we had a jump discontinuity, then the limit wouldn't exist, and we've seen that in previous videos. Now I'm just going to manipulate this algebraically, so it essentially gives us this conclusion right over here. So this we can rewrite. It's important to realize that u of c, this is just going to be some value."}, {"video_title": "If function u is continuous at x, then _u_0 as _x_0 AP Calculus AB Khan Academy.mp3", "Sentence": "If we had a jump discontinuity, then the limit wouldn't exist, and we've seen that in previous videos. Now I'm just going to manipulate this algebraically, so it essentially gives us this conclusion right over here. So this we can rewrite. It's important to realize that u of c, this is just going to be some value. It looks like I've kind of, maybe this is a function of x or something, but no, this is just going to be some value. I've inputted c here, and I've evaluated the function of that, and so this is going to be some number. It could be five or seven or pi or negative one, but it's just going to be some value, some constant."}, {"video_title": "If function u is continuous at x, then _u_0 as _x_0 AP Calculus AB Khan Academy.mp3", "Sentence": "It's important to realize that u of c, this is just going to be some value. It looks like I've kind of, maybe this is a function of x or something, but no, this is just going to be some value. I've inputted c here, and I've evaluated the function of that, and so this is going to be some number. It could be five or seven or pi or negative one, but it's just going to be some value, some constant. So I can treat it like a constant. So this is going to be the same thing as saying the limit as x approaches c of u of x minus u of c is equal to zero. And actually, in the video where we prove that differentiability implies continuity, we started with this, and we proved that right over there."}, {"video_title": "If function u is continuous at x, then _u_0 as _x_0 AP Calculus AB Khan Academy.mp3", "Sentence": "It could be five or seven or pi or negative one, but it's just going to be some value, some constant. So I can treat it like a constant. So this is going to be the same thing as saying the limit as x approaches c of u of x minus u of c is equal to zero. And actually, in the video where we prove that differentiability implies continuity, we started with this, and we proved that right over there. We showed that these two are equivalent things. But hopefully you can even think about the intuition. This is just, if the limit as u of x approaches, the limit of u of x as x approaches c is equal to this, then when you evaluate this limit, the limit as x approaches c, well, this thing is going to approach u of c, because we saw it right up here."}, {"video_title": "If function u is continuous at x, then _u_0 as _x_0 AP Calculus AB Khan Academy.mp3", "Sentence": "And actually, in the video where we prove that differentiability implies continuity, we started with this, and we proved that right over there. We showed that these two are equivalent things. But hopefully you can even think about the intuition. This is just, if the limit as u of x approaches, the limit of u of x as x approaches c is equal to this, then when you evaluate this limit, the limit as x approaches c, well, this thing is going to approach u of c, because we saw it right up here. u of c minus u of c is indeed going to be equal to zero. So hopefully you don't feel like it's too much of a stretch, and you can just subtract u of c from both sides and apply properties of limits, and you can get this result as well. But this is interesting, because this essentially can take us to this, that the idea that as our change in x gets smaller and smaller and smaller, as it approaches zero, then our change in our function is also going to approach zero."}, {"video_title": "If function u is continuous at x, then _u_0 as _x_0 AP Calculus AB Khan Academy.mp3", "Sentence": "This is just, if the limit as u of x approaches, the limit of u of x as x approaches c is equal to this, then when you evaluate this limit, the limit as x approaches c, well, this thing is going to approach u of c, because we saw it right up here. u of c minus u of c is indeed going to be equal to zero. So hopefully you don't feel like it's too much of a stretch, and you can just subtract u of c from both sides and apply properties of limits, and you can get this result as well. But this is interesting, because this essentially can take us to this, that the idea that as our change in x gets smaller and smaller and smaller, as it approaches zero, then our change in our function is also going to approach zero. Now let's just look and take a, let's just graph this or visualize this to get a sense of that. So this is our x-axis. Whoops, that's our x-axis."}, {"video_title": "If function u is continuous at x, then _u_0 as _x_0 AP Calculus AB Khan Academy.mp3", "Sentence": "But this is interesting, because this essentially can take us to this, that the idea that as our change in x gets smaller and smaller and smaller, as it approaches zero, then our change in our function is also going to approach zero. Now let's just look and take a, let's just graph this or visualize this to get a sense of that. So this is our x-axis. Whoops, that's our x-axis. Let's call that our u-axis maybe. And I did u intentionally, because that's the variable we'll use in our proof of the chain rule video. And let's say this right over here is our function."}, {"video_title": "If function u is continuous at x, then _u_0 as _x_0 AP Calculus AB Khan Academy.mp3", "Sentence": "Whoops, that's our x-axis. Let's call that our u-axis maybe. And I did u intentionally, because that's the variable we'll use in our proof of the chain rule video. And let's say this right over here is our function. Let's say this right over here, that is c. This right over here is u of c. u of c. And then let's just take some arbitrary x over here. So some arbitrary x, and then this right over here, this right over here is u of x. is u of x. So if we define, if we define our change, if we define, let me do this."}, {"video_title": "If function u is continuous at x, then _u_0 as _x_0 AP Calculus AB Khan Academy.mp3", "Sentence": "And let's say this right over here is our function. Let's say this right over here, that is c. This right over here is u of c. u of c. And then let's just take some arbitrary x over here. So some arbitrary x, and then this right over here, this right over here is u of x. is u of x. So if we define, if we define our change, if we define, let me do this. If we define our change in, our change in u is equal to u of x minus u of c, which makes sense, because this is our change in u. So let's say this is going to be u of x minus u of c. And if we define our change in x is equal to x minus c, which it is in this case, it is x minus c. It is x minus c. Then we can rewrite this limit right over here. Instead of saying the limit as x approaches c, we could write the limit as delta x approaches zero, because if x approaches c, then delta x is going to approach zero."}, {"video_title": "If function u is continuous at x, then _u_0 as _x_0 AP Calculus AB Khan Academy.mp3", "Sentence": "So if we define, if we define our change, if we define, let me do this. If we define our change in, our change in u is equal to u of x minus u of c, which makes sense, because this is our change in u. So let's say this is going to be u of x minus u of c. And if we define our change in x is equal to x minus c, which it is in this case, it is x minus c. It is x minus c. Then we can rewrite this limit right over here. Instead of saying the limit as x approaches c, we could write the limit as delta x approaches zero, because if x approaches c, then delta x is going to approach zero. So we could write this, the limit as delta x approaches zero of delta u, of delta u, is going to be equal to zero. This, we define this as our change in u, and it is our change in u. So this is equal to zero."}, {"video_title": "If function u is continuous at x, then _u_0 as _x_0 AP Calculus AB Khan Academy.mp3", "Sentence": "Instead of saying the limit as x approaches c, we could write the limit as delta x approaches zero, because if x approaches c, then delta x is going to approach zero. So we could write this, the limit as delta x approaches zero of delta u, of delta u, is going to be equal to zero. This, we define this as our change in u, and it is our change in u. So this is equal to zero. So another way of thinking about this is as delta x approaches zero, our change in u, our change in the function, is going to approach zero. So as delta x approaches zero, as delta x approaches zero, delta u approaches zero. And that's what we wrote over here."}, {"video_title": "If function u is continuous at x, then _u_0 as _x_0 AP Calculus AB Khan Academy.mp3", "Sentence": "So this is equal to zero. So another way of thinking about this is as delta x approaches zero, our change in u, our change in the function, is going to approach zero. So as delta x approaches zero, as delta x approaches zero, delta u approaches zero. And that's what we wrote over here. Delta u approaches zero as delta x approaches zero. And in a lot of ways, this is hopefully common sense. We're dealing with a continuous function."}, {"video_title": "If function u is continuous at x, then _u_0 as _x_0 AP Calculus AB Khan Academy.mp3", "Sentence": "And that's what we wrote over here. Delta u approaches zero as delta x approaches zero. And in a lot of ways, this is hopefully common sense. We're dealing with a continuous function. As you get smaller and smaller, and you could just think of it this way, as you get smaller and smaller changes in x's, as our change in x gets smaller and smaller and smaller and smaller, well because it's continuous, and you wouldn't be able to say this for a discontinuous function, but because it's continuous, or you wouldn't be able to say this for some discontinuous functions, as our change in x gets smaller and smaller and smaller, then our change in u is going to get smaller and smaller and smaller, so it makes intuitive sense. But hopefully this makes you feel even better about it. Because we're going to use this idea to prove the chain rule in the next video."}, {"video_title": "Minimizing combined area Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And to do that, we just have to take the derivative of this business, figure out where our derivative is either undefined or 0, and then just make sure that that is a minimum value, and then we'll be all set. So let me rewrite this. So our combined area as a function of x, let me just rewrite this so it's a little bit easier to take the derivative. So this is going to be the square root of 3 times x squared over, let's see, this is 4 times 9. This is x squared over 9. So this is going to be 4 times 9 is 36. And then over here in blue, this is going to be plus 100 minus x squared over 16."}, {"video_title": "Minimizing combined area Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So this is going to be the square root of 3 times x squared over, let's see, this is 4 times 9. This is x squared over 9. So this is going to be 4 times 9 is 36. And then over here in blue, this is going to be plus 100 minus x squared over 16. Now let's take the derivative of this. So a prime, our combined area, the derivative of our combined area as a function of x, is going to be equal to, well, the derivative of this with respect to x is just going to be square root of 3x over 18. The derivative of this with respect to x was the derivative of something squared over 16 with respect to that something."}, {"video_title": "Minimizing combined area Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And then over here in blue, this is going to be plus 100 minus x squared over 16. Now let's take the derivative of this. So a prime, our combined area, the derivative of our combined area as a function of x, is going to be equal to, well, the derivative of this with respect to x is just going to be square root of 3x over 18. The derivative of this with respect to x was the derivative of something squared over 16 with respect to that something. So that's going to be that something to the first power times 2 over 16, which is just over 8. And then times, we're just doing the chain rule, times the derivative of the something with respect to x. The derivative of 100 minus x with respect to x is just negative 1."}, {"video_title": "Minimizing combined area Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "The derivative of this with respect to x was the derivative of something squared over 16 with respect to that something. So that's going to be that something to the first power times 2 over 16, which is just over 8. And then times, we're just doing the chain rule, times the derivative of the something with respect to x. The derivative of 100 minus x with respect to x is just negative 1. So times negative 1. Times, so we'll multiply negative 1 right over here. And so we can rewrite all of that as this is going to be equal to the square root of 3 over 18x plus, let's see, I could write this as positive x over 8."}, {"video_title": "Minimizing combined area Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "The derivative of 100 minus x with respect to x is just negative 1. So times negative 1. Times, so we'll multiply negative 1 right over here. And so we can rewrite all of that as this is going to be equal to the square root of 3 over 18x plus, let's see, I could write this as positive x over 8. So I could write this as 1 8th x, because negative 1 times negative x is positive x over 8. And then minus 100 over 8, which is negative 12.5. Minus 12.5."}, {"video_title": "Minimizing combined area Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And so we can rewrite all of that as this is going to be equal to the square root of 3 over 18x plus, let's see, I could write this as positive x over 8. So I could write this as 1 8th x, because negative 1 times negative x is positive x over 8. And then minus 100 over 8, which is negative 12.5. Minus 12.5. And we want to figure out an x that minimizes this area. So this derivative right over here is defined for any x. So we're not going to get our critical point by figuring out where the derivative is undefined."}, {"video_title": "Minimizing combined area Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Minus 12.5. And we want to figure out an x that minimizes this area. So this derivative right over here is defined for any x. So we're not going to get our critical point by figuring out where the derivative is undefined. But we might get a critical point by setting this derivative equal to 0 to figure out what x values make our derivative 0. When do we have a 0 slope for our original function? And then we just have to verify that this is going to be a minimum point."}, {"video_title": "Minimizing combined area Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So we're not going to get our critical point by figuring out where the derivative is undefined. But we might get a critical point by setting this derivative equal to 0 to figure out what x values make our derivative 0. When do we have a 0 slope for our original function? And then we just have to verify that this is going to be a minimum point. If we can find an x, it makes this thing equal to 0. So let's try to solve for x. So if we add 12.5 to both sides, we get 12.5 is equal to, if you add the x terms, you get square root of 3 over 18 plus 1 8th x."}, {"video_title": "Minimizing combined area Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And then we just have to verify that this is going to be a minimum point. If we can find an x, it makes this thing equal to 0. So let's try to solve for x. So if we add 12.5 to both sides, we get 12.5 is equal to, if you add the x terms, you get square root of 3 over 18 plus 1 8th x. To solve for x, divide both sides by this business, you get x is equal to 12.5 over square root of 3 over 18 plus 1 over 8. And we are done. At x equals this, our derivative is equal to 0."}, {"video_title": "Minimizing combined area Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So if we add 12.5 to both sides, we get 12.5 is equal to, if you add the x terms, you get square root of 3 over 18 plus 1 8th x. To solve for x, divide both sides by this business, you get x is equal to 12.5 over square root of 3 over 18 plus 1 over 8. And we are done. At x equals this, our derivative is equal to 0. Now, we know, I shouldn't say we're done yet. We don't know whether this is a minimum point. In order to figure out whether this is a minimum point, we have to figure out whether our function is concave upward or concave downward when x is equal to this business."}, {"video_title": "Minimizing combined area Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "At x equals this, our derivative is equal to 0. Now, we know, I shouldn't say we're done yet. We don't know whether this is a minimum point. In order to figure out whether this is a minimum point, we have to figure out whether our function is concave upward or concave downward when x is equal to this business. And to figure that out, let's take the second derivative here. So the second derivative, so let me rewrite the second derivative of all of this business. The second derivative, well, this was the same function as this right over here."}, {"video_title": "Minimizing combined area Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "In order to figure out whether this is a minimum point, we have to figure out whether our function is concave upward or concave downward when x is equal to this business. And to figure that out, let's take the second derivative here. So the second derivative, so let me rewrite the second derivative of all of this business. The second derivative, well, this was the same function as this right over here. So let me rewrite it. So a prime, the derivative of our combined area, was equal to the square root of 3 over 18x plus 1 8th x minus 12.5. The second derivative is going to be square root of 3 over 18 plus 1 over 8."}, {"video_title": "Minimizing combined area Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "The second derivative, well, this was the same function as this right over here. So let me rewrite it. So a prime, the derivative of our combined area, was equal to the square root of 3 over 18x plus 1 8th x minus 12.5. The second derivative is going to be square root of 3 over 18 plus 1 over 8. So this thing right over here is greater than 0, which means we're concave upward for all x's. Which means for all x's, we're kind of doing a situation like this. So if we find an x where the slope is 0, it's going to be over an interval where it's concave upwards."}, {"video_title": "Minimizing combined area Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "The second derivative is going to be square root of 3 over 18 plus 1 over 8. So this thing right over here is greater than 0, which means we're concave upward for all x's. Which means for all x's, we're kind of doing a situation like this. So if we find an x where the slope is 0, it's going to be over an interval where it's concave upwards. This is concave upwards for all x. So we're going to be at a minimum point. The slope is 0 right over here."}, {"video_title": "Minimizing combined area Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So if we find an x where the slope is 0, it's going to be over an interval where it's concave upwards. This is concave upwards for all x. So we're going to be at a minimum point. The slope is 0 right over here. This right over here will be a minimum point. So once again, this is going to be a minimum point. Now, if we actually had 100 meter wire, this expression isn't too valuable."}, {"video_title": "Minimizing combined area Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "The slope is 0 right over here. This right over here will be a minimum point. So once again, this is going to be a minimum point. Now, if we actually had 100 meter wire, this expression isn't too valuable. We'd want to get a pretty close approximation in terms of where to actually make the cut, an actual decimal number. So let's use a calculator to get that. So we have 12.5 divided by square root of 3 divided by 18 plus 1 divided by 8."}, {"video_title": "Minimizing combined area Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Now, if we actually had 100 meter wire, this expression isn't too valuable. We'd want to get a pretty close approximation in terms of where to actually make the cut, an actual decimal number. So let's use a calculator to get that. So we have 12.5 divided by square root of 3 divided by 18 plus 1 divided by 8. 1 divided by 8 gives us, and now we deserve our drum roll, this is 56.5. So this is approximately equal to 56.5 meters. So you make this cut roughly 56.5."}, {"video_title": "Minimizing combined area Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So we have 12.5 divided by square root of 3 divided by 18 plus 1 divided by 8. 1 divided by 8 gives us, and now we deserve our drum roll, this is 56.5. So this is approximately equal to 56.5 meters. So you make this cut roughly 56.5. I'll write roughly. I'll make it little. 56.5 meters from the left-hand side."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And what we want to do is take its first and second derivatives and use as much of our techniques as we have at our disposal to attempt to graph it without a graphing calculator. If we have time, I'll take out the graphing calculator and see if our answer matches up. So a good place to start is to take the first derivative of this. So the derivative of f, well, you take the derivative of the inside, so take the derivative of that right there, which is 4x to the third. And then multiply it times the derivative of the outside with respect to the inside. So the derivative of natural log of x is 1 over x. So the derivative of this whole thing with respect to this inside expression is going to be times 1 over x to the fourth plus 27."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So the derivative of f, well, you take the derivative of the inside, so take the derivative of that right there, which is 4x to the third. And then multiply it times the derivative of the outside with respect to the inside. So the derivative of natural log of x is 1 over x. So the derivative of this whole thing with respect to this inside expression is going to be times 1 over x to the fourth plus 27. If you found that confusing, you might want to re-watch the chain rule videos. But that's the first derivative of our function. And I could rewrite this, this is equal to 4x to the third over x to the fourth plus 27, or I could write it as 4x to the third times x to the fourth plus 27 to the negative 1."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So the derivative of this whole thing with respect to this inside expression is going to be times 1 over x to the fourth plus 27. If you found that confusing, you might want to re-watch the chain rule videos. But that's the first derivative of our function. And I could rewrite this, this is equal to 4x to the third over x to the fourth plus 27, or I could write it as 4x to the third times x to the fourth plus 27 to the negative 1. All three of these expressions are equivalent. I'm just writing, I've multiplied it out, or I could write this as a negative exponent, or I could write this as a fraction with this in the denominator. They're all equivalent."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And I could rewrite this, this is equal to 4x to the third over x to the fourth plus 27, or I could write it as 4x to the third times x to the fourth plus 27 to the negative 1. All three of these expressions are equivalent. I'm just writing, I've multiplied it out, or I could write this as a negative exponent, or I could write this as a fraction with this in the denominator. They're all equivalent. So that's our first derivative. Let's do our second derivative. Our second derivative, this looks like it'll get a little bit hairier."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "They're all equivalent. So that's our first derivative. Let's do our second derivative. Our second derivative, this looks like it'll get a little bit hairier. So our second derivative is the derivative of this. So it's equal to, we can now use the product rule, it's the derivative of this first expression times the second expression. So the derivative of this first expression, 3 times 4 is 12."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Our second derivative, this looks like it'll get a little bit hairier. So our second derivative is the derivative of this. So it's equal to, we can now use the product rule, it's the derivative of this first expression times the second expression. So the derivative of this first expression, 3 times 4 is 12. 12x squared, we just decrement the 3 by 1, times the second expression, times x to the fourth plus 27 to the minus 1. And then to that we want to add just the first expression, not its derivative, so just 4x to the third, times the derivative of the second expression. So the second expression, we could take the derivative of the inside, which is just 4x to the third, this derivative of 27 is just 0."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So the derivative of this first expression, 3 times 4 is 12. 12x squared, we just decrement the 3 by 1, times the second expression, times x to the fourth plus 27 to the minus 1. And then to that we want to add just the first expression, not its derivative, so just 4x to the third, times the derivative of the second expression. So the second expression, we could take the derivative of the inside, which is just 4x to the third, this derivative of 27 is just 0. So times 4x to the third, times the derivative of this whole thing with respect to the inside. So times, so you take this exponent, put it out front. So times minus 1, times this whole thing, x to the fourth plus 27, to the, we decrement this by 1 more, so minus 2."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So the second expression, we could take the derivative of the inside, which is just 4x to the third, this derivative of 27 is just 0. So times 4x to the third, times the derivative of this whole thing with respect to the inside. So times, so you take this exponent, put it out front. So times minus 1, times this whole thing, x to the fourth plus 27, to the, we decrement this by 1 more, so minus 2. So let's see if I can simplify this expression a little bit. So this is equal to, so this right here is equal to 12x squared over this thing, x to the fourth plus 27. And then let's see, we're going to have a minus here, so it's minus, you multiply these two guys, 4 times 4 is 16."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So times minus 1, times this whole thing, x to the fourth plus 27, to the, we decrement this by 1 more, so minus 2. So let's see if I can simplify this expression a little bit. So this is equal to, so this right here is equal to 12x squared over this thing, x to the fourth plus 27. And then let's see, we're going to have a minus here, so it's minus, you multiply these two guys, 4 times 4 is 16. 16x to the third times x to the third is x to the sixth, over this thing squared, over x to the fourth plus 27 squared. That's just another way to rewrite that expression right there, right? To the minus 2, you can just put it in the denominator and make it to a positive 2 in the denominator."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And then let's see, we're going to have a minus here, so it's minus, you multiply these two guys, 4 times 4 is 16. 16x to the third times x to the third is x to the sixth, over this thing squared, over x to the fourth plus 27 squared. That's just another way to rewrite that expression right there, right? To the minus 2, you can just put it in the denominator and make it to a positive 2 in the denominator. Same thing. Now, if you've seen these problems in the past, we always want to set these things equal to 0. We want to solve for x equals 0, so it'll be useful to have this expressed as just one fraction, instead of the difference or the sum of two fractions."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "To the minus 2, you can just put it in the denominator and make it to a positive 2 in the denominator. Same thing. Now, if you've seen these problems in the past, we always want to set these things equal to 0. We want to solve for x equals 0, so it'll be useful to have this expressed as just one fraction, instead of the difference or the sum of two fractions. So what we could do is to have, we could have a common denominator, so we could multiply both the numerator and denominator of this expression by x to the fourth plus 27, and what do we get? So this is equal to, so if we multiply this first expression times x to the fourth plus 27, we get 12x squared times x to the fourth plus 27, and then in the denominator, you have x to the fourth plus 27 squared. All I did, I multiplied this numerator and this denominator by x to the fourth plus 27, I didn't change it."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "We want to solve for x equals 0, so it'll be useful to have this expressed as just one fraction, instead of the difference or the sum of two fractions. So what we could do is to have, we could have a common denominator, so we could multiply both the numerator and denominator of this expression by x to the fourth plus 27, and what do we get? So this is equal to, so if we multiply this first expression times x to the fourth plus 27, we get 12x squared times x to the fourth plus 27, and then in the denominator, you have x to the fourth plus 27 squared. All I did, I multiplied this numerator and this denominator by x to the fourth plus 27, I didn't change it. And then we have that second term, minus 16x to the sixth over x to the fourth plus 27 squared. The whole reason why I did that, now I have a common denominator, now I can just add the numerators. So this is going to be equal to, this is going to be equal to, let's see, let's multiply, well, the denominator, we know what the denominator is."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "All I did, I multiplied this numerator and this denominator by x to the fourth plus 27, I didn't change it. And then we have that second term, minus 16x to the sixth over x to the fourth plus 27 squared. The whole reason why I did that, now I have a common denominator, now I can just add the numerators. So this is going to be equal to, this is going to be equal to, let's see, let's multiply, well, the denominator, we know what the denominator is. It is x to the fourth plus 27 squared, that's our denominator, and then we can multiply this out. This is 12x squared times x to the fourth, so that's 12x to the sixth plus 27 times 12, I don't even feel like multiplying 27 times 12, so I'll just write that out. So plus 27 times 12x squared, I just multiply the 12x squared times 27, and then minus 16x to the sixth."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So this is going to be equal to, this is going to be equal to, let's see, let's multiply, well, the denominator, we know what the denominator is. It is x to the fourth plus 27 squared, that's our denominator, and then we can multiply this out. This is 12x squared times x to the fourth, so that's 12x to the sixth plus 27 times 12, I don't even feel like multiplying 27 times 12, so I'll just write that out. So plus 27 times 12x squared, I just multiply the 12x squared times 27, and then minus 16x to the sixth. And this simplifies to, let's see if I can simplify this even further. So I have an x to the sixth here, x to the sixth here, so this is equal to, I'll do this in pink, this is equal to the 27 times 12x squared, I don't feel like figuring that out right now, times 12x squared, and then you have minus 16x to the sixth and plus 12x to the sixth. So you add those two, you get minus 4, 12 minus 16 is minus 4, x to the sixth, all of that over x to the fourth plus 27 squared."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So plus 27 times 12x squared, I just multiply the 12x squared times 27, and then minus 16x to the sixth. And this simplifies to, let's see if I can simplify this even further. So I have an x to the sixth here, x to the sixth here, so this is equal to, I'll do this in pink, this is equal to the 27 times 12x squared, I don't feel like figuring that out right now, times 12x squared, and then you have minus 16x to the sixth and plus 12x to the sixth. So you add those two, you get minus 4, 12 minus 16 is minus 4, x to the sixth, all of that over x to the fourth plus 27 squared. And that is our second derivative. Now, we've done all of the derivatives, and this was actually a pretty hairy problem, and now we can solve for when the first and the second derivatives equal 0, and we'll have our candidate, well we'll know our critical points, and then we'll have our candidate inflection points and see if we can make any headway from there. So first let's see where our first derivative is equal to 0 and get our critical points, or at least maybe, also maybe where it's undefined."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So you add those two, you get minus 4, 12 minus 16 is minus 4, x to the sixth, all of that over x to the fourth plus 27 squared. And that is our second derivative. Now, we've done all of the derivatives, and this was actually a pretty hairy problem, and now we can solve for when the first and the second derivatives equal 0, and we'll have our candidate, well we'll know our critical points, and then we'll have our candidate inflection points and see if we can make any headway from there. So first let's see where our first derivative is equal to 0 and get our critical points, or at least maybe, also maybe where it's undefined. So this is equal to 0, if we want to set, the only place that this can equal to 0 is if this numerator is equal to 0. This denominator actually, if we are assuming we're dealing with real numbers, this term right here is always going to be greater than or equal to 0 for any value of x, because it's an even exponent, so this thing can never equal 0, right, because you're adding 27 to something that's non-negative, so this will never equal 0, so this will also never be undefined. So there's no undefined critical points here, but we can set the numerator equal to 0 pretty easily."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So first let's see where our first derivative is equal to 0 and get our critical points, or at least maybe, also maybe where it's undefined. So this is equal to 0, if we want to set, the only place that this can equal to 0 is if this numerator is equal to 0. This denominator actually, if we are assuming we're dealing with real numbers, this term right here is always going to be greater than or equal to 0 for any value of x, because it's an even exponent, so this thing can never equal 0, right, because you're adding 27 to something that's non-negative, so this will never equal 0, so this will also never be undefined. So there's no undefined critical points here, but we can set the numerator equal to 0 pretty easily. If we wanted to set this equal to 0, we just say 4x to the third is equal to 0, and we know what x value will make that equal to 0, x has to be equal to 0. 4 times something to the third is equal to 0, that something has to be 0. x to the third has to be 0, x has to be 0. So we can write f prime of 0 is equal to 0."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So there's no undefined critical points here, but we can set the numerator equal to 0 pretty easily. If we wanted to set this equal to 0, we just say 4x to the third is equal to 0, and we know what x value will make that equal to 0, x has to be equal to 0. 4 times something to the third is equal to 0, that something has to be 0. x to the third has to be 0, x has to be 0. So we can write f prime of 0 is equal to 0. So 0 is a critical point. The slope at 0 is 0, we don't know if it's a maximum or a minimum or an inflection point yet, or it could be, you know, we'll explore it a little bit more. And actually just so we get the coordinate, what's the coordinate?"}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So we can write f prime of 0 is equal to 0. So 0 is a critical point. The slope at 0 is 0, we don't know if it's a maximum or a minimum or an inflection point yet, or it could be, you know, we'll explore it a little bit more. And actually just so we get the coordinate, what's the coordinate? The coordinate x is 0, and then y is the natural log of x is 0, this just turns out, so it's the natural log of 27. So it's the natural log of, let me figure out what that is and get the calculator out. I said I wouldn't use a graphing calculator, but I could use a regular calculator."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And actually just so we get the coordinate, what's the coordinate? The coordinate x is 0, and then y is the natural log of x is 0, this just turns out, so it's the natural log of 27. So it's the natural log of, let me figure out what that is and get the calculator out. I said I wouldn't use a graphing calculator, but I could use a regular calculator. 27, if I were to take the natural log of that, it's like 3 point, well for our purpose this is called 3.3, we're just trying to get the general shape of the graph. So 3.3, 3 point, well we could just say 2.9 and it kept going. So this is a critical point right here, the slope is 0 here."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "I said I wouldn't use a graphing calculator, but I could use a regular calculator. 27, if I were to take the natural log of that, it's like 3 point, well for our purpose this is called 3.3, we're just trying to get the general shape of the graph. So 3.3, 3 point, well we could just say 2.9 and it kept going. So this is a critical point right here, the slope is 0 here. Slope is equal to 0 at x is equal to 0. So this is one thing we want to block off. And let's see if we can find any candidate inflection points."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So this is a critical point right here, the slope is 0 here. Slope is equal to 0 at x is equal to 0. So this is one thing we want to block off. And let's see if we can find any candidate inflection points. And remember, candidate inflection points are where the second derivative equals 0. Now if the second derivative equals 0, that doesn't tell us that those are definitely inflection points. Let me make this very clear."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And let's see if we can find any candidate inflection points. And remember, candidate inflection points are where the second derivative equals 0. Now if the second derivative equals 0, that doesn't tell us that those are definitely inflection points. Let me make this very clear. If x is inflection, then the second derivative at x is going to be equal to 0. Because you're having a change in the concavity, I always say con-ca-ti-vity, but it's the con-cavity. You have a change in the slope goes from either increasing to decreasing or from decreasing to increasing."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Let me make this very clear. If x is inflection, then the second derivative at x is going to be equal to 0. Because you're having a change in the concavity, I always say con-ca-ti-vity, but it's the con-cavity. You have a change in the slope goes from either increasing to decreasing or from decreasing to increasing. But if the derivative is equal to 0, you cannot assume that it's an inflection point. So what we're going to do is, we're going to find all of the points at which this is true, and then see if we actually do have a sign change in the second derivative at that point. And only if you have a sign change, then you can say it's an inflection point."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "You have a change in the slope goes from either increasing to decreasing or from decreasing to increasing. But if the derivative is equal to 0, you cannot assume that it's an inflection point. So what we're going to do is, we're going to find all of the points at which this is true, and then see if we actually do have a sign change in the second derivative at that point. And only if you have a sign change, then you can say it's an inflection point. So let's see if we can do that. So just because the second derivative is 0, that by itself does not tell you it's an inflection point. It has to have a second derivative of 0, and when you go below that x, the second derivative has to actually change signs."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And only if you have a sign change, then you can say it's an inflection point. So let's see if we can do that. So just because the second derivative is 0, that by itself does not tell you it's an inflection point. It has to have a second derivative of 0, and when you go below that x, the second derivative has to actually change signs. Only then. So we can say if f' changes signs around x, then we can say that x is an inflection. And if it's changing signs around x, then it's definitely going to be 0 right at x."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "It has to have a second derivative of 0, and when you go below that x, the second derivative has to actually change signs. Only then. So we can say if f' changes signs around x, then we can say that x is an inflection. And if it's changing signs around x, then it's definitely going to be 0 right at x. But you have to actually see that if it's negative before x, it has to be positive after x, or if it's positive before x, it has to be negative after x. So let's test that out. So the first thing we need to do is find these candidate points."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And if it's changing signs around x, then it's definitely going to be 0 right at x. But you have to actually see that if it's negative before x, it has to be positive after x, or if it's positive before x, it has to be negative after x. So let's test that out. So the first thing we need to do is find these candidate points. Remember, the candidate points are where the second derivative is equal to 0. We're going to find those points and then see if this is true, that the sign actually changes. So we want to find where this thing over here is equal to 0."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So the first thing we need to do is find these candidate points. Remember, the candidate points are where the second derivative is equal to 0. We're going to find those points and then see if this is true, that the sign actually changes. So we want to find where this thing over here is equal to 0. And once again, for this to be equal to 0, the numerator has to be equal to 0. This denominator can never be equal to 0 if we're dealing with real numbers, which I think is a fair assumption. So let's see where our numerator can be equal to 0 for the second derivative."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So we want to find where this thing over here is equal to 0. And once again, for this to be equal to 0, the numerator has to be equal to 0. This denominator can never be equal to 0 if we're dealing with real numbers, which I think is a fair assumption. So let's see where our numerator can be equal to 0 for the second derivative. So let's set the numerator of the second derivative. 27 times 12 x squared minus 4x to the 6th is equal to 0. Remember, that's just the numerator of our second derivative."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So let's see where our numerator can be equal to 0 for the second derivative. So let's set the numerator of the second derivative. 27 times 12 x squared minus 4x to the 6th is equal to 0. Remember, that's just the numerator of our second derivative. Any x that makes the numerator 0 is making the second derivative 0. So let's factor out a 4x squared. Now we'll have 27 times, if we factor a 4 out of the 12, we'll just get a 3, and we factored out the x squared, minus, we factored out the 4, we factored out an x squared, so we have x to the 4th is equal to 0."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Remember, that's just the numerator of our second derivative. Any x that makes the numerator 0 is making the second derivative 0. So let's factor out a 4x squared. Now we'll have 27 times, if we factor a 4 out of the 12, we'll just get a 3, and we factored out the x squared, minus, we factored out the 4, we factored out an x squared, so we have x to the 4th is equal to 0. So the x's that will make this equal to 0 will satisfy either 4x squared is equal to 0, or, now 27 times 3, I can do that in my head, that's 81, 20 times 3 is 60, 7 times 3 is 21, 60 plus 21 is 81. Or 81 minus x to the 4th is equal to 0. Any x that satisfies either of these will make this entire expression equal to 0, so if this thing is 0, the whole thing is going to be equal to 0."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Now we'll have 27 times, if we factor a 4 out of the 12, we'll just get a 3, and we factored out the x squared, minus, we factored out the 4, we factored out an x squared, so we have x to the 4th is equal to 0. So the x's that will make this equal to 0 will satisfy either 4x squared is equal to 0, or, now 27 times 3, I can do that in my head, that's 81, 20 times 3 is 60, 7 times 3 is 21, 60 plus 21 is 81. Or 81 minus x to the 4th is equal to 0. Any x that satisfies either of these will make this entire expression equal to 0, so if this thing is 0, the whole thing is going to be equal to 0. If this thing is 0, the whole thing is going to be equal to 0. Let me be clear, this is 81 right there. So let's solve this."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Any x that satisfies either of these will make this entire expression equal to 0, so if this thing is 0, the whole thing is going to be equal to 0. If this thing is 0, the whole thing is going to be equal to 0. Let me be clear, this is 81 right there. So let's solve this. This is going to be 0 when x is equal to 0 itself. This is going to be equal to 0 when, let's see, if we add x to the 4th to both sides, you get x to the 4th is equal to 81. If we take the square root of both sides of this, you get x squared is equal to 9, or so you get x is plus or minus 3. x is equal to plus or minus 3."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So let's solve this. This is going to be 0 when x is equal to 0 itself. This is going to be equal to 0 when, let's see, if we add x to the 4th to both sides, you get x to the 4th is equal to 81. If we take the square root of both sides of this, you get x squared is equal to 9, or so you get x is plus or minus 3. x is equal to plus or minus 3. These are our candidate inflection points. x is equal to 0, x is equal to plus 3, or x is equal to minus 3. What we have to do now is to see whether the second derivative changes signs around these points in order to be able to label them inflection points."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "If we take the square root of both sides of this, you get x squared is equal to 9, or so you get x is plus or minus 3. x is equal to plus or minus 3. These are our candidate inflection points. x is equal to 0, x is equal to plus 3, or x is equal to minus 3. What we have to do now is to see whether the second derivative changes signs around these points in order to be able to label them inflection points. What happens when x is slightly below 0? Let's take the situation, let's do all the scenarios. What happens when x is slightly below 0?"}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "What we have to do now is to see whether the second derivative changes signs around these points in order to be able to label them inflection points. What happens when x is slightly below 0? Let's take the situation, let's do all the scenarios. What happens when x is slightly below 0? Not all of them necessarily, but if x is like 0.1, what is the second derivative going to be doing? If x is minus 0.1, this term right here is going to be positive, and then this is going to be 81 minus 0.1 to the 4th. That's going to be a very small number."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "What happens when x is slightly below 0? Not all of them necessarily, but if x is like 0.1, what is the second derivative going to be doing? If x is minus 0.1, this term right here is going to be positive, and then this is going to be 81 minus 0.1 to the 4th. That's going to be a very small number. It's going to be some positive number times 81 minus a small number, so it's going to be a positive number. When x is less than 0, or just slightly less than 0, our second derivative is positive. What happens when x is slightly larger?"}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "That's going to be a very small number. It's going to be some positive number times 81 minus a small number, so it's going to be a positive number. When x is less than 0, or just slightly less than 0, our second derivative is positive. What happens when x is slightly larger? When I write this notation, I want to be careful. I mean really just right below 0. When x is right above 0, what happens?"}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "What happens when x is slightly larger? When I write this notation, I want to be careful. I mean really just right below 0. When x is right above 0, what happens? Let's say x was 0.01, or 0.1, positive 0.1. It's going to be the same thing, because in both cases we're squaring and we're taking the 4th, so you're kind of losing your sign information. If x is 0.1, this thing is going to be a small positive number."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "When x is right above 0, what happens? Let's say x was 0.01, or 0.1, positive 0.1. It's going to be the same thing, because in both cases we're squaring and we're taking the 4th, so you're kind of losing your sign information. If x is 0.1, this thing is going to be a small positive number. You're going to be subtracting a very small positive number from 81, but 81 minus a small number is still going to be positive. You're going to have a positive times a positive, so your second derivative is still going to be greater than 0. Something interesting here."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "If x is 0.1, this thing is going to be a small positive number. You're going to be subtracting a very small positive number from 81, but 81 minus a small number is still going to be positive. You're going to have a positive times a positive, so your second derivative is still going to be greater than 0. Something interesting here. Your second derivative is 0 when x is equal to 0, but it is not an inflection point, because notice the concavity did not change around 0. Our second derivative is positive as we approach 0 from the left, and it's positive as we approach 0 from the right. In general, at 0, as we're near 0 from either direction, we're going to be concave upwards."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Something interesting here. Your second derivative is 0 when x is equal to 0, but it is not an inflection point, because notice the concavity did not change around 0. Our second derivative is positive as we approach 0 from the left, and it's positive as we approach 0 from the right. In general, at 0, as we're near 0 from either direction, we're going to be concave upwards. The fact that 0 is a critical point, and that we're always concave upward as we approach 0 from either side, this tells us that this is a minimum point. Because we're concave upwards all around 0. 0 is not an inflection point."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "In general, at 0, as we're near 0 from either direction, we're going to be concave upwards. The fact that 0 is a critical point, and that we're always concave upward as we approach 0 from either side, this tells us that this is a minimum point. Because we're concave upwards all around 0. 0 is not an inflection point. Let's see if positive and negative 3 are inflection points. If you study this equation, let me write our... Actually, I just want to be clear. I've just been using the numerator of the second derivative."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "0 is not an inflection point. Let's see if positive and negative 3 are inflection points. If you study this equation, let me write our... Actually, I just want to be clear. I've just been using the numerator of the second derivative. The whole second derivative is this thing right here, but I've been ignoring the denominator, because the denominator is always positive. If we're trying to understand whether things are positive or negative, we just really have to determine whether the numerator is positive or negative, because this expression right here is always positive. It's something to the second power."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "I've just been using the numerator of the second derivative. The whole second derivative is this thing right here, but I've been ignoring the denominator, because the denominator is always positive. If we're trying to understand whether things are positive or negative, we just really have to determine whether the numerator is positive or negative, because this expression right here is always positive. It's something to the second power. Let's test whether we have a change in concavity around x is equal to positive or negative 3. Remember, the numerator of our... Let me just rewrite our second derivative, just so you see it here. f prime prime of x."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "It's something to the second power. Let's test whether we have a change in concavity around x is equal to positive or negative 3. Remember, the numerator of our... Let me just rewrite our second derivative, just so you see it here. f prime prime of x. The numerator is this thing right here. It's 4x squared times 81 minus x to the fourth. The denominator was up here, x to the fourth plus 27 squared."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "f prime prime of x. The numerator is this thing right here. It's 4x squared times 81 minus x to the fourth. The denominator was up here, x to the fourth plus 27 squared. x to the fourth plus 27 squared. That was our second derivative. Let's see if this changes signs around positive or negative 3."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "The denominator was up here, x to the fourth plus 27 squared. x to the fourth plus 27 squared. That was our second derivative. Let's see if this changes signs around positive or negative 3. Actually, we should get the same answer, because regardless of whether we put positive or negative 3 here, you lose all your sign information, because you're taking it to the fourth power. You're taking it to the second power. Obviously, anything to the fourth power is always going to be positive."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Let's see if this changes signs around positive or negative 3. Actually, we should get the same answer, because regardless of whether we put positive or negative 3 here, you lose all your sign information, because you're taking it to the fourth power. You're taking it to the second power. Obviously, anything to the fourth power is always going to be positive. Anything to the second power is always going to be negative. When we do our test, if it's true for positive 3, it's probably going to be true for negative 3 as well. Let's just try it out."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Obviously, anything to the fourth power is always going to be positive. Anything to the second power is always going to be negative. When we do our test, if it's true for positive 3, it's probably going to be true for negative 3 as well. Let's just try it out. When x is just a little bit less than positive 3, what's the sign of f prime prime of x? It's going to be 4 times 9. It's going to be 4 times a positive number."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Let's just try it out. When x is just a little bit less than positive 3, what's the sign of f prime prime of x? It's going to be 4 times 9. It's going to be 4 times a positive number. It might be like 2.999, but this is still going to be positive. This is going to be positive when x is approaching 3. Then this is going to be, well, if x is 3, this is 0."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "It's going to be 4 times a positive number. It might be like 2.999, but this is still going to be positive. This is going to be positive when x is approaching 3. Then this is going to be, well, if x is 3, this is 0. If x is a little bit less than 3, if it's like 2.9999, this number is going to be less than 81. This is also going to be positive. Of course, the denominator is always positive."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Then this is going to be, well, if x is 3, this is 0. If x is a little bit less than 3, if it's like 2.9999, this number is going to be less than 81. This is also going to be positive. Of course, the denominator is always positive. As x is less than 3, as it's approaching from the left, we are concave upwards. This thing is going to be a positive. f prime prime is greater than 0."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Of course, the denominator is always positive. As x is less than 3, as it's approaching from the left, we are concave upwards. This thing is going to be a positive. f prime prime is greater than 0. We are upwards, concave upwards. When x is just larger than 3, what's going to happen? This first term is still going to be positive, but if x is just larger than 3, x to the fourth is going to be just larger than 81."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "f prime prime is greater than 0. We are upwards, concave upwards. When x is just larger than 3, what's going to happen? This first term is still going to be positive, but if x is just larger than 3, x to the fourth is going to be just larger than 81. This second term is going to be negative in that situation. It's going to be negative. Let me do it in a new color."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "This first term is still going to be positive, but if x is just larger than 3, x to the fourth is going to be just larger than 81. This second term is going to be negative in that situation. It's going to be negative. Let me do it in a new color. It's going to be negative when x is larger than 3 because this is going to be larger than 81. If this is negative and this is positive, then the whole thing is going to be negative because this denominator is still going to be positive. Then f prime prime is going to be less than 0."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Let me do it in a new color. It's going to be negative when x is larger than 3 because this is going to be larger than 81. If this is negative and this is positive, then the whole thing is going to be negative because this denominator is still going to be positive. Then f prime prime is going to be less than 0. We're going to be concave downwards. One last one. What happens when x is just greater than minus 3?"}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Then f prime prime is going to be less than 0. We're going to be concave downwards. One last one. What happens when x is just greater than minus 3? Just being greater than minus 3, that's like minus 2.9999. When you take minus 2.99 and square it, you're going to get a positive number. This is going to be positive."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "What happens when x is just greater than minus 3? Just being greater than minus 3, that's like minus 2.9999. When you take minus 2.99 and square it, you're going to get a positive number. This is going to be positive. If you take minus 2.99 to the fourth, that's going to be a little bit less than 81 because 2.99 to the fourth is a little bit less than 81. This is still going to be positive. You have positive times a positive divided by a positive."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "This is going to be positive. If you take minus 2.99 to the fourth, that's going to be a little bit less than 81 because 2.99 to the fourth is a little bit less than 81. This is still going to be positive. You have positive times a positive divided by a positive. You're going to be concave upwards because your second derivative is going to be greater than 0. Concave upwards. Finally, when x is just less than negative 3, remember when I write this, I don't mean for all x's larger than negative 3 or all x's smaller than negative 3."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "You have positive times a positive divided by a positive. You're going to be concave upwards because your second derivative is going to be greater than 0. Concave upwards. Finally, when x is just less than negative 3, remember when I write this, I don't mean for all x's larger than negative 3 or all x's smaller than negative 3. There's actually no notation that would say just as we just approach 3, in this case, from the left. What happens if we just go to minus 3.11 or 3.01? This term right here is going to be positive."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Finally, when x is just less than negative 3, remember when I write this, I don't mean for all x's larger than negative 3 or all x's smaller than negative 3. There's actually no notation that would say just as we just approach 3, in this case, from the left. What happens if we just go to minus 3.11 or 3.01? This term right here is going to be positive. If we take minus 3.1 to the fourth, that's going to be larger than positive 81. The sign will become positive. It will be larger than 81, so this will become negative."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "This term right here is going to be positive. If we take minus 3.1 to the fourth, that's going to be larger than positive 81. The sign will become positive. It will be larger than 81, so this will become negative. In that case as well, we'll have a positive times a negative divided by a positive. Then our second derivative is going to be negative. We're going to be downwards."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "It will be larger than 81, so this will become negative. In that case as well, we'll have a positive times a negative divided by a positive. Then our second derivative is going to be negative. We're going to be downwards. I think we're ready to plot. First of all, is x plus or minus 3 inflection points? Sure."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "We're going to be downwards. I think we're ready to plot. First of all, is x plus or minus 3 inflection points? Sure. As we approach x is equal to 3 from the left, we are concave upwards. Then as we cross 3, the second derivative is 0. The second derivative is 0."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Sure. As we approach x is equal to 3 from the left, we are concave upwards. Then as we cross 3, the second derivative is 0. The second derivative is 0. I lost it up here. The second derivative is 0. Then as we go to the right of 3, we become concave downwards."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "The second derivative is 0. I lost it up here. The second derivative is 0. Then as we go to the right of 3, we become concave downwards. We got our sign change in the second derivative. x is equal to 3. This 3 is definitely an inflection point."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Then as we go to the right of 3, we become concave downwards. We got our sign change in the second derivative. x is equal to 3. This 3 is definitely an inflection point. The same argument can be made for negative 3. We switch signs as we cross 3. These definitely are inflection points."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "This 3 is definitely an inflection point. The same argument can be made for negative 3. We switch signs as we cross 3. These definitely are inflection points. Just so we get the exact coordinates, let's figure out what f of 3 is, or f of positive and negative 3. Then we're ready to graph. First of all, we know that the point 0,3.29, that this was a minimum."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "These definitely are inflection points. Just so we get the exact coordinates, let's figure out what f of 3 is, or f of positive and negative 3. Then we're ready to graph. First of all, we know that the point 0,3.29, that this was a minimum. Because 0 is a critical point, the slope is 0 there. Because it's concave upwards, all around 0. So 0 is definitely not an inflection point."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "First of all, we know that the point 0,3.29, that this was a minimum. Because 0 is a critical point, the slope is 0 there. Because it's concave upwards, all around 0. So 0 is definitely not an inflection point. Then we know that the points positive 3 and minus 3 are inflection points. In order to figure out their y coordinates, we can just evaluate them. They're actually going to have the same y coordinates."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So 0 is definitely not an inflection point. Then we know that the points positive 3 and minus 3 are inflection points. In order to figure out their y coordinates, we can just evaluate them. They're actually going to have the same y coordinates. Because if you put a minus 3 or positive 3 and take it to the fourth power, you're going to get the same thing. Let's figure out what they are. If we take 3 to the fourth power, that's 81."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "They're actually going to have the same y coordinates. Because if you put a minus 3 or positive 3 and take it to the fourth power, you're going to get the same thing. Let's figure out what they are. If we take 3 to the fourth power, that's 81. 81 plus 27 is equal to 108. Then we want to take the natural log of it. We want to take the natural log."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "If we take 3 to the fourth power, that's 81. 81 plus 27 is equal to 108. Then we want to take the natural log of it. We want to take the natural log. It's like 4.7, just to get a rough idea. That's true of whether we do positive or negative 3, because we took to the fourth power. So it's 4.7."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "We want to take the natural log. It's like 4.7, just to get a rough idea. That's true of whether we do positive or negative 3, because we took to the fourth power. So it's 4.7. These are both inflection points. We should be ready to graph it. Let's graph it."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So it's 4.7. These are both inflection points. We should be ready to graph it. Let's graph it. Let me draw my axis. Just like that. This is my y-axis."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Let's graph it. Let me draw my axis. Just like that. This is my y-axis. This is my x-axis. This is y. You can even call it the f of x axis, if you like."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "This is my y-axis. This is my x-axis. This is y. You can even call it the f of x axis, if you like. This is x. The point 0, 3.29. Let's say this is 1, 2, 3, 4, 5."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "You can even call it the f of x axis, if you like. This is x. The point 0, 3.29. Let's say this is 1, 2, 3, 4, 5. The point 0, 3.29. That's 0, 1, 2, 3. A little bit above 3 is right there."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Let's say this is 1, 2, 3, 4, 5. The point 0, 3.29. That's 0, 1, 2, 3. A little bit above 3 is right there. That's a minimum point. Then we're concave. The slope is 0 right there."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "A little bit above 3 is right there. That's a minimum point. Then we're concave. The slope is 0 right there. We figured that out, because the first derivative was 0 there. It's a critical point. It's concave upwards around there."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "The slope is 0 right there. We figured that out, because the first derivative was 0 there. It's a critical point. It's concave upwards around there. That told us we're at a minimum point right there. Then at positive 3. 1, 2, 3."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "It's concave upwards around there. That told us we're at a minimum point right there. Then at positive 3. 1, 2, 3. At positive 3, 4.7. 4.7 will look something like that. We have an inflection point."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "1, 2, 3. At positive 3, 4.7. 4.7 will look something like that. We have an inflection point. Before that, we're concave upwards. Then after that, we're concave downwards. It looks something like this."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "We have an inflection point. Before that, we're concave upwards. Then after that, we're concave downwards. It looks something like this. We're concave upwards up to that point. Maybe, actually, you should let me ignore that yellow thing I drew before. Let me get rid of that."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "It looks something like this. We're concave upwards up to that point. Maybe, actually, you should let me ignore that yellow thing I drew before. Let me get rid of that. Let me draw it like this. 1, 2, 3. 3, 4.7 looks like that."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Let me get rid of that. Let me draw it like this. 1, 2, 3. 3, 4.7 looks like that. Minus 3, 4.7. 1, 2, 3. 4.7 looks like that."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "3, 4.7 looks like that. Minus 3, 4.7. 1, 2, 3. 4.7 looks like that. We know at 0, we have a slope of 0. We're concave upwards. We look like this."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "4.7 looks like that. We know at 0, we have a slope of 0. We're concave upwards. We look like this. We're concave upwards until x is equal to 3. Then at x equals 3, we become concave downwards. We become concave downwards."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "We look like this. We're concave upwards until x is equal to 3. Then at x equals 3, we become concave downwards. We become concave downwards. Let me try my best to draw it well. We go off like that. Then we're concave upwards around 0."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "We become concave downwards. Let me try my best to draw it well. We go off like that. Then we're concave upwards around 0. Until we get, we're concave upwards as long as x is greater than minus 3. Then at minus 3, we become concave downwards again. Maybe, I should do it in that color."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Then we're concave upwards around 0. Until we get, we're concave upwards as long as x is greater than minus 3. Then at minus 3, we become concave downwards again. Maybe, I should do it in that color. This concave downwards right here, that's this right here. That's that right there. This concave downwards right here."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Maybe, I should do it in that color. This concave downwards right here, that's this right here. That's that right there. This concave downwards right here. Sorry, I meant to do it in that red color. This concave downwards right here is this right there. Then the concave upwards around 0 is right there."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "This concave downwards right here. Sorry, I meant to do it in that red color. This concave downwards right here is this right there. Then the concave upwards around 0 is right there. You could even imagine this concave upwards that we measured. That's this concave upwards. This concave upwards is that."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Then the concave upwards around 0 is right there. You could even imagine this concave upwards that we measured. That's this concave upwards. This concave upwards is that. Then around 0, we're always upwards. This is my sense of what the graph will look like. Maybe, it turns into, you could think about what it does."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "This concave upwards is that. Then around 0, we're always upwards. This is my sense of what the graph will look like. Maybe, it turns into, you could think about what it does. As x approaches positive or negative infinity, some of the terms. I won't go into that. Let's test whether we've graphed it correctly using a graphing calculator."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Maybe, it turns into, you could think about what it does. As x approaches positive or negative infinity, some of the terms. I won't go into that. Let's test whether we've graphed it correctly using a graphing calculator. Let me get out my trusty TI-85 and let's graph this sucker. Press graph, y equals, it was the natural log of x to the fourth plus 27. I want to get that graph there."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Let's test whether we've graphed it correctly using a graphing calculator. Let me get out my trusty TI-85 and let's graph this sucker. Press graph, y equals, it was the natural log of x to the fourth plus 27. I want to get that graph there. I do second graph. Let's cross our fingers. It looks pretty good."}, {"video_title": "Another example graphing with derivatives Differential Calculus Khan Academy.mp3", "Sentence": "I want to get that graph there. I do second graph. Let's cross our fingers. It looks pretty good. It looks almost exactly like what we drew. I think our mathematics was correct. This was actually very satisfying."}, {"video_title": "Definite integral of trig function AP Calculus AB Khan Academy.mp3", "Sentence": "So the first thing, let's see if we can take the antiderivative of nine sine of x. So we could use some of our integration properties to simplify this a little bit. So this is going to be equal to, this is the same thing as nine times the integral from 11 pi over two to six pi of sine of x dx. And what's the antiderivative of sine of x? Well we know from our derivatives that the derivative with respect to x of cosine of x is equal to negative sine of x. Negative sine of x. So can we construct this in some way so this is a negative sine of x?"}, {"video_title": "Definite integral of trig function AP Calculus AB Khan Academy.mp3", "Sentence": "And what's the antiderivative of sine of x? Well we know from our derivatives that the derivative with respect to x of cosine of x is equal to negative sine of x. Negative sine of x. So can we construct this in some way so this is a negative sine of x? Well what if I multiplied on the inside, what if I multiplied it by negative one? Well I can't just multiply it only one place by negative one I need to multiply by negative one twice so I'm not changing its value. So what if I said negative nine times negative sine of x?"}, {"video_title": "Definite integral of trig function AP Calculus AB Khan Academy.mp3", "Sentence": "So can we construct this in some way so this is a negative sine of x? Well what if I multiplied on the inside, what if I multiplied it by negative one? Well I can't just multiply it only one place by negative one I need to multiply by negative one twice so I'm not changing its value. So what if I said negative nine times negative sine of x? Well this is still gonna be nine sine of x. If you took negative nine times negative sine of x it is nine sine of x. And I did it this way because now negative sine of x it matches the derivative of cosine of x."}, {"video_title": "Definite integral of trig function AP Calculus AB Khan Academy.mp3", "Sentence": "So what if I said negative nine times negative sine of x? Well this is still gonna be nine sine of x. If you took negative nine times negative sine of x it is nine sine of x. And I did it this way because now negative sine of x it matches the derivative of cosine of x. So we could say that this is all going to be equal to, it's all going to be equal to, you have your negative nine out front, negative nine times, times, and I'll put it in brackets, negative nine times the antiderivative of negative sine of x. Well that is just going to be cosine of x. Cosine of x. And we're going to evaluate it at its bounds."}, {"video_title": "Definite integral of trig function AP Calculus AB Khan Academy.mp3", "Sentence": "And I did it this way because now negative sine of x it matches the derivative of cosine of x. So we could say that this is all going to be equal to, it's all going to be equal to, you have your negative nine out front, negative nine times, times, and I'll put it in brackets, negative nine times the antiderivative of negative sine of x. Well that is just going to be cosine of x. Cosine of x. And we're going to evaluate it at its bounds. We're going to evaluate it at six pi. Let me do that in a color I haven't used yet. We're gonna do that at six pi."}, {"video_title": "Definite integral of trig function AP Calculus AB Khan Academy.mp3", "Sentence": "And we're going to evaluate it at its bounds. We're going to evaluate it at six pi. Let me do that in a color I haven't used yet. We're gonna do that at six pi. And we're also going to do that at 11 pi over two. 11 pi over two. And so this is going to be equal to, this is equal to negative nine times, I'm gonna create some space here."}, {"video_title": "Definite integral of trig function AP Calculus AB Khan Academy.mp3", "Sentence": "We're gonna do that at six pi. And we're also going to do that at 11 pi over two. 11 pi over two. And so this is going to be equal to, this is equal to negative nine times, I'm gonna create some space here. So, actually that's probably more space than I need. It's going to be cosine of six pi. Cosine of six, six pi."}, {"video_title": "Definite integral of trig function AP Calculus AB Khan Academy.mp3", "Sentence": "And so this is going to be equal to, this is equal to negative nine times, I'm gonna create some space here. So, actually that's probably more space than I need. It's going to be cosine of six pi. Cosine of six, six pi. Cosine of six pi minus, minus cosine of 11 pi over two. Cosine of 11 pi over two. Well what is cosine of six pi going to be?"}, {"video_title": "Definite integral of trig function AP Calculus AB Khan Academy.mp3", "Sentence": "Cosine of six, six pi. Cosine of six pi minus, minus cosine of 11 pi over two. Cosine of 11 pi over two. Well what is cosine of six pi going to be? Well, cosine of any multiple of two pi is going to be equal to one. You could use six pi as we're going around the unit circle three times. So, this is the same thing as cosine of two pi or the same thing as cosine of zero."}, {"video_title": "Definite integral of trig function AP Calculus AB Khan Academy.mp3", "Sentence": "Well what is cosine of six pi going to be? Well, cosine of any multiple of two pi is going to be equal to one. You could use six pi as we're going around the unit circle three times. So, this is the same thing as cosine of two pi or the same thing as cosine of zero. So that is going to be equal to one. If that seems unfamiliar to you, I encourage you to review the unit circle definition of cosine. And what is cosine of 11 pi over two?"}, {"video_title": "Definite integral of trig function AP Calculus AB Khan Academy.mp3", "Sentence": "So, this is the same thing as cosine of two pi or the same thing as cosine of zero. So that is going to be equal to one. If that seems unfamiliar to you, I encourage you to review the unit circle definition of cosine. And what is cosine of 11 pi over two? Let's see, let's subtract some, let's subtract some multiple of two pi here to put it in values that we can understand better. So this is, so let me write it here. Cosine of 11 pi over two."}, {"video_title": "Definite integral of trig function AP Calculus AB Khan Academy.mp3", "Sentence": "And what is cosine of 11 pi over two? Let's see, let's subtract some, let's subtract some multiple of two pi here to put it in values that we can understand better. So this is, so let me write it here. Cosine of 11 pi over two. That is the same thing as, let's see, if we were to subtract, this is the same thing as cosine of 11 pi over two minus, let's see, this is the same thing as five and 1 1\u20442 pi. Right? Yeah, so this is, so we could view this as, we could subtract, let's subtract four pi, which is going to be, we could write that as eight pi over two."}, {"video_title": "Definite integral of trig function AP Calculus AB Khan Academy.mp3", "Sentence": "Cosine of 11 pi over two. That is the same thing as, let's see, if we were to subtract, this is the same thing as cosine of 11 pi over two minus, let's see, this is the same thing as five and 1 1\u20442 pi. Right? Yeah, so this is, so we could view this as, we could subtract, let's subtract four pi, which is going to be, we could write that as eight pi over two. In fact, no, let's subtract five. Let's subtract, no, let's subtract four pi, which is eight pi over two. So once again, I'm just subtracting a multiple of two pi, which isn't gonna change the value of cosine."}, {"video_title": "Definite integral of trig function AP Calculus AB Khan Academy.mp3", "Sentence": "Yeah, so this is, so we could view this as, we could subtract, let's subtract four pi, which is going to be, we could write that as eight pi over two. In fact, no, let's subtract five. Let's subtract, no, let's subtract four pi, which is eight pi over two. So once again, I'm just subtracting a multiple of two pi, which isn't gonna change the value of cosine. And so this is going to be equal to cosine of three pi over two. And if we imagine the unit circle, let me draw the unit circle here. So it's my y-axis, my x-axis, and then I have the unit circle."}, {"video_title": "Definite integral of trig function AP Calculus AB Khan Academy.mp3", "Sentence": "So once again, I'm just subtracting a multiple of two pi, which isn't gonna change the value of cosine. And so this is going to be equal to cosine of three pi over two. And if we imagine the unit circle, let me draw the unit circle here. So it's my y-axis, my x-axis, and then I have the unit circle. So, whoops, all right, the unit circle, just like that. So if we start at, this is zero, then you go to pi over two, then you go to pi, then you go to three pi over two. So that's this point on the unit circle."}, {"video_title": "Definite integral of trig function AP Calculus AB Khan Academy.mp3", "Sentence": "So it's my y-axis, my x-axis, and then I have the unit circle. So, whoops, all right, the unit circle, just like that. So if we start at, this is zero, then you go to pi over two, then you go to pi, then you go to three pi over two. So that's this point on the unit circle. So the cosine is the x-coordinate, so this is going to be zero. This is zero, so this is zero. And so we get one minus zero, so everything in the brackets evaluates out to one."}, {"video_title": "Definite integral of trig function AP Calculus AB Khan Academy.mp3", "Sentence": "So that's this point on the unit circle. So the cosine is the x-coordinate, so this is going to be zero. This is zero, so this is zero. And so we get one minus zero, so everything in the brackets evaluates out to one. And so we are left with, so let me do that. So all of this is equal to one. And so you have negative nine times one, which of course is just negative nine, is what this definite integral evaluates to."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "This is going to be the same thing as the limit as x approaches negative one of x plus one over, over the limit, the limit as x approaches negative one of square root of x plus five minus two. And then we could say, all right, this thing up here, x plus one, if we think about the graph y equals x plus one, it's continuous everywhere, especially at x equals negative one, and so to evaluate this limit, we just have to evaluate this expression at x equals negative one. So this numerator is just going to evaluate to negative one plus one. And then our denominator, square root of x plus five minus two isn't continuous everywhere, but it is continuous at x equals negative one, and so we can do the same thing. We can just substitute negative one for x. So this is going to be the square root of negative one plus five minus two. Now what does this evaluate to?"}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And then our denominator, square root of x plus five minus two isn't continuous everywhere, but it is continuous at x equals negative one, and so we can do the same thing. We can just substitute negative one for x. So this is going to be the square root of negative one plus five minus two. Now what does this evaluate to? Well, in the numerator we get a zero, and in the denominator, negative one plus five is four, take the principal root is two minus two, we get zero again. So we get, we got zero over zero. Now when you see that, you might be tempted to give up."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Now what does this evaluate to? Well, in the numerator we get a zero, and in the denominator, negative one plus five is four, take the principal root is two minus two, we get zero again. So we get, we got zero over zero. Now when you see that, you might be tempted to give up. You say, oh look, there's a zero in the denominator, maybe this limit doesn't exist, maybe I'm done here, what do I do? And if this was non-zero up here in the numerator, if you're taking a non-zero value and dividing it by zero, that is undefined, and your limit would not exist. But when you have zero over zero, this is indeterminate form, and it doesn't mean necessarily that your limit does not exist."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Now when you see that, you might be tempted to give up. You say, oh look, there's a zero in the denominator, maybe this limit doesn't exist, maybe I'm done here, what do I do? And if this was non-zero up here in the numerator, if you're taking a non-zero value and dividing it by zero, that is undefined, and your limit would not exist. But when you have zero over zero, this is indeterminate form, and it doesn't mean necessarily that your limit does not exist. And as we'll see in this video and many future ones, there are tools at our disposal to address this, and we will look at one of them. Now the tool that we're gonna look at is, is there another way of rewriting this expression so that we can evaluate its limit without getting the zero over zero? Well let's just rewrite, let's just take this, let me give it, so let's take this thing right over here, and let's say this is g of x."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "But when you have zero over zero, this is indeterminate form, and it doesn't mean necessarily that your limit does not exist. And as we'll see in this video and many future ones, there are tools at our disposal to address this, and we will look at one of them. Now the tool that we're gonna look at is, is there another way of rewriting this expression so that we can evaluate its limit without getting the zero over zero? Well let's just rewrite, let's just take this, let me give it, so let's take this thing right over here, and let's say this is g of x. So essentially what we're trying to do is find the limit of g of x as x approaches negative one. So we could write g of x is equal to x plus one, and the whole reason why I'm defining it as g of x is just to be able to think of it more clearly as a function and manipulate the function, and then think about similar functions, over x plus five minus two, or x plus one over the square root of x plus five minus two. Now the technique we're gonna use is, when you get this indeterminate form, and if you have a square root in either the numerator or the denominator, it might help to get rid of that square root."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Well let's just rewrite, let's just take this, let me give it, so let's take this thing right over here, and let's say this is g of x. So essentially what we're trying to do is find the limit of g of x as x approaches negative one. So we could write g of x is equal to x plus one, and the whole reason why I'm defining it as g of x is just to be able to think of it more clearly as a function and manipulate the function, and then think about similar functions, over x plus five minus two, or x plus one over the square root of x plus five minus two. Now the technique we're gonna use is, when you get this indeterminate form, and if you have a square root in either the numerator or the denominator, it might help to get rid of that square root. And this is often called rationalizing expression. In this case you have a square root in the denominator, so it would be rationalizing the denominator. And so this would be, the way we would do it, is we'd be leveraging our knowledge of difference of squares."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Now the technique we're gonna use is, when you get this indeterminate form, and if you have a square root in either the numerator or the denominator, it might help to get rid of that square root. And this is often called rationalizing expression. In this case you have a square root in the denominator, so it would be rationalizing the denominator. And so this would be, the way we would do it, is we'd be leveraging our knowledge of difference of squares. We know, we know that a plus b times a minus b is equal to a squared minus b squared, you learned that in algebra a little while ago. Or, if we had the square root of a plus b, and we were to multiply that times the square root of a minus b, well that'd be the square root of a squared, which is just going to be a minus b squared. So we can just leverage these ideas to get rid of this radical down here."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And so this would be, the way we would do it, is we'd be leveraging our knowledge of difference of squares. We know, we know that a plus b times a minus b is equal to a squared minus b squared, you learned that in algebra a little while ago. Or, if we had the square root of a plus b, and we were to multiply that times the square root of a minus b, well that'd be the square root of a squared, which is just going to be a minus b squared. So we can just leverage these ideas to get rid of this radical down here. The way we're going to do it, is we're gonna multiply the numerator and the denominator by the square root of x plus five plus two, right? We have the minus two, so we're gonna multiply it times the plus two. So let's do that."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So we can just leverage these ideas to get rid of this radical down here. The way we're going to do it, is we're gonna multiply the numerator and the denominator by the square root of x plus five plus two, right? We have the minus two, so we're gonna multiply it times the plus two. So let's do that. So we have square root of x plus five plus two, and we're gonna multiply the numerator times the same thing, because we don't want to change the value of the expression. This is one, so if we take the expression divided by the same expression, it's going to be one. So this is, so square root of x plus five plus two."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So let's do that. So we have square root of x plus five plus two, and we're gonna multiply the numerator times the same thing, because we don't want to change the value of the expression. This is one, so if we take the expression divided by the same expression, it's going to be one. So this is, so square root of x plus five plus two. And so this is going to be equal to, this is going to be equal to x plus one times the square root, times the square root of x plus five plus two. And then the denominator is going to be, well, it's going to be the square root of x plus five squared, which would be just x plus five, and then minus two squared, minus four. And so this down here simplifies to x plus five minus four, it's just x plus one."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So this is, so square root of x plus five plus two. And so this is going to be equal to, this is going to be equal to x plus one times the square root, times the square root of x plus five plus two. And then the denominator is going to be, well, it's going to be the square root of x plus five squared, which would be just x plus five, and then minus two squared, minus four. And so this down here simplifies to x plus five minus four, it's just x plus one. So this is just, this is just x plus one. And it probably jumps out at you that both the numerator and the denominator have an x plus one in it, so maybe we can simplify. So we could simplify and just say, well, g of x is equal to the square root of x plus five plus two."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And so this down here simplifies to x plus five minus four, it's just x plus one. So this is just, this is just x plus one. And it probably jumps out at you that both the numerator and the denominator have an x plus one in it, so maybe we can simplify. So we could simplify and just say, well, g of x is equal to the square root of x plus five plus two. Now some of you might be feeling a little off here, and you would be correct. Your spider senses would be, say, is this, is this definitely the same thing as what we originally had before we canceled out the x plus ones? And the answer is the way I just wrote it, it is not the exact same thing."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So we could simplify and just say, well, g of x is equal to the square root of x plus five plus two. Now some of you might be feeling a little off here, and you would be correct. Your spider senses would be, say, is this, is this definitely the same thing as what we originally had before we canceled out the x plus ones? And the answer is the way I just wrote it, it is not the exact same thing. It is the exact same thing everywhere, except at x equals negative one. This thing right over here is defined at x equals negative one. This thing right over here is not defined at x equals negative one."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And the answer is the way I just wrote it, it is not the exact same thing. It is the exact same thing everywhere, except at x equals negative one. This thing right over here is defined at x equals negative one. This thing right over here is not defined at x equals negative one. And g of x was not, was not, so g of x right over here, you don't get a good result when you try x equals negative one. And so in order for this to truly be the same thing as g of x, the same function, we have to say for x not equal to negative one. Now this is a simplified version of g of x."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "This thing right over here is not defined at x equals negative one. And g of x was not, was not, so g of x right over here, you don't get a good result when you try x equals negative one. And so in order for this to truly be the same thing as g of x, the same function, we have to say for x not equal to negative one. Now this is a simplified version of g of x. It is the same thing. For any input x that g of x is defined, this is going to give you the same output. And this has the exact same domain now, now that we've put this constraint in, as g of x."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Now this is a simplified version of g of x. It is the same thing. For any input x that g of x is defined, this is going to give you the same output. And this has the exact same domain now, now that we've put this constraint in, as g of x. Now you might say, okay, well, how does this help us? Because we want to find the limit as x approaches negative one. And even here, I had to put this little constraint here that x cannot be equal to negative one."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And this has the exact same domain now, now that we've put this constraint in, as g of x. Now you might say, okay, well, how does this help us? Because we want to find the limit as x approaches negative one. And even here, I had to put this little constraint here that x cannot be equal to negative one. How do we think about this limit? Well, lucky for us, we know, lucky for us, we know that if we just take another function, f of x, if we say f of x is equal to the square root of x plus five plus two, well, then we know that f of x is equal to g of x for all x not equal to negative one, because f of x does not have that constraint. And we know if this is true of two, if this is true of two functions, then the limit as x approaches, the limit, let me write this down, is since we know this, because of this, we know that the limit of f of x as x approaches negative one is going to be equal to the limit of g of x as x approaches negative one."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And even here, I had to put this little constraint here that x cannot be equal to negative one. How do we think about this limit? Well, lucky for us, we know, lucky for us, we know that if we just take another function, f of x, if we say f of x is equal to the square root of x plus five plus two, well, then we know that f of x is equal to g of x for all x not equal to negative one, because f of x does not have that constraint. And we know if this is true of two, if this is true of two functions, then the limit as x approaches, the limit, let me write this down, is since we know this, because of this, we know that the limit of f of x as x approaches negative one is going to be equal to the limit of g of x as x approaches negative one. And this, of course, is what we want to figure out, what was the beginning of the problem. And but we can now use f of x here, because only at x equals negative one that they are not the same. And if you were to graph g of x, it just has a point discontinuity, or a removable, or I should just say, yeah, a point discontinuity right over here at x equals negative one."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And we know if this is true of two, if this is true of two functions, then the limit as x approaches, the limit, let me write this down, is since we know this, because of this, we know that the limit of f of x as x approaches negative one is going to be equal to the limit of g of x as x approaches negative one. And this, of course, is what we want to figure out, what was the beginning of the problem. And but we can now use f of x here, because only at x equals negative one that they are not the same. And if you were to graph g of x, it just has a point discontinuity, or a removable, or I should just say, yeah, a point discontinuity right over here at x equals negative one. And so what is the limit? And we are in the home stretch now. What is the limit of f of x, or we could say the limit of the square root of x plus five plus two as x approaches negative one?"}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And if you were to graph g of x, it just has a point discontinuity, or a removable, or I should just say, yeah, a point discontinuity right over here at x equals negative one. And so what is the limit? And we are in the home stretch now. What is the limit of f of x, or we could say the limit of the square root of x plus five plus two as x approaches negative one? Well, this expression is continuous, or this function is continuous at x equals negative one, so we can just evaluate it at x equals negative one. So this is going to be the square root of negative one plus five plus two. So this is four, square root, principle root of four is two."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "What is the limit of f of x, or we could say the limit of the square root of x plus five plus two as x approaches negative one? Well, this expression is continuous, or this function is continuous at x equals negative one, so we can just evaluate it at x equals negative one. So this is going to be the square root of negative one plus five plus two. So this is four, square root, principle root of four is two. Two plus two is equal to four. So since the limit of f of x as x approaches negative one is four, the limit of g of x as x approaches negative one is also four. And if this little, this little, I guess you could say, leap that I just made right over here doesn't make sense to you, think about it visually."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So this is four, square root, principle root of four is two. Two plus two is equal to four. So since the limit of f of x as x approaches negative one is four, the limit of g of x as x approaches negative one is also four. And if this little, this little, I guess you could say, leap that I just made right over here doesn't make sense to you, think about it visually. Think about it visually. So if this is my y-axis, and this is my x-axis, g of x looked something like this. The g of x, the g of x, let me draw it, g of x looked something, something like this."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And if this little, this little, I guess you could say, leap that I just made right over here doesn't make sense to you, think about it visually. Think about it visually. So if this is my y-axis, and this is my x-axis, g of x looked something like this. The g of x, the g of x, let me draw it, g of x looked something, something like this. And it had a gap at negative one. So it had a gap right over there. While f of x, f of x would have the same graph, except it wouldn't have, it wouldn't have the gap."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "The g of x, the g of x, let me draw it, g of x looked something, something like this. And it had a gap at negative one. So it had a gap right over there. While f of x, f of x would have the same graph, except it wouldn't have, it wouldn't have the gap. And so if you're trying to find the limit, it seems completely reasonable. Well, let's just use f of x and evaluate what f of x would be to kind of fill that gap at x equals negative one. So hopefully this graphical version helps a little bit, or if it confuses you, ignore it."}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_ AP Calculus AB Khan Academy.mp3", "Sentence": "Here we have a function of t, and we're taking the antiderivative with respect to t. And so you would not write a dx here. That is not the notation. You'll see why when we focus on definite integrals. So what's the antiderivative of this business right over here? Well, it's going to be the same thing as the antiderivative of sine of t. It's going to be the antiderivative of sine of t, or the indefinite integral of sine of t, plus the indefinite integral, or the antiderivative of cosine of t. Plus the antiderivative of cosine of t. So let's think about what these antiderivatives are. And we already know a little bit about taking the derivatives of trig functions. We know that the derivative with respect to t of cosine of t is equal to negative sine of t. So if we want a sine of t here, we would just have to take the derivative of negative cosine t. If we take the derivative of negative cosine t, then we get positive sine of t. Derivative with respect to t of cosine t is negative sine of t. We have the negative out front."}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_ AP Calculus AB Khan Academy.mp3", "Sentence": "So what's the antiderivative of this business right over here? Well, it's going to be the same thing as the antiderivative of sine of t. It's going to be the antiderivative of sine of t, or the indefinite integral of sine of t, plus the indefinite integral, or the antiderivative of cosine of t. Plus the antiderivative of cosine of t. So let's think about what these antiderivatives are. And we already know a little bit about taking the derivatives of trig functions. We know that the derivative with respect to t of cosine of t is equal to negative sine of t. So if we want a sine of t here, we would just have to take the derivative of negative cosine t. If we take the derivative of negative cosine t, then we get positive sine of t. Derivative with respect to t of cosine t is negative sine of t. We have the negative out front. It becomes positive sine of t. So the antiderivative of sine of t is negative cosine of t. So this is going to be equal to negative cosine of t. And then what's the antiderivative of cosine of t? Well, we already know that the derivative with respect to t of sine of t is equal to cosine of t. So cosine of t's antiderivative is just sine of t. So plus sine of t. And we're done. We found the antiderivative."}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_ AP Calculus AB Khan Academy.mp3", "Sentence": "We know that the derivative with respect to t of cosine of t is equal to negative sine of t. So if we want a sine of t here, we would just have to take the derivative of negative cosine t. If we take the derivative of negative cosine t, then we get positive sine of t. Derivative with respect to t of cosine t is negative sine of t. We have the negative out front. It becomes positive sine of t. So the antiderivative of sine of t is negative cosine of t. So this is going to be equal to negative cosine of t. And then what's the antiderivative of cosine of t? Well, we already know that the derivative with respect to t of sine of t is equal to cosine of t. So cosine of t's antiderivative is just sine of t. So plus sine of t. And we're done. We found the antiderivative. Now let's tackle this. Now we don't have a t. We're taking the indefinite integral with respect to, actually, this is a mistake. This should be with respect to a."}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_ AP Calculus AB Khan Academy.mp3", "Sentence": "We found the antiderivative. Now let's tackle this. Now we don't have a t. We're taking the indefinite integral with respect to, actually, this is a mistake. This should be with respect to a. Let me clean this up. This should be a dA. If we were taking this with respect to t, then we would treat all of these things as just constants."}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_ AP Calculus AB Khan Academy.mp3", "Sentence": "This should be with respect to a. Let me clean this up. This should be a dA. If we were taking this with respect to t, then we would treat all of these things as just constants. But I don't want to confuse you right now. Let me make it clear. This is going to be dA."}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_ AP Calculus AB Khan Academy.mp3", "Sentence": "If we were taking this with respect to t, then we would treat all of these things as just constants. But I don't want to confuse you right now. Let me make it clear. This is going to be dA. That's what we are integrating. We're taking the antiderivative with respect to. So what is this going to be equal to?"}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_ AP Calculus AB Khan Academy.mp3", "Sentence": "This is going to be dA. That's what we are integrating. We're taking the antiderivative with respect to. So what is this going to be equal to? Well, once again, we can rewrite it as the sum of integrals. This is the indefinite integral of e to the a dA. So this one right over here."}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_ AP Calculus AB Khan Academy.mp3", "Sentence": "So what is this going to be equal to? Well, once again, we can rewrite it as the sum of integrals. This is the indefinite integral of e to the a dA. So this one right over here. I'll do it in green. Plus the indefinite integral, or the antiderivative, of 1 over a dA. Now, what is the antiderivative of e to the a?"}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_ AP Calculus AB Khan Academy.mp3", "Sentence": "So this one right over here. I'll do it in green. Plus the indefinite integral, or the antiderivative, of 1 over a dA. Now, what is the antiderivative of e to the a? Well, we already know a little bit about exponentials. The derivative with respect to x of e to the x is equal to e to the x. That's one of the reasons why e and the exponential function in general is so amazing."}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_ AP Calculus AB Khan Academy.mp3", "Sentence": "Now, what is the antiderivative of e to the a? Well, we already know a little bit about exponentials. The derivative with respect to x of e to the x is equal to e to the x. That's one of the reasons why e and the exponential function in general is so amazing. And if we just replaced a with x, or x with a, you get the derivative with respect to a of e to the a is equal to e to the a. So the antiderivative here, the derivative of e to the a is e to the a. The antiderivative is going to be e to the a."}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_ AP Calculus AB Khan Academy.mp3", "Sentence": "That's one of the reasons why e and the exponential function in general is so amazing. And if we just replaced a with x, or x with a, you get the derivative with respect to a of e to the a is equal to e to the a. So the antiderivative here, the derivative of e to the a is e to the a. The antiderivative is going to be e to the a. And maybe you could shift it by some type of a constant. Oh, and let me not forget, I have to put my constant right over here. I could have a constant factor."}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_ AP Calculus AB Khan Academy.mp3", "Sentence": "The antiderivative is going to be e to the a. And maybe you could shift it by some type of a constant. Oh, and let me not forget, I have to put my constant right over here. I could have a constant factor. So let me, always important, remember the constant. So you have a constant factor right over here. Never forget that."}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_ AP Calculus AB Khan Academy.mp3", "Sentence": "I could have a constant factor. So let me, always important, remember the constant. So you have a constant factor right over here. Never forget that. I almost did. So once again, over here, what's the antiderivative of e to the a? It is e to the a."}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_ AP Calculus AB Khan Academy.mp3", "Sentence": "Never forget that. I almost did. So once again, over here, what's the antiderivative of e to the a? It is e to the a. What's the antiderivative of 1 over a? Well, we've seen that in the last video. It is going to be the natural log of the absolute value of a."}, {"video_title": "Indefinite integrals of sin(x), cos(x), and e_ AP Calculus AB Khan Academy.mp3", "Sentence": "It is e to the a. What's the antiderivative of 1 over a? Well, we've seen that in the last video. It is going to be the natural log of the absolute value of a. And then we want to have the most general antiderivative. So there could be a constant factor out here as well. And we are done."}, {"video_title": "One-sided limits from graphs asymptote Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And what I want to do is I want to figure out the limit of g of x as x approaches positive six from values that are less than positive six, or you could say from the left, from the, you could say the negative direction. So what is this going to be equal to? And if you have a sense of it, pause the video and give a go at it. Well, to think about this, let's just take different x values that approach six from the left and look at what the values of the function are. So g of two looks like it's a little bit more than one. G of three, it's a little bit more than that. G of four looks like it's a little under two."}, {"video_title": "One-sided limits from graphs asymptote Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Well, to think about this, let's just take different x values that approach six from the left and look at what the values of the function are. So g of two looks like it's a little bit more than one. G of three, it's a little bit more than that. G of four looks like it's a little under two. G of five, it looks like it's around three. G of 5.5 looks like it's around five. G of, let's say 5.75, looks like it's like nine."}, {"video_title": "One-sided limits from graphs asymptote Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "G of four looks like it's a little under two. G of five, it looks like it's around three. G of 5.5 looks like it's around five. G of, let's say 5.75, looks like it's like nine. And so as x gets closer and closer to six from the left, it looks like the value of our function just becomes unbounded. It's just getting infinitely large. And so in some context, you might see someone write that."}, {"video_title": "One-sided limits from graphs asymptote Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "G of, let's say 5.75, looks like it's like nine. And so as x gets closer and closer to six from the left, it looks like the value of our function just becomes unbounded. It's just getting infinitely large. And so in some context, you might see someone write that. Maybe this is equal to infinity. But infinity isn't, we're not talking about a specific number. And if we're talking technically about limits, the way that we've looked at it, what is, you'll sometimes see this in some classes, but in this context, especially on the exercises on Khan Academy, we'll say that this does not exist."}, {"video_title": "One-sided limits from graphs asymptote Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And so in some context, you might see someone write that. Maybe this is equal to infinity. But infinity isn't, we're not talking about a specific number. And if we're talking technically about limits, the way that we've looked at it, what is, you'll sometimes see this in some classes, but in this context, especially on the exercises on Khan Academy, we'll say that this does not exist. Not exist. This thing right over here is unbounded. And this is interesting because the left-handed limit here doesn't exist, but the right-handed limit does."}, {"video_title": "One-sided limits from graphs asymptote Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And if we're talking technically about limits, the way that we've looked at it, what is, you'll sometimes see this in some classes, but in this context, especially on the exercises on Khan Academy, we'll say that this does not exist. Not exist. This thing right over here is unbounded. And this is interesting because the left-handed limit here doesn't exist, but the right-handed limit does. If I were to say the limit of g of x as x approaches six from the right-hand side, well, let's see. We have g of eight is there, g of seven is there, g of 6.5, looks like it's a little less than negative three, g of 6.01, a little even closer to negative three, g of 6.0000001 is very close to negative three. So it looks like this limit right over here, at least looking at it graphically, it looks like when we approach six from the right, it looks like the function is approaching negative three, but from the left, it's just unbounded, so we'll say it doesn't exist."}, {"video_title": "Derivative of __ Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And once again, it reinforces the idea that e is really this somewhat magical number. So we're gonna do a little bit of an exploration. Let's just pick some points on this curve of y is equal to e to the x, and think about what the slope of the tangent line is, or what the derivative looks like. And so let's say when y is equal to one, or when e to the x is equal to one, this is the case when x is equal to zero, well, the slope of the tangent line looks like it is one, which is curious, because that's exactly the value of the function at that point. What about when e to the x is equal to two, right over here? Well, here, let me do it in another color, the slope of the tangent line sure looks pretty close, sure looks pretty close to two. What about, what about when e to the x is equal to 1 1\u20442?"}, {"video_title": "Derivative of __ Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And so let's say when y is equal to one, or when e to the x is equal to one, this is the case when x is equal to zero, well, the slope of the tangent line looks like it is one, which is curious, because that's exactly the value of the function at that point. What about when e to the x is equal to two, right over here? Well, here, let me do it in another color, the slope of the tangent line sure looks pretty close, sure looks pretty close to two. What about, what about when e to the x is equal to 1 1\u20442? So that's happening right about here. Well, it sure looks like the slope of the tangent line is about 1 1\u20442. We could try, what happens when e to the x is equal to five?"}, {"video_title": "Derivative of __ Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "What about, what about when e to the x is equal to 1 1\u20442? So that's happening right about here. Well, it sure looks like the slope of the tangent line is about 1 1\u20442. We could try, what happens when e to the x is equal to five? Well, the slope of the tangent line here sure does look pretty close, sure does look pretty close to five. And so, just eyeballing it, is it the case that the slope of the tangent line of e to the x is the same thing, is e to the x? And I will tell you, and this is an amazing thing, that that is indeed true."}, {"video_title": "Derivative of __ Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We could try, what happens when e to the x is equal to five? Well, the slope of the tangent line here sure does look pretty close, sure does look pretty close to five. And so, just eyeballing it, is it the case that the slope of the tangent line of e to the x is the same thing, is e to the x? And I will tell you, and this is an amazing thing, that that is indeed true. That if I have some function, f of x, that is equal to e to the x, and if I were to take the derivative of this, this is going to be equal to e to the x as well. Or another way of saying it, the derivative with respect to x of e to the x is equal to e to the x. And that is an amazing thing."}, {"video_title": "Derivative of __ Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And I will tell you, and this is an amazing thing, that that is indeed true. That if I have some function, f of x, that is equal to e to the x, and if I were to take the derivative of this, this is going to be equal to e to the x as well. Or another way of saying it, the derivative with respect to x of e to the x is equal to e to the x. And that is an amazing thing. In previous lessons or courses, you've learned about ways to define e. And this could be a new one. E is the number that where if you take that number to the power x, if you define a function or expression as e to the x, it's that number where if you take the derivative of that, it's still going to be e to the x. And what you're looking here, this curve, it's a curve where the value, that's y value, at any point is the same as the slope of the tangent line."}, {"video_title": "Derivative of __ Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And that is an amazing thing. In previous lessons or courses, you've learned about ways to define e. And this could be a new one. E is the number that where if you take that number to the power x, if you define a function or expression as e to the x, it's that number where if you take the derivative of that, it's still going to be e to the x. And what you're looking here, this curve, it's a curve where the value, that's y value, at any point is the same as the slope of the tangent line. If that doesn't strike you as mysterious and magical and amazing just yet, it will. Maybe tonight, you'll wake up in the middle of the night and you'll realize just what's going on. Now some of you might be saying, okay, this is cool, you're telling me this, but how do I know it's true?"}, {"video_title": "Disc method rotating around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "And this time we're going to rotate our function. We're going to rotate it around a vertical line that is not the y-axis. And if we do that, so we're going to rotate y is equal to x squared minus 1, or at least this part of it, we're going to rotate it around the vertical line x is equal to negative 2. And if we do that, we get this gumball shape that looks something like this. So what I want to do is I want to find the volume of this using the disk method. So what I want to do is construct some disks. So construct some disks."}, {"video_title": "Disc method rotating around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "And if we do that, we get this gumball shape that looks something like this. So what I want to do is I want to find the volume of this using the disk method. So what I want to do is construct some disks. So construct some disks. So let's take this one of the disks right over here. It's going to have some depth. And that depth is going to be dy."}, {"video_title": "Disc method rotating around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "So construct some disks. So let's take this one of the disks right over here. It's going to have some depth. And that depth is going to be dy. It's going to be dy right over there. And it's going to have some area on top of it that is a function of any given y that I have. So the volume of a given disk is going to be the area as a function of y times the depth of the disk times dy."}, {"video_title": "Disc method rotating around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "And that depth is going to be dy. It's going to be dy right over there. And it's going to have some area on top of it that is a function of any given y that I have. So the volume of a given disk is going to be the area as a function of y times the depth of the disk times dy. And then we just have to integrate it over the interval that we care about. And we're doing it in terms of the volume of the disk. And then we're going to integrate it over the interval that we care about."}, {"video_title": "Disc method rotating around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "So the volume of a given disk is going to be the area as a function of y times the depth of the disk times dy. And then we just have to integrate it over the interval that we care about. And we're doing it in terms of the volume of the disk. And then we're going to integrate it over the interval that we care about. And we're doing it all in terms of y. And in this case, we're going to integrate from y is equal to... Well, this is going to hit this y intercept right over here. So y is equal to negative 1."}, {"video_title": "Disc method rotating around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "And then we're going to integrate it over the interval that we care about. And we're doing it all in terms of y. And in this case, we're going to integrate from y is equal to... Well, this is going to hit this y intercept right over here. So y is equal to negative 1. And let's go all the way to y is equal to... Let's say y is equal to 3. y is equal to 3 right over here. So from y equals negative 1 to y equals 3. So y equals negative 1 to y is equal to 3."}, {"video_title": "Disc method rotating around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "So y is equal to negative 1. And let's go all the way to y is equal to... Let's say y is equal to 3. y is equal to 3 right over here. So from y equals negative 1 to y equals 3. So y equals negative 1 to y is equal to 3. And that's going to give us the volume of our upside-down gumdrop-type-looking thing. So the key here is so that we can start evaluating the double integral is to just figure out what the area of each of these disks are as a function of y. And we know that area is just... Area as a function of y is just going to be pi times radius as a function of y squared."}, {"video_title": "Disc method rotating around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "So y equals negative 1 to y is equal to 3. And that's going to give us the volume of our upside-down gumdrop-type-looking thing. So the key here is so that we can start evaluating the double integral is to just figure out what the area of each of these disks are as a function of y. And we know that area is just... Area as a function of y is just going to be pi times radius as a function of y squared. So the real key is, what is the radius as a function of y for any one of these y's? The radius as a function of y. So let's think about that a little bit."}, {"video_title": "Disc method rotating around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "And we know that area is just... Area as a function of y is just going to be pi times radius as a function of y squared. So the real key is, what is the radius as a function of y for any one of these y's? The radius as a function of y. So let's think about that a little bit. What is this curve? Well, let's write it as a function of y. If you add 1 to both sides, and I'm going to swap sides, you'll get x squared is equal to y plus 1."}, {"video_title": "Disc method rotating around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "So let's think about that a little bit. What is this curve? Well, let's write it as a function of y. If you add 1 to both sides, and I'm going to swap sides, you'll get x squared is equal to y plus 1. I just added 1 to both sides and then swapped sides. And then you get x is equal to the principal root of the square root of y plus 1. So this we can write as x or we can even write it as f of y if we want."}, {"video_title": "Disc method rotating around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "If you add 1 to both sides, and I'm going to swap sides, you'll get x squared is equal to y plus 1. I just added 1 to both sides and then swapped sides. And then you get x is equal to the principal root of the square root of y plus 1. So this we can write as x or we can even write it as f of y if we want. f of y is equal to the square root of y plus 1. Or we could say x is equal to a function of y, which is the square root of y plus 1. So what's the distance?"}, {"video_title": "Disc method rotating around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "So this we can write as x or we can even write it as f of y if we want. f of y is equal to the square root of y plus 1. Or we could say x is equal to a function of y, which is the square root of y plus 1. So what's the distance? What's the distance here at any point? Well, this distance... Let me make it very clear. So it's going to be our total distance in the horizontal direction."}, {"video_title": "Disc method rotating around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "So what's the distance? What's the distance here at any point? Well, this distance... Let me make it very clear. So it's going to be our total distance in the horizontal direction. So this first part, as we're... I'm going to do it in a different color that you can't see. So this part right over here is just going to be the value of the function."}, {"video_title": "Disc method rotating around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "So it's going to be our total distance in the horizontal direction. So this first part, as we're... I'm going to do it in a different color that you can't see. So this part right over here is just going to be the value of the function. It's going to give you an x value. But then you have to add another 2 to go all the way over here. So your entire radius is a function of y."}, {"video_title": "Disc method rotating around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "So this part right over here is just going to be the value of the function. It's going to give you an x value. But then you have to add another 2 to go all the way over here. So your entire radius is a function of y. Your radius as a function of y is going to be equal to the square root of y plus 1. This essentially will give you one of these x values when you're sitting on this curve. It's x as a function of y."}, {"video_title": "Disc method rotating around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "So your entire radius is a function of y. Your radius as a function of y is going to be equal to the square root of y plus 1. This essentially will give you one of these x values when you're sitting on this curve. It's x as a function of y. It'll give you one of these x values. And then from that you add another 2. So plus 2."}, {"video_title": "Disc method rotating around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "It's x as a function of y. It'll give you one of these x values. And then from that you add another 2. So plus 2. Another way of thinking about it, you get an x value here and from that x value you subtract out x is equal to negative 2. And when you subtract x is equal to negative 2 you're adding 2 here. But hopefully this makes intuitive sense."}, {"video_title": "Disc method rotating around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "So plus 2. Another way of thinking about it, you get an x value here and from that x value you subtract out x is equal to negative 2. And when you subtract x is equal to negative 2 you're adding 2 here. But hopefully this makes intuitive sense. This is the x value... Let me do this in a better color. This right over here, this distance right over here is the x value you get when you just evaluate the function of y. But then if you wanted the full radius, you have to go another 2 to go to the center of our axis of rotation."}, {"video_title": "Disc method rotating around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "But hopefully this makes intuitive sense. This is the x value... Let me do this in a better color. This right over here, this distance right over here is the x value you get when you just evaluate the function of y. But then if you wanted the full radius, you have to go another 2 to go to the center of our axis of rotation. Once again, if you just take a given y right over there, you evaluate the y you get an x value. That x value will just give you this distance. If you want the full distance, you have to subtract negative 2 from that x value, which is essentially the same thing as adding 2 to get our full radius."}, {"video_title": "Disc method rotating around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "But then if you wanted the full radius, you have to go another 2 to go to the center of our axis of rotation. Once again, if you just take a given y right over there, you evaluate the y you get an x value. That x value will just give you this distance. If you want the full distance, you have to subtract negative 2 from that x value, which is essentially the same thing as adding 2 to get our full radius. So our radius as a function of y is this thing right over here. So substituting back into this, we can now write our definite integral for our volume. The volume is going to be equal to the definite integral from negative 1 to 3 of pi times our radius squared dy."}, {"video_title": "Disc method rotating around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "If you want the full distance, you have to subtract negative 2 from that x value, which is essentially the same thing as adding 2 to get our full radius. So our radius as a function of y is this thing right over here. So substituting back into this, we can now write our definite integral for our volume. The volume is going to be equal to the definite integral from negative 1 to 3 of pi times our radius squared dy. So I can write the pi out here. We've done this multiple times. Times radius squared, so it's going to be square root of y plus 1 plus 2 squared, that's our radius, times dy."}, {"video_title": "Disc method rotating around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "The volume is going to be equal to the definite integral from negative 1 to 3 of pi times our radius squared dy. So I can write the pi out here. We've done this multiple times. Times radius squared, so it's going to be square root of y plus 1 plus 2 squared, that's our radius, times dy. So we've set up the definite integral and now we just have to evaluate this thing. I'll save that for the next video. I encourage you to try this out on your own."}, {"video_title": "Mistakes when finding inflection points second derivative undefined AP Calculus AB Khan Academy.mp3", "Sentence": "And then later we are asked, is Robert's work correct? If not, what's his mistake? So pause this video and try to figure it out on your own. All right, now let's work through this together. So our original g of x is equal to the cube root of x, which is the same thing as x to the 1 3rd. So in step one, it looks like Robert's trying to find the first and second derivative. So the first derivative, we just do the power rule."}, {"video_title": "Mistakes when finding inflection points second derivative undefined AP Calculus AB Khan Academy.mp3", "Sentence": "All right, now let's work through this together. So our original g of x is equal to the cube root of x, which is the same thing as x to the 1 3rd. So in step one, it looks like Robert's trying to find the first and second derivative. So the first derivative, we just do the power rule. So it'll be 1 3rd x to the decrement the exponent. So this is looking good. Second derivative, we take this, multiply this times 1 3rd, which would be negative 2 9ths, and then decrement negative 2 3rds, which would indeed be negative 5 3rds."}, {"video_title": "Mistakes when finding inflection points second derivative undefined AP Calculus AB Khan Academy.mp3", "Sentence": "So the first derivative, we just do the power rule. So it'll be 1 3rd x to the decrement the exponent. So this is looking good. Second derivative, we take this, multiply this times 1 3rd, which would be negative 2 9ths, and then decrement negative 2 3rds, which would indeed be negative 5 3rds. So that looks right. And then it looks like Robert's trying to rewrite it. So we have the negative 2 9ths still."}, {"video_title": "Mistakes when finding inflection points second derivative undefined AP Calculus AB Khan Academy.mp3", "Sentence": "Second derivative, we take this, multiply this times 1 3rd, which would be negative 2 9ths, and then decrement negative 2 3rds, which would indeed be negative 5 3rds. So that looks right. And then it looks like Robert's trying to rewrite it. So we have the negative 2 9ths still. But then he recognized that this is the same thing as x to the 5 3rds in the denominator. And x to the 5 3rds is the same thing as x, as the cube root of x to the 5th. So this is all looking good."}, {"video_title": "Mistakes when finding inflection points second derivative undefined AP Calculus AB Khan Academy.mp3", "Sentence": "So we have the negative 2 9ths still. But then he recognized that this is the same thing as x to the 5 3rds in the denominator. And x to the 5 3rds is the same thing as x, as the cube root of x to the 5th. So this is all looking good. Step one looks good. And then step two, it looks like he's trying to find the solution, or he's trying to find x values where the second derivative is equal to zero. And it is indeed true that this has no solution, that you can never make this second derivative equal to zero."}, {"video_title": "Mistakes when finding inflection points second derivative undefined AP Calculus AB Khan Academy.mp3", "Sentence": "So this is all looking good. Step one looks good. And then step two, it looks like he's trying to find the solution, or he's trying to find x values where the second derivative is equal to zero. And it is indeed true that this has no solution, that you can never make this second derivative equal to zero. In order to be zero, the numerator would have to be zero, and well, two is never going to be equal to zero. So this is correct. And then step three, he says g doesn't have any inflection points."}, {"video_title": "Mistakes when finding inflection points second derivative undefined AP Calculus AB Khan Academy.mp3", "Sentence": "And it is indeed true that this has no solution, that you can never make this second derivative equal to zero. In order to be zero, the numerator would have to be zero, and well, two is never going to be equal to zero. So this is correct. And then step three, he says g doesn't have any inflection points. Now this is a little bit suspect. It is, in many cases, our inflection point is a situation where our second derivative is equal to zero. And even then, we don't know it's an inflection point."}, {"video_title": "Mistakes when finding inflection points second derivative undefined AP Calculus AB Khan Academy.mp3", "Sentence": "And then step three, he says g doesn't have any inflection points. Now this is a little bit suspect. It is, in many cases, our inflection point is a situation where our second derivative is equal to zero. And even then, we don't know it's an inflection point. It would be a candidate inflection point. We would have to confirm that our second derivative crosses signs, or switches signs, as we cross that x value. But here we can't find a situation where our second derivative is equal to zero."}, {"video_title": "Mistakes when finding inflection points second derivative undefined AP Calculus AB Khan Academy.mp3", "Sentence": "And even then, we don't know it's an inflection point. It would be a candidate inflection point. We would have to confirm that our second derivative crosses signs, or switches signs, as we cross that x value. But here we can't find a situation where our second derivative is equal to zero. But we have to remind ourselves that other candidate inflection points are where our second derivative is undefined. And so he can't make this statement without seeing where our second derivative could be undefined. So for example, he could say that g prime prime is undefined, undefined when what?"}, {"video_title": "Mistakes when finding inflection points second derivative undefined AP Calculus AB Khan Academy.mp3", "Sentence": "But here we can't find a situation where our second derivative is equal to zero. But we have to remind ourselves that other candidate inflection points are where our second derivative is undefined. And so he can't make this statement without seeing where our second derivative could be undefined. So for example, he could say that g prime prime is undefined, undefined when what? Well, this is going to be undefined when x is equal to zero. X zero to the fifth, cube root of that, that's gonna be zero but then you're dividing by zero. So g prime prime undefined when x is equal to zero."}, {"video_title": "Mistakes when finding inflection points second derivative undefined AP Calculus AB Khan Academy.mp3", "Sentence": "So for example, he could say that g prime prime is undefined, undefined when what? Well, this is going to be undefined when x is equal to zero. X zero to the fifth, cube root of that, that's gonna be zero but then you're dividing by zero. So g prime prime undefined when x is equal to zero. So therefore, x equals, so we could say, candidate, candidate, candidate inflection point when x equals zero. And so then we would want to test it. And we could set up a traditional table that you might have seen before where we have our interval or intervals."}, {"video_title": "Mistakes when finding inflection points second derivative undefined AP Calculus AB Khan Academy.mp3", "Sentence": "So g prime prime undefined when x is equal to zero. So therefore, x equals, so we could say, candidate, candidate, candidate inflection point when x equals zero. And so then we would want to test it. And we could set up a traditional table that you might have seen before where we have our interval or intervals. We could have test values in our intervals. We have to be careful with those, make sure that they are indicative. And then we would say the sine of our second derivative of g prime prime."}, {"video_title": "Mistakes when finding inflection points second derivative undefined AP Calculus AB Khan Academy.mp3", "Sentence": "And we could set up a traditional table that you might have seen before where we have our interval or intervals. We could have test values in our intervals. We have to be careful with those, make sure that they are indicative. And then we would say the sine of our second derivative of g prime prime. And then we would have our concavity, concavity of g. And in order for x equals zero to be an inflection point, we would have to switch sines as, or our second derivative would have to switch sines as we cross x equals zero, and which would mean our concavity of g switches sines as we go, as we cross x equals zero. So let's do values less than zero, negative infinity to zero, and then values greater than zero, infinity, zero to infinity. I could do test values, let's say, I'll use negative one and one."}, {"video_title": "Mistakes when finding inflection points second derivative undefined AP Calculus AB Khan Academy.mp3", "Sentence": "And then we would say the sine of our second derivative of g prime prime. And then we would have our concavity, concavity of g. And in order for x equals zero to be an inflection point, we would have to switch sines as, or our second derivative would have to switch sines as we cross x equals zero, and which would mean our concavity of g switches sines as we go, as we cross x equals zero. So let's do values less than zero, negative infinity to zero, and then values greater than zero, infinity, zero to infinity. I could do test values, let's say, I'll use negative one and one. And you have to be careful when you use these, you have to make sure that we are close enough that nothing unusual happens between these test values up until we get to that candidate inflection point. And now what's the sine of our second derivative when x is equal to negative one? When x equals negative one, so let's see, negative one to the fifth power is negative one cube root of negative one is negative one."}, {"video_title": "Mistakes when finding inflection points second derivative undefined AP Calculus AB Khan Academy.mp3", "Sentence": "I could do test values, let's say, I'll use negative one and one. And you have to be careful when you use these, you have to make sure that we are close enough that nothing unusual happens between these test values up until we get to that candidate inflection point. And now what's the sine of our second derivative when x is equal to negative one? When x equals negative one, so let's see, negative one to the fifth power is negative one cube root of negative one is negative one. And so we're gonna have negative 2 9ths divided by negative one is gonna be positive 2 9ths. So our sine right over here is gonna be positive. And when, and this is gonna be in general when we're dealing with any negative value, because if you take any negative value to the fifth power it's gonna be negative."}, {"video_title": "Mistakes when finding inflection points second derivative undefined AP Calculus AB Khan Academy.mp3", "Sentence": "When x equals negative one, so let's see, negative one to the fifth power is negative one cube root of negative one is negative one. And so we're gonna have negative 2 9ths divided by negative one is gonna be positive 2 9ths. So our sine right over here is gonna be positive. And when, and this is gonna be in general when we're dealing with any negative value, because if you take any negative value to the fifth power it's gonna be negative. And you take that, the cube root of that, you're gonna have negative, but then when you have a negative value divided by that, you're gonna get a positive value. So you can feel good that this test values is indicative of actually this entire interval. And if you're dealing with a positive value, well, that to the fifth power is gonna be positive, cube root of that is still going to be positive, but then your gonna have negative 2 9ths divided by that positive value."}, {"video_title": "Mistakes when finding inflection points second derivative undefined AP Calculus AB Khan Academy.mp3", "Sentence": "And when, and this is gonna be in general when we're dealing with any negative value, because if you take any negative value to the fifth power it's gonna be negative. And you take that, the cube root of that, you're gonna have negative, but then when you have a negative value divided by that, you're gonna get a positive value. So you can feel good that this test values is indicative of actually this entire interval. And if you're dealing with a positive value, well, that to the fifth power is gonna be positive, cube root of that is still going to be positive, but then your gonna have negative 2 9ths divided by that positive value. So this is going to be negative. So it is indeed the case that our concavity of g switches as we cross x equals zero. We're concave upwards when x is less than zero, our second derivative is positive, and we are concave downwards when x is greater than zero."}, {"video_title": "Mistakes when finding inflection points second derivative undefined AP Calculus AB Khan Academy.mp3", "Sentence": "And if you're dealing with a positive value, well, that to the fifth power is gonna be positive, cube root of that is still going to be positive, but then your gonna have negative 2 9ths divided by that positive value. So this is going to be negative. So it is indeed the case that our concavity of g switches as we cross x equals zero. We're concave upwards when x is less than zero, our second derivative is positive, and we are concave downwards when x is greater than zero. Let me write that a little bit. Downwards, downwards when x is greater than zero. So we are switching concavity as we cross x equals zero, and so this tells us that x, so let's see, we are switching signs, switching, let me say g prime prime switching signs as we cross x equals zero, and our function is defined at x equals zero, and function defined at x equals zero."}, {"video_title": "Mistakes when finding inflection points second derivative undefined AP Calculus AB Khan Academy.mp3", "Sentence": "We're concave upwards when x is less than zero, our second derivative is positive, and we are concave downwards when x is greater than zero. Let me write that a little bit. Downwards, downwards when x is greater than zero. So we are switching concavity as we cross x equals zero, and so this tells us that x, so let's see, we are switching signs, switching, let me say g prime prime switching signs as we cross x equals zero, and our function is defined at x equals zero, and function defined at x equals zero. So we have an inflection point at x equals zero. So inflection point at x is equal to zero. And if you're familiar with the graph of the cube root, you would indeed see an inflection point at that point."}, {"video_title": "Mistakes when finding inflection points second derivative undefined AP Calculus AB Khan Academy.mp3", "Sentence": "So we are switching concavity as we cross x equals zero, and so this tells us that x, so let's see, we are switching signs, switching, let me say g prime prime switching signs as we cross x equals zero, and our function is defined at x equals zero, and function defined at x equals zero. So we have an inflection point at x equals zero. So inflection point at x is equal to zero. And if you're familiar with the graph of the cube root, you would indeed see an inflection point at that point. So there we go. He was wrong in step three. There actually is an inflection point."}, {"video_title": "Finding specific antiderivatives exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "We're told that f of seven is equal to 40 plus 5e to the seventh power. And f prime of x is equal to 5e to the x. What is f of zero? So to evaluate f of zero, let's take the antiderivative of f prime of x, and then we're going to have a constant of integration there, so we can use the information that they gave us up here, that f of seven is equal to this. This might look like an expression, well it is an expression, but it's really just a number. There's no variables in this. And so we can use that to solve for our constant of integration, and then we will have fully known what f of x is, and we can use that to evaluate f of zero."}, {"video_title": "Finding specific antiderivatives exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "So to evaluate f of zero, let's take the antiderivative of f prime of x, and then we're going to have a constant of integration there, so we can use the information that they gave us up here, that f of seven is equal to this. This might look like an expression, well it is an expression, but it's really just a number. There's no variables in this. And so we can use that to solve for our constant of integration, and then we will have fully known what f of x is, and we can use that to evaluate f of zero. So let's just do it. So if f prime of x is equal to 5e to the x, then f of x is going to be equal to the antiderivative of f prime of x, so the antiderivative of 5e to the x, dx. And this is the thing that I always find amazing about exponentials, and actually let me just take a step."}, {"video_title": "Finding specific antiderivatives exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "And so we can use that to solve for our constant of integration, and then we will have fully known what f of x is, and we can use that to evaluate f of zero. So let's just do it. So if f prime of x is equal to 5e to the x, then f of x is going to be equal to the antiderivative of f prime of x, so the antiderivative of 5e to the x, dx. And this is the thing that I always find amazing about exponentials, and actually let me just take a step. I'll take that five out of the integral so it becomes a little bit more obvious. And so the antiderivative of e to the x, well that's just e to the x, because the derivative of e to the x is e to the x, which I find amazing every time I have to manipulate or take the derivative or antiderivative of e to the x. So this is going to be 5e to the x plus c. And you can verify, take the derivative of 5e to the x plus c. The derivative of 5e to the x, well that's 5e to the x, so that works out, and the derivative of c is zero, so you wouldn't see it over here."}, {"video_title": "Finding specific antiderivatives exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "And this is the thing that I always find amazing about exponentials, and actually let me just take a step. I'll take that five out of the integral so it becomes a little bit more obvious. And so the antiderivative of e to the x, well that's just e to the x, because the derivative of e to the x is e to the x, which I find amazing every time I have to manipulate or take the derivative or antiderivative of e to the x. So this is going to be 5e to the x plus c. And you can verify, take the derivative of 5e to the x plus c. The derivative of 5e to the x, well that's 5e to the x, so that works out, and the derivative of c is zero, so you wouldn't see it over here. So now let's use this information to figure out what c is so that we know exactly what f of x is, and then we can evaluate f of zero. So we know that f of seven, so when x is equal to seven, we're going to, this expression is going to evaluate to this thing, 40 plus 5e to the seven. So five times e to the seventh power plus c is equal to 40, is equal to 40 plus 5e to the seventh power."}, {"video_title": "Finding specific antiderivatives exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be 5e to the x plus c. And you can verify, take the derivative of 5e to the x plus c. The derivative of 5e to the x, well that's 5e to the x, so that works out, and the derivative of c is zero, so you wouldn't see it over here. So now let's use this information to figure out what c is so that we know exactly what f of x is, and then we can evaluate f of zero. So we know that f of seven, so when x is equal to seven, we're going to, this expression is going to evaluate to this thing, 40 plus 5e to the seven. So five times e to the seventh power plus c is equal to 40, is equal to 40 plus 5e to the seventh power. And notice all I did is said okay, f of seven. Well if this is f of x, f of, let me write this down. So if this is f of seven, if this is f of x, I just replaced the x with a seven to find f of seven, and we know that f of seven is also going to be equal to that, they gave us that information."}, {"video_title": "Finding specific antiderivatives exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "So five times e to the seventh power plus c is equal to 40, is equal to 40 plus 5e to the seventh power. And notice all I did is said okay, f of seven. Well if this is f of x, f of, let me write this down. So if this is f of seven, if this is f of x, I just replaced the x with a seven to find f of seven, and we know that f of seven is also going to be equal to that, they gave us that information. But when you just look at this, it's pretty easy to figure out what c is going to be. You can subtract 5e to the seventh from both sides, and you see that c is equal to 40. And so we can rewrite f of x, we can say that f of x is equal to 5e to the x plus c, which is 40."}, {"video_title": "Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3", "Sentence": "So we've got the differential equation, the derivative of y with respect to x is equal to three times y. And we want to find the particular solution that gives us y being equal to two when x is equal to one. So I encourage you to pause this video and see if you can figure this out on your own. All right, now let's work through it together. So some of you might have immediately said, hey, this is the form of a differential equation where the solution is going to be an exponential and you just got right to it. But I'm not gonna go straight to that. I'm just gonna recognize that this is a separable differential equation and then I'm gonna solve it that way."}, {"video_title": "Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3", "Sentence": "All right, now let's work through it together. So some of you might have immediately said, hey, this is the form of a differential equation where the solution is going to be an exponential and you just got right to it. But I'm not gonna go straight to that. I'm just gonna recognize that this is a separable differential equation and then I'm gonna solve it that way. So when I say it's separable, that means we can separate all the y's, d y's on one side and all the x's, d x's on the other side. And so what I could do is if I divide both sides of this equation by y and multiply both sides by d x, I get one over y, d y, is equal to three d x. One, two, three, d x."}, {"video_title": "Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3", "Sentence": "I'm just gonna recognize that this is a separable differential equation and then I'm gonna solve it that way. So when I say it's separable, that means we can separate all the y's, d y's on one side and all the x's, d x's on the other side. And so what I could do is if I divide both sides of this equation by y and multiply both sides by d x, I get one over y, d y, is equal to three d x. One, two, three, d x. Now on the left and right hand sides, I have these clean things that I can now integrate. That's what people talk about when they say separable differential equations. Now here on the left, if I wanted to write it in a fairly general form, I could write, well, the antiderivative of one over y is gonna be the natural log of the absolute value of y. I'm taking the antiderivative with respect to y here."}, {"video_title": "Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3", "Sentence": "One, two, three, d x. Now on the left and right hand sides, I have these clean things that I can now integrate. That's what people talk about when they say separable differential equations. Now here on the left, if I wanted to write it in a fairly general form, I could write, well, the antiderivative of one over y is gonna be the natural log of the absolute value of y. I'm taking the antiderivative with respect to y here. Now I could add a constant, but I'm gonna add a constant on the right hand side so there's no reason to add two arbitrary constants on both sides. I could just add one on one side. So that is going to be equal to, the antiderivative here is going to be three x and I'll add the promised constant plus c right over there."}, {"video_title": "Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3", "Sentence": "Now here on the left, if I wanted to write it in a fairly general form, I could write, well, the antiderivative of one over y is gonna be the natural log of the absolute value of y. I'm taking the antiderivative with respect to y here. Now I could add a constant, but I'm gonna add a constant on the right hand side so there's no reason to add two arbitrary constants on both sides. I could just add one on one side. So that is going to be equal to, the antiderivative here is going to be three x and I'll add the promised constant plus c right over there. And now let's think about it a little bit. Well, we can rewrite this in exponential form. We could say, we could write that e to the three x plus c is equal to the natural log of y. I could write the natural log of y is equal to e to the three x plus c. Now I could rewrite this as equal to e to the three x times e to the c. Now e to the c is just going to be some other arbitrary constant, which I could still denote by c. They're gonna be different values, but we're just trying to just get a sense of what the structure of this thing looks like."}, {"video_title": "Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3", "Sentence": "So that is going to be equal to, the antiderivative here is going to be three x and I'll add the promised constant plus c right over there. And now let's think about it a little bit. Well, we can rewrite this in exponential form. We could say, we could write that e to the three x plus c is equal to the natural log of y. I could write the natural log of y is equal to e to the three x plus c. Now I could rewrite this as equal to e to the three x times e to the c. Now e to the c is just going to be some other arbitrary constant, which I could still denote by c. They're gonna be different values, but we're just trying to just get a sense of what the structure of this thing looks like. So we could say this is going to be some constant times e to the three x. So another way of thinking about it, saying the absolute value of y is equal to this, this isn't a function yet. We're trying to find a function solution to this differential equation."}, {"video_title": "Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3", "Sentence": "We could say, we could write that e to the three x plus c is equal to the natural log of y. I could write the natural log of y is equal to e to the three x plus c. Now I could rewrite this as equal to e to the three x times e to the c. Now e to the c is just going to be some other arbitrary constant, which I could still denote by c. They're gonna be different values, but we're just trying to just get a sense of what the structure of this thing looks like. So we could say this is going to be some constant times e to the three x. So another way of thinking about it, saying the absolute value of y is equal to this, this isn't a function yet. We're trying to find a function solution to this differential equation. So this would tell us that either y is equal to c e to the three x or y is equal to negative c e to the three x. Well, we've kept it in general terms. I haven't put any, we don't know what c is, so what we could do instead is just pick this one, and then we can solve for c, assuming this one right over here."}, {"video_title": "Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3", "Sentence": "We're trying to find a function solution to this differential equation. So this would tell us that either y is equal to c e to the three x or y is equal to negative c e to the three x. Well, we've kept it in general terms. I haven't put any, we don't know what c is, so what we could do instead is just pick this one, and then we can solve for c, assuming this one right over here. And so we will see if we can meet these constraints using this, and it'll essentially take the other one into consideration, whether we're going positive or negative. So let's do that. So when y is equal to two, when y is equal to two, I'm not going to solve for c to find the particular solution, x is equal to one, or when x is equal to one, y is equal to two."}, {"video_title": "Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3", "Sentence": "I haven't put any, we don't know what c is, so what we could do instead is just pick this one, and then we can solve for c, assuming this one right over here. And so we will see if we can meet these constraints using this, and it'll essentially take the other one into consideration, whether we're going positive or negative. So let's do that. So when y is equal to two, when y is equal to two, I'm not going to solve for c to find the particular solution, x is equal to one, or when x is equal to one, y is equal to two. So I could write it like that. And we get two is equal to c times e to the third power, three times one. And so to solve for c, I can just divide both sides by e to the third, and so I could, or I could multiply both sides times e to the negative third, and I could get two e to the negative third power is equal to c. And so let's now substitute it back in."}, {"video_title": "Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3", "Sentence": "So when y is equal to two, when y is equal to two, I'm not going to solve for c to find the particular solution, x is equal to one, or when x is equal to one, y is equal to two. So I could write it like that. And we get two is equal to c times e to the third power, three times one. And so to solve for c, I can just divide both sides by e to the third, and so I could, or I could multiply both sides times e to the negative third, and I could get two e to the negative third power is equal to c. And so let's now substitute it back in. And our particular solution is going to be y is equal to c. C is two e to the negative third power times e to the three x. Now I have, I'm taking the product of two things with the same base. I can add the exponents."}, {"video_title": "Worked example exponential solution to differential equation AP Calculus AB Khan Academy.mp3", "Sentence": "And so to solve for c, I can just divide both sides by e to the third, and so I could, or I could multiply both sides times e to the negative third, and I could get two e to the negative third power is equal to c. And so let's now substitute it back in. And our particular solution is going to be y is equal to c. C is two e to the negative third power times e to the three x. Now I have, I'm taking the product of two things with the same base. I can add the exponents. So I could say y is equal to two times e to the three x, e to the three x, and then I'll add the exponents to three x minus three. And there you go. This is one way that you could write the particular solution that meets these constraints for this separable differential equation."}, {"video_title": "The derivative & tangent line equations Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So whenever you see something like this, it doesn't hurt to try to visualize it. You might want to draw it out or just visualize it in your head, but since you can't get in my head, I will draw it out. So let me draw the information that they are giving us. So that's x-axis, that is the y-axis. Let's see, the relevant points here are two comma three and seven comma six. So let me go one, two, three, four, five, six, seven along the x-axis. And I'm gonna go one, two, three, four, five, and six along the y-axis."}, {"video_title": "The derivative & tangent line equations Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So that's x-axis, that is the y-axis. Let's see, the relevant points here are two comma three and seven comma six. So let me go one, two, three, four, five, six, seven along the x-axis. And I'm gonna go one, two, three, four, five, and six along the y-axis. And now this point, so we have the point two comma three. So let me mark that. So two comma three is right over there."}, {"video_title": "The derivative & tangent line equations Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And I'm gonna go one, two, three, four, five, and six along the y-axis. And now this point, so we have the point two comma three. So let me mark that. So two comma three is right over there. So that's two comma three. And we also have the point seven comma six. Seven comma six is going to be right over there."}, {"video_title": "The derivative & tangent line equations Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So two comma three is right over there. So that's two comma three. And we also have the point seven comma six. Seven comma six is going to be right over there. Seven comma six. Now let's remind ourselves what they're saying. They're saying the tangent line to the graph of function f at this point passes through the point seven comma six."}, {"video_title": "The derivative & tangent line equations Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Seven comma six is going to be right over there. Seven comma six. Now let's remind ourselves what they're saying. They're saying the tangent line to the graph of function f at this point passes through the point seven comma six. So if it's the tangent line to the graph at that point, it must go through two comma three. That's the only place where it intersects our graph, and it goes through seven comma six. So you only need two points to define a line."}, {"video_title": "The derivative & tangent line equations Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "They're saying the tangent line to the graph of function f at this point passes through the point seven comma six. So if it's the tangent line to the graph at that point, it must go through two comma three. That's the only place where it intersects our graph, and it goes through seven comma six. So you only need two points to define a line. And so the tangent line is going to look like, it's going to look like, let me see if I can, no, that's not right. Let me draw it. Like it's going to look, oh, that's not exactly right."}, {"video_title": "The derivative & tangent line equations Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So you only need two points to define a line. And so the tangent line is going to look like, it's going to look like, let me see if I can, no, that's not right. Let me draw it. Like it's going to look, oh, that's not exactly right. Let me try one more time. Okay, there you go. So the tangent line is going to look like that."}, {"video_title": "The derivative & tangent line equations Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Like it's going to look, oh, that's not exactly right. Let me try one more time. Okay, there you go. So the tangent line is going to look like that. It goes, it's tangent to f right at two comma three, and it goes through the point seven comma six. And so we don't know anything other than f, but we can imagine what f looks like. Our function f could, so our function f, it could look something like this."}, {"video_title": "The derivative & tangent line equations Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So the tangent line is going to look like that. It goes, it's tangent to f right at two comma three, and it goes through the point seven comma six. And so we don't know anything other than f, but we can imagine what f looks like. Our function f could, so our function f, it could look something like this. It just has to be tangent. So that line has to be tangent to our function right at that point. So our function f could look something like that."}, {"video_title": "The derivative & tangent line equations Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Our function f could, so our function f, it could look something like this. It just has to be tangent. So that line has to be tangent to our function right at that point. So our function f could look something like that. So when they say find f prime of two, they're really saying what is the slope of the tangent line when x is equal to two? So when x is equal to two, well, the slope of the tangent line is the slope of this line. They gave us, they gave us the two points that sit on the tangent line."}, {"video_title": "The derivative & tangent line equations Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So our function f could look something like that. So when they say find f prime of two, they're really saying what is the slope of the tangent line when x is equal to two? So when x is equal to two, well, the slope of the tangent line is the slope of this line. They gave us, they gave us the two points that sit on the tangent line. So we just have to figure out its slope, because that is going to be the rate of change of that function right over there. It's derivative. It's going to be the slope of the tangent line, because this is the tangent line."}, {"video_title": "The derivative & tangent line equations Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "They gave us, they gave us the two points that sit on the tangent line. So we just have to figure out its slope, because that is going to be the rate of change of that function right over there. It's derivative. It's going to be the slope of the tangent line, because this is the tangent line. So let's do that. So as we know, slope is change in y over change in x. So if we change our, to go from two comma three to seven comma six, our change in x, change in x, we go from x equals two to x equals seven, so our change in x is equal to five."}, {"video_title": "The derivative & tangent line equations Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "It's going to be the slope of the tangent line, because this is the tangent line. So let's do that. So as we know, slope is change in y over change in x. So if we change our, to go from two comma three to seven comma six, our change in x, change in x, we go from x equals two to x equals seven, so our change in x is equal to five. And our change in y, our change in y, we go from y equals three to y equals six, so our change in y is equal to three. So our change in y over change in x is going to be three over five, which is the slope of this line, which is the derivative of the function at two, because this is the tangent line at x equals two. Let's do another one of these."}, {"video_title": "The derivative & tangent line equations Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So if we change our, to go from two comma three to seven comma six, our change in x, change in x, we go from x equals two to x equals seven, so our change in x is equal to five. And our change in y, our change in y, we go from y equals three to y equals six, so our change in y is equal to three. So our change in y over change in x is going to be three over five, which is the slope of this line, which is the derivative of the function at two, because this is the tangent line at x equals two. Let's do another one of these. So, for a function g, we are given that g of negative one equals three, and g prime of negative one is equal to negative two. What is the equation of the tangent line to the graph of g at x equals negative one? All right, so once again, I think it will be helpful to graph this."}, {"video_title": "The derivative & tangent line equations Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Let's do another one of these. So, for a function g, we are given that g of negative one equals three, and g prime of negative one is equal to negative two. What is the equation of the tangent line to the graph of g at x equals negative one? All right, so once again, I think it will be helpful to graph this. So, we have our y-axis. We have our x-axis. And let's see."}, {"video_title": "The derivative & tangent line equations Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "All right, so once again, I think it will be helpful to graph this. So, we have our y-axis. We have our x-axis. And let's see. We say for a function g, we are given that g of negative one is equal to three. So the point negative one comma three is on our function. So this is negative one, and then we have one, two, and three."}, {"video_title": "The derivative & tangent line equations Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And let's see. We say for a function g, we are given that g of negative one is equal to three. So the point negative one comma three is on our function. So this is negative one, and then we have one, two, and three. So that's that right over there. That is the point, that is the point negative one comma three. It's going to be on our function."}, {"video_title": "The derivative & tangent line equations Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So this is negative one, and then we have one, two, and three. So that's that right over there. That is the point, that is the point negative one comma three. It's going to be on our function. And we also know that g prime of negative one is equal to negative two. So the slope of the tangent line right at that point on our function is going to be negative two. That's what that tells us."}, {"video_title": "The derivative & tangent line equations Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "It's going to be on our function. And we also know that g prime of negative one is equal to negative two. So the slope of the tangent line right at that point on our function is going to be negative two. That's what that tells us. The slope of the tangent line when x is equal to negative one is equal to negative two. So I could use that information to actually draw the tangent line. So let me see if I can, let me see if I can do this."}, {"video_title": "The derivative & tangent line equations Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "That's what that tells us. The slope of the tangent line when x is equal to negative one is equal to negative two. So I could use that information to actually draw the tangent line. So let me see if I can, let me see if I can do this. So it will look, so I think it will, let me just draw it like this. So it's gonna go, so it's a slope of negative two is going to look something like that. So as we can see, if we move positive one in the x direction, we go down two in the y direction."}, {"video_title": "The derivative & tangent line equations Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So let me see if I can, let me see if I can do this. So it will look, so I think it will, let me just draw it like this. So it's gonna go, so it's a slope of negative two is going to look something like that. So as we can see, if we move positive one in the x direction, we go down two in the y direction. So that has a slope of negative two. And so you might say, well where is g? Well we could draw what g could look like."}, {"video_title": "The derivative & tangent line equations Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So as we can see, if we move positive one in the x direction, we go down two in the y direction. So that has a slope of negative two. And so you might say, well where is g? Well we could draw what g could look like. G might look something like this. Might look something like that right over there where that is the tangent line. We could make g do all sorts of crazy things after that."}, {"video_title": "The derivative & tangent line equations Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Well we could draw what g could look like. G might look something like this. Might look something like that right over there where that is the tangent line. We could make g do all sorts of crazy things after that. But all we really care about is the equation for this green line. And there's a couple of ways that you could do this. You could say, well look, a line is generally, there's a bunch of different ways where you can define the equation for a line."}, {"video_title": "The derivative & tangent line equations Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "We could make g do all sorts of crazy things after that. But all we really care about is the equation for this green line. And there's a couple of ways that you could do this. You could say, well look, a line is generally, there's a bunch of different ways where you can define the equation for a line. You could say a line has a form y is equal to mx plus b where m is the slope and b is the y intercept. Well we already know what the slope of this line is. It is negative two."}, {"video_title": "The derivative & tangent line equations Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "You could say, well look, a line is generally, there's a bunch of different ways where you can define the equation for a line. You could say a line has a form y is equal to mx plus b where m is the slope and b is the y intercept. Well we already know what the slope of this line is. It is negative two. So we could say y is equal to negative two, negative two times x, times x plus b. And then to solve for b, we know that the point negative one comma three is on this line. And this goes back to some of your algebra one that you might have learned a few years ago."}, {"video_title": "The derivative & tangent line equations Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "It is negative two. So we could say y is equal to negative two, negative two times x, times x plus b. And then to solve for b, we know that the point negative one comma three is on this line. And this goes back to some of your algebra one that you might have learned a few years ago. So let's substitute negative one and three for x and y. So when y is equal to three, so three, three is equal to, is equal to negative two, negative two times x times negative one, times negative one plus b, plus b. And so let's see, this is negative two times negative one is positive two."}, {"video_title": "The derivative & tangent line equations Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And this goes back to some of your algebra one that you might have learned a few years ago. So let's substitute negative one and three for x and y. So when y is equal to three, so three, three is equal to, is equal to negative two, negative two times x times negative one, times negative one plus b, plus b. And so let's see, this is negative two times negative one is positive two. And so if you subtract two from both sides, you get one is equal to b. And there you have it. That is the equation of our line."}, {"video_title": "The derivative & tangent line equations Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And so let's see, this is negative two times negative one is positive two. And so if you subtract two from both sides, you get one is equal to b. And there you have it. That is the equation of our line. Y is equal to negative two x plus one. And there's other ways that you could have done this. You could have written the line in point slope form or you could have done it this way."}, {"video_title": "Local linearity and differentiability Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "What we're going to do in this video is explore the relationship between local linearity at a point and differentiability at a point. So local linearity is this idea that if we zoom in sufficiently on a point, that even a nonlinear function that is differentiable at that point will actually look linear. So let me show some examples of that. So let's say we had y is equal to x squared. So that's that there, clearly a nonlinear function. But we can zoom in on a point and if we zoom sufficiently in, we will see that it looks roughly linear. So let's say we want to zoom in on the point one comma one."}, {"video_title": "Local linearity and differentiability Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say we had y is equal to x squared. So that's that there, clearly a nonlinear function. But we can zoom in on a point and if we zoom sufficiently in, we will see that it looks roughly linear. So let's say we want to zoom in on the point one comma one. So let's do that. So zooming in on the point one comma one, already it is looking roughly linear at that point. And this property of local linearity is very helpful when trying to approximate a function around a point."}, {"video_title": "Local linearity and differentiability Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say we want to zoom in on the point one comma one. So let's do that. So zooming in on the point one comma one, already it is looking roughly linear at that point. And this property of local linearity is very helpful when trying to approximate a function around a point. So for example, we could figure out, we could take the derivative at the point one one, use that as the slope of our tangent line, find the equation of the tangent line, and use that equation to approximate values of our function around x equals one. And you might not need to do that for y is equal to x squared, but it could actually be very, very useful for a more complex function. But the big takeaway here, at the point one one, it is displaying this idea of local linearity."}, {"video_title": "Local linearity and differentiability Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And this property of local linearity is very helpful when trying to approximate a function around a point. So for example, we could figure out, we could take the derivative at the point one one, use that as the slope of our tangent line, find the equation of the tangent line, and use that equation to approximate values of our function around x equals one. And you might not need to do that for y is equal to x squared, but it could actually be very, very useful for a more complex function. But the big takeaway here, at the point one one, it is displaying this idea of local linearity. And it is also differentiable at that point. Now let's look at another example of a point on a function where we aren't differentiable and we also don't see the local linearity. So for example, let's do the absolute value of x."}, {"video_title": "Local linearity and differentiability Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "But the big takeaway here, at the point one one, it is displaying this idea of local linearity. And it is also differentiable at that point. Now let's look at another example of a point on a function where we aren't differentiable and we also don't see the local linearity. So for example, let's do the absolute value of x. And let me shift it over a little bit just so that we don't overlap as much. All right. So the absolute value of x minus one, it actually is differentiable as long as we're not at this corner right over here, as long as we're not at the point one comma zero."}, {"video_title": "Local linearity and differentiability Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So for example, let's do the absolute value of x. And let me shift it over a little bit just so that we don't overlap as much. All right. So the absolute value of x minus one, it actually is differentiable as long as we're not at this corner right over here, as long as we're not at the point one comma zero. For any other x value, it is differentiable. But right at x equals one, we've talked in other videos how we aren't differentiable there. And then we can use this local linearity idea to test it as well."}, {"video_title": "Local linearity and differentiability Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So the absolute value of x minus one, it actually is differentiable as long as we're not at this corner right over here, as long as we're not at the point one comma zero. For any other x value, it is differentiable. But right at x equals one, we've talked in other videos how we aren't differentiable there. And then we can use this local linearity idea to test it as well. And once again, this is not rigorous mathematics, but it is to give you an intuition. No matter how far we zoom in, we still see this sharp corner. It would be hard to construct the only tangent line, a unique line that goes through this point one comma zero."}, {"video_title": "Local linearity and differentiability Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And then we can use this local linearity idea to test it as well. And once again, this is not rigorous mathematics, but it is to give you an intuition. No matter how far we zoom in, we still see this sharp corner. It would be hard to construct the only tangent line, a unique line that goes through this point one comma zero. I can construct an actual infinite number of lines that go through one comma zero, but that do not go through the rest of the curve. And so notice, wherever you see a hard corner, like we're seeing at one comma zero in this absolute value function, that's a pretty good indication that we are not going to be differentiable at that point. Now let's zoom out a little bit and let's take another function."}, {"video_title": "Local linearity and differentiability Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "It would be hard to construct the only tangent line, a unique line that goes through this point one comma zero. I can construct an actual infinite number of lines that go through one comma zero, but that do not go through the rest of the curve. And so notice, wherever you see a hard corner, like we're seeing at one comma zero in this absolute value function, that's a pretty good indication that we are not going to be differentiable at that point. Now let's zoom out a little bit and let's take another function. Let's take a function where the differentiability or the lack of differentiability is not because of a corner, but it's because as we zoom in, it starts to look linear, but it starts to look like a vertical line. So a good example of that would be square root of, let's say, four minus x squared. So that's the top half of a circle of radius two."}, {"video_title": "Local linearity and differentiability Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's zoom out a little bit and let's take another function. Let's take a function where the differentiability or the lack of differentiability is not because of a corner, but it's because as we zoom in, it starts to look linear, but it starts to look like a vertical line. So a good example of that would be square root of, let's say, four minus x squared. So that's the top half of a circle of radius two. And let's focus on the point two comma zero because right over there, we actually are not differentiable. And if we zoom in far enough, we see right at two comma zero that we are approaching what looks like a vertical line. So once again, we would not be differentiable at two comma zero."}, {"video_title": "Local linearity and differentiability Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So that's the top half of a circle of radius two. And let's focus on the point two comma zero because right over there, we actually are not differentiable. And if we zoom in far enough, we see right at two comma zero that we are approaching what looks like a vertical line. So once again, we would not be differentiable at two comma zero. Now another thing I wanna point out, all of these, you really didn't have to zoom in too much to appreciate that, hey, I got a corner here on this absolute value function, or at two comma zero or at negative two comma zero, something a little bit stranger than normal is happening there, so maybe I'm not differentiable. But there are some functions that we don't see as typically in a algebra or precalculus or calculus class, but it can look like a hard corner from a zoomed out perspective. But as we zoom in, once again, we'll see the local linearity, and they're also differentiable at those points."}, {"video_title": "Local linearity and differentiability Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So once again, we would not be differentiable at two comma zero. Now another thing I wanna point out, all of these, you really didn't have to zoom in too much to appreciate that, hey, I got a corner here on this absolute value function, or at two comma zero or at negative two comma zero, something a little bit stranger than normal is happening there, so maybe I'm not differentiable. But there are some functions that we don't see as typically in a algebra or precalculus or calculus class, but it can look like a hard corner from a zoomed out perspective. But as we zoom in, once again, we'll see the local linearity, and they're also differentiable at those points. So a good example of that, let me actually get rid of some of these just so we can really zoom in. Let's say y is equal to x to the, and I'm gonna make a very large exponent here. So x to the 10th power, it's starting to look a little bit like a corner there."}, {"video_title": "Local linearity and differentiability Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "But as we zoom in, once again, we'll see the local linearity, and they're also differentiable at those points. So a good example of that, let me actually get rid of some of these just so we can really zoom in. Let's say y is equal to x to the, and I'm gonna make a very large exponent here. So x to the 10th power, it's starting to look a little bit like a corner there. Let's make it to the 100th power. Well now, it's looking even more like a corner there. Let me go to the 1,000th power just for good measure."}, {"video_title": "Local linearity and differentiability Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So x to the 10th power, it's starting to look a little bit like a corner there. Let's make it to the 100th power. Well now, it's looking even more like a corner there. Let me go to the 1,000th power just for good measure. So at this scale, it looks like we have a corner at the point one comma zero. Now this curve actually does not go to the point one comma zero. If x is one, then y is going to be one."}, {"video_title": "Local linearity and differentiability Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Let me go to the 1,000th power just for good measure. So at this scale, it looks like we have a corner at the point one comma zero. Now this curve actually does not go to the point one comma zero. If x is one, then y is going to be one. And we'll see that as we zoom in, this, what looks like a hard corner, is going to soften. And that's good because this function is actually differentiable at every value of x. It's a little bit more exotic than what we typically see."}, {"video_title": "Local linearity and differentiability Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "If x is one, then y is going to be one. And we'll see that as we zoom in, this, what looks like a hard corner, is going to soften. And that's good because this function is actually differentiable at every value of x. It's a little bit more exotic than what we typically see. But as we zoom in, we'll actually see that. Let's just zoom in on what looks like a fairly hard corner. But if we zoom sufficiently enough, even at the part that looks like the hardest part of the corner, the real corner, we'll see that it starts to soften and it curves."}, {"video_title": "Local linearity and differentiability Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "It's a little bit more exotic than what we typically see. But as we zoom in, we'll actually see that. Let's just zoom in on what looks like a fairly hard corner. But if we zoom sufficiently enough, even at the part that looks like the hardest part of the corner, the real corner, we'll see that it starts to soften and it curves. And if we zoom in sufficiently, it will actually look like a line. It's hard to believe when you're really zoomed out. And I'm going at the point that really looked like a corner from a distance."}, {"video_title": "Local linearity and differentiability Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "But if we zoom sufficiently enough, even at the part that looks like the hardest part of the corner, the real corner, we'll see that it starts to soften and it curves. And if we zoom in sufficiently, it will actually look like a line. It's hard to believe when you're really zoomed out. And I'm going at the point that really looked like a corner from a distance. But as we zoom in, we see once again this local linearity. And it's a non-vertical line. And so once again, this is true at any point on this curve, that we are going to be differentiable."}, {"video_title": "Local linearity and differentiability Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And I'm going at the point that really looked like a corner from a distance. But as we zoom in, we see once again this local linearity. And it's a non-vertical line. And so once again, this is true at any point on this curve, that we are going to be differentiable. So the whole point here is sometimes you might have to zoom in a lot. A tool like Desmos, which I'm using right now, is very helpful for doing that. And this isn't rigorous mathematics, but it's to give you an intuitive sense that if you zoom in sufficiently and you start to see a curve looking more and more like a line, good indication that you are differentiable."}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And that is just a fancy way of saying does the function have a defined derivative at a point? So let's just remind ourselves a definition of a derivative. And there's multiple ways of writing this. For the sake of this video, I'll write it as the derivative of our function at point C, this is Lagrange notation with this F prime, the derivative of our function F at C, is going to be equal to the limit as X approaches C of F of X minus F of C over X minus C. And at first when you see this formula, and we've seen it before, it looks a little bit strange, but all it is is it's calculating the slope. This is our change in the value of our function, or you could think of it as our change in Y if Y is equal to F of X. And this is our change in X. And we're just trying to see, well, what is that slope as X gets closer and closer to C, as our change in X gets closer and closer to zero?"}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "For the sake of this video, I'll write it as the derivative of our function at point C, this is Lagrange notation with this F prime, the derivative of our function F at C, is going to be equal to the limit as X approaches C of F of X minus F of C over X minus C. And at first when you see this formula, and we've seen it before, it looks a little bit strange, but all it is is it's calculating the slope. This is our change in the value of our function, or you could think of it as our change in Y if Y is equal to F of X. And this is our change in X. And we're just trying to see, well, what is that slope as X gets closer and closer to C, as our change in X gets closer and closer to zero? And we talk about that in other videos. So I'm now going to make a few claims in this video. And I'm not going to prove them rigorously."}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And we're just trying to see, well, what is that slope as X gets closer and closer to C, as our change in X gets closer and closer to zero? And we talk about that in other videos. So I'm now going to make a few claims in this video. And I'm not going to prove them rigorously. There is another video that will go a little bit more into the proof direction. But this is more to get an intuition. And so the first claim that I'm going to make is if F is differentiable at X equals C, at X equals C, then F is continuous at X equals C. So I'm saying if we know it's differentiable, if we can find this limit, if we can find this derivative at X equals C, then our function is also continuous at X equals C. It doesn't necessarily mean the other way around."}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And I'm not going to prove them rigorously. There is another video that will go a little bit more into the proof direction. But this is more to get an intuition. And so the first claim that I'm going to make is if F is differentiable at X equals C, at X equals C, then F is continuous at X equals C. So I'm saying if we know it's differentiable, if we can find this limit, if we can find this derivative at X equals C, then our function is also continuous at X equals C. It doesn't necessarily mean the other way around. And actually, we'll look at a case where it's not necessarily the case the other way around, that if you're continuous, then you're definitely differentiable. But another way to interpret what I just wrote down is if you are not continuous, then you definitely will not be differentiable. If F not continuous at X equals C, at X equals C, then F is not differentiable."}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And so the first claim that I'm going to make is if F is differentiable at X equals C, at X equals C, then F is continuous at X equals C. So I'm saying if we know it's differentiable, if we can find this limit, if we can find this derivative at X equals C, then our function is also continuous at X equals C. It doesn't necessarily mean the other way around. And actually, we'll look at a case where it's not necessarily the case the other way around, that if you're continuous, then you're definitely differentiable. But another way to interpret what I just wrote down is if you are not continuous, then you definitely will not be differentiable. If F not continuous at X equals C, at X equals C, then F is not differentiable. Differentiable at X is equal to C. So let me give a few examples of a non-continuous function. And then think about would we be able to find this limit? So the first is where you have a discontinuity."}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "If F not continuous at X equals C, at X equals C, then F is not differentiable. Differentiable at X is equal to C. So let me give a few examples of a non-continuous function. And then think about would we be able to find this limit? So the first is where you have a discontinuity. Our function is defined at C. It's equal to this value. But you can see as X becomes larger than C, it just jumps down and shifts right over here. So what would happen if you were trying to find this limit?"}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So the first is where you have a discontinuity. Our function is defined at C. It's equal to this value. But you can see as X becomes larger than C, it just jumps down and shifts right over here. So what would happen if you were trying to find this limit? Well, remember, all this is is a slope of a line between when X is some arbitrary value. Let's say it's out here. So that would be X."}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So what would happen if you were trying to find this limit? Well, remember, all this is is a slope of a line between when X is some arbitrary value. Let's say it's out here. So that would be X. This would be the point X comma F of X. And then this is the point C comma F of C right over here. So this is C comma F of C. So if you find the left-sided limit right over here, you're essentially saying, okay, let's find this slope."}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So that would be X. This would be the point X comma F of X. And then this is the point C comma F of C right over here. So this is C comma F of C. So if you find the left-sided limit right over here, you're essentially saying, okay, let's find this slope. And then let me get a little bit closer. And let's get X a little bit closer, and then let's find this slope. And then let's get X even closer than that and find this slope."}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So this is C comma F of C. So if you find the left-sided limit right over here, you're essentially saying, okay, let's find this slope. And then let me get a little bit closer. And let's get X a little bit closer, and then let's find this slope. And then let's get X even closer than that and find this slope. And in all of those cases, it would be zero. The slope is zero. So one way to think about it, the derivative, or this limit as we approach from the left, seems to be approaching zero."}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And then let's get X even closer than that and find this slope. And in all of those cases, it would be zero. The slope is zero. So one way to think about it, the derivative, or this limit as we approach from the left, seems to be approaching zero. But what about if we were to take Xs to the right? So instead of our Xs being there, what if we were to take Xs right over here? Well, for this point, X comma F of X, our slope, if we take F of X minus F of C over X minus C, that would be the slope of this line."}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So one way to think about it, the derivative, or this limit as we approach from the left, seems to be approaching zero. But what about if we were to take Xs to the right? So instead of our Xs being there, what if we were to take Xs right over here? Well, for this point, X comma F of X, our slope, if we take F of X minus F of C over X minus C, that would be the slope of this line. If we get X to be even closer, let's say right over here, then this would be the slope of this line. If we get even closer, then this expression would be the slope of this line. And so as we get closer and closer to X being equal to C, we see that our slope is actually approaching negative infinity."}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Well, for this point, X comma F of X, our slope, if we take F of X minus F of C over X minus C, that would be the slope of this line. If we get X to be even closer, let's say right over here, then this would be the slope of this line. If we get even closer, then this expression would be the slope of this line. And so as we get closer and closer to X being equal to C, we see that our slope is actually approaching negative infinity. And most importantly, it's approaching a very different value from the right. This expression is approaching a very different value from the right as it is from the left. And so in this case, this limit up here won't exist."}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And so as we get closer and closer to X being equal to C, we see that our slope is actually approaching negative infinity. And most importantly, it's approaching a very different value from the right. This expression is approaching a very different value from the right as it is from the left. And so in this case, this limit up here won't exist. So we can clearly say this is not differentiable. So once again, not a proof here. I'm just getting an intuition for, if something isn't continuous, it's pretty clear, at least in this case, that it's not going to be differentiable."}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And so in this case, this limit up here won't exist. So we can clearly say this is not differentiable. So once again, not a proof here. I'm just getting an intuition for, if something isn't continuous, it's pretty clear, at least in this case, that it's not going to be differentiable. Let's look at another case. Let's look at a case where we have what's sometimes called a removable discontinuity or a point discontinuity. So once again, let's say we're approaching from the left."}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "I'm just getting an intuition for, if something isn't continuous, it's pretty clear, at least in this case, that it's not going to be differentiable. Let's look at another case. Let's look at a case where we have what's sometimes called a removable discontinuity or a point discontinuity. So once again, let's say we're approaching from the left. This is X, this is the point X comma F of X. Now what's interesting is, where as this expression is the slope of the line connecting X comma F of X and C comma F of C, which is this point, not that point. Remember, we have this removable discontinuity right over here."}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So once again, let's say we're approaching from the left. This is X, this is the point X comma F of X. Now what's interesting is, where as this expression is the slope of the line connecting X comma F of X and C comma F of C, which is this point, not that point. Remember, we have this removable discontinuity right over here. And so this would be, this expression is calculating the slope of that line. And then if X gets even closer to C, well then we're gonna be calculating the slope of that line. If X gets even closer to C, we're gonna be calculating the slope of that line."}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Remember, we have this removable discontinuity right over here. And so this would be, this expression is calculating the slope of that line. And then if X gets even closer to C, well then we're gonna be calculating the slope of that line. If X gets even closer to C, we're gonna be calculating the slope of that line. And so as we approach from the left, as X approaches C from the left, we actually have a situation where this expression right over here is going to approach negative infinity. And if we approach from the right, if we approach with X as large as in C, well this is our X comma F of X. So we have a positive slope."}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "If X gets even closer to C, we're gonna be calculating the slope of that line. And so as we approach from the left, as X approaches C from the left, we actually have a situation where this expression right over here is going to approach negative infinity. And if we approach from the right, if we approach with X as large as in C, well this is our X comma F of X. So we have a positive slope. And then as we get closer, it gets more positive. More positive approaches positive infinity. But either way, it's not approaching a finite value."}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So we have a positive slope. And then as we get closer, it gets more positive. More positive approaches positive infinity. But either way, it's not approaching a finite value. And one side is approaching positive infinity and the other side is approaching negative infinity. This, the limit of this expression is not going to exist. So once again, I'm not doing a rigorous proof here."}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "But either way, it's not approaching a finite value. And one side is approaching positive infinity and the other side is approaching negative infinity. This, the limit of this expression is not going to exist. So once again, I'm not doing a rigorous proof here. But try to construct a discontinuous function where you will be able to find this. It is very, very hard. And you might say, well what about the situations where F is not even defined at C?"}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So once again, I'm not doing a rigorous proof here. But try to construct a discontinuous function where you will be able to find this. It is very, very hard. And you might say, well what about the situations where F is not even defined at C? Which for sure, you're not gonna be continuous if F is not defined at C. Well if F is not defined at C, then this part of the expression wouldn't even make sense. So you definitely wouldn't be differentiable. But now let's ask another thing."}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And you might say, well what about the situations where F is not even defined at C? Which for sure, you're not gonna be continuous if F is not defined at C. Well if F is not defined at C, then this part of the expression wouldn't even make sense. So you definitely wouldn't be differentiable. But now let's ask another thing. I've just given you good arguments for when you're not continuous, you're not going to be differentiable. But can we make another claim that if you are continuous, then you definitely will be differentiable? Well it turns out that there are for sure many functions, an infinite number of functions, that can be continuous at C but not differentiable."}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "But now let's ask another thing. I've just given you good arguments for when you're not continuous, you're not going to be differentiable. But can we make another claim that if you are continuous, then you definitely will be differentiable? Well it turns out that there are for sure many functions, an infinite number of functions, that can be continuous at C but not differentiable. So for example, this could be an absolute value function. It doesn't have to be an absolute value function, but this could be Y is equal to the absolute value of X minus C. And why is this one not differentiable at C? Well think about what's happening."}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Well it turns out that there are for sure many functions, an infinite number of functions, that can be continuous at C but not differentiable. So for example, this could be an absolute value function. It doesn't have to be an absolute value function, but this could be Y is equal to the absolute value of X minus C. And why is this one not differentiable at C? Well think about what's happening. Think about this expression. Remember, this expression, all it's doing is calculating the slope between the point X comma F of X and the point C comma F of C. So if X is say out here, this is X comma F of X, it's going to be calculated, and so as we take the limit as X approaches C from the left, we'll be looking at this slope. And then as we get closer, we'll be looking at this slope, which is actually going to be the same."}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Well think about what's happening. Think about this expression. Remember, this expression, all it's doing is calculating the slope between the point X comma F of X and the point C comma F of C. So if X is say out here, this is X comma F of X, it's going to be calculated, and so as we take the limit as X approaches C from the left, we'll be looking at this slope. And then as we get closer, we'll be looking at this slope, which is actually going to be the same. In this case it would be a negative one. So as X approaches C from the left, this expression would be negative one. But as X approaches C from the right, this expression is going to be one."}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And then as we get closer, we'll be looking at this slope, which is actually going to be the same. In this case it would be a negative one. So as X approaches C from the left, this expression would be negative one. But as X approaches C from the right, this expression is going to be one. The slope of the line that connects these points is one. The slope of the line that connects these points is one. So the limit of this expression, or I would say the value of this expression, is approaching two different values as X approaches C from the left or the right."}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "But as X approaches C from the right, this expression is going to be one. The slope of the line that connects these points is one. The slope of the line that connects these points is one. So the limit of this expression, or I would say the value of this expression, is approaching two different values as X approaches C from the left or the right. From the left, it's approaching negative one, or it's constantly negative one, and so it's approaching negative one, you could say. And from the right, it's one, and it's approaching one the entire time. And so we know if you're approaching two different values from on the left-sided or the right-sided limit, then this limit will not exist."}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So the limit of this expression, or I would say the value of this expression, is approaching two different values as X approaches C from the left or the right. From the left, it's approaching negative one, or it's constantly negative one, and so it's approaching negative one, you could say. And from the right, it's one, and it's approaching one the entire time. And so we know if you're approaching two different values from on the left-sided or the right-sided limit, then this limit will not exist. So here, this is not differentiable. And even intuitively, we think of the derivative as the slope of the tangent line, and you could actually draw an infinite number of tangent lines here. It's one way to think about it."}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And so we know if you're approaching two different values from on the left-sided or the right-sided limit, then this limit will not exist. So here, this is not differentiable. And even intuitively, we think of the derivative as the slope of the tangent line, and you could actually draw an infinite number of tangent lines here. It's one way to think about it. You could say, well, maybe this is the tangent line right over there, but then why can't I make something like this the tangent line? That only intersects at the point C comma zero, and then you could keep doing things like that. Why can't that be the tangent line?"}, {"video_title": "Differentiability and continuity Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "It's one way to think about it. You could say, well, maybe this is the tangent line right over there, but then why can't I make something like this the tangent line? That only intersects at the point C comma zero, and then you could keep doing things like that. Why can't that be the tangent line? And you could go on and on and on. So the big takeaways here, at least intuitively, in a future video, I'm going to prove to you that if F is differentiable at C, then it is continuous at C, which can also be interpreted that if you're not continuous at C, then you're not gonna be differentiable. These two examples will hopefully give you some intuition for that, but it's not the case that if something is continuous, that it has to be differentiable."}, {"video_title": "Proof of special case of l'H\u00f4pital's rule Differential Calculus Khan Academy.mp3", "Sentence": "And it's a more constrained version of the general case we've been looking at, but it's still very powerful and very applicable. And the reason why we're going to go over this special case is because its proof is fairly straightforward and will give you an intuition for why L'Hospital's Rule works at all. So the special case of L'Hospital's Rule is a situation where f of a is equal to 0, f prime of a exists, g of a is equal to 0, g prime of a exists. If these constraints are met, then the limit as x approaches a of f of x over g of x is going to be equal to f prime of a over g prime of a. So it's very similar to the general case, it's a little bit more constrained. We're assuming that f prime of a exists, we're not just taking the limit now. We're assuming f prime of a and g prime of a actually exist."}, {"video_title": "Proof of special case of l'H\u00f4pital's rule Differential Calculus Khan Academy.mp3", "Sentence": "If these constraints are met, then the limit as x approaches a of f of x over g of x is going to be equal to f prime of a over g prime of a. So it's very similar to the general case, it's a little bit more constrained. We're assuming that f prime of a exists, we're not just taking the limit now. We're assuming f prime of a and g prime of a actually exist. But notice if we substitute a right over here, we get 0 over 0. But then if the derivatives exist, we can just evaluate the derivatives at a and then we get the limit. So this is very close to the general case of L'Hospital's Rule."}, {"video_title": "Proof of special case of l'H\u00f4pital's rule Differential Calculus Khan Academy.mp3", "Sentence": "We're assuming f prime of a and g prime of a actually exist. But notice if we substitute a right over here, we get 0 over 0. But then if the derivatives exist, we can just evaluate the derivatives at a and then we get the limit. So this is very close to the general case of L'Hospital's Rule. Now let's actually prove it. And to prove it, we're going to start with the right hand and then show that if we use the definition of derivatives, we get the left hand right over here. So let me do that."}, {"video_title": "Proof of special case of l'H\u00f4pital's rule Differential Calculus Khan Academy.mp3", "Sentence": "So this is very close to the general case of L'Hospital's Rule. Now let's actually prove it. And to prove it, we're going to start with the right hand and then show that if we use the definition of derivatives, we get the left hand right over here. So let me do that. So I'll do it right over here. So f prime of a is equal to what by the definition of derivatives? Well, we could view that as the limit as x approaches a of f of x minus f of a over x minus a."}, {"video_title": "Proof of special case of l'H\u00f4pital's rule Differential Calculus Khan Academy.mp3", "Sentence": "So let me do that. So I'll do it right over here. So f prime of a is equal to what by the definition of derivatives? Well, we could view that as the limit as x approaches a of f of x minus f of a over x minus a. So this is literally just the slope between two points. So if you have your function f of x like this, this is the point a, f of a right over here. This right over here is the point x, f of x."}, {"video_title": "Proof of special case of l'H\u00f4pital's rule Differential Calculus Khan Academy.mp3", "Sentence": "Well, we could view that as the limit as x approaches a of f of x minus f of a over x minus a. So this is literally just the slope between two points. So if you have your function f of x like this, this is the point a, f of a right over here. This right over here is the point x, f of x. This expression right over here is the slope between these two points. The change in our y value is f of x minus f of a. The change in our x value is x minus a."}, {"video_title": "Proof of special case of l'H\u00f4pital's rule Differential Calculus Khan Academy.mp3", "Sentence": "This right over here is the point x, f of x. This expression right over here is the slope between these two points. The change in our y value is f of x minus f of a. The change in our x value is x minus a. So this expression is just the slope of this line. And we're just taking the line that connects these two points. That's the slope of it."}, {"video_title": "Proof of special case of l'H\u00f4pital's rule Differential Calculus Khan Academy.mp3", "Sentence": "The change in our x value is x minus a. So this expression is just the slope of this line. And we're just taking the line that connects these two points. That's the slope of it. I'll do that in white. The slope of the line that connects those two points. And we're taking the limit as x gets closer and closer and closer to a."}, {"video_title": "Proof of special case of l'H\u00f4pital's rule Differential Calculus Khan Academy.mp3", "Sentence": "That's the slope of it. I'll do that in white. The slope of the line that connects those two points. And we're taking the limit as x gets closer and closer and closer to a. So this is just another way of writing the definition of the derivative. So that's fine. Let's do the same thing for g prime of a."}, {"video_title": "Proof of special case of l'H\u00f4pital's rule Differential Calculus Khan Academy.mp3", "Sentence": "And we're taking the limit as x gets closer and closer and closer to a. So this is just another way of writing the definition of the derivative. So that's fine. Let's do the same thing for g prime of a. So f prime of a over g prime of a is going to be this business, which is in orange, f prime of a over g prime of a, which we can write as the limit as x approaches a of g of x minus g of a over x minus a. Well, in the numerator we're taking the limit as x approaches a. In the denominator we're taking the limit as x approaches a."}, {"video_title": "Proof of special case of l'H\u00f4pital's rule Differential Calculus Khan Academy.mp3", "Sentence": "Let's do the same thing for g prime of a. So f prime of a over g prime of a is going to be this business, which is in orange, f prime of a over g prime of a, which we can write as the limit as x approaches a of g of x minus g of a over x minus a. Well, in the numerator we're taking the limit as x approaches a. In the denominator we're taking the limit as x approaches a. So we can just rewrite this. This we can rewrite as the limit as x approaches a of all this business in orange, f of x minus f of a over x minus a over all the business in green, g of x minus g of a, all of that over x minus a. Now to simplify this, we can multiply the numerator and the denominator by x minus a to get rid of these x minus a."}, {"video_title": "Proof of special case of l'H\u00f4pital's rule Differential Calculus Khan Academy.mp3", "Sentence": "In the denominator we're taking the limit as x approaches a. So we can just rewrite this. This we can rewrite as the limit as x approaches a of all this business in orange, f of x minus f of a over x minus a over all the business in green, g of x minus g of a, all of that over x minus a. Now to simplify this, we can multiply the numerator and the denominator by x minus a to get rid of these x minus a. So let's do that. Let's multiply by x minus a over x minus a. So the numerator x minus a and we're dividing by x minus a, those cancel out, and then these two cancel out, and we're left with this thing over here is equal to the limit as x approaches a of, in the numerator we have f of x minus f of a, and in the denominator we have g of x minus g of a. I think you see where this is going."}, {"video_title": "Proof of special case of l'H\u00f4pital's rule Differential Calculus Khan Academy.mp3", "Sentence": "Now to simplify this, we can multiply the numerator and the denominator by x minus a to get rid of these x minus a. So let's do that. Let's multiply by x minus a over x minus a. So the numerator x minus a and we're dividing by x minus a, those cancel out, and then these two cancel out, and we're left with this thing over here is equal to the limit as x approaches a of, in the numerator we have f of x minus f of a, and in the denominator we have g of x minus g of a. I think you see where this is going. What is f of a equal to? Well, we assumed f of a is equal to zero. That's why we're using L'Hospital's Rule from the get-go."}, {"video_title": "Proof of special case of l'H\u00f4pital's rule Differential Calculus Khan Academy.mp3", "Sentence": "So the numerator x minus a and we're dividing by x minus a, those cancel out, and then these two cancel out, and we're left with this thing over here is equal to the limit as x approaches a of, in the numerator we have f of x minus f of a, and in the denominator we have g of x minus g of a. I think you see where this is going. What is f of a equal to? Well, we assumed f of a is equal to zero. That's why we're using L'Hospital's Rule from the get-go. f of a is equal to zero, g of a is equal to zero. f of a is equal to zero, g of a is equal to zero, and this simplifies to the limit as x approaches a of f prime of x, sorry, of f of x, we've got to be careful, of f of x over g of x. So we just showed that if f of a equals zero, g of a equals zero, and these two derivatives exist, then the derivatives evaluated a over each other are going to be equal to the limit as x approaches a of f of x over g of x, or the limit as x approaches a of f of x over g of x is going to be equal to f prime of a over g prime of a."}, {"video_title": "Second derivatives (vector-valued functions) Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "When I say vector-valued, it means you give me a t, it's a function of t, and so you give me a t, I'm not just gonna give you a number, I'm gonna give you a vector. And as we'll see, you're gonna get a two-dimensional vector. You could view this as the x component of the vector and the y component of the vector. And you are probably familiar by now that there's multiple notations for even a two-dimensional vector. For example, you could use what's often viewed as engineering notation here, where the x component is being multiplied by the horizontal unit vector. So you might see something like that, where that's the unit vector, plus the y component, four t to the fourth plus two t plus one, is multiplied by the vertical unit vector. So these are both representing the same thing, it just has a different notation."}, {"video_title": "Second derivatives (vector-valued functions) Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "And you are probably familiar by now that there's multiple notations for even a two-dimensional vector. For example, you could use what's often viewed as engineering notation here, where the x component is being multiplied by the horizontal unit vector. So you might see something like that, where that's the unit vector, plus the y component, four t to the fourth plus two t plus one, is multiplied by the vertical unit vector. So these are both representing the same thing, it just has a different notation. And sometimes you'll see vector-valued functions with an arrow on top to make it explicit that this is a vector-valued function. Sometimes you'll just hear people say, well, let h be a vector-valued function, and they might not write that arrow on top. So now that we have that out of the way, what we are interested in is, well, let's find the first and second derivatives of h with respect to t. So let's take the first derivative, h prime of t. Well, as you'll see, that's actually quite straightforward."}, {"video_title": "Second derivatives (vector-valued functions) Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "So these are both representing the same thing, it just has a different notation. And sometimes you'll see vector-valued functions with an arrow on top to make it explicit that this is a vector-valued function. Sometimes you'll just hear people say, well, let h be a vector-valued function, and they might not write that arrow on top. So now that we have that out of the way, what we are interested in is, well, let's find the first and second derivatives of h with respect to t. So let's take the first derivative, h prime of t. Well, as you'll see, that's actually quite straightforward. You're just gonna take the respective components, take the derivative of the respective components with respect to t. So the x component with respect to t, if you were to take the derivative, so with respect to t, what are you going to get? Well, we're gonna use the power rule right over here. Five times the negative one, or times the negative, you're gonna get negative five, times t to the five minus one power, so t to the fourth power."}, {"video_title": "Second derivatives (vector-valued functions) Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "So now that we have that out of the way, what we are interested in is, well, let's find the first and second derivatives of h with respect to t. So let's take the first derivative, h prime of t. Well, as you'll see, that's actually quite straightforward. You're just gonna take the respective components, take the derivative of the respective components with respect to t. So the x component with respect to t, if you were to take the derivative, so with respect to t, what are you going to get? Well, we're gonna use the power rule right over here. Five times the negative one, or times the negative, you're gonna get negative five, times t to the five minus one power, so t to the fourth power. The derivative with respect to t of negative six, well, that's just zero. So that's the rate of change of the x component with respect to t. And now we go to the y component. So we're gonna do the same thing."}, {"video_title": "Second derivatives (vector-valued functions) Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "Five times the negative one, or times the negative, you're gonna get negative five, times t to the five minus one power, so t to the fourth power. The derivative with respect to t of negative six, well, that's just zero. So that's the rate of change of the x component with respect to t. And now we go to the y component. So we're gonna do the same thing. Derivative with respect to t is going to be, and once again, we just use the power rule, four times four is 16t to the third power. Derivative of 2t is just two. And then derivative of a constant, well, that's zero, we've already seen that."}, {"video_title": "Second derivatives (vector-valued functions) Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "So we're gonna do the same thing. Derivative with respect to t is going to be, and once again, we just use the power rule, four times four is 16t to the third power. Derivative of 2t is just two. And then derivative of a constant, well, that's zero, we've already seen that. So there you have it. So this is the rate of change of the x component with respect to t. This is the rate of change of the y component with respect to t. And one way to do it, and vectors can represent many, many, many different things, but the type of a two-dimensional vector like this, you could imagine this being, h of t being a position vector in two dimensions. And then if you're looking at the rate of change of position with respect to time, well, then this would be the velocity vector."}, {"video_title": "Second derivatives (vector-valued functions) Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "And then derivative of a constant, well, that's zero, we've already seen that. So there you have it. So this is the rate of change of the x component with respect to t. This is the rate of change of the y component with respect to t. And one way to do it, and vectors can represent many, many, many different things, but the type of a two-dimensional vector like this, you could imagine this being, h of t being a position vector in two dimensions. And then if you're looking at the rate of change of position with respect to time, well, then this would be the velocity vector. And then if we were to take the derivative of this with respect to time, well, we're going to get the acceleration vector. So if we say h prime prime of t, what is that going to be equal to? H prime prime of t?"}, {"video_title": "Second derivatives (vector-valued functions) Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "And then if you're looking at the rate of change of position with respect to time, well, then this would be the velocity vector. And then if we were to take the derivative of this with respect to time, well, we're going to get the acceleration vector. So if we say h prime prime of t, what is that going to be equal to? H prime prime of t? Well, we just apply the power rule again. So four times negative five is equal to negative 20 t to the four minus one, so t to the third power. And then we have three times 16 is 48t squared, and then the derivative of two is just zero."}, {"video_title": "Second derivatives (vector-valued functions) Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "H prime prime of t? Well, we just apply the power rule again. So four times negative five is equal to negative 20 t to the four minus one, so t to the third power. And then we have three times 16 is 48t squared, and then the derivative of two is just zero. And so there you have it. For any, if you view t as time, for any time, if you view this one as position, this one as velocity, and this as acceleration, you could, this would now give you the position, velocity, and acceleration. But it's important to realize that these vectors could represent anything of a two-dimensional nature."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "It would look something like this. It would look something like that. So that is e to the x. And what I want to do is I want to approximate f of x is equal to e to the x using a Taylor series approximation, or Taylor series expansion. And I want to do it not around x is equal to 0. I want to do it around x is equal to 3, just to pick another arbitrary value. So we're going to do it around x is equal to 3."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "And what I want to do is I want to approximate f of x is equal to e to the x using a Taylor series approximation, or Taylor series expansion. And I want to do it not around x is equal to 0. I want to do it around x is equal to 3, just to pick another arbitrary value. So we're going to do it around x is equal to 3. This is x is equal to 3. This right there, that is f of 3. f of 3 is e to the third power. So this is e to the third power right over there."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "So we're going to do it around x is equal to 3. This is x is equal to 3. This right there, that is f of 3. f of 3 is e to the third power. So this is e to the third power right over there. So when we take the Taylor series expansion, if we have a 0 degree polynomial approximating it, the best we could probably do is have a constant function going straight through e to the third. If we do a first order approximation, so we have a first degree term, then it will be the tangent line. And as we add more and more degrees to it, we should hopefully be able to kind of contour or converge with the curve better and better and better."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "So this is e to the third power right over there. So when we take the Taylor series expansion, if we have a 0 degree polynomial approximating it, the best we could probably do is have a constant function going straight through e to the third. If we do a first order approximation, so we have a first degree term, then it will be the tangent line. And as we add more and more degrees to it, we should hopefully be able to kind of contour or converge with the curve better and better and better. And in the future, we'll talk a little bit more about how we can test for convergences and how well are we converging and all of that type of thing. But with that said, let's just apply the formula that hopefully we got the intuition for in the last video. So the Taylor series expansion for f of x is equal to e to the x will be the polynomial."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "And as we add more and more degrees to it, we should hopefully be able to kind of contour or converge with the curve better and better and better. And in the future, we'll talk a little bit more about how we can test for convergences and how well are we converging and all of that type of thing. But with that said, let's just apply the formula that hopefully we got the intuition for in the last video. So the Taylor series expansion for f of x is equal to e to the x will be the polynomial. So what's f of c? Well, if x is equal to 3, we're saying that c is 3 in this situation. So if c is 3, f of 3 is e to the third power."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "So the Taylor series expansion for f of x is equal to e to the x will be the polynomial. So what's f of c? Well, if x is equal to 3, we're saying that c is 3 in this situation. So if c is 3, f of 3 is e to the third power. So it's e to the third power plus. What's f prime of c? Well, f prime of x is also going to be e to the x."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "So if c is 3, f of 3 is e to the third power. So it's e to the third power plus. What's f prime of c? Well, f prime of x is also going to be e to the x. You take the derivative of e to the x, you get e to the x. That's one of the super cool things about e to the x. So this is also f prime of x."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "Well, f prime of x is also going to be e to the x. You take the derivative of e to the x, you get e to the x. That's one of the super cool things about e to the x. So this is also f prime of x. Frankly, this is the same thing as f, the nth derivative of x. You could just keep taking the derivative of this and you'll get e to the x. So f prime of x is e to the x."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "So this is also f prime of x. Frankly, this is the same thing as f, the nth derivative of x. You could just keep taking the derivative of this and you'll get e to the x. So f prime of x is e to the x. You evaluate that at 3. You get e to the third power again times x minus 3. c is 3. Plus the second derivative of our function is still e to the x."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "So f prime of x is e to the x. You evaluate that at 3. You get e to the third power again times x minus 3. c is 3. Plus the second derivative of our function is still e to the x. Evaluate that at 3. You get e to the third power over 2 factorial times x minus 3 to the second power. And then we could keep going."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "Plus the second derivative of our function is still e to the x. Evaluate that at 3. You get e to the third power over 2 factorial times x minus 3 to the second power. And then we could keep going. The third derivative is still e to the x. Evaluate that at 3. c is 3 in this situation. So you get e to the third power over 3 factorial times x minus 3 to the third power."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "And then we could keep going. The third derivative is still e to the x. Evaluate that at 3. c is 3 in this situation. So you get e to the third power over 3 factorial times x minus 3 to the third power. And we can keep going with this, but I think you get the general idea. But what's even more interesting than just kind of going through the mechanics of finding the expansion is seeing how, as we add more and more terms, it starts to approximate e to the x better and better and better. And our approximation gets good further and further away from x is equal to 3."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "So you get e to the third power over 3 factorial times x minus 3 to the third power. And we can keep going with this, but I think you get the general idea. But what's even more interesting than just kind of going through the mechanics of finding the expansion is seeing how, as we add more and more terms, it starts to approximate e to the x better and better and better. And our approximation gets good further and further away from x is equal to 3. And to do that, I used Wolfram Alpha, available at wolframalpha.com. And I think I typed in like Taylor series expansion e to the x and x equals 3. And I just knew what I wanted and gave me all of this business right over here."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "And our approximation gets good further and further away from x is equal to 3. And to do that, I used Wolfram Alpha, available at wolframalpha.com. And I think I typed in like Taylor series expansion e to the x and x equals 3. And I just knew what I wanted and gave me all of this business right over here. And I actually calculated the expansion. You can see it's the exact same thing that we have over here. e to the third plus e to the third times x minus 3."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "And I just knew what I wanted and gave me all of this business right over here. And I actually calculated the expansion. You can see it's the exact same thing that we have over here. e to the third plus e to the third times x minus 3. We have e to the third plus e to the third times x minus 3 plus 1 half. They actually expanded out the factorial. So instead of 3 factorial, they wrote a 6 over here."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "e to the third plus e to the third times x minus 3. We have e to the third plus e to the third times x minus 3 plus 1 half. They actually expanded out the factorial. So instead of 3 factorial, they wrote a 6 over here. And they did a bunch of terms up here. But what's even more interesting is that they actually graph each of these polynomials with more and more terms. So in orange, we have e to the x."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "So instead of 3 factorial, they wrote a 6 over here. And they did a bunch of terms up here. But what's even more interesting is that they actually graph each of these polynomials with more and more terms. So in orange, we have e to the x. We have f of x is equal to e to the x. And then they tell us order n approximation shown with n dots. So the order 1 approximation, so that should be the situation where we have a first degree polynomial."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "So in orange, we have e to the x. We have f of x is equal to e to the x. And then they tell us order n approximation shown with n dots. So the order 1 approximation, so that should be the situation where we have a first degree polynomial. So that's literally a first degree polynomial would be these two terms right over here. Because this is a 0th degree. This is a first degree."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "So the order 1 approximation, so that should be the situation where we have a first degree polynomial. So that's literally a first degree polynomial would be these two terms right over here. Because this is a 0th degree. This is a first degree. We just have x to the first power involved here. If we just were to plot this, if this was our polynomial, that is plotted with one dot. And that is this one right over here with one dot."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "This is a first degree. We just have x to the first power involved here. If we just were to plot this, if this was our polynomial, that is plotted with one dot. And that is this one right over here with one dot. And they plot it right over here. And we can see that it's just a tangent line at x is equal to 3. That is x is equal to 3 right over there."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "And that is this one right over here with one dot. And they plot it right over here. And we can see that it's just a tangent line at x is equal to 3. That is x is equal to 3 right over there. So this is the tangent line. If we add a term, now we're getting to a second degree polynomial. Because we're adding an x squared."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "That is x is equal to 3 right over there. So this is the tangent line. If we add a term, now we're getting to a second degree polynomial. Because we're adding an x squared. If you expand this out, you'll have an x squared term. And you'll have another x term. But the degree of the polynomial will now be a second degree."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "Because we're adding an x squared. If you expand this out, you'll have an x squared term. And you'll have another x term. But the degree of the polynomial will now be a second degree. So let's look for two dots. So that's this one right over here. So let's see, two dots coming in."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "But the degree of the polynomial will now be a second degree. So let's look for two dots. So that's this one right over here. So let's see, two dots coming in. So you'll notice one, two dots. So you have two dots. And it comes in."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "So let's see, two dots coming in. So you'll notice one, two dots. So you have two dots. And it comes in. And this is a parabola. It's a second degree polynomial. And then it comes back like this."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "And it comes in. And this is a parabola. It's a second degree polynomial. And then it comes back like this. But notice, it does a better job, especially around x equals 3, of approximating e to the x. It stays with the curve a little bit longer. You add another term."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "And then it comes back like this. But notice, it does a better job, especially around x equals 3, of approximating e to the x. It stays with the curve a little bit longer. You add another term. Let me do this in a new color, a color that I have not used. You add another term. Now you have a third degree polynomial."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "You add another term. Let me do this in a new color, a color that I have not used. You add another term. Now you have a third degree polynomial. If you have all of these combined, if this is your polynomial, and you were to graph that. So let's look for the three dots right over here. So one, two, three."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "Now you have a third degree polynomial. If you have all of these combined, if this is your polynomial, and you were to graph that. So let's look for the three dots right over here. So one, two, three. So it's this curve. Third degree polynomial is this curve right over here. And notice, it starts contouring e to the x a little bit sooner than the second degree version."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "So one, two, three. So it's this curve. Third degree polynomial is this curve right over here. And notice, it starts contouring e to the x a little bit sooner than the second degree version. And it stays with it a little bit longer. And so you have it just like that. You add another term to it."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "And notice, it starts contouring e to the x a little bit sooner than the second degree version. And it stays with it a little bit longer. And so you have it just like that. You add another term to it. You add the fourth degree term to it. So now we have all of this plus all of this. If this is your polynomial, now you have this curve right over here."}, {"video_title": "Visualizing Taylor polynomial approximations AP Calculus BC Khan Academy.mp3", "Sentence": "You add another term to it. You add the fourth degree term to it. So now we have all of this plus all of this. If this is your polynomial, now you have this curve right over here. Notice, every time you add a term, it's getting better and better at approximating e to the x further and further away from x is equal to 3. And then if you add another term, you get this one up here. But hopefully that satisfies you that we are getting closer and closer the more terms we add."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "Position as a function of time. And let me graph a potential s of t right over here. We have a horizontal axis as the time axis. And let me just graph something. I'll draw it kind of parabola looking. Although I could have done it general, but just to make things a little bit simpler for me. So I'll draw it kind of parabola looking."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "And let me just graph something. I'll draw it kind of parabola looking. Although I could have done it general, but just to make things a little bit simpler for me. So I'll draw it kind of parabola looking. So that is, if we call this a y-axis, we could even call this y equals s of t as a reasonable way to graph our position as a function of time function. And now let's think about what happens if we want to think about the change in position between two times. Let's say between time a."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "So I'll draw it kind of parabola looking. So that is, if we call this a y-axis, we could even call this y equals s of t as a reasonable way to graph our position as a function of time function. And now let's think about what happens if we want to think about the change in position between two times. Let's say between time a. Let's say that's time a right over there. And then this right over here is time b. So time b is right over here."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say between time a. Let's say that's time a right over there. And then this right over here is time b. So time b is right over here. So what would be the change in position between time a and between time b? Well, at time b, we are at s of b. We are at s of b position."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "So time b is right over here. So what would be the change in position between time a and between time b? Well, at time b, we are at s of b. We are at s of b position. And at time a, we were at s of a position. So the change in position between time a and time b, let me write this down, the change in position between, and this might be obvious to you, but I'll write it down, between times a and b is going to be equal to s of b, this position, s of b minus this position, minus s of a. So nothing earth shattering so far."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "We are at s of b position. And at time a, we were at s of a position. So the change in position between time a and time b, let me write this down, the change in position between, and this might be obvious to you, but I'll write it down, between times a and b is going to be equal to s of b, this position, s of b minus this position, minus s of a. So nothing earth shattering so far. But now let's think about what happens if we take the derivative of this function right over here. So what happens when we take the derivative of a position as a function of time? So remember, the derivative gives us the slope of the tangent line at any point."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "So nothing earth shattering so far. But now let's think about what happens if we take the derivative of this function right over here. So what happens when we take the derivative of a position as a function of time? So remember, the derivative gives us the slope of the tangent line at any point. So let's say we're looking at a point right over there. The slope of the tangent line, it tells us for a very small change in t, I'm exaggerating it visually, for a very, very small change in t, how much are we changing in position? How much are we changing in position?"}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "So remember, the derivative gives us the slope of the tangent line at any point. So let's say we're looking at a point right over there. The slope of the tangent line, it tells us for a very small change in t, I'm exaggerating it visually, for a very, very small change in t, how much are we changing in position? How much are we changing in position? So we write that as ds dt is the derivative of our position function at any given time. So when we're talking about how the rate at which position changes with respect to time, what is that? Well, that is equal to velocity."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "How much are we changing in position? So we write that as ds dt is the derivative of our position function at any given time. So when we're talking about how the rate at which position changes with respect to time, what is that? Well, that is equal to velocity. So this is equal to velocity. But let me write this in different notations. So this itself is going to be a function of time."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "Well, that is equal to velocity. So this is equal to velocity. But let me write this in different notations. So this itself is going to be a function of time. So we could write this, this is equal to s prime of t. These are just two different ways of writing the derivative of s with respect to t. This makes it a little bit clearer that this itself is a function of time. And we know that this is the exact same thing as velocity as function of time, which we will write as v of t. So let's graph what v of t might look like down here. Let's graph it."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "So this itself is going to be a function of time. So we could write this, this is equal to s prime of t. These are just two different ways of writing the derivative of s with respect to t. This makes it a little bit clearer that this itself is a function of time. And we know that this is the exact same thing as velocity as function of time, which we will write as v of t. So let's graph what v of t might look like down here. Let's graph it. So let me put another axis down here that looks pretty close to the original. Give myself some real estate. So that looks pretty good."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "Let's graph it. So let me put another axis down here that looks pretty close to the original. Give myself some real estate. So that looks pretty good. And then let me try to graph v of t. So once again, if this is my y-axis, this is my t-axis. And I'm going to graph y is equal to v of t. And if this really is a parabola, then the slope over here is 0. The slope, the rate of change is 0."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "So that looks pretty good. And then let me try to graph v of t. So once again, if this is my y-axis, this is my t-axis. And I'm going to graph y is equal to v of t. And if this really is a parabola, then the slope over here is 0. The slope, the rate of change is 0. And then it keeps increasing. The slope gets steeper and steeper and steeper. And so v of t might look something like this."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "The slope, the rate of change is 0. And then it keeps increasing. The slope gets steeper and steeper and steeper. And so v of t might look something like this. So this is the graph of y is equal to v of t. Now, using this graph, let's think if we can conceptualize the distance or the change in position between time a and between time b. Well, let's go back to our Riemann sums. Let's think about what an area of a very small rectangle would represent."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "And so v of t might look something like this. So this is the graph of y is equal to v of t. Now, using this graph, let's think if we can conceptualize the distance or the change in position between time a and between time b. Well, let's go back to our Riemann sums. Let's think about what an area of a very small rectangle would represent. So let's divide this into a bunch of rectangles. So I'll do fairly large rectangles so we have some space to work with. You can imagine much smaller ones."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "Let's think about what an area of a very small rectangle would represent. So let's divide this into a bunch of rectangles. So I'll do fairly large rectangles so we have some space to work with. You can imagine much smaller ones. And I'm going to do a left Riemann sum here, just because we've done those a bunch. But we could do a right Riemann sum. We could do a trapezoidal sum."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "You can imagine much smaller ones. And I'm going to do a left Riemann sum here, just because we've done those a bunch. But we could do a right Riemann sum. We could do a trapezoidal sum. We could do anything we want. So then we could keep going all the way. Actually, let me just do 3 right now."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "We could do a trapezoidal sum. We could do anything we want. So then we could keep going all the way. Actually, let me just do 3 right now. Let me just do 3 right over here. And so this is actually a very rough approximation. But you can imagine it might get closer."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "Actually, let me just do 3 right now. Let me just do 3 right over here. And so this is actually a very rough approximation. But you can imagine it might get closer. But what is the area of each of these rectangles trying? What is it an approximation for? Well, this one right over here, you have f of a, or I should say v of a."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "But you can imagine it might get closer. But what is the area of each of these rectangles trying? What is it an approximation for? Well, this one right over here, you have f of a, or I should say v of a. So your velocity at time a is the height right over here. And then this distance right over here is a change in time, times delta t. So the area for that rectangle is your velocity at that moment times your change in time. What is the velocity at that moment times your change in time?"}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "Well, this one right over here, you have f of a, or I should say v of a. So your velocity at time a is the height right over here. And then this distance right over here is a change in time, times delta t. So the area for that rectangle is your velocity at that moment times your change in time. What is the velocity at that moment times your change in time? Well, that's going to be your change in position. So this will tell you this is an approximation of your change in position over this time. Then this rectangle, the area of this rectangle, is another approximation for your change in position over the next delta t. And then you can imagine this right over here is an approximation for your change in position for the next delta t. So if you really wanted to figure out your change in position between a and b, you might want to just do a Riemann sum if you wanted to approximate it."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "What is the velocity at that moment times your change in time? Well, that's going to be your change in position. So this will tell you this is an approximation of your change in position over this time. Then this rectangle, the area of this rectangle, is another approximation for your change in position over the next delta t. And then you can imagine this right over here is an approximation for your change in position for the next delta t. So if you really wanted to figure out your change in position between a and b, you might want to just do a Riemann sum if you wanted to approximate it. You would want to take the sum from i equals 1 to i equals n of v of, and I'll do a left Riemann sum. But once again, we could use a midpoint. We could do trapezoids."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "Then this rectangle, the area of this rectangle, is another approximation for your change in position over the next delta t. And then you can imagine this right over here is an approximation for your change in position for the next delta t. So if you really wanted to figure out your change in position between a and b, you might want to just do a Riemann sum if you wanted to approximate it. You would want to take the sum from i equals 1 to i equals n of v of, and I'll do a left Riemann sum. But once again, we could use a midpoint. We could do trapezoids. We could do the right Riemann sum. But I'll just do a left one because that's what I depicted right here. v of t of i minus 1."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "We could do trapezoids. We could do the right Riemann sum. But I'll just do a left one because that's what I depicted right here. v of t of i minus 1. So if this would be t0, it would be a. So this is the first rectangle. So the first rectangle, you use the function evaluated at t0."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "v of t of i minus 1. So if this would be t0, it would be a. So this is the first rectangle. So the first rectangle, you use the function evaluated at t0. For the second rectangle, you use the function evaluated at t1. We've done this in multiple videos already. And then we multiply it times each of the changes in time."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "So the first rectangle, you use the function evaluated at t0. For the second rectangle, you use the function evaluated at t1. We've done this in multiple videos already. And then we multiply it times each of the changes in time. This will be an approximation for our total. And let me make it clear. Where delta t is equal to b minus a over the number of intervals we have."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "And then we multiply it times each of the changes in time. This will be an approximation for our total. And let me make it clear. Where delta t is equal to b minus a over the number of intervals we have. We already know from many, many videos when we looked at Riemann sums that this will be an approximation. Well, it will be an approximation for two things. We just talked about it will be an approximation for our change in position."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "Where delta t is equal to b minus a over the number of intervals we have. We already know from many, many videos when we looked at Riemann sums that this will be an approximation. Well, it will be an approximation for two things. We just talked about it will be an approximation for our change in position. But it's also an approximation for our area. So this right over here. So we're trying to approximate change in position."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "We just talked about it will be an approximation for our change in position. But it's also an approximation for our area. So this right over here. So we're trying to approximate change in position. And this is also approximate of the area under the curve. So hopefully this satisfies you that if you are able to calculate the area under the curve. And actually, this one's pretty easy because it's a trapezoid."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "So we're trying to approximate change in position. And this is also approximate of the area under the curve. So hopefully this satisfies you that if you are able to calculate the area under the curve. And actually, this one's pretty easy because it's a trapezoid. But even if this was a function, if it was kind of a wacky function, it would still apply. That when you're calculating the area under the curve of the velocity function, you are actually figuring out the change in position. These are the two things."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "And actually, this one's pretty easy because it's a trapezoid. But even if this was a function, if it was kind of a wacky function, it would still apply. That when you're calculating the area under the curve of the velocity function, you are actually figuring out the change in position. These are the two things. Well, we already know what could we do to get the exact area under the curve or to get the exact change in position. Well, we just have a ton of rectangles. We take the limit as the number of rectangles we have approaches infinity."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "These are the two things. Well, we already know what could we do to get the exact area under the curve or to get the exact change in position. Well, we just have a ton of rectangles. We take the limit as the number of rectangles we have approaches infinity. We take the limit as n approaches infinity. And as n approaches infinity, because delta t is b minus a divided by n, delta t is going to become infinitely small. It's going to turn into dt."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "We take the limit as the number of rectangles we have approaches infinity. We take the limit as n approaches infinity. And as n approaches infinity, because delta t is b minus a divided by n, delta t is going to become infinitely small. It's going to turn into dt. This is one way to think about it. And we already have notation for this. This is one way to think about a Riemann integral."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "It's going to turn into dt. This is one way to think about it. And we already have notation for this. This is one way to think about a Riemann integral. We just use the left Riemann sum. Once again, we could use the right Riemann sum, et cetera, et cetera. We could have used a more general Riemann sum, but this one will work."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "This is one way to think about a Riemann integral. We just use the left Riemann sum. Once again, we could use the right Riemann sum, et cetera, et cetera. We could have used a more general Riemann sum, but this one will work. So this will be equal to the definite integral from a to b of v of t dt. So this right over here is one way of saying, look, if we want the exact area under the curve of the velocity curve, which is going to be the exact change in position between a and b, we can denote it this way. It's the limit of this Riemann sum as n approaches infinity or the definite integral from a to b of v of t dt."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "We could have used a more general Riemann sum, but this one will work. So this will be equal to the definite integral from a to b of v of t dt. So this right over here is one way of saying, look, if we want the exact area under the curve of the velocity curve, which is going to be the exact change in position between a and b, we can denote it this way. It's the limit of this Riemann sum as n approaches infinity or the definite integral from a to b of v of t dt. But what did we just figure out? So remember, this is another. We could call this the exact change in position between times a and b."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "It's the limit of this Riemann sum as n approaches infinity or the definite integral from a to b of v of t dt. But what did we just figure out? So remember, this is another. We could call this the exact change in position between times a and b. But we already figured out what the exact change in positions between times a and b are. It's this thing right over here. And so this gets interesting."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "We could call this the exact change in position between times a and b. But we already figured out what the exact change in positions between times a and b are. It's this thing right over here. And so this gets interesting. We now have a way of evaluating this definite integral. Conceptually, we knew that this is the exact change in position between a and b, but we already figured out a way to figure out the exact change in position between a and b. So let me write all this down."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "And so this gets interesting. We now have a way of evaluating this definite integral. Conceptually, we knew that this is the exact change in position between a and b, but we already figured out a way to figure out the exact change in position between a and b. So let me write all this down. We have that the definite integral between a and b of v of t dt is equal to s of b minus s of a, where s of t is the anti-derivative of v of t. And this notion, although I've written in a very nontraditional used position velocity, this is the second fundamental theorem of calculus. And you're probably wondering about the first. We'll talk about that in another video."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "So let me write all this down. We have that the definite integral between a and b of v of t dt is equal to s of b minus s of a, where s of t is the anti-derivative of v of t. And this notion, although I've written in a very nontraditional used position velocity, this is the second fundamental theorem of calculus. And you're probably wondering about the first. We'll talk about that in another video. But this is a super useful way of evaluating definite integrals and finding the area under a curve. Second fundamental theorem of calculus, very closely tied to the first fundamental theorem, which we won't talk about now. So why is this such a big deal?"}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "We'll talk about that in another video. But this is a super useful way of evaluating definite integrals and finding the area under a curve. Second fundamental theorem of calculus, very closely tied to the first fundamental theorem, which we won't talk about now. So why is this such a big deal? Well, let me write it in a more general notation, the way that you might be used to seeing it in your calculus book. It's telling us that if we want the area under the curve between two points, a and b, between two x points, a and b, of f of x. And so this is how we would denote the area under the curve between those two intervals."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "So why is this such a big deal? Well, let me write it in a more general notation, the way that you might be used to seeing it in your calculus book. It's telling us that if we want the area under the curve between two points, a and b, between two x points, a and b, of f of x. And so this is how we would denote the area under the curve between those two intervals. So let me draw that just to make it clear what I'm talking about in general terms. So this right over here could be f of x. And we care about the area under the curve between a and b."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "And so this is how we would denote the area under the curve between those two intervals. So let me draw that just to make it clear what I'm talking about in general terms. So this right over here could be f of x. And we care about the area under the curve between a and b. If we want to find the exact area under the curve, we can figure it out by taking the antiderivative of f. And let's just say that capital F of x is the antiderivative, or is an antiderivative, because you can have multiple that are shifted by constants, is an antiderivative of f. Then you just have to evaluate the antiderivative at the end points and take the difference. So you take the end point first. You subtract the antiderivative evaluated at the starting point from the antiderivative evaluated at the end point."}, {"video_title": "Intuition for second part of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "And we care about the area under the curve between a and b. If we want to find the exact area under the curve, we can figure it out by taking the antiderivative of f. And let's just say that capital F of x is the antiderivative, or is an antiderivative, because you can have multiple that are shifted by constants, is an antiderivative of f. Then you just have to evaluate the antiderivative at the end points and take the difference. So you take the end point first. You subtract the antiderivative evaluated at the starting point from the antiderivative evaluated at the end point. So you get capital F of b minus capital F of a. So if you want to figure out the exact area under the curve, you take the antiderivative of it and evaluate that at the end point. And from that, you subtract the starting point."}, {"video_title": "Dividing expressions to evaluate integral AP Calculus BC Khan Academy.mp3", "Sentence": "See if you can evaluate this integral right over here. So I'm assuming you've had a go at it, so let's work through this together. So you probably realize that some of the traditional techniques that we've already had in our toolkits don't seem to be directly applicable. You substitution and others. And the key here to realize is we have a rational expression here where the numerator has the same degree or higher than the denominator. In this case, the numerator and the denominator have the same degree. And whenever you see something like that, it's probably a good idea to divide the denominator into the numerator."}, {"video_title": "Dividing expressions to evaluate integral AP Calculus BC Khan Academy.mp3", "Sentence": "You substitution and others. And the key here to realize is we have a rational expression here where the numerator has the same degree or higher than the denominator. In this case, the numerator and the denominator have the same degree. And whenever you see something like that, it's probably a good idea to divide the denominator into the numerator. That's what this rational expression could be interpreted as, x minus five divided by negative two x plus two. So let's do a little bit of algebraic long division to actually divide negative two x plus two into x minus five to see if we can rewrite this in a way that where we can evaluate the integral. So let's do that."}, {"video_title": "Dividing expressions to evaluate integral AP Calculus BC Khan Academy.mp3", "Sentence": "And whenever you see something like that, it's probably a good idea to divide the denominator into the numerator. That's what this rational expression could be interpreted as, x minus five divided by negative two x plus two. So let's do a little bit of algebraic long division to actually divide negative two x plus two into x minus five to see if we can rewrite this in a way that where we can evaluate the integral. So let's do that. We're gonna take x minus five. So x minus five. And divide negative two x plus two into that."}, {"video_title": "Dividing expressions to evaluate integral AP Calculus BC Khan Academy.mp3", "Sentence": "So let's do that. We're gonna take x minus five. So x minus five. And divide negative two x plus two into that. So negative two x plus two. So look at the highest degree terms. How many times does negative two x go into x?"}, {"video_title": "Dividing expressions to evaluate integral AP Calculus BC Khan Academy.mp3", "Sentence": "And divide negative two x plus two into that. So negative two x plus two. So look at the highest degree terms. How many times does negative two x go into x? Well, it's gonna go negative 1 1 2 times. Negative 1 1 2 times two is negative one. Negative 1 1 2 times negative two x is just going to be positive x, just like that."}, {"video_title": "Dividing expressions to evaluate integral AP Calculus BC Khan Academy.mp3", "Sentence": "How many times does negative two x go into x? Well, it's gonna go negative 1 1 2 times. Negative 1 1 2 times two is negative one. Negative 1 1 2 times negative two x is just going to be positive x, just like that. And now we want to subtract this yellow expression from this blue expression. And so let's just, let me just take the negative of this and then add. So I'm just gonna take the negative of it and add."}, {"video_title": "Dividing expressions to evaluate integral AP Calculus BC Khan Academy.mp3", "Sentence": "Negative 1 1 2 times negative two x is just going to be positive x, just like that. And now we want to subtract this yellow expression from this blue expression. And so let's just, let me just take the negative of this and then add. So I'm just gonna take the negative of it and add. And so we are left with negative five plus one is negative four. So you could say negative two x plus two goes into x minus five negative 1 1 2 times with negative four left over. And we can rewrite this integral, our original integral, as, we can rewrite it as negative 1 1 2 minus four over negative two x plus two dx."}, {"video_title": "Dividing expressions to evaluate integral AP Calculus BC Khan Academy.mp3", "Sentence": "So I'm just gonna take the negative of it and add. And so we are left with negative five plus one is negative four. So you could say negative two x plus two goes into x minus five negative 1 1 2 times with negative four left over. And we can rewrite this integral, our original integral, as, we can rewrite it as negative 1 1 2 minus four over negative two x plus two dx. Now let's see, it looks like we can simplify this expression a little bit more. The numerator and the denominator, they're both divisible by two, all of these terms are divisible by two. Actually we have all these negatives, that always unnecessarily complicates things."}, {"video_title": "Dividing expressions to evaluate integral AP Calculus BC Khan Academy.mp3", "Sentence": "And we can rewrite this integral, our original integral, as, we can rewrite it as negative 1 1 2 minus four over negative two x plus two dx. Now let's see, it looks like we can simplify this expression a little bit more. The numerator and the denominator, they're both divisible by two, all of these terms are divisible by two. Actually we have all these negatives, that always unnecessarily complicates things. So let's actually divide the numerator and the denominator by negative two. So what are we gonna have then? So if we divide the numerator by negative two, if this is negative four, this is going to become positive two."}, {"video_title": "Dividing expressions to evaluate integral AP Calculus BC Khan Academy.mp3", "Sentence": "Actually we have all these negatives, that always unnecessarily complicates things. So let's actually divide the numerator and the denominator by negative two. So what are we gonna have then? So if we divide the numerator by negative two, if this is negative four, this is going to become positive two. Then this, if we divide negative two x by negative two, that's just going to become x. And then two divided by negative two is going to be minus one. So our original integral, once again, this is just algebra, everything we've done so far is algebra."}, {"video_title": "Dividing expressions to evaluate integral AP Calculus BC Khan Academy.mp3", "Sentence": "So if we divide the numerator by negative two, if this is negative four, this is going to become positive two. Then this, if we divide negative two x by negative two, that's just going to become x. And then two divided by negative two is going to be minus one. So our original integral, once again, this is just algebra, everything we've done so far is algebra. We've just rewritten it using a little bit of algebraic long division. Our original integral has simplified to negative 1 1 2. And some might argue it's not simplified, but it's actually much more useful for finding the integral."}, {"video_title": "Dividing expressions to evaluate integral AP Calculus BC Khan Academy.mp3", "Sentence": "So our original integral, once again, this is just algebra, everything we've done so far is algebra. We've just rewritten it using a little bit of algebraic long division. Our original integral has simplified to negative 1 1 2. And some might argue it's not simplified, but it's actually much more useful for finding the integral. Negative 1 1 2 plus two over x minus one dx. Now, how do we evaluate this? Well, the antiderivative of negative 1 1 2 is pretty straightforward."}, {"video_title": "Dividing expressions to evaluate integral AP Calculus BC Khan Academy.mp3", "Sentence": "And some might argue it's not simplified, but it's actually much more useful for finding the integral. Negative 1 1 2 plus two over x minus one dx. Now, how do we evaluate this? Well, the antiderivative of negative 1 1 2 is pretty straightforward. That's just going to be negative 1 1 2 x. Negative 1 1 2 x. What's the antiderivative of two over x minus one?"}, {"video_title": "Dividing expressions to evaluate integral AP Calculus BC Khan Academy.mp3", "Sentence": "Well, the antiderivative of negative 1 1 2 is pretty straightforward. That's just going to be negative 1 1 2 x. Negative 1 1 2 x. What's the antiderivative of two over x minus one? And you might be able to do this in your head. The derivative of x minus one is just one. So you could say that the derivative is sitting there."}, {"video_title": "Dividing expressions to evaluate integral AP Calculus BC Khan Academy.mp3", "Sentence": "What's the antiderivative of two over x minus one? And you might be able to do this in your head. The derivative of x minus one is just one. So you could say that the derivative is sitting there. And so we can essentially do u substitution in our heads and say, okay, let's just take the antiderivative, I guess you could say with respect to x minus one, which would be the natural log of the absolute value of x minus one. If all of that sounds really confusing, I'll actually do the u substitution. So if I were just trying to evaluate, if I were just trying to evaluate the integral two over x minus one dx, I could see, okay, the derivative of x minus one is just one."}, {"video_title": "Dividing expressions to evaluate integral AP Calculus BC Khan Academy.mp3", "Sentence": "So you could say that the derivative is sitting there. And so we can essentially do u substitution in our heads and say, okay, let's just take the antiderivative, I guess you could say with respect to x minus one, which would be the natural log of the absolute value of x minus one. If all of that sounds really confusing, I'll actually do the u substitution. So if I were just trying to evaluate, if I were just trying to evaluate the integral two over x minus one dx, I could see, okay, the derivative of x minus one is just one. So I could say u is equal to x minus one. And then du is going to be equal to dx. And so this is going to be, we can rewrite in terms of u as two, I'll just take the constant out, two times the integral of one over u du, which we know is two times the natural log of the absolute value of u plus c. And in this case, we know that u is x minus one."}, {"video_title": "Dividing expressions to evaluate integral AP Calculus BC Khan Academy.mp3", "Sentence": "So if I were just trying to evaluate, if I were just trying to evaluate the integral two over x minus one dx, I could see, okay, the derivative of x minus one is just one. So I could say u is equal to x minus one. And then du is going to be equal to dx. And so this is going to be, we can rewrite in terms of u as two, I'll just take the constant out, two times the integral of one over u du, which we know is two times the natural log of the absolute value of u plus c. And in this case, we know that u is x minus one. So this is equal to two times the natural log of x minus one plus c. And so that's what we're gonna have right over here. So plus two times the natural log of the absolute value of x minus one plus c. And the plus c doesn't just come from this one. This, in general, when we're taking the integral of the whole thing, there could be some constant, because obviously if we go the other way, we take the derivative, the constant will go away."}, {"video_title": "Product rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "What we will talk about in this video is the product rule, which is one of the fundamental ways of evaluating derivatives. And we won't prove it in this video, but we will learn how to apply it. And all it tells us is that if we have a function that can be expressed as the product of two functions, so let's say it can be expressed as f of x times g of x, and we want to take the derivative of this function, we want to take the derivative of it, that it's going to be equal to the derivative of one of these functions, f prime of x, let's say the derivative of the first one, times the second function, plus the first function, not taking its derivative, times the derivative of the second function. So here we have two terms. In each term, we took the derivative of one of the functions and not the other. And we multiply the derivative of the first function times the second function plus just the first function times the derivative of the second function. Now let's see if we can actually apply this to actually find the derivative of something."}, {"video_title": "Product rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So here we have two terms. In each term, we took the derivative of one of the functions and not the other. And we multiply the derivative of the first function times the second function plus just the first function times the derivative of the second function. Now let's see if we can actually apply this to actually find the derivative of something. So let's say we are dealing with x squared times cosine of x. Or let's say, well, yeah, sure. Let's do x squared times sine of x."}, {"video_title": "Product rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's see if we can actually apply this to actually find the derivative of something. So let's say we are dealing with x squared times cosine of x. Or let's say, well, yeah, sure. Let's do x squared times sine of x. We could have done it either way. And we are curious about taking the derivative of this. We are curious about what its derivative is."}, {"video_title": "Product rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Let's do x squared times sine of x. We could have done it either way. And we are curious about taking the derivative of this. We are curious about what its derivative is. Well, we might immediately recognize that this can be expressed as a product of two functions. We could set f of x is equal to x squared. So that is f of x right over there."}, {"video_title": "Product rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "We are curious about what its derivative is. Well, we might immediately recognize that this can be expressed as a product of two functions. We could set f of x is equal to x squared. So that is f of x right over there. And we could set g of x to be equal to sine of x. And there we have it. We have our f of x times g of x."}, {"video_title": "Product rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So that is f of x right over there. And we could set g of x to be equal to sine of x. And there we have it. We have our f of x times g of x. And we could think about what these individual derivatives are. The derivative of f of x is just going to be equal to 2x by the power rule. And the derivative of g of x is just the derivative of sine of x."}, {"video_title": "Product rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "We have our f of x times g of x. And we could think about what these individual derivatives are. The derivative of f of x is just going to be equal to 2x by the power rule. And the derivative of g of x is just the derivative of sine of x. And we covered this when we just talked about common derivatives. Derivative of sine of x is cosine of x. And so now we're ready to apply the product rule."}, {"video_title": "Product rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And the derivative of g of x is just the derivative of sine of x. And we covered this when we just talked about common derivatives. Derivative of sine of x is cosine of x. And so now we're ready to apply the product rule. This is going to be equal to f prime of x times g of x. So f prime of x, the derivative of f, is 2x times g of x, which is sine of x, plus just our function f, which is x squared, times the derivative of g, times cosine of x. And we're done."}, {"video_title": "Definite integral of rational function AP Calculus AB Khan Academy.mp3", "Sentence": "Now at first this might seem daunting. I have this rational expression. I have x's in the numerators and x's in the denominators, but we just have to remember, we just have to do some algebraic manipulation and this is going to seem a lot more tractable. This is the same thing as the definite integral from negative one to negative two of 16 over x to the third minus x to the third over x to the third. Minus x to the third over x to the third dx. And now what is that going to be equal to? That is going to be equal to the definite integral from negative one to negative two of, I could write this first term right over here."}, {"video_title": "Definite integral of rational function AP Calculus AB Khan Academy.mp3", "Sentence": "This is the same thing as the definite integral from negative one to negative two of 16 over x to the third minus x to the third over x to the third. Minus x to the third over x to the third dx. And now what is that going to be equal to? That is going to be equal to the definite integral from negative one to negative two of, I could write this first term right over here. Let me do this in a different color. I could write this as 16x to the negative three. x to the negative three."}, {"video_title": "Definite integral of rational function AP Calculus AB Khan Academy.mp3", "Sentence": "That is going to be equal to the definite integral from negative one to negative two of, I could write this first term right over here. Let me do this in a different color. I could write this as 16x to the negative three. x to the negative three. And this second one, we have minus x to the third over x to the third. Well, x to the third is just over, x to the third over x to the third is just going to be equal to one. So this is going to be minus one dx."}, {"video_title": "Definite integral of rational function AP Calculus AB Khan Academy.mp3", "Sentence": "x to the negative three. And this second one, we have minus x to the third over x to the third. Well, x to the third is just over, x to the third over x to the third is just going to be equal to one. So this is going to be minus one dx. So dx. And so what is this going to be equal to? Well, let's take the antiderivative of each of these parts and then we're going to have to evaluate them at the different bounds."}, {"video_title": "Definite integral of rational function AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be minus one dx. So dx. And so what is this going to be equal to? Well, let's take the antiderivative of each of these parts and then we're going to have to evaluate them at the different bounds. So let's see. The antiderivative of 16x to the negative three. We're just going to do the power rule for derivatives in reverse."}, {"video_title": "Definite integral of rational function AP Calculus AB Khan Academy.mp3", "Sentence": "Well, let's take the antiderivative of each of these parts and then we're going to have to evaluate them at the different bounds. So let's see. The antiderivative of 16x to the negative three. We're just going to do the power rule for derivatives in reverse. You could view this as the power rule of integration or power rule of taking the antiderivative where what you do is you're going to increase our exponent by one. So you're going to go from negative three to negative two. And then you're going to divide by that amount, by negative two."}, {"video_title": "Definite integral of rational function AP Calculus AB Khan Academy.mp3", "Sentence": "We're just going to do the power rule for derivatives in reverse. You could view this as the power rule of integration or power rule of taking the antiderivative where what you do is you're going to increase our exponent by one. So you're going to go from negative three to negative two. And then you're going to divide by that amount, by negative two. So it's going to be 16 divided by negative two times x to the negative two. All I did is I increased the exponent and I divided by that amount. So that's the antiderivative here."}, {"video_title": "Definite integral of rational function AP Calculus AB Khan Academy.mp3", "Sentence": "And then you're going to divide by that amount, by negative two. So it's going to be 16 divided by negative two times x to the negative two. All I did is I increased the exponent and I divided by that amount. So that's the antiderivative here. And 16 divided by negative two, that is just negative eight. So we have negative eight x to the negative two. And then the antiderivative of negative one, well, that's just negative x."}, {"video_title": "Definite integral of rational function AP Calculus AB Khan Academy.mp3", "Sentence": "So that's the antiderivative here. And 16 divided by negative two, that is just negative eight. So we have negative eight x to the negative two. And then the antiderivative of negative one, well, that's just negative x. Negative, negative x. Negative x. And actually you could, you might just know that."}, {"video_title": "Definite integral of rational function AP Calculus AB Khan Academy.mp3", "Sentence": "And then the antiderivative of negative one, well, that's just negative x. Negative, negative x. Negative x. And actually you could, you might just know that. And hey, if I take the derivative of negative x, I get negative one. Or if you viewed this as negative x to the zero power, because that's what one is, well, it's the same thing. You increase the exponent by one to get x to the first power."}, {"video_title": "Definite integral of rational function AP Calculus AB Khan Academy.mp3", "Sentence": "And actually you could, you might just know that. And hey, if I take the derivative of negative x, I get negative one. Or if you viewed this as negative x to the zero power, because that's what one is, well, it's the same thing. You increase the exponent by one to get x to the first power. And then you divide by one. And so, I mean, you could view it as that right over there. But either way, you get to negative or minus x."}, {"video_title": "Definite integral of rational function AP Calculus AB Khan Academy.mp3", "Sentence": "You increase the exponent by one to get x to the first power. And then you divide by one. And so, I mean, you could view it as that right over there. But either way, you get to negative or minus x. And so now we want to evaluate that. We're going to evaluate that at the bounds and take the difference. So we're going to evaluate that at negative two and then subtract from that, this evaluated at negative one."}, {"video_title": "Definite integral of rational function AP Calculus AB Khan Academy.mp3", "Sentence": "But either way, you get to negative or minus x. And so now we want to evaluate that. We're going to evaluate that at the bounds and take the difference. So we're going to evaluate that at negative two and then subtract from that, this evaluated at negative one. And let me do those in two different colors so we can see what's going on. So we're gonna evaluate it at negative two and we're going to evaluate it at negative one. So let's first evaluate it at negative two."}, {"video_title": "Definite integral of rational function AP Calculus AB Khan Academy.mp3", "Sentence": "So we're going to evaluate that at negative two and then subtract from that, this evaluated at negative one. And let me do those in two different colors so we can see what's going on. So we're gonna evaluate it at negative two and we're going to evaluate it at negative one. So let's first evaluate it at negative two. So this is going to be equal to, this is going to be equal to, when you evaluate it at negative two, it's going to be negative eight, negative eight times x to the negative two. So negative two to the negative two power minus negative two. And from that, we're going to subtract it, evaluate it at negative one."}, {"video_title": "Definite integral of rational function AP Calculus AB Khan Academy.mp3", "Sentence": "So let's first evaluate it at negative two. So this is going to be equal to, this is going to be equal to, when you evaluate it at negative two, it's going to be negative eight, negative eight times x to the negative two. So negative two to the negative two power minus negative two. And from that, we're going to subtract it, evaluate it at negative one. So it's gonna be negative eight times negative one to the negative two power minus negative one. All right, so what is this going to be? So negative two to the negative two, so negative two to the negative two is equal to one over negative two squared, which is equal to 1 4th."}, {"video_title": "Definite integral of rational function AP Calculus AB Khan Academy.mp3", "Sentence": "And from that, we're going to subtract it, evaluate it at negative one. So it's gonna be negative eight times negative one to the negative two power minus negative one. All right, so what is this going to be? So negative two to the negative two, so negative two to the negative two is equal to one over negative two squared, which is equal to 1 4th. So this is equal to positive 1 4th, but then negative eight times positive 1 4th is going to be equal to negative two. And then we have negative two minus negative two, so that's negative two plus two. And so everything I've just done in this purplish color, that is just going to be zero."}, {"video_title": "Definite integral of rational function AP Calculus AB Khan Academy.mp3", "Sentence": "So negative two to the negative two, so negative two to the negative two is equal to one over negative two squared, which is equal to 1 4th. So this is equal to positive 1 4th, but then negative eight times positive 1 4th is going to be equal to negative two. And then we have negative two minus negative two, so that's negative two plus two. And so everything I've just done in this purplish color, that is just going to be zero. And then if we look at what's going on in the orange, when we evaluate at negative one, let's see, negative one to the negative two power, well, that's one over negative one squared. Well, this is all just going to be one. And so we're gonna have negative eight plus one, which is equal to negative seven."}, {"video_title": "Area bounded by polar curves Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "We now have a lot of experience finding the areas under curves when we're dealing with things in rectangular coordinates. And we saw, we took the Riemann sums, a bunch of rectangles, we took the limit as we had an infinite number of infinitely thin rectangles, and we were able to find the area. But now let's move on to polar coordinates. And polar coordinates, I won't say we're finding the area under a curve, but really, in this example right over here, we have a part of the graph of r is equal to f of theta, and we've graphed it between theta is equal to alpha and theta is equal to beta. What I wanna do in this video is come up with a general expression for this area in blue, this area that is bounded, I guess you could say, by those angles, and the graph of r is equal to, r is equal to f of theta. And I want you to come up, or at least attempt to come up with an expression on your own, but I'll give you a little bit of a hint here. When we did it in rectangular coordinates, we divided things into rectangles."}, {"video_title": "Area bounded by polar curves Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "And polar coordinates, I won't say we're finding the area under a curve, but really, in this example right over here, we have a part of the graph of r is equal to f of theta, and we've graphed it between theta is equal to alpha and theta is equal to beta. What I wanna do in this video is come up with a general expression for this area in blue, this area that is bounded, I guess you could say, by those angles, and the graph of r is equal to, r is equal to f of theta. And I want you to come up, or at least attempt to come up with an expression on your own, but I'll give you a little bit of a hint here. When we did it in rectangular coordinates, we divided things into rectangles. Over here, rectangles don't seem as obvious, because they're all kind of coming to this point. But what if we could divide things into, if we could divide things into sector, I guess we could say, little pie pieces. Someone's doing some serious drilling downstairs."}, {"video_title": "Area bounded by polar curves Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "When we did it in rectangular coordinates, we divided things into rectangles. Over here, rectangles don't seem as obvious, because they're all kind of coming to this point. But what if we could divide things into, if we could divide things into sector, I guess we could say, little pie pieces. Someone's doing some serious drilling downstairs. I don't know if it's picking up on the microphone, but anyway, I will continue. So what would happen if we could divide this into a whole series of kind of pie pieces? If we could divide it into a whole series of pie pieces, and then take the limit as if we had an infinite number, an infinite number of pie pieces."}, {"video_title": "Area bounded by polar curves Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "Someone's doing some serious drilling downstairs. I don't know if it's picking up on the microphone, but anyway, I will continue. So what would happen if we could divide this into a whole series of kind of pie pieces? If we could divide it into a whole series of pie pieces, and then take the limit as if we had an infinite number, an infinite number of pie pieces. So we want to find the area of each of these pie pieces, and then take the limit as the pie pieces, I guess you could say, become infinitely thin, and we have an infinite number of them. And I'll give you one more hint. I'll give you one more hint."}, {"video_title": "Area bounded by polar curves Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "If we could divide it into a whole series of pie pieces, and then take the limit as if we had an infinite number, an infinite number of pie pieces. So we want to find the area of each of these pie pieces, and then take the limit as the pie pieces, I guess you could say, become infinitely thin, and we have an infinite number of them. And I'll give you one more hint. I'll give you one more hint. For thinking about the area of these pie, I guess you could say pie, the area of these pie wedges. I'll give you another hint. So if I have a circle, I'm doing my best attempt at a circle."}, {"video_title": "Area bounded by polar curves Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "I'll give you one more hint. For thinking about the area of these pie, I guess you could say pie, the area of these pie wedges. I'll give you another hint. So if I have a circle, I'm doing my best attempt at a circle. Luckily the plumbing or whatever is going on downstairs has stopped for now, allowing me to focus more on the calculus, which is obviously more important. All right, so if I have a circle, that's my best attempt at a circle, and it's of radius r, it's of radius r, and let's say, let me draw a sector of this circle. It's a sector of a circle, so that's obviously r as well."}, {"video_title": "Area bounded by polar curves Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So if I have a circle, I'm doing my best attempt at a circle. Luckily the plumbing or whatever is going on downstairs has stopped for now, allowing me to focus more on the calculus, which is obviously more important. All right, so if I have a circle, that's my best attempt at a circle, and it's of radius r, it's of radius r, and let's say, let me draw a sector of this circle. It's a sector of a circle, so that's obviously r as well. And if this angle right here is theta, what is going to be the area, what is going to be the area of this sector right over here? So that's my hint for you. Think about what this area is going to be, and we're assuming theta is in radians."}, {"video_title": "Area bounded by polar curves Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "It's a sector of a circle, so that's obviously r as well. And if this angle right here is theta, what is going to be the area, what is going to be the area of this sector right over here? So that's my hint for you. Think about what this area is going to be, and we're assuming theta is in radians. Think about what this area is going to be, and then see if you can extend that to what we're trying to do here to figure out a, somehow, I'm giving you a hint again, using integration, finding an expression for this area. So I'm assuming you've had a go at it, so first let's think about this. So what's the area of the entire, the area of the entire circle?"}, {"video_title": "Area bounded by polar curves Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "Think about what this area is going to be, and we're assuming theta is in radians. Think about what this area is going to be, and then see if you can extend that to what we're trying to do here to figure out a, somehow, I'm giving you a hint again, using integration, finding an expression for this area. So I'm assuming you've had a go at it, so first let's think about this. So what's the area of the entire, the area of the entire circle? Well, we already know that. That's going to be pi r squared, formula for the area of a circle. And then what's going to be the area of this?"}, {"video_title": "Area bounded by polar curves Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So what's the area of the entire, the area of the entire circle? Well, we already know that. That's going to be pi r squared, formula for the area of a circle. And then what's going to be the area of this? Well, it's going to be a fraction of the circle. If this is pi, sorry, if this is theta, where if we went two pi radians, that would be the whole circle. So this is going to be pi, sorry, this is going to be theta over two pi of the circle."}, {"video_title": "Area bounded by polar curves Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "And then what's going to be the area of this? Well, it's going to be a fraction of the circle. If this is pi, sorry, if this is theta, where if we went two pi radians, that would be the whole circle. So this is going to be pi, sorry, this is going to be theta over two pi of the circle. So times theta over, over two pi, times theta over two pi, would be the area of this sector right over here, area of the whole circle times the proportion of the circle that we've kind of, we have defined, or that this sector is made up of. And so this would give us, this would give us, the pi's cancel out, it would give us 1 1 2 r squared times theta. Now, what happens if instead of theta, so let's look at each of these over here."}, {"video_title": "Area bounded by polar curves Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So this is going to be pi, sorry, this is going to be theta over two pi of the circle. So times theta over, over two pi, times theta over two pi, would be the area of this sector right over here, area of the whole circle times the proportion of the circle that we've kind of, we have defined, or that this sector is made up of. And so this would give us, this would give us, the pi's cancel out, it would give us 1 1 2 r squared times theta. Now, what happens if instead of theta, so let's look at each of these over here. So each of these things that I've drawn. So let's focus just on one of these, one of these wedges. I will highlight it in orange."}, {"video_title": "Area bounded by polar curves Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "Now, what happens if instead of theta, so let's look at each of these over here. So each of these things that I've drawn. So let's focus just on one of these, one of these wedges. I will highlight it in orange. I'll highlight it in orange. So instead of the angle being theta, let's just assume it's a really, really, really small angle. We'll use a differential, although this is a little bit of loosey-goosey mathematics, but the important here is to give you the conceptual understanding."}, {"video_title": "Area bounded by polar curves Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "I will highlight it in orange. I'll highlight it in orange. So instead of the angle being theta, let's just assume it's a really, really, really small angle. We'll use a differential, although this is a little bit of loosey-goosey mathematics, but the important here is to give you the conceptual understanding. I could call it a delta theta, and then eventually take the limit as our delta thetas approach zero, but I'm just gonna, just for conceptual purposes, assume we have a infinitely small or super small change in theta. So let's call that d theta, and d theta, and this, this, the radius here, or I guess we could say this length right over here, you could view it as the radius of at least the arc right at that point. It's going to be r, it's going to be r as a function of the thetas that were around right over here, but we're just gonna call that our r right over there."}, {"video_title": "Area bounded by polar curves Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "We'll use a differential, although this is a little bit of loosey-goosey mathematics, but the important here is to give you the conceptual understanding. I could call it a delta theta, and then eventually take the limit as our delta thetas approach zero, but I'm just gonna, just for conceptual purposes, assume we have a infinitely small or super small change in theta. So let's call that d theta, and d theta, and this, this, the radius here, or I guess we could say this length right over here, you could view it as the radius of at least the arc right at that point. It's going to be r, it's going to be r as a function of the thetas that were around right over here, but we're just gonna call that our r right over there. And so what is going to be the area of this little sector? Well, the area of this little sector is going to, instead of my angle being theta, I'm calling my angle d theta, this little differential. So instead of 1 1\u20442 r squared, it's going to be, let me do that in a color you can see, this, this, this area is going to be 1 1\u20442 r squared d theta."}, {"video_title": "Area bounded by polar curves Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "It's going to be r, it's going to be r as a function of the thetas that were around right over here, but we're just gonna call that our r right over there. And so what is going to be the area of this little sector? Well, the area of this little sector is going to, instead of my angle being theta, I'm calling my angle d theta, this little differential. So instead of 1 1\u20442 r squared, it's going to be, let me do that in a color you can see, this, this, this area is going to be 1 1\u20442 r squared d theta. Notice here the angle was theta, here the angle was just d theta, super, super small angle. Now, if I wanted to take the sum of all of these, from theta is equal to alpha to theta is equal to beta, and there really is an infinite number of these, this is an infinitely small angle, well then for the entire, for the entire area right over here, I could just integrate, I could just integrate all of these. So that's going to be the integral from alpha to beta of 1 1\u20442 r squared, 1 1\u20442 r squared d theta."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "In the last video, we tried to figure out the slope of a point, or the slope of a curve at a certain point. And the way we did it, we said, OK, well, let's find the slope between that point and then another point that's not too far away from that point. And we got the slope of the secant line. And it looks all fancy, but this is just the y value of the point that's not too far away. And this is just the y value of the point in question. So this is just your change in y. And then you divide that by your change in x."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "And it looks all fancy, but this is just the y value of the point that's not too far away. And this is just the y value of the point in question. So this is just your change in y. And then you divide that by your change in x. So in the example we did, h was the difference between our two x values. This distance was h. And that gave us the point, the slope of that line. We said, hey, what if we take the limit as this point right here gets closer and closer to this point?"}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "And then you divide that by your change in x. So in the example we did, h was the difference between our two x values. This distance was h. And that gave us the point, the slope of that line. We said, hey, what if we take the limit as this point right here gets closer and closer to this point? If this point essentially almost becomes this point, then our slope is going to be the slope of our tangent line. And we define that as the derivative of our function. We said that's equal to f prime of x."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "We said, hey, what if we take the limit as this point right here gets closer and closer to this point? If this point essentially almost becomes this point, then our slope is going to be the slope of our tangent line. And we define that as the derivative of our function. We said that's equal to f prime of x. So let's see if we can apply this in this video to maybe make things a little bit more concrete in your head. So let me do one. First I'll do a particular case where I want to find the slope at exactly some point."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "We said that's equal to f prime of x. So let's see if we can apply this in this video to maybe make things a little bit more concrete in your head. So let me do one. First I'll do a particular case where I want to find the slope at exactly some point. So let's say, let me draw my axes again. Let's draw some axes right there. And let's say I have the curve."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "First I'll do a particular case where I want to find the slope at exactly some point. So let's say, let me draw my axes again. Let's draw some axes right there. And let's say I have the curve. This is the curve. y is equal to x squared. So this is my y-axis."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "And let's say I have the curve. This is the curve. y is equal to x squared. So this is my y-axis. This is my x-axis. And I want to know the slope at the point x is equal to 3. When I say the slope, you can imagine a tangent line here."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "So this is my y-axis. This is my x-axis. And I want to know the slope at the point x is equal to 3. When I say the slope, you can imagine a tangent line here. Do it in a light yellow. You can imagine a tangent line that goes just like that. And it would just barely graze the curve at that point."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "When I say the slope, you can imagine a tangent line here. Do it in a light yellow. You can imagine a tangent line that goes just like that. And it would just barely graze the curve at that point. But what is the slope of that tangent line? What is the slope of that tangent line? Which is the same as the slope of the curve right at that point."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "And it would just barely graze the curve at that point. But what is the slope of that tangent line? What is the slope of that tangent line? Which is the same as the slope of the curve right at that point. So to do it, I'm going to actually do this exact technique that we did before. And then we'll generalize it so you don't have to do it every time for a particular number. So let's take some other point here."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "Which is the same as the slope of the curve right at that point. So to do it, I'm going to actually do this exact technique that we did before. And then we'll generalize it so you don't have to do it every time for a particular number. So let's take some other point here. Let's call this 3 plus delta x. I'm changing the notation because in some books you'll see an h, some books you'll see a delta x. Doesn't hurt to be exposed to both of them. So this is 3 plus delta x."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "So let's take some other point here. Let's call this 3 plus delta x. I'm changing the notation because in some books you'll see an h, some books you'll see a delta x. Doesn't hurt to be exposed to both of them. So this is 3 plus delta x. So first of all, what is this point right here? This is a curve y is equal to x squared. So f of x is 3 squared."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "So this is 3 plus delta x. So first of all, what is this point right here? This is a curve y is equal to x squared. So f of x is 3 squared. This is the point 9. This is the point 3, 9 right here. Let me draw that out."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "So f of x is 3 squared. This is the point 9. This is the point 3, 9 right here. Let me draw that out. Well, I'll draw it out in a second. And what is this point right here? So if we were to go all the way up here, what is that point?"}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "Let me draw that out. Well, I'll draw it out in a second. And what is this point right here? So if we were to go all the way up here, what is that point? Well, here our x is now 3 plus delta x. It's the same thing as this one right here. It's x naught plus h. I could have called this 3 plus h just as easily."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "So if we were to go all the way up here, what is that point? Well, here our x is now 3 plus delta x. It's the same thing as this one right here. It's x naught plus h. I could have called this 3 plus h just as easily. So it's 3 plus delta x up there. So what's the y value going to be? Well, whatever x value is, it's on the curve, it's going to be that squared."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "It's x naught plus h. I could have called this 3 plus h just as easily. So it's 3 plus delta x up there. So what's the y value going to be? Well, whatever x value is, it's on the curve, it's going to be that squared. So it's going to be, this is going to be the point 3 plus delta x squared. So let's figure out the slope of this secant line. Let me zoom in a little bit because that might help."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "Well, whatever x value is, it's on the curve, it's going to be that squared. So it's going to be, this is going to be the point 3 plus delta x squared. So let's figure out the slope of this secant line. Let me zoom in a little bit because that might help. So if I zoom in on just this part of the curve, it might look like that. And then I have one point here, and then I have the other point is up here. Let me see if I can draw that."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "Let me zoom in a little bit because that might help. So if I zoom in on just this part of the curve, it might look like that. And then I have one point here, and then I have the other point is up here. Let me see if I can draw that. That's the secant line. Just like that. This was the point over here."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "Let me see if I can draw that. That's the secant line. Just like that. This was the point over here. This point is the point 3, 9. And then this point up here is the point 3 plus delta x. So just some larger number than 3."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "This was the point over here. This point is the point 3, 9. And then this point up here is the point 3 plus delta x. So just some larger number than 3. And then it's going to be that number squared. So it's going to be 3 plus delta x squared. What is that?"}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "So just some larger number than 3. And then it's going to be that number squared. So it's going to be 3 plus delta x squared. What is that? That's going to be 9. I'm just FOILing this out, or you do the distributive property twice. If a plus b squared is a squared plus 2ab plus b squared, so it's going to be 9 plus 2 times the product of these things."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "What is that? That's going to be 9. I'm just FOILing this out, or you do the distributive property twice. If a plus b squared is a squared plus 2ab plus b squared, so it's going to be 9 plus 2 times the product of these things. So plus 6 delta x, and then plus delta x squared. Plus delta x plus delta x squared. That's the coordinate of the second line."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "If a plus b squared is a squared plus 2ab plus b squared, so it's going to be 9 plus 2 times the product of these things. So plus 6 delta x, and then plus delta x squared. Plus delta x plus delta x squared. That's the coordinate of the second line. This looks complicated, but I just took this x value and I squared it, because it's on the line y is equal to x squared. So the slope of this line, the slope of the secant line is going to be the change in y divided by the change in x. So the change in y is just going to be this guy's y value, which is 9 plus 6 delta x plus delta x squared."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "That's the coordinate of the second line. This looks complicated, but I just took this x value and I squared it, because it's on the line y is equal to x squared. So the slope of this line, the slope of the secant line is going to be the change in y divided by the change in x. So the change in y is just going to be this guy's y value, which is 9 plus 6 delta x plus delta x squared. That's this guy's y value. Minus this guy's y value. I'll do it in green."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "So the change in y is just going to be this guy's y value, which is 9 plus 6 delta x plus delta x squared. That's this guy's y value. Minus this guy's y value. I'll do it in green. So minus 9. That's your change in y. And you want to divide that by your change in x."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "I'll do it in green. So minus 9. That's your change in y. And you want to divide that by your change in x. Well, what is your change in x? This is actually going to be pretty convenient. This larger x value, we started with this point on the top."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "And you want to divide that by your change in x. Well, what is your change in x? This is actually going to be pretty convenient. This larger x value, we started with this point on the top. So we have to start with this point on the bottom. So it's going to be 3 plus delta x. And then what's this x value?"}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "This larger x value, we started with this point on the top. So we have to start with this point on the bottom. So it's going to be 3 plus delta x. And then what's this x value? Well, it's minus 3. That's his x value. So what does this simplify to?"}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "And then what's this x value? Well, it's minus 3. That's his x value. So what does this simplify to? The numerator, this 9 and that 9 cancel out. We get a 9 minus 9. And the denominator, what happens?"}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "So what does this simplify to? The numerator, this 9 and that 9 cancel out. We get a 9 minus 9. And the denominator, what happens? This 3 and minus 3 cancel out. So the change in x actually end up becoming this delta x, which makes sense, because this delta x is essentially how much more this guy is than that guy. So that should be the change in x, delta x."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "And the denominator, what happens? This 3 and minus 3 cancel out. So the change in x actually end up becoming this delta x, which makes sense, because this delta x is essentially how much more this guy is than that guy. So that should be the change in x, delta x. So the slope of my secant line has simplified to 6 times my change in x plus my change in x squared, all of that over my change in x. Now we can simplify this even more. Let's divide the numerator and the denominator by our change in x."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "So that should be the change in x, delta x. So the slope of my secant line has simplified to 6 times my change in x plus my change in x squared, all of that over my change in x. Now we can simplify this even more. Let's divide the numerator and the denominator by our change in x. And I will switch colors just to ease the monotony. So my slope of my tangent of my secant line, the one that goes through both of these, is going to be equal if you divide the numerator and denominator. This becomes 6."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "Let's divide the numerator and the denominator by our change in x. And I will switch colors just to ease the monotony. So my slope of my tangent of my secant line, the one that goes through both of these, is going to be equal if you divide the numerator and denominator. This becomes 6. I'm just dividing numerator and denominator by delta x. Plus 6 plus delta x. So that is the slope of the secant line."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "This becomes 6. I'm just dividing numerator and denominator by delta x. Plus 6 plus delta x. So that is the slope of the secant line. So slope is equal to 6 plus delta x. That's this one right here. That's this reddish line that I've drawn right there."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "So that is the slope of the secant line. So slope is equal to 6 plus delta x. That's this one right here. That's this reddish line that I've drawn right there. So if this number right here, if the delta x was 1, if these were the points 3 and 4, then my slope would be 6 plus 1. Because I'm picking a point 4 where the delta x here would have to be 1. So the slope would be 7."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "That's this reddish line that I've drawn right there. So if this number right here, if the delta x was 1, if these were the points 3 and 4, then my slope would be 6 plus 1. Because I'm picking a point 4 where the delta x here would have to be 1. So the slope would be 7. So we have a general formula for no matter what my delta x is, I can find the slope between 3 and 3 plus delta x, between those two points. Now, we wanted to find the slope at exactly that point right there. So let's see what happens when delta x gets smaller and smaller."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "So the slope would be 7. So we have a general formula for no matter what my delta x is, I can find the slope between 3 and 3 plus delta x, between those two points. Now, we wanted to find the slope at exactly that point right there. So let's see what happens when delta x gets smaller and smaller. This is what delta x is right now. It's this distance. But if delta x got a little bit smaller, then it would start, the secant line would look like that."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "So let's see what happens when delta x gets smaller and smaller. This is what delta x is right now. It's this distance. But if delta x got a little bit smaller, then it would start, the secant line would look like that. Got even smaller, the secant line would look like that. It gets even smaller, then we're getting pretty close to the slope of the tangent line. The tangent line is this thing right here that I want to find the slope of."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "But if delta x got a little bit smaller, then it would start, the secant line would look like that. Got even smaller, the secant line would look like that. It gets even smaller, then we're getting pretty close to the slope of the tangent line. The tangent line is this thing right here that I want to find the slope of. So let's find the limit as our delta x approaches 0. So the limit as delta x approaches 0 of our slope of the secant line of 6 plus delta x is equal to what? This is pretty straightforward."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "The tangent line is this thing right here that I want to find the slope of. So let's find the limit as our delta x approaches 0. So the limit as delta x approaches 0 of our slope of the secant line of 6 plus delta x is equal to what? This is pretty straightforward. You could just set this equal to 0 and it's equal to 6. So the slope of our tangent line at the point x is equal to 3 right there is equal to 6. And another way we could write this, if we wrote that f of x is equal to x squared, we now know that the derivative, or the slope, of the tangent line of this function at the point 3."}, {"video_title": "Calculating slope of tangent line using derivative definition Differential Calculus Khan Academy.mp3", "Sentence": "This is pretty straightforward. You could just set this equal to 0 and it's equal to 6. So the slope of our tangent line at the point x is equal to 3 right there is equal to 6. And another way we could write this, if we wrote that f of x is equal to x squared, we now know that the derivative, or the slope, of the tangent line of this function at the point 3. I just only evaluated it at the point 3 right there. That that is equal to 6. I haven't yet come up with a general formula for the slope of this line at any point."}, {"video_title": "Worked example limit comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let me underline that, the limit comparison test, in order to determine whether S converges. So let's just remind ourselves about the limit comparison test. If we say, if we say that we have two series, and I'll just use this notation, A sub n, and then another series, B sub n, and we know that A sub n and B sub n are greater than or equal to zero for all n, for all n, if we know this, then if, then if the limit as n approaches infinity of A sub n over B sub n is equal to some positive constant, so zero is less than that constant is less than infinity, then either both converge, then both converge, both converge, both converge, or both diverge, both diverge. And it really makes a lot of sense, because it's saying, look, as we get into our really large values of n, as we go really far out there in terms of the terms, if our behavior starts to look the same, then it makes sense that both these series would converge or diverge, and we have an introductory video on this in another video. So let's think about what, if we say that this is our A sub n, if we say that this is A sub n right over here, what is a series that we can really compare to that seems to have the same behavior as n gets really large? Well, this one just seems to get unbounded. This one doesn't look that similar."}, {"video_title": "Worked example limit comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And it really makes a lot of sense, because it's saying, look, as we get into our really large values of n, as we go really far out there in terms of the terms, if our behavior starts to look the same, then it makes sense that both these series would converge or diverge, and we have an introductory video on this in another video. So let's think about what, if we say that this is our A sub n, if we say that this is A sub n right over here, what is a series that we can really compare to that seems to have the same behavior as n gets really large? Well, this one just seems to get unbounded. This one doesn't look that similar. It has a three to the n minus one in the denominator, but the numerator doesn't behave the same. This one over here is interesting, because we could write this, this is the same thing as the sum n equals one to infinity. We could write this as two to the n over three to the n, and these are very similar."}, {"video_title": "Worked example limit comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "This one doesn't look that similar. It has a three to the n minus one in the denominator, but the numerator doesn't behave the same. This one over here is interesting, because we could write this, this is the same thing as the sum n equals one to infinity. We could write this as two to the n over three to the n, and these are very similar. The only difference between this and this is that in the denominator here, or in the denominator up here, we have a minus one, and down here, we don't have that minus one. And so it makes sense, given that that's just a constant, that as n gets very large, that these might behave the same. So let's try it out."}, {"video_title": "Worked example limit comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "We could write this as two to the n over three to the n, and these are very similar. The only difference between this and this is that in the denominator here, or in the denominator up here, we have a minus one, and down here, we don't have that minus one. And so it makes sense, given that that's just a constant, that as n gets very large, that these might behave the same. So let's try it out. Let's find the limit, and we also know that the A sub n's and the B sub n's, if we say that this right over here is B sub n, if we say that's B sub n, that this is going to be positive, or this is going to be greater than or equal to zero for n equals one, two, three, so for any values, this is going to be greater than or equal to zero, and the same thing right over here. It's gonna be greater than or equal to zero for all the n's that we care about. So we meet these first constraints, and so let's find the limit as n approaches infinity of A sub n, which is, I'll write it in that red color, which is two to the nth power over three to the n minus one over B sub n, over two to the nth, over three to the nth."}, {"video_title": "Worked example limit comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's try it out. Let's find the limit, and we also know that the A sub n's and the B sub n's, if we say that this right over here is B sub n, if we say that's B sub n, that this is going to be positive, or this is going to be greater than or equal to zero for n equals one, two, three, so for any values, this is going to be greater than or equal to zero, and the same thing right over here. It's gonna be greater than or equal to zero for all the n's that we care about. So we meet these first constraints, and so let's find the limit as n approaches infinity of A sub n, which is, I'll write it in that red color, which is two to the nth power over three to the n minus one over B sub n, over two to the nth, over three to the nth. So let me actually do a little algebraic manipulation right over here. This is going to be the same thing as two to the nth over three to the n minus one times three to the n over two to the n. Divide the numerator and denominators by two to the n. Those cancel out. And so this will give us three to the n over three to the n minus one."}, {"video_title": "Worked example limit comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So we meet these first constraints, and so let's find the limit as n approaches infinity of A sub n, which is, I'll write it in that red color, which is two to the nth power over three to the n minus one over B sub n, over two to the nth, over three to the nth. So let me actually do a little algebraic manipulation right over here. This is going to be the same thing as two to the nth over three to the n minus one times three to the n over two to the n. Divide the numerator and denominators by two to the n. Those cancel out. And so this will give us three to the n over three to the n minus one. We can divide the numerator and denominator by three to the n, and that'll give us one over one minus one over three to the n. So we could say this is the same thing as the limit as n approaches infinity of one over one minus one over three to the n. Well, what's this going to be equal to? Well, as n approaches infinity, this thing, one over three to the n, that's gonna go to zero. So this whole thing is just going to approach one."}, {"video_title": "Worked example limit comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so this will give us three to the n over three to the n minus one. We can divide the numerator and denominator by three to the n, and that'll give us one over one minus one over three to the n. So we could say this is the same thing as the limit as n approaches infinity of one over one minus one over three to the n. Well, what's this going to be equal to? Well, as n approaches infinity, this thing, one over three to the n, that's gonna go to zero. So this whole thing is just going to approach one. And one is clearly between zero and infinity. So the destinies of these two series are tied. They either both converge or they both diverge."}, {"video_title": "Worked example limit comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So this whole thing is just going to approach one. And one is clearly between zero and infinity. So the destinies of these two series are tied. They either both converge or they both diverge. And so this is a good one to use the limit comparison test with. And so let's think about it. Do they either both converge or do they both diverge?"}, {"video_title": "Worked example limit comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "They either both converge or they both diverge. And so this is a good one to use the limit comparison test with. And so let's think about it. Do they either both converge or do they both diverge? Well, this is a geometric series. Our common ratio here is less than one. So this is going to converge."}, {"video_title": "Worked example limit comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Do they either both converge or do they both diverge? Well, this is a geometric series. Our common ratio here is less than one. So this is going to converge. This is going to converge. And because this one converges, by the limit comparison test, our original series S converges. Converges."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "So let's start with a little bit of a geometric or trigonometric construction that I have here. So this white circle, this is a unit circle. Let me label it as such. So it has radius one, unit circle. So what does the length of this salmon colored line represent? Well, the height of this line would be the y coordinate of where this radius intersects the unit circle. And so by definition, by the unit circle definition of trig functions, the length of this line is going to be sine of theta."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "So it has radius one, unit circle. So what does the length of this salmon colored line represent? Well, the height of this line would be the y coordinate of where this radius intersects the unit circle. And so by definition, by the unit circle definition of trig functions, the length of this line is going to be sine of theta. If we wanted to make sure that it also worked for thetas that end up in the fourth quadrant, that will be useful, we can just ensure that it's the absolute value of the sine of theta. Now what about this blue line over here? Can I express that in terms of a trigonometric function?"}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "And so by definition, by the unit circle definition of trig functions, the length of this line is going to be sine of theta. If we wanted to make sure that it also worked for thetas that end up in the fourth quadrant, that will be useful, we can just ensure that it's the absolute value of the sine of theta. Now what about this blue line over here? Can I express that in terms of a trigonometric function? Well, let's think about it. What would tangent of theta be? Let me write it over here."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "Can I express that in terms of a trigonometric function? Well, let's think about it. What would tangent of theta be? Let me write it over here. Tangent of theta is equal to opposite over adjacent. So if we look at this broader triangle right over here, this is our angle theta in radians. This is the opposite side."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "Let me write it over here. Tangent of theta is equal to opposite over adjacent. So if we look at this broader triangle right over here, this is our angle theta in radians. This is the opposite side. The adjacent side down here, this just has length one. Remember, this is a unit circle. So this just has length one."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "This is the opposite side. The adjacent side down here, this just has length one. Remember, this is a unit circle. So this just has length one. So the tangent of theta is the opposite side. The opposite side is equal to the tangent of theta. And just like before, this is going to be a positive value if we're sitting here in the first quadrant, but I want things to work in both the first and the fourth quadrant for the sake of our proof, so I'm just gonna put an absolute value here."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "So this just has length one. So the tangent of theta is the opposite side. The opposite side is equal to the tangent of theta. And just like before, this is going to be a positive value if we're sitting here in the first quadrant, but I want things to work in both the first and the fourth quadrant for the sake of our proof, so I'm just gonna put an absolute value here. So now that we've done that, I'm gonna think about some triangles and their respective areas. So first, I'm gonna draw a triangle that sits in this wedge, in this pie piece, this pie slice within this circle. So I can construct this triangle."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "And just like before, this is going to be a positive value if we're sitting here in the first quadrant, but I want things to work in both the first and the fourth quadrant for the sake of our proof, so I'm just gonna put an absolute value here. So now that we've done that, I'm gonna think about some triangles and their respective areas. So first, I'm gonna draw a triangle that sits in this wedge, in this pie piece, this pie slice within this circle. So I can construct this triangle. And so let's think about the area of what I am shading in right over here. How can I express that area? Well, it's a triangle."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "So I can construct this triangle. And so let's think about the area of what I am shading in right over here. How can I express that area? Well, it's a triangle. We know that the area of a triangle is 1 1 2 base times height. We know the height is the absolute value of the sine of theta, and we know that the base is equal to one. So the area here is going to be equal to 1 1 2 times our base, which is one, times our height, which is the absolute value of the sine of theta."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "Well, it's a triangle. We know that the area of a triangle is 1 1 2 base times height. We know the height is the absolute value of the sine of theta, and we know that the base is equal to one. So the area here is going to be equal to 1 1 2 times our base, which is one, times our height, which is the absolute value of the sine of theta. I'll rewrite it over here. I could just rewrite that as the absolute value of the sine of theta over two. Now let's think about the area of this wedge that I am highlighting in this yellow color."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "So the area here is going to be equal to 1 1 2 times our base, which is one, times our height, which is the absolute value of the sine of theta. I'll rewrite it over here. I could just rewrite that as the absolute value of the sine of theta over two. Now let's think about the area of this wedge that I am highlighting in this yellow color. So what fraction of the entire circle is this going to be? If I were to go all the way around the circle, it would be two pi radians. So this is theta over two piths of the entire circle, and we know the area of the circle."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "Now let's think about the area of this wedge that I am highlighting in this yellow color. So what fraction of the entire circle is this going to be? If I were to go all the way around the circle, it would be two pi radians. So this is theta over two piths of the entire circle, and we know the area of the circle. This is a unit circle. It has a radius one. So it would be times the area of the circle, which would be pi times the radius squared."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "So this is theta over two piths of the entire circle, and we know the area of the circle. This is a unit circle. It has a radius one. So it would be times the area of the circle, which would be pi times the radius squared. The radius is one, so it's just gonna be times pi. And so the area of this wedge right over here, theta over two. And if we wanted to make this work for thetas in the fourth quadrant, we could just write an absolute value sign right over there because we're talking about positive area."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "So it would be times the area of the circle, which would be pi times the radius squared. The radius is one, so it's just gonna be times pi. And so the area of this wedge right over here, theta over two. And if we wanted to make this work for thetas in the fourth quadrant, we could just write an absolute value sign right over there because we're talking about positive area. And now let's think about this larger triangle in this blue color. And this is pretty straightforward. The area here is gonna be 1 1\u20442 times base times height."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "And if we wanted to make this work for thetas in the fourth quadrant, we could just write an absolute value sign right over there because we're talking about positive area. And now let's think about this larger triangle in this blue color. And this is pretty straightforward. The area here is gonna be 1 1\u20442 times base times height. So the area, and once again, this is this entire area, that's going to be 1 1\u20442 times our base, which is one, times our height, which is our absolute value of tangent of theta. And so I can just write that down as the absolute value of the tangent of theta over two. Now how would you compare the areas of this pink or this salmon-colored triangle, which sits inside of this wedge, and how would you compare that area of the wedge to the bigger triangle?"}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "The area here is gonna be 1 1\u20442 times base times height. So the area, and once again, this is this entire area, that's going to be 1 1\u20442 times our base, which is one, times our height, which is our absolute value of tangent of theta. And so I can just write that down as the absolute value of the tangent of theta over two. Now how would you compare the areas of this pink or this salmon-colored triangle, which sits inside of this wedge, and how would you compare that area of the wedge to the bigger triangle? Well, it's clear that the area of the salmon triangle is less than or equal to the area of the wedge, and the area of the wedge is less than or equal to the area of the big blue triangle. The wedge includes the salmon triangle plus this area right over here. And then the blue triangle includes the wedge plus it has this area right over here."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "Now how would you compare the areas of this pink or this salmon-colored triangle, which sits inside of this wedge, and how would you compare that area of the wedge to the bigger triangle? Well, it's clear that the area of the salmon triangle is less than or equal to the area of the wedge, and the area of the wedge is less than or equal to the area of the big blue triangle. The wedge includes the salmon triangle plus this area right over here. And then the blue triangle includes the wedge plus it has this area right over here. So I think we can feel good visually that this statement right over here is true. And now I'm just going to do a little bit of algebraic manipulation. Let me multiply everything by two."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "And then the blue triangle includes the wedge plus it has this area right over here. So I think we can feel good visually that this statement right over here is true. And now I'm just going to do a little bit of algebraic manipulation. Let me multiply everything by two. So I can rewrite that the absolute value of sine of theta is less than or equal to the absolute value of theta, which is less than or equal to the absolute value of tangent of theta. And let's see, actually, instead of writing the absolute value of tangent of theta, I'm gonna rewrite that as the absolute value of sine of theta over the absolute value of cosine of theta. That's going to be the same thing as the absolute value of tangent of theta."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "Let me multiply everything by two. So I can rewrite that the absolute value of sine of theta is less than or equal to the absolute value of theta, which is less than or equal to the absolute value of tangent of theta. And let's see, actually, instead of writing the absolute value of tangent of theta, I'm gonna rewrite that as the absolute value of sine of theta over the absolute value of cosine of theta. That's going to be the same thing as the absolute value of tangent of theta. And the reason why I did that is we can now divide everything by the absolute value of sine of theta. Since we're dividing by a positive quantity, it's not going to change the direction of the inequalities. So let's do that."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "That's going to be the same thing as the absolute value of tangent of theta. And the reason why I did that is we can now divide everything by the absolute value of sine of theta. Since we're dividing by a positive quantity, it's not going to change the direction of the inequalities. So let's do that. I'm gonna divide this by an absolute value of sine of theta. I'm gonna divide this by an absolute value of the sine of theta. And then I'm gonna divide this by an absolute value of the sine of theta."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "So let's do that. I'm gonna divide this by an absolute value of sine of theta. I'm gonna divide this by an absolute value of the sine of theta. And then I'm gonna divide this by an absolute value of the sine of theta. And what do I get? Well, over here, I get a one. And on the right-hand side, I get a one over the absolute value of cosine theta."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "And then I'm gonna divide this by an absolute value of the sine of theta. And what do I get? Well, over here, I get a one. And on the right-hand side, I get a one over the absolute value of cosine theta. These two cancel out. So the next step I'm gonna do is take the reciprocal of everything. And so when I take the reciprocal of everything, that actually will switch the inequalities."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "And on the right-hand side, I get a one over the absolute value of cosine theta. These two cancel out. So the next step I'm gonna do is take the reciprocal of everything. And so when I take the reciprocal of everything, that actually will switch the inequalities. The reciprocal of one is still going to be one. But now, since I'm taking the reciprocal of this here, it's going to be greater than or equal to the absolute value of the sine of theta over the absolute value of theta. And that's going to be greater than or equal to the reciprocal of one over the absolute value of cosine of theta is the absolute value of cosine of theta."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "And so when I take the reciprocal of everything, that actually will switch the inequalities. The reciprocal of one is still going to be one. But now, since I'm taking the reciprocal of this here, it's going to be greater than or equal to the absolute value of the sine of theta over the absolute value of theta. And that's going to be greater than or equal to the reciprocal of one over the absolute value of cosine of theta is the absolute value of cosine of theta. We really just care about the first and fourth quadrants. You can think about this theta approaching zero from that direction or from that direction there. So that would be the first and fourth quadrants."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "And that's going to be greater than or equal to the reciprocal of one over the absolute value of cosine of theta is the absolute value of cosine of theta. We really just care about the first and fourth quadrants. You can think about this theta approaching zero from that direction or from that direction there. So that would be the first and fourth quadrants. So if we're in the first quadrant and theta is positive, sine of theta is gonna be positive as well. And if we're in the fourth quadrant and theta's negative, well, sine of theta's gonna have the same sign. It's going to be negative as well."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "So that would be the first and fourth quadrants. So if we're in the first quadrant and theta is positive, sine of theta is gonna be positive as well. And if we're in the fourth quadrant and theta's negative, well, sine of theta's gonna have the same sign. It's going to be negative as well. And so these absolute value signs aren't necessary. In the first quadrant, sine of theta and theta are both positive. In the fourth quadrant, they're both negative, but when you divide them, you're going to get a positive value."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "It's going to be negative as well. And so these absolute value signs aren't necessary. In the first quadrant, sine of theta and theta are both positive. In the fourth quadrant, they're both negative, but when you divide them, you're going to get a positive value. So I can erase those. If we're in the first or fourth quadrant, our x value is not negative, and so cosine of theta, which is the x coordinate on our unit circle, is not going to be negative. And so we don't need the absolute value signs over there."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "In the fourth quadrant, they're both negative, but when you divide them, you're going to get a positive value. So I can erase those. If we're in the first or fourth quadrant, our x value is not negative, and so cosine of theta, which is the x coordinate on our unit circle, is not going to be negative. And so we don't need the absolute value signs over there. Now, we should pause a second because we're actually almost done. We have just set up three functions. You could think of this as f of x is equal to, you could view this as f of theta is equal to one, g of theta is equal to this, and h of theta is equal to that."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "And so we don't need the absolute value signs over there. Now, we should pause a second because we're actually almost done. We have just set up three functions. You could think of this as f of x is equal to, you could view this as f of theta is equal to one, g of theta is equal to this, and h of theta is equal to that. And over the interval that we care about, we could say four, negative pi over two is less than theta, is less than pi over two. But over this interval, this is true for any theta over which these functions are defined. Sine of theta over theta is defined over this interval except where theta is equal to zero."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "You could think of this as f of x is equal to, you could view this as f of theta is equal to one, g of theta is equal to this, and h of theta is equal to that. And over the interval that we care about, we could say four, negative pi over two is less than theta, is less than pi over two. But over this interval, this is true for any theta over which these functions are defined. Sine of theta over theta is defined over this interval except where theta is equal to zero. But since we're defined everywhere else, we can now find the limit. So what we can say is, well, by the squeeze theorem or by the sandwich theorem, if this is true over the interval, then we also know that the following is true. And this we deserve a little bit of a drum roll."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "Sine of theta over theta is defined over this interval except where theta is equal to zero. But since we're defined everywhere else, we can now find the limit. So what we can say is, well, by the squeeze theorem or by the sandwich theorem, if this is true over the interval, then we also know that the following is true. And this we deserve a little bit of a drum roll. The limit as theta approaches zero of this is going to be greater than or equal to the limit as theta approaches zero of this, which is the one that we care about, sine of theta over theta, which is going to be greater than or equal to the limit as theta approaches zero of this. Now, this is clearly going to be just equal to one. This is what we care about."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "And this we deserve a little bit of a drum roll. The limit as theta approaches zero of this is going to be greater than or equal to the limit as theta approaches zero of this, which is the one that we care about, sine of theta over theta, which is going to be greater than or equal to the limit as theta approaches zero of this. Now, this is clearly going to be just equal to one. This is what we care about. And this, what's the limit as theta approaches zero of cosine of theta? Well, cosine of zero is just one, and it's a continuous function, so this is just going to be one. So let's see."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "In the last video, we tried to come up with a somewhat rigorous definition of what a limit is, where we say when you say that the limit of f of x as x approaches c is equal to l, you're really saying, and this is a somewhat rigorous definition, that you can get f of x as close as you want to l by making x sufficiently close to c. So let's see if we can put a little bit meat on it. So instead of saying as close as you want, let's call that some positive number epsilon. So I'm just going to use the Greek letter epsilon right over there. So it really turns into a game. So you tell me how close you want. So this is the game. You tell me how close you want f of x to be to l. And you do this by giving me a positive number that we call epsilon, which is really how close you want f of x to be to l. So you give a positive number epsilon."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "So it really turns into a game. So you tell me how close you want. So this is the game. You tell me how close you want f of x to be to l. And you do this by giving me a positive number that we call epsilon, which is really how close you want f of x to be to l. So you give a positive number epsilon. And epsilon is how close do you want to be. How close. So for example, if epsilon is 0.01, that says that you want f of x to be within 0.01 of epsilon."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "You tell me how close you want f of x to be to l. And you do this by giving me a positive number that we call epsilon, which is really how close you want f of x to be to l. So you give a positive number epsilon. And epsilon is how close do you want to be. How close. So for example, if epsilon is 0.01, that says that you want f of x to be within 0.01 of epsilon. And so what I then do is I say, well, OK, you've given me that epsilon. I'm going to find you. I will find you another number, find another positive number, another number which we'll call delta, the lowercase delta, the Greek letter delta, such that, so I'll say where, if x is within delta of c, then f of x will be within epsilon of our limit."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "So for example, if epsilon is 0.01, that says that you want f of x to be within 0.01 of epsilon. And so what I then do is I say, well, OK, you've given me that epsilon. I'm going to find you. I will find you another number, find another positive number, another number which we'll call delta, the lowercase delta, the Greek letter delta, such that, so I'll say where, if x is within delta of c, then f of x will be within epsilon of our limit. So let's see if these are really saying the same thing. And this yellow definition right over here, we said you can get f of x as close as you want to l by making x sufficiently close to c. This second definition, which I kind of made as a little bit more of a game, is doing the same thing. Someone is saying how close they want f of x to be to l. And the burden is then to find a delta where as long as x is within delta of c, then f of x will be within epsilon of the limit."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "I will find you another number, find another positive number, another number which we'll call delta, the lowercase delta, the Greek letter delta, such that, so I'll say where, if x is within delta of c, then f of x will be within epsilon of our limit. So let's see if these are really saying the same thing. And this yellow definition right over here, we said you can get f of x as close as you want to l by making x sufficiently close to c. This second definition, which I kind of made as a little bit more of a game, is doing the same thing. Someone is saying how close they want f of x to be to l. And the burden is then to find a delta where as long as x is within delta of c, then f of x will be within epsilon of the limit. So that is doing it. It's saying, look, if we were constraining x in such a way that if x is in that range to c, then f of x will be as close as you want. So let's make this a little bit clearer by diagramming right over here."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "Someone is saying how close they want f of x to be to l. And the burden is then to find a delta where as long as x is within delta of c, then f of x will be within epsilon of the limit. So that is doing it. It's saying, look, if we were constraining x in such a way that if x is in that range to c, then f of x will be as close as you want. So let's make this a little bit clearer by diagramming right over here. You show up and you say, well, I want f of x to be within epsilon of our limit. So this right over here, this point right over here is our limit plus epsilon. And this right over here might be our limit minus."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "So let's make this a little bit clearer by diagramming right over here. You show up and you say, well, I want f of x to be within epsilon of our limit. So this right over here, this point right over here is our limit plus epsilon. And this right over here might be our limit minus. This right over here is the limit minus epsilon. And you say, OK, sure. I think I can get your f of x within this range of our limit."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "And this right over here might be our limit minus. This right over here is the limit minus epsilon. And you say, OK, sure. I think I can get your f of x within this range of our limit. And I can do that by defining a range around c. And really, I could visually look at this boundary. But I could even go narrower than boundary. I can go right over here."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "I think I can get your f of x within this range of our limit. And I can do that by defining a range around c. And really, I could visually look at this boundary. But I could even go narrower than boundary. I can go right over here. Says, OK, I meet your challenge. I will find another number, delta. So this right over here is c plus delta."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "I can go right over here. Says, OK, I meet your challenge. I will find another number, delta. So this right over here is c plus delta. This right over here is c minus. Let me write this down. Is c minus delta."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "So this right over here is c plus delta. This right over here is c minus. Let me write this down. Is c minus delta. So I'll find you some delta so that if you take any x in the range c minus delta to c plus delta, and maybe the function's not even defined at c. So we think of ones that maybe aren't c, but are getting very close. If you find any x in that range, f of those x's are going to be as close as you want to your limit. They're going to be within the range l plus epsilon or l minus epsilon."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "Is c minus delta. So I'll find you some delta so that if you take any x in the range c minus delta to c plus delta, and maybe the function's not even defined at c. So we think of ones that maybe aren't c, but are getting very close. If you find any x in that range, f of those x's are going to be as close as you want to your limit. They're going to be within the range l plus epsilon or l minus epsilon. So what's another way of saying this? Another way of saying this is you give me an epsilon. Then I will find you a delta."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "They're going to be within the range l plus epsilon or l minus epsilon. So what's another way of saying this? Another way of saying this is you give me an epsilon. Then I will find you a delta. So let me write this in a little bit more math notation. So I'll write the same exact statements with a little bit more math here, but it's the exact same thing. So you give me, or let me write it this way."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "Then I will find you a delta. So let me write this in a little bit more math notation. So I'll write the same exact statements with a little bit more math here, but it's the exact same thing. So you give me, or let me write it this way. Given an epsilon greater than 0, we can find, so that's kind of the first part of the game, we can find a delta greater than 0 such that if x is within delta of c. So what's another way of saying that x is within delta of c? Well, one way you could say, well, what's the distance between x and c is going to be less than delta. This statement is true for any x that's within delta of c. The difference between the two is going to be less than delta."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "So you give me, or let me write it this way. Given an epsilon greater than 0, we can find, so that's kind of the first part of the game, we can find a delta greater than 0 such that if x is within delta of c. So what's another way of saying that x is within delta of c? Well, one way you could say, well, what's the distance between x and c is going to be less than delta. This statement is true for any x that's within delta of c. The difference between the two is going to be less than delta. So that if you pick an x that is in this range between c minus delta and c plus delta, and these are the x's that satisfy that right over here, then the distance between your f of x and your limit, and this is just the distance between the f of x and the limit, it's going to be less than epsilon. So all this is saying is if the limit truly does exist, it truly is L, is if you give me any positive number epsilon, it could be a super, super small one, we can find a delta. So we can define a range around c so that if we take any x value that is within delta of c, that's all this statement is saying, that the distance between x and c is less than delta."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "This statement is true for any x that's within delta of c. The difference between the two is going to be less than delta. So that if you pick an x that is in this range between c minus delta and c plus delta, and these are the x's that satisfy that right over here, then the distance between your f of x and your limit, and this is just the distance between the f of x and the limit, it's going to be less than epsilon. So all this is saying is if the limit truly does exist, it truly is L, is if you give me any positive number epsilon, it could be a super, super small one, we can find a delta. So we can define a range around c so that if we take any x value that is within delta of c, that's all this statement is saying, that the distance between x and c is less than delta. So it's within delta c. So that's these points right over here. That f of those x's, the function evaluated at those x's is going to be within the range that you are specifying. It's going to be within epsilon of our limit."}, {"video_title": "Epsilon-delta definition of limits.mp3", "Sentence": "So we can define a range around c so that if we take any x value that is within delta of c, that's all this statement is saying, that the distance between x and c is less than delta. So it's within delta c. So that's these points right over here. That f of those x's, the function evaluated at those x's is going to be within the range that you are specifying. It's going to be within epsilon of our limit. The f of x, the difference between f of x and your limit will be less than epsilon. Your f of x is going to sit someplace over there. So that's all the epsilon delta definition is telling us."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "So first, differentiability. Differentiability. So let's think about that first. It's always helpful to draw ourselves a function. So that's our y-axis. This is our x-axis. And let's just draw some function here."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "It's always helpful to draw ourselves a function. So that's our y-axis. This is our x-axis. And let's just draw some function here. So let's say my function looks like this. And we care about the point x equals C, which is right over here. So that's the point x equals C. And then this value, of course, is going to be f of C. And one way that we can find the derivative at x equals C, or the slope of the tangent line at x equals C, is we could start with some other point, say some arbitrary x out here."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "And let's just draw some function here. So let's say my function looks like this. And we care about the point x equals C, which is right over here. So that's the point x equals C. And then this value, of course, is going to be f of C. And one way that we can find the derivative at x equals C, or the slope of the tangent line at x equals C, is we could start with some other point, say some arbitrary x out here. So let's say this is some arbitrary x out here. So then this point right over there, this value, this y value, would be f of x. This graph, of course, is a graph of y equals f of x."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "So that's the point x equals C. And then this value, of course, is going to be f of C. And one way that we can find the derivative at x equals C, or the slope of the tangent line at x equals C, is we could start with some other point, say some arbitrary x out here. So let's say this is some arbitrary x out here. So then this point right over there, this value, this y value, would be f of x. This graph, of course, is a graph of y equals f of x. And we can think about finding the slope of this line, this secant line between these two points. But then we can find the limit as x approaches C. And as x approaches C, this secant, the slope of the secant line is going to approach the slope of the tangent line, or it's going to be the derivative. And so we could take the limit as x approaches C, of the slope of this secant line."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "This graph, of course, is a graph of y equals f of x. And we can think about finding the slope of this line, this secant line between these two points. But then we can find the limit as x approaches C. And as x approaches C, this secant, the slope of the secant line is going to approach the slope of the tangent line, or it's going to be the derivative. And so we could take the limit as x approaches C, of the slope of this secant line. So what's the slope? Well, it's going to be change in y over change in x. The change in y is f of x minus f of C. That's our change in y right over here."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "And so we could take the limit as x approaches C, of the slope of this secant line. So what's the slope? Well, it's going to be change in y over change in x. The change in y is f of x minus f of C. That's our change in y right over here. And this is all a review. This is just one definition of the derivative, or one way to think about the derivative. So it's going to be f of x minus f of C, that's our change in y, over our change in x, which is x minus C. It is x minus C. So if this limit exists, then we're able to find the slope of the tangent line at this point."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "The change in y is f of x minus f of C. That's our change in y right over here. And this is all a review. This is just one definition of the derivative, or one way to think about the derivative. So it's going to be f of x minus f of C, that's our change in y, over our change in x, which is x minus C. It is x minus C. So if this limit exists, then we're able to find the slope of the tangent line at this point. And we call that slope of the tangent line, we call that the derivative at x equals C. We say that this is going to be equal to f prime of C. All of this is review. So if we're saying, one way to think about it, if we're saying that the function f is differentiable at x equals C, we're really just saying that this limit right over here actually exists. And if this limit actually exists, we just call that value f prime of C. So that's just a review of differentiability."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "So it's going to be f of x minus f of C, that's our change in y, over our change in x, which is x minus C. It is x minus C. So if this limit exists, then we're able to find the slope of the tangent line at this point. And we call that slope of the tangent line, we call that the derivative at x equals C. We say that this is going to be equal to f prime of C. All of this is review. So if we're saying, one way to think about it, if we're saying that the function f is differentiable at x equals C, we're really just saying that this limit right over here actually exists. And if this limit actually exists, we just call that value f prime of C. So that's just a review of differentiability. Now let's give ourselves a review of continuity. Con-ti-nuity. So the definition for continuity is if the limit as x approaches C of f of x is equal to f of C. Now this might seem a little bit, you know, well, it might pop out to you as being intuitive, or it might seem a little, well, where did this come from?"}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "And if this limit actually exists, we just call that value f prime of C. So that's just a review of differentiability. Now let's give ourselves a review of continuity. Con-ti-nuity. So the definition for continuity is if the limit as x approaches C of f of x is equal to f of C. Now this might seem a little bit, you know, well, it might pop out to you as being intuitive, or it might seem a little, well, where did this come from? Well, let's visualize it, and then hopefully it'll make some intuitive sense. So if you have a function, so let's actually look at some cases where you're not continuous. And that actually might make it a little bit more clear."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "So the definition for continuity is if the limit as x approaches C of f of x is equal to f of C. Now this might seem a little bit, you know, well, it might pop out to you as being intuitive, or it might seem a little, well, where did this come from? Well, let's visualize it, and then hopefully it'll make some intuitive sense. So if you have a function, so let's actually look at some cases where you're not continuous. And that actually might make it a little bit more clear. So if you had a point discontinuity at x equals C, so this is x equals C. So if you had a point discontinuity, so let me draw it like this actually. So you have a gap here, and x equals, when x equals C, f of C is actually way up here. So this is f of C, and then the function continues like this."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "And that actually might make it a little bit more clear. So if you had a point discontinuity at x equals C, so this is x equals C. So if you had a point discontinuity, so let me draw it like this actually. So you have a gap here, and x equals, when x equals C, f of C is actually way up here. So this is f of C, and then the function continues like this. The limit as x approaches C of f of x is going to be this value, which is clearly different than f of C, this value right over here. If you take the limit, if you take the limit as x approaches C of f of x, you're approaching this value. This right over here is the limit as x approaches C of f of x, which is different than f of C. So this definition of continuity seems to be good at least for this case, because this is not a continuous function."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "So this is f of C, and then the function continues like this. The limit as x approaches C of f of x is going to be this value, which is clearly different than f of C, this value right over here. If you take the limit, if you take the limit as x approaches C of f of x, you're approaching this value. This right over here is the limit as x approaches C of f of x, which is different than f of C. So this definition of continuity seems to be good at least for this case, because this is not a continuous function. You have a point discontinuity. So for at least in this case, this definition of continuity would properly identify this as not a continuous function. Now you could also think about a jump discontinuity."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "This right over here is the limit as x approaches C of f of x, which is different than f of C. So this definition of continuity seems to be good at least for this case, because this is not a continuous function. You have a point discontinuity. So for at least in this case, this definition of continuity would properly identify this as not a continuous function. Now you could also think about a jump discontinuity. You could also think about a jump discontinuity. So let's look at this. And all of this is hopefully a little bit of review."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "Now you could also think about a jump discontinuity. You could also think about a jump discontinuity. So let's look at this. And all of this is hopefully a little bit of review. So a jump discontinuity at C, at x equals C, might look like this. Might look like this. So this is at x equals C. So this is x equals C right over here."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "And all of this is hopefully a little bit of review. So a jump discontinuity at C, at x equals C, might look like this. Might look like this. So this is at x equals C. So this is x equals C right over here. This would be f of C. But if you tried to evaluate the limit as x approaches C of f of x, you'd get a different value as you approach C from the negative side. You would approach this value. And as you approach C from the positive side, you would approach f of C. And so the limit wouldn't exist."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "So this is at x equals C. So this is x equals C right over here. This would be f of C. But if you tried to evaluate the limit as x approaches C of f of x, you'd get a different value as you approach C from the negative side. You would approach this value. And as you approach C from the positive side, you would approach f of C. And so the limit wouldn't exist. So this limit right over here wouldn't exist in the case of this type of a jump discontinuity. So once again, this definition would properly say that this is not, this one right over here is not continuous. This limit actually would not even exist."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "And as you approach C from the positive side, you would approach f of C. And so the limit wouldn't exist. So this limit right over here wouldn't exist in the case of this type of a jump discontinuity. So once again, this definition would properly say that this is not, this one right over here is not continuous. This limit actually would not even exist. And then you could even look at a, you could look at a function that is truly continuous. If you look at a function that is truly continuous, so something like this. Something like this."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "This limit actually would not even exist. And then you could even look at a, you could look at a function that is truly continuous. If you look at a function that is truly continuous, so something like this. Something like this. That is x equals C. Well, this is f of C. This is f of C. And if you were to take the limit as x approaches C, as x approaches C from either side of f of x, you're going to approach f of C. So here you have the limit as x approaches C of f of x indeed is equal to f of C. So it's what you would expect for a continuous function. So now that we've done that review of differentiability and continuity, let's prove that differentiability actually implies continuity. And I think it's important to kind of do this review just so that you can really visualize things."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "Something like this. That is x equals C. Well, this is f of C. This is f of C. And if you were to take the limit as x approaches C, as x approaches C from either side of f of x, you're going to approach f of C. So here you have the limit as x approaches C of f of x indeed is equal to f of C. So it's what you would expect for a continuous function. So now that we've done that review of differentiability and continuity, let's prove that differentiability actually implies continuity. And I think it's important to kind of do this review just so that you can really visualize things. So differentiability implies this, this limit right over here exists. So let's start with a slightly different limit. Let me draw a line here actually."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "And I think it's important to kind of do this review just so that you can really visualize things. So differentiability implies this, this limit right over here exists. So let's start with a slightly different limit. Let me draw a line here actually. Let me draw a line just so we're doing something different. So let's take, let us take the limit as x approaches C of f of x, of f of x minus f of C. Of f of x minus f of C. Well can we rewrite this? Well we could rewrite this as the limit as x approaches C. And we can essentially take this expression and multiply and divide it by x minus C. So let's multiply it times x minus C. x minus C and divide it by x minus C. So we have f of x minus f of C. All of that over x minus C. So all I did is I multiplied and I divided by x minus C. Well what's this limit going to be equal to?"}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "Let me draw a line here actually. Let me draw a line just so we're doing something different. So let's take, let us take the limit as x approaches C of f of x, of f of x minus f of C. Of f of x minus f of C. Well can we rewrite this? Well we could rewrite this as the limit as x approaches C. And we can essentially take this expression and multiply and divide it by x minus C. So let's multiply it times x minus C. x minus C and divide it by x minus C. So we have f of x minus f of C. All of that over x minus C. So all I did is I multiplied and I divided by x minus C. Well what's this limit going to be equal to? This is going to be equal to, it's going to be the limit, and I'm just applying the property of limit, property, I'm applying a property of limits here. So the limit of the product is equal to the same thing as the product of the limits. So it's the limit as x approaches C of x minus C times the limit, let me write it this way, times the limit as x approaches C of f of x minus f of C. All of that over x minus C. Now what is this thing right over here?"}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "Well we could rewrite this as the limit as x approaches C. And we can essentially take this expression and multiply and divide it by x minus C. So let's multiply it times x minus C. x minus C and divide it by x minus C. So we have f of x minus f of C. All of that over x minus C. So all I did is I multiplied and I divided by x minus C. Well what's this limit going to be equal to? This is going to be equal to, it's going to be the limit, and I'm just applying the property of limit, property, I'm applying a property of limits here. So the limit of the product is equal to the same thing as the product of the limits. So it's the limit as x approaches C of x minus C times the limit, let me write it this way, times the limit as x approaches C of f of x minus f of C. All of that over x minus C. Now what is this thing right over here? Well if we assume that f is differentiable at C, and we're going to do that, actually I should have started off there. Let's assume, let's assume, because we wanted to show that differentiability improves continuity. If we assume f differentiable, differentiable at C, well then this right over here is just going to be f prime of C. This right over here, we just saw it right over here, that's this exact same thing."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "So it's the limit as x approaches C of x minus C times the limit, let me write it this way, times the limit as x approaches C of f of x minus f of C. All of that over x minus C. Now what is this thing right over here? Well if we assume that f is differentiable at C, and we're going to do that, actually I should have started off there. Let's assume, let's assume, because we wanted to show that differentiability improves continuity. If we assume f differentiable, differentiable at C, well then this right over here is just going to be f prime of C. This right over here, we just saw it right over here, that's this exact same thing. This is f prime, f prime of C. And what is this thing right over here? The limit as x approaches C of x minus C? Well that's just going to be zero."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "If we assume f differentiable, differentiable at C, well then this right over here is just going to be f prime of C. This right over here, we just saw it right over here, that's this exact same thing. This is f prime, f prime of C. And what is this thing right over here? The limit as x approaches C of x minus C? Well that's just going to be zero. As x approaches C, it's going to approach C minus C, it's just going to be zero. So what's zero times f prime of C? Well f prime of C is just going to be some value, so zero times anything is just going to be zero."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "Well that's just going to be zero. As x approaches C, it's going to approach C minus C, it's just going to be zero. So what's zero times f prime of C? Well f prime of C is just going to be some value, so zero times anything is just going to be zero. So I did all that work to get a zero. Now why is this interesting? Well we just said, we just assumed that if f is differentiable at C, and we evaluate this limit, we get zero."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "Well f prime of C is just going to be some value, so zero times anything is just going to be zero. So I did all that work to get a zero. Now why is this interesting? Well we just said, we just assumed that if f is differentiable at C, and we evaluate this limit, we get zero. So if we assume f is differentiable at C, we can write, we can write the limit, I'm just rewriting it, the limit as x approaches C of f of x minus f of C, and I could even put parentheses around it like that, which I already did up here, is equal to zero. Well this is the same thing, I could use limit properties again, this is the same thing as saying, and I'll do it over here, actually let me do it down here. The limit as x approaches C of f of x minus the limit as x approaches C of f of C, of f of C, is equal to zero."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "Well we just said, we just assumed that if f is differentiable at C, and we evaluate this limit, we get zero. So if we assume f is differentiable at C, we can write, we can write the limit, I'm just rewriting it, the limit as x approaches C of f of x minus f of C, and I could even put parentheses around it like that, which I already did up here, is equal to zero. Well this is the same thing, I could use limit properties again, this is the same thing as saying, and I'll do it over here, actually let me do it down here. The limit as x approaches C of f of x minus the limit as x approaches C of f of C, of f of C, is equal to zero. The limit of the difference is the same thing as the difference of the limits. Well what's this thing over here going to be? Well f of C is just a number, it's not a function of x anymore, it's just f of C is going to evaluate to something."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "The limit as x approaches C of f of x minus the limit as x approaches C of f of C, of f of C, is equal to zero. The limit of the difference is the same thing as the difference of the limits. Well what's this thing over here going to be? Well f of C is just a number, it's not a function of x anymore, it's just f of C is going to evaluate to something. So this is just going to be f of C. This is just going to be f of C. So if the limit of f of x as x approaches C minus f of C is equal to zero. Well just add f of C to both sides and what do you get? Well you get the limit as x approaches C of f of x is equal to f of C. And this is the definition of continuity, the limit of my function as x approaches C is equal to the function, is equal to the value of the function at C. This is, this means that our function is continuous."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "Well f of C is just a number, it's not a function of x anymore, it's just f of C is going to evaluate to something. So this is just going to be f of C. This is just going to be f of C. So if the limit of f of x as x approaches C minus f of C is equal to zero. Well just add f of C to both sides and what do you get? Well you get the limit as x approaches C of f of x is equal to f of C. And this is the definition of continuity, the limit of my function as x approaches C is equal to the function, is equal to the value of the function at C. This is, this means that our function is continuous. Continuous at C. So just a reminder, we started assuming f differentiable at C, we use that fact to evaluate this limit right over here, which we got to be equal to zero. And if that limit is equal to zero, then it just follows, just doing a little bit of algebra and using properties of limits, that the limit as x approaches C of f of x is equal to f of C. And that's our definition of being continuous, continuous at the point C. So hopefully that satisfies you. If we know that the derivative exists at a point, if it's differentiable at a point C, that means it's also continuous at that point C. The function is also continuous at that point."}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And for the sake of this video, we can assume that the graph of this function just keeps getting lower and lower and lower as x becomes more and more negative, and lower and lower and lower as x goes beyond the interval that I've depicted right over here. So what is the maximum value that this function takes on? Well, we can eyeball that. It looks like it's at this point. It looks like it's at that point right over there. So we would call this a global maximum. The function never takes on a value larger than this."}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "It looks like it's at this point. It looks like it's at that point right over there. So we would call this a global maximum. The function never takes on a value larger than this. So we could say that we have a global maximum at the point x naught, because f of x naught is greater than or equal to f of x for any other x in the domain. And that's pretty obvious when you look at it like this. Now, do we have a global minimum point the way that I've drawn it?"}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "The function never takes on a value larger than this. So we could say that we have a global maximum at the point x naught, because f of x naught is greater than or equal to f of x for any other x in the domain. And that's pretty obvious when you look at it like this. Now, do we have a global minimum point the way that I've drawn it? Well, no. This function can take on arbitrarily negative values. It approaches negative infinity as x approaches negative infinity."}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Now, do we have a global minimum point the way that I've drawn it? Well, no. This function can take on arbitrarily negative values. It approaches negative infinity as x approaches negative infinity. It approaches negative infinity as x approaches positive infinity. So we have, let me write this down, we have no global minimum. Now, let me ask you a question."}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "It approaches negative infinity as x approaches negative infinity. It approaches negative infinity as x approaches positive infinity. So we have, let me write this down, we have no global minimum. Now, let me ask you a question. Do we have local minima or local maximum? When I say minima, it's just the plural of minimum. And maxima is just the plural of maximum."}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Now, let me ask you a question. Do we have local minima or local maximum? When I say minima, it's just the plural of minimum. And maxima is just the plural of maximum. So do we have a local minima here or a local minimum here? Well, a local minimum, you could imagine, means that that value of the function at that point is lower than the points around it. So right over here, it looks like we have a local minimum."}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And maxima is just the plural of maximum. So do we have a local minima here or a local minimum here? Well, a local minimum, you could imagine, means that that value of the function at that point is lower than the points around it. So right over here, it looks like we have a local minimum. And I'm not giving you a very rigorous definition here, but one way to think about it is we can say that we have a local minimum point at x1 is if we have a region around x1 where f of x1 is less than an f of x for any x in this region right over here. And it's pretty easy to eyeball, too. This is a low point for any of the values of f around it right over there."}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So right over here, it looks like we have a local minimum. And I'm not giving you a very rigorous definition here, but one way to think about it is we can say that we have a local minimum point at x1 is if we have a region around x1 where f of x1 is less than an f of x for any x in this region right over here. And it's pretty easy to eyeball, too. This is a low point for any of the values of f around it right over there. Now, do we have any other local minima? Well, it doesn't look like we do. Now, what about local maxima?"}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "This is a low point for any of the values of f around it right over there. Now, do we have any other local minima? Well, it doesn't look like we do. Now, what about local maxima? Well, this one right over here, let me do it in purple. I don't want to get people confused, actually. Let me do it in this color."}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Now, what about local maxima? Well, this one right over here, let me do it in purple. I don't want to get people confused, actually. Let me do it in this color. This point right over here looks like a local maximum. Local, not lox. That would have to deal with salmon."}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Let me do it in this color. This point right over here looks like a local maximum. Local, not lox. That would have to deal with salmon. Local maximum right over there. So we could say at the point x2, we have a local maximum point at x2 because f of x2 is larger than f of x for any x around a neighborhood around x2. I'm not being very rigorous, but you can see it just by looking at it."}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "That would have to deal with salmon. Local maximum right over there. So we could say at the point x2, we have a local maximum point at x2 because f of x2 is larger than f of x for any x around a neighborhood around x2. I'm not being very rigorous, but you can see it just by looking at it. So that's fair enough. We've identified all of the maxima and minima, often called the extrema, for this function. Now, how can we identify those if we knew something about the derivative of the function?"}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "I'm not being very rigorous, but you can see it just by looking at it. So that's fair enough. We've identified all of the maxima and minima, often called the extrema, for this function. Now, how can we identify those if we knew something about the derivative of the function? Well, let's look at the derivative at each of these points. So at this first point right over here, if I were to try to visualize the tangent line, let me do that in a better color than brown. If I were to try to visualize the tangent line, it would look something like that."}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Now, how can we identify those if we knew something about the derivative of the function? Well, let's look at the derivative at each of these points. So at this first point right over here, if I were to try to visualize the tangent line, let me do that in a better color than brown. If I were to try to visualize the tangent line, it would look something like that. So the slope here is 0. So we would say that f prime of x0 is equal to 0. So the slope of the tangent line at this point is 0."}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "If I were to try to visualize the tangent line, it would look something like that. So the slope here is 0. So we would say that f prime of x0 is equal to 0. So the slope of the tangent line at this point is 0. What about over here? Well, once again, the tangent line would look something like that. So once again, we would say f prime at x1 is equal to 0."}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So the slope of the tangent line at this point is 0. What about over here? Well, once again, the tangent line would look something like that. So once again, we would say f prime at x1 is equal to 0. What about over here? Well, here, the tangent line is actually not well-defined. We have a positive slope going into it, and then it immediately jumps to being a negative slope."}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So once again, we would say f prime at x1 is equal to 0. What about over here? Well, here, the tangent line is actually not well-defined. We have a positive slope going into it, and then it immediately jumps to being a negative slope. So over here, f prime of x2 is not defined. Let me just write undefined. So we have an interesting, and once again, I'm not rigorously proving it to you."}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "We have a positive slope going into it, and then it immediately jumps to being a negative slope. So over here, f prime of x2 is not defined. Let me just write undefined. So we have an interesting, and once again, I'm not rigorously proving it to you. I just want you to get the intuition here. We see that if we have some type of an extrema, and we're not talking about when x is at an endpoint of an interval. Just to be clear what I'm talking about when I'm talking about x as an endpoint of an interval."}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So we have an interesting, and once again, I'm not rigorously proving it to you. I just want you to get the intuition here. We see that if we have some type of an extrema, and we're not talking about when x is at an endpoint of an interval. Just to be clear what I'm talking about when I'm talking about x as an endpoint of an interval. We're saying, let's say that the function is, let's say we have an interval from there. So let's say a function starts right over there and then keeps going. This would be a maximum point, but it would be an endpoint."}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Just to be clear what I'm talking about when I'm talking about x as an endpoint of an interval. We're saying, let's say that the function is, let's say we have an interval from there. So let's say a function starts right over there and then keeps going. This would be a maximum point, but it would be an endpoint. We're not talking about endpoints right now. We're talking about when we have points in between, or when our interval is infinite. So we're not talking about points like that or points like this."}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "This would be a maximum point, but it would be an endpoint. We're not talking about endpoints right now. We're talking about when we have points in between, or when our interval is infinite. So we're not talking about points like that or points like this. We're talking about the points in between. So if you have a point inside of an interval, it's going to be a minimum or maximum, and we see the intuition here. If you have non-endpoint min or max at, let's say, x is equal to a."}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So we're not talking about points like that or points like this. We're talking about the points in between. So if you have a point inside of an interval, it's going to be a minimum or maximum, and we see the intuition here. If you have non-endpoint min or max at, let's say, x is equal to a. So if you know that you have a minimum or maximum point at some point x is equal to a and x isn't the endpoint of some interval, this tells you something interesting, or at least we have the intuition. We see that the derivative at x is equal to a is going to be equal to 0, or the derivative at x is equal to a is going to be undefined. And we see that in each of these cases."}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "If you have non-endpoint min or max at, let's say, x is equal to a. So if you know that you have a minimum or maximum point at some point x is equal to a and x isn't the endpoint of some interval, this tells you something interesting, or at least we have the intuition. We see that the derivative at x is equal to a is going to be equal to 0, or the derivative at x is equal to a is going to be undefined. And we see that in each of these cases. Derivative is 0. Derivative is 0. Derivative is undefined."}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And we see that in each of these cases. Derivative is 0. Derivative is 0. Derivative is undefined. And we have a word for these points where the derivative is either 0 or the derivative is undefined. We call them critical points. So for the sake of this function, the critical points are, if we could include x sub 0, we could include x sub 1, at x sub 0 and x sub 1 the derivative is 0, and x sub 2 where the function is undefined."}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Derivative is undefined. And we have a word for these points where the derivative is either 0 or the derivative is undefined. We call them critical points. So for the sake of this function, the critical points are, if we could include x sub 0, we could include x sub 1, at x sub 0 and x sub 1 the derivative is 0, and x sub 2 where the function is undefined. Now, so if we have a non-endpoint minimum or maximum point, then it's going to be a critical point. But can we say it the other way around? If we find a critical point where the derivative is 0 or the derivative is undefined, is that going to be a maximum or minimum point?"}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So for the sake of this function, the critical points are, if we could include x sub 0, we could include x sub 1, at x sub 0 and x sub 1 the derivative is 0, and x sub 2 where the function is undefined. Now, so if we have a non-endpoint minimum or maximum point, then it's going to be a critical point. But can we say it the other way around? If we find a critical point where the derivative is 0 or the derivative is undefined, is that going to be a maximum or minimum point? And to think about that, let's imagine this point right over here. So let's call this x sub 3. If we look at the tangent line right over here, if we look at the slope right over here, it looks like f prime of x sub 3 is equal to 0."}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "If we find a critical point where the derivative is 0 or the derivative is undefined, is that going to be a maximum or minimum point? And to think about that, let's imagine this point right over here. So let's call this x sub 3. If we look at the tangent line right over here, if we look at the slope right over here, it looks like f prime of x sub 3 is equal to 0. So based on our definition of a critical point, x sub 3 would also be a critical point. But it does not appear to be a minimum or a maximum point. So a minimum or a maximum point, that's not an endpoint."}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "If we look at the tangent line right over here, if we look at the slope right over here, it looks like f prime of x sub 3 is equal to 0. So based on our definition of a critical point, x sub 3 would also be a critical point. But it does not appear to be a minimum or a maximum point. So a minimum or a maximum point, that's not an endpoint. It's definitely going to be a critical point. But being a critical point by itself does not mean you're at a minimum or maximum point. So just to be clear, at all of these points, we're at a minimum or maximum point."}, {"video_title": "Critical points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So a minimum or a maximum point, that's not an endpoint. It's definitely going to be a critical point. But being a critical point by itself does not mean you're at a minimum or maximum point. So just to be clear, at all of these points, we're at a minimum or maximum point. This, we're at a critical point. All of these are critical points, but this is not a minimum or maximum point. In the next video, we'll start to think about how you can differentiate or how you can tell whether you have a minimum or maximum at a critical point."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And we saw it's really just the difference between the function and our approximation of the function. So for example, this distance right over here, that is our error at x is equal to b. And what we really care about is the absolute value of it, because at some points, f of x might be larger than the polynomial. Sometimes the polynomial might be larger than f of x. What we care is the absolute distance between them. And so what I want to do in this video is try to bound our error at some b. Try to bound our error."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Sometimes the polynomial might be larger than f of x. What we care is the absolute distance between them. And so what I want to do in this video is try to bound our error at some b. Try to bound our error. So say it's less than or equal to some constant value. Try to bound it at b for some b is greater than a. We're just going to assume that b is greater than a."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Try to bound our error. So say it's less than or equal to some constant value. Try to bound it at b for some b is greater than a. We're just going to assume that b is greater than a. And we saw some tantalizing. We got to a bit of a tantalizing result that seems like we might be able to bound it in the last video. We saw that the n plus 1th derivative of our error function is equal to the n plus 1th derivative of our function, or that their absolute values would also be equal to."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "We're just going to assume that b is greater than a. And we saw some tantalizing. We got to a bit of a tantalizing result that seems like we might be able to bound it in the last video. We saw that the n plus 1th derivative of our error function is equal to the n plus 1th derivative of our function, or that their absolute values would also be equal to. So if we can somehow bound the n plus 1th derivative of our function over some interval, an interval that matters to us, an interval that maybe has b in it, then we can at least bound the n plus 1th derivative of our error function. And then maybe we can do a little bit of integration to bound the error itself at some value b. So let's see if we can do that."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "We saw that the n plus 1th derivative of our error function is equal to the n plus 1th derivative of our function, or that their absolute values would also be equal to. So if we can somehow bound the n plus 1th derivative of our function over some interval, an interval that matters to us, an interval that maybe has b in it, then we can at least bound the n plus 1th derivative of our error function. And then maybe we can do a little bit of integration to bound the error itself at some value b. So let's see if we can do that. Well, let's just assume that we're in a reality where we do know something about the n plus 1th derivative of f of x. Let's say we do know that this, we do it in a color that I haven't used yet. I'll do it in white."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's see if we can do that. Well, let's just assume that we're in a reality where we do know something about the n plus 1th derivative of f of x. Let's say we do know that this, we do it in a color that I haven't used yet. I'll do it in white. So let's say that thing right over there looks something like that. So that is f, the n plus 1th derivative. And I only care about it over this interval right over here."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "I'll do it in white. So let's say that thing right over there looks something like that. So that is f, the n plus 1th derivative. And I only care about it over this interval right over here. Who cares what it does later? I just want to bound it over the interval because at the end of the day, I just want to bound b right over here. So let's say that the absolute value of this, let's say that we know, let me write it over here."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And I only care about it over this interval right over here. Who cares what it does later? I just want to bound it over the interval because at the end of the day, I just want to bound b right over here. So let's say that the absolute value of this, let's say that we know, let me write it over here. Let's say that we know that the absolute value of the n plus 1th derivative, the n plus 1th, and I apologize, I actually switched between the capital n and the lowercase n, and I did that in the last video. I shouldn't have, but now that you know that I did that, hopefully it doesn't confuse you. n plus 1th."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's say that the absolute value of this, let's say that we know, let me write it over here. Let's say that we know that the absolute value of the n plus 1th derivative, the n plus 1th, and I apologize, I actually switched between the capital n and the lowercase n, and I did that in the last video. I shouldn't have, but now that you know that I did that, hopefully it doesn't confuse you. n plus 1th. So let's say we know that the n plus 1th derivative of f of x, the absolute value of it, let's say it's bounded, let's say it's less than or equal to some m over the interval, because we only care about the interval. It might not be bounded in general, but all we care is that it takes some maximum value over this interval. So over the interval x, I could write it this way, over the interval x is a member between a and b."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "n plus 1th. So let's say we know that the n plus 1th derivative of f of x, the absolute value of it, let's say it's bounded, let's say it's less than or equal to some m over the interval, because we only care about the interval. It might not be bounded in general, but all we care is that it takes some maximum value over this interval. So over the interval x, I could write it this way, over the interval x is a member between a and b. And this includes both of them. It's a closed interval. x could be a, x could be b, or x could be anything in between."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So over the interval x, I could write it this way, over the interval x is a member between a and b. And this includes both of them. It's a closed interval. x could be a, x could be b, or x could be anything in between. And we can say this generally, that this derivative will have some maximum value. So this is the absolute value is maximum value, max value, m for max. We know that it will have a maximum value if this thing is continuous."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "x could be a, x could be b, or x could be anything in between. And we can say this generally, that this derivative will have some maximum value. So this is the absolute value is maximum value, max value, m for max. We know that it will have a maximum value if this thing is continuous. So once again, we're going to assume that it is continuous, that it has some maximum value over this interval right over here. Well, this thing, this thing right over here, we know is the same thing as the n plus 1th derivative of the error function. So then we know, so then that implies, that implies that the, that's a new color."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "We know that it will have a maximum value if this thing is continuous. So once again, we're going to assume that it is continuous, that it has some maximum value over this interval right over here. Well, this thing, this thing right over here, we know is the same thing as the n plus 1th derivative of the error function. So then we know, so then that implies, that implies that the, that's a new color. Let me do that blue or that green. That implies that the n plus 1th derivative of the error function, the absolute value of it, because these are the same thing, is also bounded by m. So that's a little bit of an interesting result, but it gets us nowhere near there. It might look similar, but this is the n plus 1th derivative of the error function."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So then we know, so then that implies, that implies that the, that's a new color. Let me do that blue or that green. That implies that the n plus 1th derivative of the error function, the absolute value of it, because these are the same thing, is also bounded by m. So that's a little bit of an interesting result, but it gets us nowhere near there. It might look similar, but this is the n plus 1th derivative of the error function. And we'll have to think about how we can get an m in the future. We're assuming that we somehow know it, and maybe we'll do some example problems where we figure that out. But this is the n plus 1th derivative."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "It might look similar, but this is the n plus 1th derivative of the error function. And we'll have to think about how we can get an m in the future. We're assuming that we somehow know it, and maybe we'll do some example problems where we figure that out. But this is the n plus 1th derivative. We bounded its absolute value, but we really want to bound the actual error function, the 0th derivative, you could say, the actual function itself. Well, we could try to integrate both sides of this and see if we can eventually get to e, to get to e of x, to get to our error function or our remainder function. So let's do that."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "But this is the n plus 1th derivative. We bounded its absolute value, but we really want to bound the actual error function, the 0th derivative, you could say, the actual function itself. Well, we could try to integrate both sides of this and see if we can eventually get to e, to get to e of x, to get to our error function or our remainder function. So let's do that. Let's take the integral, let's take the integral of both sides of this. Now the integral on this left-hand side, it's a little interesting. We take the integral of the absolute value."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's do that. Let's take the integral, let's take the integral of both sides of this. Now the integral on this left-hand side, it's a little interesting. We take the integral of the absolute value. It would be easier if we were taking the absolute value of the integral. And lucky for us, the way it's set up, so let me just write a little aside here. We know generally that if I take, and it's something for you to think about."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "We take the integral of the absolute value. It would be easier if we were taking the absolute value of the integral. And lucky for us, the way it's set up, so let me just write a little aside here. We know generally that if I take, and it's something for you to think about. If I take, so if I have two options, this option versus, and I know they look the same right now. So over here I'm going to have the integral of the absolute value, and over here I'm going to have the absolute value of the integral. Which of these is going to be, which of these can be larger?"}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "We know generally that if I take, and it's something for you to think about. If I take, so if I have two options, this option versus, and I know they look the same right now. So over here I'm going to have the integral of the absolute value, and over here I'm going to have the absolute value of the integral. Which of these is going to be, which of these can be larger? Well, you just have to think about the scenarios. If f of x is always positive over the interval that you're taking the integration, then they're going to be the same thing. So you're going to get positive values, take the absolute value of a positive value, it doesn't make a difference."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Which of these is going to be, which of these can be larger? Well, you just have to think about the scenarios. If f of x is always positive over the interval that you're taking the integration, then they're going to be the same thing. So you're going to get positive values, take the absolute value of a positive value, it doesn't make a difference. What matters is if f of x is negative. If f of x is negative the entire time, so if this is our x-axis, that is our y-axis. If f of x is, well we saw if it's positive the entire time, you're taking the absolute value of a positive, absolute value of a positive, it's not going to matter, these two things are going to be equal."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So you're going to get positive values, take the absolute value of a positive value, it doesn't make a difference. What matters is if f of x is negative. If f of x is negative the entire time, so if this is our x-axis, that is our y-axis. If f of x is, well we saw if it's positive the entire time, you're taking the absolute value of a positive, absolute value of a positive, it's not going to matter, these two things are going to be equal. If f of x is negative the whole time, then you're going to get, then this integral is going to evaluate to a negative value, but then you're going to take the absolute value of it, and then over here you're just going to, the integral is going to evaluate to a positive value, but it's still going to be the same thing. The interesting case is when f of x is both positive and negative. So you can imagine a situation like this."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "If f of x is, well we saw if it's positive the entire time, you're taking the absolute value of a positive, absolute value of a positive, it's not going to matter, these two things are going to be equal. If f of x is negative the whole time, then you're going to get, then this integral is going to evaluate to a negative value, but then you're going to take the absolute value of it, and then over here you're just going to, the integral is going to evaluate to a positive value, but it's still going to be the same thing. The interesting case is when f of x is both positive and negative. So you can imagine a situation like this. If f of x looks something like that, then this right over here, the integral, you'd have positive, this would be positive, and then this would be negative right over here, and so they would cancel each other out. So this would be a smaller value than if you took the integral of the absolute value. So the integral, the absolute value of f would look something like this."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So you can imagine a situation like this. If f of x looks something like that, then this right over here, the integral, you'd have positive, this would be positive, and then this would be negative right over here, and so they would cancel each other out. So this would be a smaller value than if you took the integral of the absolute value. So the integral, the absolute value of f would look something like this. So all of the areas are going to be, if you view the integral, if you view this as maybe a definite integral, all of the areas would be positive. So you're going to get a bigger value when you take the integral of the absolute value, then you will, especially when f of x goes both positive and negative over the interval, then you would if you took the integral first and then the absolute value. Because once again, if you took the integral first for something like this, you would get a low value because this stuff would cancel out with this stuff right over here, and then you would take the absolute value of just a lower magnitude number."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So the integral, the absolute value of f would look something like this. So all of the areas are going to be, if you view the integral, if you view this as maybe a definite integral, all of the areas would be positive. So you're going to get a bigger value when you take the integral of the absolute value, then you will, especially when f of x goes both positive and negative over the interval, then you would if you took the integral first and then the absolute value. Because once again, if you took the integral first for something like this, you would get a low value because this stuff would cancel out with this stuff right over here, and then you would take the absolute value of just a lower magnitude number. And so in general, the absolute value of the integral is going to be less than or equal to the integral of the absolute value. So we can say, so this right here is the integral of the absolute value, which is going to be greater than or equal, what we have written over here is just this, that's going to be greater than or equal to, and I think you'll see why I'm doing this in a second, greater than or equal to the absolute value of the integral of the n plus 1th derivative of x dx. And the reason why this is useful is that we can still keep the inequality that this is less than or equal to this, but now this is a pretty straightforward integral to evaluate."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Because once again, if you took the integral first for something like this, you would get a low value because this stuff would cancel out with this stuff right over here, and then you would take the absolute value of just a lower magnitude number. And so in general, the absolute value of the integral is going to be less than or equal to the integral of the absolute value. So we can say, so this right here is the integral of the absolute value, which is going to be greater than or equal, what we have written over here is just this, that's going to be greater than or equal to, and I think you'll see why I'm doing this in a second, greater than or equal to the absolute value of the integral of the n plus 1th derivative of x dx. And the reason why this is useful is that we can still keep the inequality that this is less than or equal to this, but now this is a pretty straightforward integral to evaluate. The anti-derivative of the n plus 1th derivative is going to be the nth derivative. So this business right over here is just going to be the absolute value of the nth derivative. The absolute value of the nth derivative of our error function."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And the reason why this is useful is that we can still keep the inequality that this is less than or equal to this, but now this is a pretty straightforward integral to evaluate. The anti-derivative of the n plus 1th derivative is going to be the nth derivative. So this business right over here is just going to be the absolute value of the nth derivative. The absolute value of the nth derivative of our error function. Did I say expected value? I shouldn't. See, it even confuses me."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "The absolute value of the nth derivative of our error function. Did I say expected value? I shouldn't. See, it even confuses me. This is the error function. I should have used r, r for remainder, but this is all error. Nothing about probability or expected value in this video."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "See, it even confuses me. This is the error function. I should have used r, r for remainder, but this is all error. Nothing about probability or expected value in this video. This is e for error. So anyway, this is going to be the nth derivative of our error function, which is going to be less than or equal to this, which is less than or equal to the anti-derivative of m. Well, that's a constant, so that's going to be mx. And since we're just taking indefinite integrals, we can't forget the idea that we have a constant over here."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Nothing about probability or expected value in this video. This is e for error. So anyway, this is going to be the nth derivative of our error function, which is going to be less than or equal to this, which is less than or equal to the anti-derivative of m. Well, that's a constant, so that's going to be mx. And since we're just taking indefinite integrals, we can't forget the idea that we have a constant over here. And in general, when you're trying to create an upper bound, you want as low of an upper bound as possible. So we want to minimize what this constant is. And lucky for us, we do know what value this function takes on at a point."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And since we're just taking indefinite integrals, we can't forget the idea that we have a constant over here. And in general, when you're trying to create an upper bound, you want as low of an upper bound as possible. So we want to minimize what this constant is. And lucky for us, we do know what value this function takes on at a point. We know that the nth derivative of our error function at a is equal to 0. I think we wrote it over here. The nth derivative at a is equal to 0, and that's because the nth derivative of the function and the approximation at a are going to be the same exact thing."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And lucky for us, we do know what value this function takes on at a point. We know that the nth derivative of our error function at a is equal to 0. I think we wrote it over here. The nth derivative at a is equal to 0, and that's because the nth derivative of the function and the approximation at a are going to be the same exact thing. And so if we evaluate both sides of this at a, and I'll do it over here on the side, we know that the absolute value of the nth derivative at a, we know that this thing is going to be equal to the absolute value of 0, which is 0, which needs to be less than or equal to when you evaluate this thing at a, which is less than or equal to ma plus c. And so if you look at this part of the inequality, you subtract ma from both sides, you get negative ma is less than or equal to c. So our constant here, based on that little condition that we were able to get in the last video, our constant is going to be greater than or equal to negative ma. So if we want to minimize the constant, if we want to get this as low of a bound as possible, we would want to pick c is equal to negative ma. That is the lowest possible c that will meet these constraints that we know are true."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "The nth derivative at a is equal to 0, and that's because the nth derivative of the function and the approximation at a are going to be the same exact thing. And so if we evaluate both sides of this at a, and I'll do it over here on the side, we know that the absolute value of the nth derivative at a, we know that this thing is going to be equal to the absolute value of 0, which is 0, which needs to be less than or equal to when you evaluate this thing at a, which is less than or equal to ma plus c. And so if you look at this part of the inequality, you subtract ma from both sides, you get negative ma is less than or equal to c. So our constant here, based on that little condition that we were able to get in the last video, our constant is going to be greater than or equal to negative ma. So if we want to minimize the constant, if we want to get this as low of a bound as possible, we would want to pick c is equal to negative ma. That is the lowest possible c that will meet these constraints that we know are true. So we will actually pick c to be negative ma. And then we can rewrite this whole thing as the absolute value of the nth derivative of the error function, not the expected value. I have a strange suspicion I might have said expected value, but this is the error function."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "That is the lowest possible c that will meet these constraints that we know are true. So we will actually pick c to be negative ma. And then we can rewrite this whole thing as the absolute value of the nth derivative of the error function, not the expected value. I have a strange suspicion I might have said expected value, but this is the error function. The absolute value of the nth derivative of the error function is less than or equal to m times x minus a. And once again, all of the constraints hold. This is for x as part of the interval, the closed interval between a and b."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "I have a strange suspicion I might have said expected value, but this is the error function. The absolute value of the nth derivative of the error function is less than or equal to m times x minus a. And once again, all of the constraints hold. This is for x as part of the interval, the closed interval between a and b. But it looks like we're making progress. We at least went from the n plus 1th derivative to the nth derivative. Let's see if we can keep going."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "This is for x as part of the interval, the closed interval between a and b. But it looks like we're making progress. We at least went from the n plus 1th derivative to the nth derivative. Let's see if we can keep going. So same general idea. If we know this, then we know that we can take the integral of both sides of this. So we can take the integral of both sides of this, the antiderivative of both sides."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let's see if we can keep going. So same general idea. If we know this, then we know that we can take the integral of both sides of this. So we can take the integral of both sides of this, the antiderivative of both sides. And we know from what we figured out up here that something that's even smaller than this right over here is the absolute value of the integral of the expected value. See, I said it. Of the error function, not the expected value."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So we can take the integral of both sides of this, the antiderivative of both sides. And we know from what we figured out up here that something that's even smaller than this right over here is the absolute value of the integral of the expected value. See, I said it. Of the error function, not the expected value. Of the error function, the nth derivative of the error function of x dx. So we know that this is less than or equal to, based on the exact same logic there. And this is useful because this is just going to be the n minus 1th derivative of our error function of x."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Of the error function, not the expected value. Of the error function, the nth derivative of the error function of x dx. So we know that this is less than or equal to, based on the exact same logic there. And this is useful because this is just going to be the n minus 1th derivative of our error function of x. And of course, we have the absolute value outside of it. And now this is going to be less than or equal to. It's less than or equal to this, which is less than or equal to this right over here."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And this is useful because this is just going to be the n minus 1th derivative of our error function of x. And of course, we have the absolute value outside of it. And now this is going to be less than or equal to. It's less than or equal to this, which is less than or equal to this right over here. The antiderivative of this right over here is going to be m times x minus a squared over 2. You can do u substitution if you want. Or you can just say, hey, look, I have a little expression here."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "It's less than or equal to this, which is less than or equal to this right over here. The antiderivative of this right over here is going to be m times x minus a squared over 2. You can do u substitution if you want. Or you can just say, hey, look, I have a little expression here. Its derivative is 1, so it's implicitly there. So I can just treat it as kind of a u. So raise it to an exponent and then divide that exponent."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Or you can just say, hey, look, I have a little expression here. Its derivative is 1, so it's implicitly there. So I can just treat it as kind of a u. So raise it to an exponent and then divide that exponent. But once again, I'm taking indefinite integrals. So I'm going to say a plus c over here. But let's use that same exact logic."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So raise it to an exponent and then divide that exponent. But once again, I'm taking indefinite integrals. So I'm going to say a plus c over here. But let's use that same exact logic. If we evaluate this at a, you're going to have it. If we evaluate this whole, let's evaluate both sides of this at a. The left side evaluated at a we know is going to be 0."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "But let's use that same exact logic. If we evaluate this at a, you're going to have it. If we evaluate this whole, let's evaluate both sides of this at a. The left side evaluated at a we know is going to be 0. We figured that out up here in the last video. So you get 0 when you evaluate the left side at a. The right side evaluated at a, you get m times a minus a squared over 2."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "The left side evaluated at a we know is going to be 0. We figured that out up here in the last video. So you get 0 when you evaluate the left side at a. The right side evaluated at a, you get m times a minus a squared over 2. So you're going to get 0 plus c. So you're going to get 0 is less than or equal to c. Once again, we want to minimize our constant. We want to minimize our upper bound over here. So we want to pick the lowest possible c that meets our constraints."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "The right side evaluated at a, you get m times a minus a squared over 2. So you're going to get 0 plus c. So you're going to get 0 is less than or equal to c. Once again, we want to minimize our constant. We want to minimize our upper bound over here. So we want to pick the lowest possible c that meets our constraints. So the lowest possible c that meets our constraint is 0. So the general idea here is that we can keep doing this. We can keep doing exactly what we're doing all the way until we keep integrating it."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So we want to pick the lowest possible c that meets our constraints. So the lowest possible c that meets our constraint is 0. So the general idea here is that we can keep doing this. We can keep doing exactly what we're doing all the way until we keep integrating it. The exact same way that I've done it, all the way that we get, and using this exact same property here, all the way until we get the bound on the error function of x. So you could view this as the 0th derivative. You're going all the way to the 0th derivative, which is really just the error function."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "We can keep doing exactly what we're doing all the way until we keep integrating it. The exact same way that I've done it, all the way that we get, and using this exact same property here, all the way until we get the bound on the error function of x. So you could view this as the 0th derivative. You're going all the way to the 0th derivative, which is really just the error function. The bound on the error function of x is going to be less than or equal to. What's it going to be? You can already see the pattern here."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "You're going all the way to the 0th derivative, which is really just the error function. The bound on the error function of x is going to be less than or equal to. What's it going to be? You can already see the pattern here. It's going to be m times x minus a. The one way to think about it, this exponent plus this derivative is going to be equal to n plus 1. Now this derivative is 0, so this exponent is going to be n plus 1."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "You can already see the pattern here. It's going to be m times x minus a. The one way to think about it, this exponent plus this derivative is going to be equal to n plus 1. Now this derivative is 0, so this exponent is going to be n plus 1. Whatever that exponent is, you're going to have, and maybe I should have done it, you're going to have n plus 1 factorial over here. If you say, wait, where does this n plus 1 factorial come from? I just had a 2 here."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Now this derivative is 0, so this exponent is going to be n plus 1. Whatever that exponent is, you're going to have, and maybe I should have done it, you're going to have n plus 1 factorial over here. If you say, wait, where does this n plus 1 factorial come from? I just had a 2 here. Well, think about what happens when we integrate this again. You're going to raise this to the 3rd power and then divide by 3. So your denominator is going to have 2 times 3."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "I just had a 2 here. Well, think about what happens when we integrate this again. You're going to raise this to the 3rd power and then divide by 3. So your denominator is going to have 2 times 3. Then when you integrate it again, you're going to raise it to the 4th power and then divide by 4. So then your denominator is going to be 2 times 3 times 4, or 4 factorial. So whatever power you're raising to, the denominator is going to be that power factorial."}, {"video_title": "Taylor polynomial remainder (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So your denominator is going to have 2 times 3. Then when you integrate it again, you're going to raise it to the 4th power and then divide by 4. So then your denominator is going to be 2 times 3 times 4, or 4 factorial. So whatever power you're raising to, the denominator is going to be that power factorial. What's really interesting now is if we are able to figure out that maximum value of our function, if we're able to figure out that maximum value of our function right there, we now have a way of bounding our error function over that interval, over that interval between a and b. So for example, the error function at b, we can now bound it if we know what an m is. We can say the error function at b is going to be less than or equal to m times b minus a to the n plus 1th power over n plus 1 factorial."}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "Let's say we have the repeating decimal 0.4008, where the digits 4, 0, 0, 8 keep on repeating. So if we were to write it out, it would look something like this, 0.4008, 4, 0, 0, 8, 4, 0, 0, 8, and keeps on going forever. What I want you to do right now is pause the video and think about whether you can represent this repeating decimal as an infinite sum, as an infinite series, and then think about whether that infinite series is a geometric series. So I'm assuming you've given a go at it, so let's think about it. So for each term of my infinite series, I'm going to represent one of these repeating patterns of 4008. So for example, I will make this 4008 my first term. So this could be viewed as 0, and this 4008 represents 0.4008."}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "So I'm assuming you've given a go at it, so let's think about it. So for each term of my infinite series, I'm going to represent one of these repeating patterns of 4008. So for example, I will make this 4008 my first term. So this could be viewed as 0, and this 4008 represents 0.4008. And then I could make this 4008 my next term, or my next term will represent this 4008. And that means this 4008 is the same thing as 0.00004008. And then this next 4008, well, that represents 0."}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "So this could be viewed as 0, and this 4008 represents 0.4008. And then I could make this 4008 my next term, or my next term will represent this 4008. And that means this 4008 is the same thing as 0.00004008. And then this next 4008, well, that represents 0. And we have eight 0's, 1, 2, 3, 4, 5, 6, 7, 8, 4008. And then we would just keep on going like that forever. So we're just going to keep on going like that forever."}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "And then this next 4008, well, that represents 0. And we have eight 0's, 1, 2, 3, 4, 5, 6, 7, 8, 4008. And then we would just keep on going like that forever. So we're just going to keep on going like that forever. So hopefully there's a pattern here. We're essentially throwing four 0's before the decimal every time, and we could just keep on going like that forever. So this is an infinite sum."}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "So we're just going to keep on going like that forever. So hopefully there's a pattern here. We're essentially throwing four 0's before the decimal every time, and we could just keep on going like that forever. So this is an infinite sum. It's an infinite series. The next question is, is this a geometric series? Well, in order for it to be a geometric series, to go from one term to the next, you must be multiplying by the same value, by the same common ratio."}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "So this is an infinite sum. It's an infinite series. The next question is, is this a geometric series? Well, in order for it to be a geometric series, to go from one term to the next, you must be multiplying by the same value, by the same common ratio. So what are we multiplying when we go from 0.4008 to this one right over here, where we add four 0's before the 4008? What are we multiplying? Well, we move the decimal four spots to the left, so we're multiplying by 10 to the negative fourth."}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "Well, in order for it to be a geometric series, to go from one term to the next, you must be multiplying by the same value, by the same common ratio. So what are we multiplying when we go from 0.4008 to this one right over here, where we add four 0's before the 4008? What are we multiplying? Well, we move the decimal four spots to the left, so we're multiplying by 10 to the negative fourth. Or you could view it as we're multiplying by 0.001. 10 to the negative 1, 2, 3, 4. To go from here to here, well, same thing."}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "Well, we move the decimal four spots to the left, so we're multiplying by 10 to the negative fourth. Or you could view it as we're multiplying by 0.001. 10 to the negative 1, 2, 3, 4. To go from here to here, well, same thing. Move the decimal four places to the left. So once again, we're multiplying by 0.0001. And so it looks pretty clear that we have a common ratio of 10 to the negative fourth power."}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "To go from here to here, well, same thing. Move the decimal four places to the left. So once again, we're multiplying by 0.0001. And so it looks pretty clear that we have a common ratio of 10 to the negative fourth power. So we can rewrite all of this business as 0.4008 times our common ratio for this first term, times our common ratio of 10 to the negative fourth to the zeroth power. So that gives us that right over there. Plus 0.4008 times 10 to the negative fourth to the first power."}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "And so it looks pretty clear that we have a common ratio of 10 to the negative fourth power. So we can rewrite all of this business as 0.4008 times our common ratio for this first term, times our common ratio of 10 to the negative fourth to the zeroth power. So that gives us that right over there. Plus 0.4008 times 10 to the negative fourth to the first power. And that gives us that value right over there. Plus 0.4008 times 10 to the negative fourth to the second power. And we keep on going."}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "Plus 0.4008 times 10 to the negative fourth to the first power. And that gives us that value right over there. Plus 0.4008 times 10 to the negative fourth to the second power. And we keep on going. And so in this form, it looks a little bit clearer like a geometric series, an infinite geometric series. And if we wanted to write that out with sigma notation, we could write this as the sum from k equals 0 to infinity. To infinity of, well, what's our first term going to be?"}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "And we keep on going. And so in this form, it looks a little bit clearer like a geometric series, an infinite geometric series. And if we wanted to write that out with sigma notation, we could write this as the sum from k equals 0 to infinity. To infinity of, well, what's our first term going to be? It's going to be 0.4008 times our common ratio, which we could write out as either 10 to the negative fourth or 0.0001. I'll just write it as 10 to the negative fourth to the k-th power. So the next interesting question, this clearly can be represented as a geometric series, is, well, what is this sum?"}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "To infinity of, well, what's our first term going to be? It's going to be 0.4008 times our common ratio, which we could write out as either 10 to the negative fourth or 0.0001. I'll just write it as 10 to the negative fourth to the k-th power. So the next interesting question, this clearly can be represented as a geometric series, is, well, what is this sum? You might say, oh, that's just going to be 4008 repeating over and over. But I want to express it as a fraction. And so I want you to pause this video."}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "So the next interesting question, this clearly can be represented as a geometric series, is, well, what is this sum? You might say, oh, that's just going to be 4008 repeating over and over. But I want to express it as a fraction. And so I want you to pause this video. Use what you already know about finding the sum of an infinite geometric series to try to express this thing right over here as a fraction. So I'm assuming you've had a go at it. So let's think about it."}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "And so I want you to pause this video. Use what you already know about finding the sum of an infinite geometric series to try to express this thing right over here as a fraction. So I'm assuming you've had a go at it. So let's think about it. We've already seen, we've already derived in previous videos, that the sum of an infinite geometric series, now let me do this in a neutral color. If I have a series like this, k equals 0 to infinity of ar to the k power, that this sum is going to be equal to a over 1 minus r. We've derived this, actually, in several other videos. So in this case, this is going to be, well, our a here is 0.4008."}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "So let's think about it. We've already seen, we've already derived in previous videos, that the sum of an infinite geometric series, now let me do this in a neutral color. If I have a series like this, k equals 0 to infinity of ar to the k power, that this sum is going to be equal to a over 1 minus r. We've derived this, actually, in several other videos. So in this case, this is going to be, well, our a here is 0.4008. And it's going to be that over 1 minus our common ratio. Minus, and I'll write it like this, 0.0001, 1 ten thousandth. So what's this going to be?"}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "So in this case, this is going to be, well, our a here is 0.4008. And it's going to be that over 1 minus our common ratio. Minus, and I'll write it like this, 0.0001, 1 ten thousandth. So what's this going to be? Well, this is going to be the same thing as 0.4008. If you take 1 minus 1 ten thousandth, or you could use this 10,000 ten thousandths, minus 1 ten thousandth, you're going to have 9,999 over, or 9,999 ten thousandths. Once again, let me write this out, just so this doesn't look confusing."}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "So what's this going to be? Well, this is going to be the same thing as 0.4008. If you take 1 minus 1 ten thousandth, or you could use this 10,000 ten thousandths, minus 1 ten thousandth, you're going to have 9,999 over, or 9,999 ten thousandths. Once again, let me write this out, just so this doesn't look confusing. 1 is the same thing as 10,000 over 10,000. And you're subtracting 1 over 10,000. And so you're going to get 9,999 over 10,000."}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "Once again, let me write this out, just so this doesn't look confusing. 1 is the same thing as 10,000 over 10,000. And you're subtracting 1 over 10,000. And so you're going to get 9,999 over 10,000. And so this is going to be the same thing as 0.4008 times 10,000 over 9,999. Well, what's this top number times 10,000? Well, that's just going to give us 4,008 over 9,999."}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "And so you're going to get 9,999 over 10,000. And so this is going to be the same thing as 0.4008 times 10,000 over 9,999. Well, what's this top number times 10,000? Well, that's just going to give us 4,008 over 9,999. And we've just expressed that repeating decimal as a fraction. So we have succeeded. And you might say, well, maybe we can simplify this thing."}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "Well, that's just going to give us 4,008 over 9,999. And we've just expressed that repeating decimal as a fraction. So we have succeeded. And you might say, well, maybe we can simplify this thing. And so let's see. This is already a fraction, so we've already kind of achieved it. But if we want to get a little bit simpler, if we add the digits up here, 4 plus 8 is 12."}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "And you might say, well, maybe we can simplify this thing. And so let's see. This is already a fraction, so we've already kind of achieved it. But if we want to get a little bit simpler, if we add the digits up here, 4 plus 8 is 12. So 1 plus 2 is 3. So this up here is divisible by 3. And this down here is clearly divisible by 3."}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "But if we want to get a little bit simpler, if we add the digits up here, 4 plus 8 is 12. So 1 plus 2 is 3. So this up here is divisible by 3. And this down here is clearly divisible by 3. So let's divide both of them by 3. So 3 goes into 4,008. Let's see."}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "And this down here is clearly divisible by 3. So let's divide both of them by 3. So 3 goes into 4,008. Let's see. It goes into 4 one time. Subtract. You get a 10."}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "Let's see. It goes into 4 one time. Subtract. You get a 10. 3 times 3 is 9. Subtract. You get another 10."}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "You get a 10. 3 times 3 is 9. Subtract. You get another 10. Goes into 3 times. 3 times 3 is 9. Subtract."}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "You get another 10. Goes into 3 times. 3 times 3 is 9. Subtract. Bring down an 8. 3 goes into 18 exactly 6 times. So our numerator is 1,336."}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "Subtract. Bring down an 8. 3 goes into 18 exactly 6 times. So our numerator is 1,336. This is no longer divisible by 3. The sum of the digits is not divisible by 3. It's not a multiple of 3."}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "So our numerator is 1,336. This is no longer divisible by 3. The sum of the digits is not divisible by 3. It's not a multiple of 3. And if you divide this bottom number by 3, you get 3,333. And I think we have simplified it. I think we have simplified it about as well as we can."}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "It's not a multiple of 3. And if you divide this bottom number by 3, you get 3,333. And I think we have simplified it. I think we have simplified it about as well as we can. Although we could check more. Let me know if I didn't. But either way, we have now written this."}, {"video_title": "Repeating decimal as infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "I think we have simplified it about as well as we can. Although we could check more. Let me know if I didn't. But either way, we have now written this. This was pretty neat. We saw that a repeating decimal can be represented not just as an infinite series, but as an infinite geometric series. And then we were able to use the formula that we derived for the sum of an infinite geometric series to actually express it as a fraction."}, {"video_title": "Worked example cosine function from power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "And I encourage you to pause this video and give it a go on your own. And I will give you a hint. The key to this is to figure out, well, what function is this the power series for, and then use that function to evaluate this. And there's another clue here is that, hey, this is kind of a mysterious or suspicious-looking number here, pi over two, that looks like something I would use a trig function to evaluate. That might be a little bit more straightforward. So I'll let you have a go at it. So I'm assuming you have tried, so let's try to work through this together."}, {"video_title": "Worked example cosine function from power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "And there's another clue here is that, hey, this is kind of a mysterious or suspicious-looking number here, pi over two, that looks like something I would use a trig function to evaluate. That might be a little bit more straightforward. So I'll let you have a go at it. So I'm assuming you have tried, so let's try to work through this together. And any of these types of problems, I like to at least expand out this power series so I get a better sense of what it's like. So this right over here, if I were to expand it out, this is going to be equal to, when n is zero, this is one. Actually, all of these are one, so it's just gonna be one."}, {"video_title": "Worked example cosine function from power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "So I'm assuming you have tried, so let's try to work through this together. And any of these types of problems, I like to at least expand out this power series so I get a better sense of what it's like. So this right over here, if I were to expand it out, this is going to be equal to, when n is zero, this is one. Actually, all of these are one, so it's just gonna be one. When n is one, it's gonna be negative one x to the sixth, x to the sixth over two factorial. When n is two, it's going to be positive. Negative one squared is positive one times x to the 12th over four factorial."}, {"video_title": "Worked example cosine function from power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "Actually, all of these are one, so it's just gonna be one. When n is one, it's gonna be negative one x to the sixth, x to the sixth over two factorial. When n is two, it's going to be positive. Negative one squared is positive one times x to the 12th over four factorial. And then let's just do one more. When x is equal to three, it's going to be negative x to the 18th, x to the 18th over six, over six factorial. And you just keep going on and on forever."}, {"video_title": "Worked example cosine function from power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "Negative one squared is positive one times x to the 12th over four factorial. And then let's just do one more. When x is equal to three, it's going to be negative x to the 18th, x to the 18th over six, over six factorial. And you just keep going on and on forever. Now, offhand, I don't know a function, especially a trigonometric function, because that was kind of our clue here. This pi over two makes me feel like I might, this might be a trigonometric function right over here. Nothing jumps out at me offhand, but this does look suspiciously familiar."}, {"video_title": "Worked example cosine function from power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "And you just keep going on and on forever. Now, offhand, I don't know a function, especially a trigonometric function, because that was kind of our clue here. This pi over two makes me feel like I might, this might be a trigonometric function right over here. Nothing jumps out at me offhand, but this does look suspiciously familiar. This looks awfully close to the power series, or the Maclaurin series for cosine of x, which we have seen multiple times. Let's just remind ourselves what that is. And if this doesn't look familiar, the previous video where I do the Maclaurin series for cosine of x goes into detail on how I get this."}, {"video_title": "Worked example cosine function from power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "Nothing jumps out at me offhand, but this does look suspiciously familiar. This looks awfully close to the power series, or the Maclaurin series for cosine of x, which we have seen multiple times. Let's just remind ourselves what that is. And if this doesn't look familiar, the previous video where I do the Maclaurin series for cosine of x goes into detail on how I get this. The Maclaurin series for cosine of x is equal to, so I'll just write a few terms. I'll write approximately equal to one minus x squared over two factorial plus x to the fourth, plus x to the fourth over four factorial minus x to the sixth over six factorial. And just like that, you're probably seeing the similarities."}, {"video_title": "Worked example cosine function from power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "And if this doesn't look familiar, the previous video where I do the Maclaurin series for cosine of x goes into detail on how I get this. The Maclaurin series for cosine of x is equal to, so I'll just write a few terms. I'll write approximately equal to one minus x squared over two factorial plus x to the fourth, plus x to the fourth over four factorial minus x to the sixth over six factorial. And just like that, you're probably seeing the similarities. Well, the first term's the same, the sine negative, positive, negative, positive, negative, positive, negative, positive, two factorial, four factorial, six factorial. The difference is the powers, the exponents on the x's. This is x squared, this is x to the sixth, this is x to the fourth, that's x to the twelfth."}, {"video_title": "Worked example cosine function from power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "And just like that, you're probably seeing the similarities. Well, the first term's the same, the sine negative, positive, negative, positive, negative, positive, negative, positive, two factorial, four factorial, six factorial. The difference is the powers, the exponents on the x's. This is x squared, this is x to the sixth, this is x to the fourth, that's x to the twelfth. This is x to the sixth, that's x to the eighteenth. Well, what if we, I guess something for you to think about is, well, how can we replace x with something here? Because anything that I change, if I take cosine of, if I change x to, I don't know, a plus b, everywhere we see an x, you would replace it with an a plus b."}, {"video_title": "Worked example cosine function from power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "This is x squared, this is x to the sixth, this is x to the fourth, that's x to the twelfth. This is x to the sixth, that's x to the eighteenth. Well, what if we, I guess something for you to think about is, well, how can we replace x with something here? Because anything that I change, if I take cosine of, if I change x to, I don't know, a plus b, everywhere we see an x, you would replace it with an a plus b. Can we put a power of x here so that these things end up like that? Well, this x to the sixth is the same thing, x to the sixth is the same thing as x to the third squared, that's x to the third squared. This right over here, x to the twelfth is the same thing as x to the third to the fourth power."}, {"video_title": "Worked example cosine function from power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "Because anything that I change, if I take cosine of, if I change x to, I don't know, a plus b, everywhere we see an x, you would replace it with an a plus b. Can we put a power of x here so that these things end up like that? Well, this x to the sixth is the same thing, x to the sixth is the same thing as x to the third squared, that's x to the third squared. This right over here, x to the twelfth is the same thing as x to the third to the fourth power. This right over here is the same thing as x to the third to the sixth power. So if we could replace each of these x's with x to the thirds, we would get this power series up here. Well, how do we do that?"}, {"video_title": "Worked example cosine function from power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "This right over here, x to the twelfth is the same thing as x to the third to the fourth power. This right over here is the same thing as x to the third to the sixth power. So if we could replace each of these x's with x to the thirds, we would get this power series up here. Well, how do we do that? Well, we would just say, well, it's the cosine of x to third. And actually, let me do that in a different color. So the cosine, and that's not a different color, so the cosine of x to the third is going to be equal to, and once again, everywhere we see an x, we replace it with x to the third."}, {"video_title": "Worked example cosine function from power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, how do we do that? Well, we would just say, well, it's the cosine of x to third. And actually, let me do that in a different color. So the cosine, and that's not a different color, so the cosine of x to the third is going to be equal to, and once again, everywhere we see an x, we replace it with x to the third. So it's one minus, and actually, I'm just going to put in parentheses squared, two factorial, which I wanted to do that in the green. Let me do all of this in the green. All right, so it's going to be equal to one minus parentheses squared over two factorial plus parentheses to the fourth power over four factorial minus parentheses to the sixth power over six factorial."}, {"video_title": "Worked example cosine function from power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "So the cosine, and that's not a different color, so the cosine of x to the third is going to be equal to, and once again, everywhere we see an x, we replace it with x to the third. So it's one minus, and actually, I'm just going to put in parentheses squared, two factorial, which I wanted to do that in the green. Let me do all of this in the green. All right, so it's going to be equal to one minus parentheses squared over two factorial plus parentheses to the fourth power over four factorial minus parentheses to the sixth power over six factorial. And now let me get changed back to that mauve color. And since I'm taking the cosine of x to the third, well, this is going to be x to the third squared. This is going to be x to the third to the fourth power."}, {"video_title": "Worked example cosine function from power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "All right, so it's going to be equal to one minus parentheses squared over two factorial plus parentheses to the fourth power over four factorial minus parentheses to the sixth power over six factorial. And now let me get changed back to that mauve color. And since I'm taking the cosine of x to the third, well, this is going to be x to the third squared. This is going to be x to the third to the fourth power. This is going to be x to the third to the sixth power, which is exactly what I have right over here. So this right over here is the power series for cosine of x to the third. So evaluating this when x is equal to the cube root of pi over two is the same thing as evaluating this when x is equal to the cube root of pi over two."}, {"video_title": "Worked example cosine function from power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "This is going to be x to the third to the fourth power. This is going to be x to the third to the sixth power, which is exactly what I have right over here. So this right over here is the power series for cosine of x to the third. So evaluating this when x is equal to the cube root of pi over two is the same thing as evaluating this when x is equal to the cube root of pi over two. Let me write that down, because this is interesting. So this, so I'll just rewrite it, from n equals zero to infinity of negative one to the n, x to the sixth n over two n factorial. This is equal to, this is the power series representation of cosine of x to the third power."}, {"video_title": "Worked example cosine function from power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "So evaluating this when x is equal to the cube root of pi over two is the same thing as evaluating this when x is equal to the cube root of pi over two. Let me write that down, because this is interesting. So this, so I'll just rewrite it, from n equals zero to infinity of negative one to the n, x to the sixth n over two n factorial. This is equal to, this is the power series representation of cosine of x to the third power. So if you want to evaluate this when x is a cube root of pi over two, we just have to evaluate this when x is a cube root of pi over two. And this does suspiciously work out nicely, because if you take the cube of the cube root, well, good things happen. So the cosine of the cube root of pi over two to the third power, well, that's just the same thing as cosine of pi over two, which, of course, is equal to zero."}, {"video_title": "Chain rule proof Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And if you've been following some of the videos on differentiability implies continuity and what happens to a continuous function as our change in x, if x is our independent variable, as that approaches zero, how the change in our function approaches zero, then this proof is actually surprisingly straightforward. So let's just get to it. And this is just one of many proofs of the chain rule. So the chain rule tells us that if y, y is a function of u, which is a function of x, and we want to figure out the derivative of this, so we want to differentiate this with respect to x. So we're gonna differentiate this with respect to x. We could write this as the derivative of y with respect to x, which is going to be equal to the derivative of y with respect to u times the derivative of u with respect to x. This is what the chain rule tells us."}, {"video_title": "Chain rule proof Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So the chain rule tells us that if y, y is a function of u, which is a function of x, and we want to figure out the derivative of this, so we want to differentiate this with respect to x. So we're gonna differentiate this with respect to x. We could write this as the derivative of y with respect to x, which is going to be equal to the derivative of y with respect to u times the derivative of u with respect to x. This is what the chain rule tells us. But how do we actually go about proving it? Well, we just have to remind ourselves that the derivative of y with respect to x, the derivative of y with respect to x, is equal to the limit as delta x approaches zero of change in y over change in x. Now we can do a little bit of algebraic manipulation here to introduce a change in u."}, {"video_title": "Chain rule proof Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "This is what the chain rule tells us. But how do we actually go about proving it? Well, we just have to remind ourselves that the derivative of y with respect to x, the derivative of y with respect to x, is equal to the limit as delta x approaches zero of change in y over change in x. Now we can do a little bit of algebraic manipulation here to introduce a change in u. So let's do that. So this is going to be the same thing as the limit as delta x approaches zero. And I'm gonna rewrite this part right over here."}, {"video_title": "Chain rule proof Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Now we can do a little bit of algebraic manipulation here to introduce a change in u. So let's do that. So this is going to be the same thing as the limit as delta x approaches zero. And I'm gonna rewrite this part right over here. I'm gonna essentially divide and multiply by a change in u. So I can rewrite this as delta y over delta u times delta u, whoops, times delta u over delta x. Change in y over change in u times change in u over change in x."}, {"video_title": "Chain rule proof Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And I'm gonna rewrite this part right over here. I'm gonna essentially divide and multiply by a change in u. So I can rewrite this as delta y over delta u times delta u, whoops, times delta u over delta x. Change in y over change in u times change in u over change in x. And you can see, these are just going to be numbers here. So our change in u, this would cancel with that, and you'd be left with change in y over change in x, which is exactly what we had here. So nothing earth-shattering just yet."}, {"video_title": "Chain rule proof Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Change in y over change in u times change in u over change in x. And you can see, these are just going to be numbers here. So our change in u, this would cancel with that, and you'd be left with change in y over change in x, which is exactly what we had here. So nothing earth-shattering just yet. But what's this going to be equal to? What's this going to be equal to? Well, the limit of the product is the same thing as the product of the limits."}, {"video_title": "Chain rule proof Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So nothing earth-shattering just yet. But what's this going to be equal to? What's this going to be equal to? Well, the limit of the product is the same thing as the product of the limits. So this is going to be the same thing as the limit as delta x approaches zero of, and I'll color-code it, of this stuff, of delta y over delta u times, maybe I'll put parentheses around it, times, times the limit, the limit as delta x approaches zero, delta x approaches zero of this business. So let me put some parentheses around it. Delta u over delta x."}, {"video_title": "Chain rule proof Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the limit of the product is the same thing as the product of the limits. So this is going to be the same thing as the limit as delta x approaches zero of, and I'll color-code it, of this stuff, of delta y over delta u times, maybe I'll put parentheses around it, times, times the limit, the limit as delta x approaches zero, delta x approaches zero of this business. So let me put some parentheses around it. Delta u over delta x. So what does this simplify to? Well, this right over here, this is the definition, and we're assuming, in order for this to even be true, we have to assume that u and y are differentiable at x. So we assume, in order for this to be true, we're assuming, we're assuming y, comma, u are differentiable, are differentiable, are differentiable at x."}, {"video_title": "Chain rule proof Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Delta u over delta x. So what does this simplify to? Well, this right over here, this is the definition, and we're assuming, in order for this to even be true, we have to assume that u and y are differentiable at x. So we assume, in order for this to be true, we're assuming, we're assuming y, comma, u are differentiable, are differentiable, are differentiable at x. And remember, also, if they're differentiable at x, that means they're continuous at x. But if u is differentiable at x, then this limit exists, and this is the derivative of, this is u prime of x, or du dx. So this right over here, we can rewrite as du dx."}, {"video_title": "Chain rule proof Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So we assume, in order for this to be true, we're assuming, we're assuming y, comma, u are differentiable, are differentiable, are differentiable at x. And remember, also, if they're differentiable at x, that means they're continuous at x. But if u is differentiable at x, then this limit exists, and this is the derivative of, this is u prime of x, or du dx. So this right over here, we can rewrite as du dx. I think you see where this is going. Now, this right over here, just looking at it the way it's written right here, we can't quite yet call this dy du, because this is the limit as delta x approaches zero, not the limit as delta u approaches zero. But we just have to remind ourselves the results from probably the previous video, depending on how you're watching it, which is, if we have a function u that is continuous at a point that as delta x approaches zero, delta u approaches zero."}, {"video_title": "Chain rule proof Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So this right over here, we can rewrite as du dx. I think you see where this is going. Now, this right over here, just looking at it the way it's written right here, we can't quite yet call this dy du, because this is the limit as delta x approaches zero, not the limit as delta u approaches zero. But we just have to remind ourselves the results from probably the previous video, depending on how you're watching it, which is, if we have a function u that is continuous at a point that as delta x approaches zero, delta u approaches zero. So we can actually rewrite this. We can rewrite this right over here. Instead of saying delta x approaches zero, that's just going to have the effect, because u is differentiable at x, which means it's continuous at x, that means that delta u is going to approach zero."}, {"video_title": "Chain rule proof Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "But we just have to remind ourselves the results from probably the previous video, depending on how you're watching it, which is, if we have a function u that is continuous at a point that as delta x approaches zero, delta u approaches zero. So we can actually rewrite this. We can rewrite this right over here. Instead of saying delta x approaches zero, that's just going to have the effect, because u is differentiable at x, which means it's continuous at x, that means that delta u is going to approach zero. As our change in x gets smaller and smaller and smaller, our change in u is going to get smaller and smaller and smaller. So we can rewrite this as our change in u approaches zero. And when we rewrite it like that, well, then this is just dy du."}, {"video_title": "Limits at infinity of quotients (Part 2) Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Let's do a few more examples of finding the limit of functions as x approaches infinity or negative infinity. So here I have this crazy function, 9x to the 7th minus 17x to the 6th plus 15 square roots of x, all of that over 3x to the 7th plus 1,000x to the 5th minus log base 2 of x. So what's going to happen as x approaches infinity? And the key here, like we've seen in other examples, is just to realize which terms will dominate. So for example, in the numerator, out of these three terms, the 9x to the 7th is going to grow much faster than any of these other terms. So this is the dominating term in the numerator. And the denominator, 3x to the 7th, is going to grow much faster than an x to the 5th term, definitely much faster than a log base 2 term."}, {"video_title": "Limits at infinity of quotients (Part 2) Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And the key here, like we've seen in other examples, is just to realize which terms will dominate. So for example, in the numerator, out of these three terms, the 9x to the 7th is going to grow much faster than any of these other terms. So this is the dominating term in the numerator. And the denominator, 3x to the 7th, is going to grow much faster than an x to the 5th term, definitely much faster than a log base 2 term. So at infinity, as we get closer and closer to infinity, this function is going to be roughly equal to 9x to the 7th over 3x to the 7th. And so we can say, especially as we get larger and larger, as we get closer and closer to infinity, these two things are going to get closer and closer to each other. We can say this limit is going to be the same thing as this limit, which is going to be equal to the limit as x approaches infinity."}, {"video_title": "Limits at infinity of quotients (Part 2) Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And the denominator, 3x to the 7th, is going to grow much faster than an x to the 5th term, definitely much faster than a log base 2 term. So at infinity, as we get closer and closer to infinity, this function is going to be roughly equal to 9x to the 7th over 3x to the 7th. And so we can say, especially as we get larger and larger, as we get closer and closer to infinity, these two things are going to get closer and closer to each other. We can say this limit is going to be the same thing as this limit, which is going to be equal to the limit as x approaches infinity. Well, we can just cancel out the x to the 7th, so it's going to be 9 thirds or just 3, which is just going to be 3. So that is our limit as x approaches infinity of all of this craziness. Now let's do the same with this function over here."}, {"video_title": "Limits at infinity of quotients (Part 2) Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "We can say this limit is going to be the same thing as this limit, which is going to be equal to the limit as x approaches infinity. Well, we can just cancel out the x to the 7th, so it's going to be 9 thirds or just 3, which is just going to be 3. So that is our limit as x approaches infinity of all of this craziness. Now let's do the same with this function over here. Once again, crazy function. We're going to negative infinity, but the same principles apply. Which terms dominate as the absolute value of x get larger and larger and larger, as x gets larger in magnitude?"}, {"video_title": "Limits at infinity of quotients (Part 2) Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's do the same with this function over here. Once again, crazy function. We're going to negative infinity, but the same principles apply. Which terms dominate as the absolute value of x get larger and larger and larger, as x gets larger in magnitude? Well, in the numerator, it's the 3x to the 3rd term. In the denominator, it's the 6x to the 4th term. So this is going to be the same thing as the limit of 3x to the 3rd over 6x to the 4th as x approaches negative infinity."}, {"video_title": "Limits at infinity of quotients (Part 2) Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Which terms dominate as the absolute value of x get larger and larger and larger, as x gets larger in magnitude? Well, in the numerator, it's the 3x to the 3rd term. In the denominator, it's the 6x to the 4th term. So this is going to be the same thing as the limit of 3x to the 3rd over 6x to the 4th as x approaches negative infinity. And if we simplify this, this is going to be equal to the limit as x approaches negative infinity of 1 over 2x. And what's this going to be? Well, if the denominator, even though it's becoming a larger and larger and larger negative number, it becomes 1 over a very, very large negative number, which is going to get us pretty darn close to 0, just as 1 over x as x approaches negative infinity gets us close to 0."}, {"video_title": "Limits at infinity of quotients (Part 2) Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be the same thing as the limit of 3x to the 3rd over 6x to the 4th as x approaches negative infinity. And if we simplify this, this is going to be equal to the limit as x approaches negative infinity of 1 over 2x. And what's this going to be? Well, if the denominator, even though it's becoming a larger and larger and larger negative number, it becomes 1 over a very, very large negative number, which is going to get us pretty darn close to 0, just as 1 over x as x approaches negative infinity gets us close to 0. So this right over here, the horizontal asymptote in this case is y is equal to 0. And I encourage you to graph it or try it out with numbers to verify that for yourself. The key realization here is to simplify the problem by just thinking about which terms are going to dominate the rest."}, {"video_title": "Limits at infinity of quotients (Part 2) Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Well, if the denominator, even though it's becoming a larger and larger and larger negative number, it becomes 1 over a very, very large negative number, which is going to get us pretty darn close to 0, just as 1 over x as x approaches negative infinity gets us close to 0. So this right over here, the horizontal asymptote in this case is y is equal to 0. And I encourage you to graph it or try it out with numbers to verify that for yourself. The key realization here is to simplify the problem by just thinking about which terms are going to dominate the rest. Now let's think about this one. What is the limit of this crazy function as x approaches infinity? Well, once again, what are the dominating terms?"}, {"video_title": "Limits at infinity of quotients (Part 2) Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "The key realization here is to simplify the problem by just thinking about which terms are going to dominate the rest. Now let's think about this one. What is the limit of this crazy function as x approaches infinity? Well, once again, what are the dominating terms? In the numerator, it's 4x to the 4th. In the denominator, it's 250x to the 3rd. These are the highest degree terms."}, {"video_title": "Limits at infinity of quotients (Part 2) Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Well, once again, what are the dominating terms? In the numerator, it's 4x to the 4th. In the denominator, it's 250x to the 3rd. These are the highest degree terms. So this is going to be the same thing as the limit as x approaches infinity of 4x to the 4th over 250x to the 3rd, which is going to be the same thing as the limit of 4. Well, this is going to be the same thing as we could divide 200 and, well, I'll just leave it like this. It's going to be the limit of 4 over 250x to the 4th divided by x to the 3rd is just x times x as x approaches infinity."}, {"video_title": "Limits at infinity of quotients (Part 2) Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "These are the highest degree terms. So this is going to be the same thing as the limit as x approaches infinity of 4x to the 4th over 250x to the 3rd, which is going to be the same thing as the limit of 4. Well, this is going to be the same thing as we could divide 200 and, well, I'll just leave it like this. It's going to be the limit of 4 over 250x to the 4th divided by x to the 3rd is just x times x as x approaches infinity. Or we could even say this is going to be 4 250ths times the limit as x approaches infinity of x. Now what's this? What's the limit of x as x approaches infinity?"}, {"video_title": "Limits at infinity of quotients (Part 2) Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "It's going to be the limit of 4 over 250x to the 4th divided by x to the 3rd is just x times x as x approaches infinity. Or we could even say this is going to be 4 250ths times the limit as x approaches infinity of x. Now what's this? What's the limit of x as x approaches infinity? Well, it's just going to keep growing forever. So this right over here is just going to be infinity. Infinity times some number right over here is going to be infinity."}, {"video_title": "Limits at infinity of quotients (Part 2) Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "What's the limit of x as x approaches infinity? Well, it's just going to keep growing forever. So this right over here is just going to be infinity. Infinity times some number right over here is going to be infinity. So the limit as x approaches infinity of all of this, it's actually unbounded. It's infinity. And a kind of obvious way of seeing that right from the get-go is to realize that the numerator has a 4th degree term, while the highest degree term in the denominator is only a 3rd degree term."}, {"video_title": "Limits at infinity of quotients (Part 2) Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Infinity times some number right over here is going to be infinity. So the limit as x approaches infinity of all of this, it's actually unbounded. It's infinity. And a kind of obvious way of seeing that right from the get-go is to realize that the numerator has a 4th degree term, while the highest degree term in the denominator is only a 3rd degree term. So the numerator is going to grow far faster than the denominator. So if the numerator is going far faster than the denominator, you're going to approach infinity in this case. If the numerator is going far slower than the denominator, if the denominator is going far faster than the numerator, like this case, you are then approaching 0."}, {"video_title": "Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "I drop a rock in the middle of that pool of water, and a little while later, a ripple has, a little wave, a ripple has formed that is moving radially outward from where I dropped the rock. So let me see how well I can draw that. So it's moving radially outward. So that is the ripple that is formed from me dropping the rock into the water. So it's a circle centered at where the rock initially hit the water. And let's say right at this moment, the radius of this circle is equal to 3 centimeters. And we also know that the radius is increasing at a rate of 1 centimeter per second."}, {"video_title": "Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So that is the ripple that is formed from me dropping the rock into the water. So it's a circle centered at where the rock initially hit the water. And let's say right at this moment, the radius of this circle is equal to 3 centimeters. And we also know that the radius is increasing at a rate of 1 centimeter per second. So radius growing at rate of 1 centimeter per second. So given this, right now our circle, our ripple circle, has a radius of 3 centimeters. And we know that the radius is growing at 1 centimeter per second."}, {"video_title": "Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And we also know that the radius is increasing at a rate of 1 centimeter per second. So radius growing at rate of 1 centimeter per second. So given this, right now our circle, our ripple circle, has a radius of 3 centimeters. And we know that the radius is growing at 1 centimeter per second. Given that, at what rate is the area of circle growing? Area of circle growing. Interesting."}, {"video_title": "Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And we know that the radius is growing at 1 centimeter per second. Given that, at what rate is the area of circle growing? Area of circle growing. Interesting. So let's think about what we know and then what we don't know, what we're trying to figure out. So if we call this radius r, we know that right now r is equal to 3 centimeters. We also know the rate at which r is changing with respect to time."}, {"video_title": "Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Interesting. So let's think about what we know and then what we don't know, what we're trying to figure out. So if we call this radius r, we know that right now r is equal to 3 centimeters. We also know the rate at which r is changing with respect to time. We also know this information right over here. dr dt, the rate at which the radius is changing with respect to time, is 1 centimeter per second. Now what do we need to figure out?"}, {"video_title": "Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We also know the rate at which r is changing with respect to time. We also know this information right over here. dr dt, the rate at which the radius is changing with respect to time, is 1 centimeter per second. Now what do we need to figure out? Well, they said, what rate is the area of the circle growing? So we need to figure out at what rate is the area of the circle, where a is the area of the circle, at what rate is this growing? This is what we need to figure out."}, {"video_title": "Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Now what do we need to figure out? Well, they said, what rate is the area of the circle growing? So we need to figure out at what rate is the area of the circle, where a is the area of the circle, at what rate is this growing? This is what we need to figure out. So what might be useful here is if we can come up with a relationship between the area of the circle and the radius of the circle, and maybe take the derivative with respect to time. And we'll have to use a little bit of the chain rule to do that. So what is the relationship at any given point in time between the area of the circle and the radius of the circle?"}, {"video_title": "Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "This is what we need to figure out. So what might be useful here is if we can come up with a relationship between the area of the circle and the radius of the circle, and maybe take the derivative with respect to time. And we'll have to use a little bit of the chain rule to do that. So what is the relationship at any given point in time between the area of the circle and the radius of the circle? Well, this is elementary geometry. The area of a circle is going to be equal to pi times the radius of the circle squared. Now, what we want to do is figure out the rate at which the area is changing with respect to time."}, {"video_title": "Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So what is the relationship at any given point in time between the area of the circle and the radius of the circle? Well, this is elementary geometry. The area of a circle is going to be equal to pi times the radius of the circle squared. Now, what we want to do is figure out the rate at which the area is changing with respect to time. So why don't we take the derivative of both sides of this with respect to time? And let me give myself a little more real estate. Actually, let me just rewrite what I just had."}, {"video_title": "Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Now, what we want to do is figure out the rate at which the area is changing with respect to time. So why don't we take the derivative of both sides of this with respect to time? And let me give myself a little more real estate. Actually, let me just rewrite what I just had. So pi r squared. Area is equal to pi r squared. Now I'm going to take the derivative of both sides with respect to time."}, {"video_title": "Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Actually, let me just rewrite what I just had. So pi r squared. Area is equal to pi r squared. Now I'm going to take the derivative of both sides with respect to time. So the derivative with respect to time. I'm not taking the derivative with respect to r. I'm taking the derivative with respect to time. So on the left-hand side, right over here, I'm going to have the derivative of our area."}, {"video_title": "Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Now I'm going to take the derivative of both sides with respect to time. So the derivative with respect to time. I'm not taking the derivative with respect to r. I'm taking the derivative with respect to time. So on the left-hand side, right over here, I'm going to have the derivative of our area. Actually, let me just write it in that green color. I'm going to have the derivative of our area with respect to time on the left-hand side. And on the right-hand side, what do I have?"}, {"video_title": "Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So on the left-hand side, right over here, I'm going to have the derivative of our area. Actually, let me just write it in that green color. I'm going to have the derivative of our area with respect to time on the left-hand side. And on the right-hand side, what do I have? Well, if I'm taking the derivative of a constant times something, I can take the constant out. So let me just do that. Pi times the derivative with respect to time of r squared."}, {"video_title": "Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And on the right-hand side, what do I have? Well, if I'm taking the derivative of a constant times something, I can take the constant out. So let me just do that. Pi times the derivative with respect to time of r squared. And to make it a little bit clearer what I'm about to do, why I'm using the chain rule, we're assuming that r is a function of time. If r wasn't a function of time, then area wouldn't be a function of time. So instead of just writing r, let me make it explicit that it is a function of time."}, {"video_title": "Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Pi times the derivative with respect to time of r squared. And to make it a little bit clearer what I'm about to do, why I'm using the chain rule, we're assuming that r is a function of time. If r wasn't a function of time, then area wouldn't be a function of time. So instead of just writing r, let me make it explicit that it is a function of time. I'll write r of t. So it's r of t, which we're squaring. We want to find the derivative of this with respect to time. And here, we just have to apply the chain rule."}, {"video_title": "Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So instead of just writing r, let me make it explicit that it is a function of time. I'll write r of t. So it's r of t, which we're squaring. We want to find the derivative of this with respect to time. And here, we just have to apply the chain rule. We're taking the derivative of something squared with respect to that something. So the derivative of that something squared with respect to the something is going to be 2 times that something to the first power. And then we're going to have, let me make it clear, this is the derivative of r of t squared with respect to r of t, the derivative of something squared with respect to that something."}, {"video_title": "Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And here, we just have to apply the chain rule. We're taking the derivative of something squared with respect to that something. So the derivative of that something squared with respect to the something is going to be 2 times that something to the first power. And then we're going to have, let me make it clear, this is the derivative of r of t squared with respect to r of t, the derivative of something squared with respect to that something. If it was the derivative of x squared with respect to x, we'd have 2x. If it's the derivative of r of t squared with respect to r of t, it's 2 r of t. But this doesn't get us just the derivative with respect to time. This is just the derivative with respect to r of t. The derivative which this changes with respect to time, we have to multiply this times the rate at which r of t changes with respect to time."}, {"video_title": "Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And then we're going to have, let me make it clear, this is the derivative of r of t squared with respect to r of t, the derivative of something squared with respect to that something. If it was the derivative of x squared with respect to x, we'd have 2x. If it's the derivative of r of t squared with respect to r of t, it's 2 r of t. But this doesn't get us just the derivative with respect to time. This is just the derivative with respect to r of t. The derivative which this changes with respect to time, we have to multiply this times the rate at which r of t changes with respect to time. So the rate at which r of t changes with respect to time, well, we could just write that as dr dt. These are equivalent expressions. And of course, we have our pi out front."}, {"video_title": "Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "This is just the derivative with respect to r of t. The derivative which this changes with respect to time, we have to multiply this times the rate at which r of t changes with respect to time. So the rate at which r of t changes with respect to time, well, we could just write that as dr dt. These are equivalent expressions. And of course, we have our pi out front. And I just want to emphasize, this is just the chain rule right over here. The derivative of something squared with respect to time is going to be the derivative of the something squared with respect to the something. So that's 2 times the something times the derivative of that something with respect to time."}, {"video_title": "Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And of course, we have our pi out front. And I just want to emphasize, this is just the chain rule right over here. The derivative of something squared with respect to time is going to be the derivative of the something squared with respect to the something. So that's 2 times the something times the derivative of that something with respect to time. I can't emphasize enough what we did right over here. This is the chain rule. So we're left with pi times this is equal to the derivative of our area with respect to time."}, {"video_title": "Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So that's 2 times the something times the derivative of that something with respect to time. I can't emphasize enough what we did right over here. This is the chain rule. So we're left with pi times this is equal to the derivative of our area with respect to time. Now, let me rewrite all of this again, just so it cleans up a little bit. So we have the derivative of our area with respect to time is equal to pi times, actually, let me put that 2 out front, is equal to 2 times pi times, I can now switch back to just calling this r. We know that r is a function of t. So I'll just write 2 pi times r times dr dt. Actually, let me make the r in blue."}, {"video_title": "Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So we're left with pi times this is equal to the derivative of our area with respect to time. Now, let me rewrite all of this again, just so it cleans up a little bit. So we have the derivative of our area with respect to time is equal to pi times, actually, let me put that 2 out front, is equal to 2 times pi times, I can now switch back to just calling this r. We know that r is a function of t. So I'll just write 2 pi times r times dr dt. Actually, let me make the r in blue. 2 pi r dr dt. Now, what do we know? We know what r is."}, {"video_title": "Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Actually, let me make the r in blue. 2 pi r dr dt. Now, what do we know? We know what r is. We know that r, at this moment right in time, is 3 centimeters. Right now, r is 3 centimeters. We know dr dt right now is 1 centimeter per second."}, {"video_title": "Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We know what r is. We know that r, at this moment right in time, is 3 centimeters. Right now, r is 3 centimeters. We know dr dt right now is 1 centimeter per second. We know this is 1 centimeter per second. So what's da dt going to be equal to? What's going to be equal to, do that same green, 2 pi times 3 times 1 times, that's purple, times 1 centimeter per second."}, {"video_title": "Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We know dr dt right now is 1 centimeter per second. We know this is 1 centimeter per second. So what's da dt going to be equal to? What's going to be equal to, do that same green, 2 pi times 3 times 1 times, that's purple, times 1 centimeter per second. And let's make sure we get the units right. So we have a centimeter times a centimeter. So it's going to be centimeters."}, {"video_title": "Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "What's going to be equal to, do that same green, 2 pi times 3 times 1 times, that's purple, times 1 centimeter per second. And let's make sure we get the units right. So we have a centimeter times a centimeter. So it's going to be centimeters. That's too dark of a color. It's going to be square centimeters, centimeters times centimeters, square centimeter per second, which is the exact units we need for a change in area. So we have da dt is equal to this."}, {"video_title": "Related rates intro Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So it's going to be centimeters. That's too dark of a color. It's going to be square centimeters, centimeters times centimeters, square centimeter per second, which is the exact units we need for a change in area. So we have da dt is equal to this. The rate at which area is changing with respect to time is equal to 6 pi. So it's going to be a little bit over 18 centimeters squared per second, right at that moment. Yep, 3 times 2 pi."}, {"video_title": "Planar motion example acceleration vector Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "A particle moves in the xy plane so that at any time t is greater than or equal to zero. Its position vector is, and they give us the x component and the y component of our position vectors, and they're both functions of time. What is the particle's acceleration vector at time t equals three? Alright, so our position, let's denote that. It's a vector-valued function. It's gonna be a function of time. It is a vector."}, {"video_title": "Planar motion example acceleration vector Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "Alright, so our position, let's denote that. It's a vector-valued function. It's gonna be a function of time. It is a vector. And they already told us that the x component of our position is negative three t to the third power plus four t squared, and the y component is t to the third power plus two. And so you give me any time greater than or equal to zero, I put it in here, and I can give you the corresponding x and y components. And this is one form of notation for a vector."}, {"video_title": "Planar motion example acceleration vector Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "It is a vector. And they already told us that the x component of our position is negative three t to the third power plus four t squared, and the y component is t to the third power plus two. And so you give me any time greater than or equal to zero, I put it in here, and I can give you the corresponding x and y components. And this is one form of notation for a vector. Another way of writing this, you might be familiar with engineering notation. It might be written like, or sometimes people write this as unit vector notation. Negative three to the third plus four t squared times the unit vector in the horizontal direction plus t to the third plus two times the unit vector in the vertical direction."}, {"video_title": "Planar motion example acceleration vector Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "And this is one form of notation for a vector. Another way of writing this, you might be familiar with engineering notation. It might be written like, or sometimes people write this as unit vector notation. Negative three to the third plus four t squared times the unit vector in the horizontal direction plus t to the third plus two times the unit vector in the vertical direction. This is just denoting the same thing. This is the x component, this is the y component. This is the component in the horizontal direction, this is the component in the vertical direction, or the y component."}, {"video_title": "Planar motion example acceleration vector Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "Negative three to the third plus four t squared times the unit vector in the horizontal direction plus t to the third plus two times the unit vector in the vertical direction. This is just denoting the same thing. This is the x component, this is the y component. This is the component in the horizontal direction, this is the component in the vertical direction, or the y component. Now the key realization is if you have the position vector, well, the velocity vector's just going to be the derivative of that. So v of t is just going to be equal to r prime of t, which is going to be equal to, well, you just have to take the corresponding derivatives of each of the components. So let's do that."}, {"video_title": "Planar motion example acceleration vector Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "This is the component in the horizontal direction, this is the component in the vertical direction, or the y component. Now the key realization is if you have the position vector, well, the velocity vector's just going to be the derivative of that. So v of t is just going to be equal to r prime of t, which is going to be equal to, well, you just have to take the corresponding derivatives of each of the components. So let's do that. So if we want to take the derivative of the x component here with respect to time, we're just gonna use the power rule a bunch. So it's three times negative three, so it's negative nine t squared, and then plus two times four is eight, so plus eight t to the first, so plus eight t. And then over here for the y component, so the derivative of t to the third with respect to t is three t squared, three t squared, and then the derivative of two is just a zero. So actually I have space to write that, three t squared even bigger."}, {"video_title": "Planar motion example acceleration vector Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "So let's do that. So if we want to take the derivative of the x component here with respect to time, we're just gonna use the power rule a bunch. So it's three times negative three, so it's negative nine t squared, and then plus two times four is eight, so plus eight t to the first, so plus eight t. And then over here for the y component, so the derivative of t to the third with respect to t is three t squared, three t squared, and then the derivative of two is just a zero. So actually I have space to write that, three t squared even bigger. Three t squared. All right, and if we want to find the acceleration function or the vector-valued function that gives us acceleration as a function of time, well, that's just going to be the derivative of the velocity function with respect to time. So this is going to be equal to, this is going to be equal to, let me give myself some space, and so the x component, well, I just take the derivative of the x component again, and let me find a color I haven't used yet."}, {"video_title": "Planar motion example acceleration vector Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "So actually I have space to write that, three t squared even bigger. Three t squared. All right, and if we want to find the acceleration function or the vector-valued function that gives us acceleration as a function of time, well, that's just going to be the derivative of the velocity function with respect to time. So this is going to be equal to, this is going to be equal to, let me give myself some space, and so the x component, well, I just take the derivative of the x component again, and let me find a color I haven't used yet. I'll use this green. So let's see, two times negative nine, negative 18 times t to the first power plus eight, derivative of eight t is just eight if we're taking the derivative with respect to t. And then, and then here in the orange, derivative of three t squared, so it's two power rule here over and over again. Two times three is six t to the first power, just six t. So this is, we've just been able to, by taking the derivative of this position vector-valued function twice, I'm able to find the acceleration function."}, {"video_title": "Planar motion example acceleration vector Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "So this is going to be equal to, this is going to be equal to, let me give myself some space, and so the x component, well, I just take the derivative of the x component again, and let me find a color I haven't used yet. I'll use this green. So let's see, two times negative nine, negative 18 times t to the first power plus eight, derivative of eight t is just eight if we're taking the derivative with respect to t. And then, and then here in the orange, derivative of three t squared, so it's two power rule here over and over again. Two times three is six t to the first power, just six t. So this is, we've just been able to, by taking the derivative of this position vector-valued function twice, I'm able to find the acceleration function. And now I just have to evaluate it at t equals three. So our acceleration at t is equal to three is equal to, so in green, it's going to be negative 18 times three plus eight, comma, eight, comma, comma, and then we're gonna have six times three, six times three. And so what does this simplify to?"}, {"video_title": "Planar motion example acceleration vector Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "Two times three is six t to the first power, just six t. So this is, we've just been able to, by taking the derivative of this position vector-valued function twice, I'm able to find the acceleration function. And now I just have to evaluate it at t equals three. So our acceleration at t is equal to three is equal to, so in green, it's going to be negative 18 times three plus eight, comma, eight, comma, comma, and then we're gonna have six times three, six times three. And so what does this simplify to? Well, this is going to be equal to, let's see, negative 18 times three is negative 54, negative 54 plus eight is negative 46, negative 46, and then six times three is 18. Did I do that arithmetic right? So this is negative 54 plus eight."}, {"video_title": "Planar motion example acceleration vector Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "And so what does this simplify to? Well, this is going to be equal to, let's see, negative 18 times three is negative 54, negative 54 plus eight is negative 46, negative 46, and then six times three is 18. Did I do that arithmetic right? So this is negative 54 plus eight. So negative 54 plus four would be negative 50, plus another four would be negative 46. Yep, there you have it. Negative 46, comma, comma, 18."}, {"video_title": "_-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "And like always, I encourage you to pause this video and see if you can figure this out on your own. So I'm assuming you've had a go at it, and so there's a couple of interesting things here. The first thing, at least that my brain does, it says, well, I'm used to taking derivatives and antiderivatives of e to the x. Not some other base to the x. So we know that the derivative with respect to x of e to the x is e to the x, or we could say that the antiderivative of e to the x is equal to e to the x plus c. So since I'm dealing with something raised to a, this particular situation, something raised to a function of x, it seems like I might wanna put some, I might wanna change the base here. But how do I do that? Well, the way I would do that is re-express two in terms of e. So what would be two in terms of e?"}, {"video_title": "_-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "Not some other base to the x. So we know that the derivative with respect to x of e to the x is e to the x, or we could say that the antiderivative of e to the x is equal to e to the x plus c. So since I'm dealing with something raised to a, this particular situation, something raised to a function of x, it seems like I might wanna put some, I might wanna change the base here. But how do I do that? Well, the way I would do that is re-express two in terms of e. So what would be two in terms of e? Well, two is equal to e, is equal to e raised to the power that you need to raise e to to get to two. Well, what's the power that you have to raise e to to get to two? Well, that's the natural log of two."}, {"video_title": "_-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the way I would do that is re-express two in terms of e. So what would be two in terms of e? Well, two is equal to e, is equal to e raised to the power that you need to raise e to to get to two. Well, what's the power that you have to raise e to to get to two? Well, that's the natural log of two. Once again, the natural log of two is the exponent that you have to raise e to to get to two. So if you actually raise e to it, you're going to get two. So this is what two is."}, {"video_title": "_-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "Well, that's the natural log of two. Once again, the natural log of two is the exponent that you have to raise e to to get to two. So if you actually raise e to it, you're going to get two. So this is what two is. Now what is two to the x to the third? Well, if we raise both sides of this to the x to the third power, if we raise both sides to the x to the third power, two to the x to the third is equal to, if I raise something to an exponent and then raise that to an exponent, it's going to be equal to e to the x to the third, x to the third times the natural log of two, times the natural log of two. So that already seems pretty interesting."}, {"video_title": "_-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "So this is what two is. Now what is two to the x to the third? Well, if we raise both sides of this to the x to the third power, if we raise both sides to the x to the third power, two to the x to the third is equal to, if I raise something to an exponent and then raise that to an exponent, it's going to be equal to e to the x to the third, x to the third times the natural log of two, times the natural log of two. So that already seems pretty interesting. So let's rewrite this. And actually, what I'm going to do is let's just focus on the indefinite integral first, see if we can figure that out, and then we can apply, and then we can evaluate the definite ones. So let's just think about this."}, {"video_title": "_-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "So that already seems pretty interesting. So let's rewrite this. And actually, what I'm going to do is let's just focus on the indefinite integral first, see if we can figure that out, and then we can apply, and then we can evaluate the definite ones. So let's just think about this. Let's think about the indefinite integral of x squared times two to the x to the third power dx. I really want to find the antiderivative of this. Well, this is going to be the exact same thing as the integral of, so I'll write my x squared still, but instead of two to the x to the third, I'm going to write all of this business."}, {"video_title": "_-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "So let's just think about this. Let's think about the indefinite integral of x squared times two to the x to the third power dx. I really want to find the antiderivative of this. Well, this is going to be the exact same thing as the integral of, so I'll write my x squared still, but instead of two to the x to the third, I'm going to write all of this business. Let me just copy and paste that. We already established this is the same thing as two to the x to the third power. Copy and paste it just like that."}, {"video_title": "_-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "Well, this is going to be the exact same thing as the integral of, so I'll write my x squared still, but instead of two to the x to the third, I'm going to write all of this business. Let me just copy and paste that. We already established this is the same thing as two to the x to the third power. Copy and paste it just like that. And then let me just close it with a dx. So I was able to get it in terms of e as a base. That makes me a little bit more comfortable, but it still seems pretty complicated."}, {"video_title": "_-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "Copy and paste it just like that. And then let me just close it with a dx. So I was able to get it in terms of e as a base. That makes me a little bit more comfortable, but it still seems pretty complicated. But you might be saying, well, okay, look, maybe u substitution could be at play here because I have this kind of crazy expression, x to the third times the natural log of two, but what's the derivative of that? Well, that's going to be three x squared times the natural log of two, or three times the natural log of two times x squared. Well, that's just a constant times x squared."}, {"video_title": "_-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "That makes me a little bit more comfortable, but it still seems pretty complicated. But you might be saying, well, okay, look, maybe u substitution could be at play here because I have this kind of crazy expression, x to the third times the natural log of two, but what's the derivative of that? Well, that's going to be three x squared times the natural log of two, or three times the natural log of two times x squared. Well, that's just a constant times x squared. We already have a x squared here, and so maybe we can engineer this a little bit to have the constant there as well. So let's think about that. So if we made this, if we defined this as u, so if we said u is equal to x to the third times the natural log of two, what is du going to be?"}, {"video_title": "_-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "Well, that's just a constant times x squared. We already have a x squared here, and so maybe we can engineer this a little bit to have the constant there as well. So let's think about that. So if we made this, if we defined this as u, so if we said u is equal to x to the third times the natural log of two, what is du going to be? Well, du is going to be, it's going to be, well, natural log of two is just a constant, so it's going to be three x squared times the natural log of two. And we can actually just change the order we're multiplying a little bit. We could say that this is the same thing as x squared times three natural log of two, which is the same thing, just using logarithm properties, as x squared times the natural log of two to the third power."}, {"video_title": "_-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "So if we made this, if we defined this as u, so if we said u is equal to x to the third times the natural log of two, what is du going to be? Well, du is going to be, it's going to be, well, natural log of two is just a constant, so it's going to be three x squared times the natural log of two. And we can actually just change the order we're multiplying a little bit. We could say that this is the same thing as x squared times three natural log of two, which is the same thing, just using logarithm properties, as x squared times the natural log of two to the third power. Three natural log of two is the same thing as the natural log of two to the third power. So this is equal to x squared times the natural log of eight. So let's see, if this is u, where is du?"}, {"video_title": "_-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "We could say that this is the same thing as x squared times three natural log of two, which is the same thing, just using logarithm properties, as x squared times the natural log of two to the third power. Three natural log of two is the same thing as the natural log of two to the third power. So this is equal to x squared times the natural log of eight. So let's see, if this is u, where is du? Oh, and of course, we can't forget the dx. This is a dx right over here. dx, dx, dx."}, {"video_title": "_-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "So let's see, if this is u, where is du? Oh, and of course, we can't forget the dx. This is a dx right over here. dx, dx, dx. So where is the du? Well, we have a dx. Let me circle things."}, {"video_title": "_-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "dx, dx, dx. So where is the du? Well, we have a dx. Let me circle things. So you have a dx here. You have a dx there. You have an x squared here."}, {"video_title": "_-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "Let me circle things. So you have a dx here. You have a dx there. You have an x squared here. You have an x squared here. So really, all we need is, all we need here is the natural log of eight. So if we, ideally, we would have a natural log of eight right over here."}, {"video_title": "_-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "You have an x squared here. You have an x squared here. So really, all we need is, all we need here is the natural log of eight. So if we, ideally, we would have a natural log of eight right over here. And we could put it there, as long as we also, we can multiply by a natural log of eight, as long as we also divide by a natural log of eight. And so we could do it right over here. We could divide by natural log of eight."}, {"video_title": "_-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "So if we, ideally, we would have a natural log of eight right over here. And we could put it there, as long as we also, we can multiply by a natural log of eight, as long as we also divide by a natural log of eight. And so we could do it right over here. We could divide by natural log of eight. But we know that the antiderivative of some constant times a function is the same thing as the constant times the antiderivative of that function. So we could just take that on the outside. So it's one over the natural log of eight."}, {"video_title": "_-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "We could divide by natural log of eight. But we know that the antiderivative of some constant times a function is the same thing as the constant times the antiderivative of that function. So we could just take that on the outside. So it's one over the natural log of eight. So let's write this in terms of u and du. This simplifies to one over the natural log of eight times the antiderivative of e, e to the u, e to the u, that's the u, du. This times this times that is du, du."}, {"video_title": "_-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "So it's one over the natural log of eight. So let's write this in terms of u and du. This simplifies to one over the natural log of eight times the antiderivative of e, e to the u, e to the u, that's the u, du. This times this times that is du, du. And this is straightforward. We know what this is going to be. This is going to be equal to, so let me just write the one over natural log of eight out here."}, {"video_title": "_-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "This times this times that is du, du. And this is straightforward. We know what this is going to be. This is going to be equal to, so let me just write the one over natural log of eight out here. One over natural log of eight times e to the u. Times e to the u, e to the u. And of course, if we're thinking in terms of just antiderivative, there would be some constant out there."}, {"video_title": "_-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "This is going to be equal to, so let me just write the one over natural log of eight out here. One over natural log of eight times e to the u. Times e to the u, e to the u. And of course, if we're thinking in terms of just antiderivative, there would be some constant out there. And then we would just reverse the substitution. We already know what u is. So this is going to be equal to the antiderivative of this expression is one over the natural log of eight times e to the, instead of u, we know that u is x to the third times the natural log of two."}, {"video_title": "_-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "And of course, if we're thinking in terms of just antiderivative, there would be some constant out there. And then we would just reverse the substitution. We already know what u is. So this is going to be equal to the antiderivative of this expression is one over the natural log of eight times e to the, instead of u, we know that u is x to the third times the natural log of two. And of course, we could put a plus c there. Now, going back to the original problem. We just need to evaluate the antiderivative of this at each of these points."}, {"video_title": "_-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be equal to the antiderivative of this expression is one over the natural log of eight times e to the, instead of u, we know that u is x to the third times the natural log of two. And of course, we could put a plus c there. Now, going back to the original problem. We just need to evaluate the antiderivative of this at each of these points. So let's rewrite this. So given what we just figured out, so let me copy and paste that. This is just going to be equal to, it's going to be equal to the antiderivative evaluated at one minus the antiderivative evaluated at zero."}, {"video_title": "_-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "We just need to evaluate the antiderivative of this at each of these points. So let's rewrite this. So given what we just figured out, so let me copy and paste that. This is just going to be equal to, it's going to be equal to the antiderivative evaluated at one minus the antiderivative evaluated at zero. We don't have to worry about the constants because those will cancel out. And so we are going to get, we are going to get one, let me evaluate it first at one. So you're gonna get one over the natural log of eight times e to the one to the third power, which is just one, times the natural log of two."}, {"video_title": "_-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "This is just going to be equal to, it's going to be equal to the antiderivative evaluated at one minus the antiderivative evaluated at zero. We don't have to worry about the constants because those will cancel out. And so we are going to get, we are going to get one, let me evaluate it first at one. So you're gonna get one over the natural log of eight times e to the one to the third power, which is just one, times the natural log of two. Natural log of two, that's evaluated at one. And then we're gonna have minus it evaluated at zero. So it's going to be one over the natural log of eight times e to the, well, when x is zero, this whole thing is going to be zero."}, {"video_title": "_-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "So you're gonna get one over the natural log of eight times e to the one to the third power, which is just one, times the natural log of two. Natural log of two, that's evaluated at one. And then we're gonna have minus it evaluated at zero. So it's going to be one over the natural log of eight times e to the, well, when x is zero, this whole thing is going to be zero. Well, e to the zero is just one. And e to the natural log of two, well, that's just going to be two. We already established that early on."}, {"video_title": "_-substitution definite integral of exponential function AP Calculus AB Khan Academy.mp3", "Sentence": "So it's going to be one over the natural log of eight times e to the, well, when x is zero, this whole thing is going to be zero. Well, e to the zero is just one. And e to the natural log of two, well, that's just going to be two. We already established that early on. This is just going to be equal to two. So we are left with two over the natural log of eight minus one over the natural log of eight, which is just going to be equal to one over the natural log of eight. And we are done."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "And also, the diameter of the top of the cup is also 4 centimeters. And I'm pouring water into this cup right now. And I'm pouring the water at a rate of 1 cubic centimeter per second. And right at this moment, there is a height of 2 centimeters of water in the cup right now. So the height right now from the bottom of the cup to this point right over here is 2 centimeters. So my question to you is, at what rate? We know the rate at which the water is flowing into the cup."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "And right at this moment, there is a height of 2 centimeters of water in the cup right now. So the height right now from the bottom of the cup to this point right over here is 2 centimeters. So my question to you is, at what rate? We know the rate at which the water is flowing into the cup. We're being given a volume per time. My question to you is, right at this moment, right when we are filling our cup at 1 cubic centimeter per second, and we have exactly 2 centimeters of water in the cup, 2 centimeters deep of water in the cup, what is the rate at which the height of the water is changing? What is the rate at which this height right over here is actually changing?"}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "We know the rate at which the water is flowing into the cup. We're being given a volume per time. My question to you is, right at this moment, right when we are filling our cup at 1 cubic centimeter per second, and we have exactly 2 centimeters of water in the cup, 2 centimeters deep of water in the cup, what is the rate at which the height of the water is changing? What is the rate at which this height right over here is actually changing? We know it's 2 centimeters, but how fast is it changing? Well, let's think about this a little bit. What are we being given?"}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "What is the rate at which this height right over here is actually changing? We know it's 2 centimeters, but how fast is it changing? Well, let's think about this a little bit. What are we being given? We're being given the rate at which the volume of the water is changing with respect to time. So let's write that down. We're being given the rate at which the volume of the water is changing with respect to time."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "What are we being given? We're being given the rate at which the volume of the water is changing with respect to time. So let's write that down. We're being given the rate at which the volume of the water is changing with respect to time. And we're being told. We're told that this is 1 cubic centimeter per second. And what are we trying to figure out?"}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "We're being given the rate at which the volume of the water is changing with respect to time. And we're being told. We're told that this is 1 cubic centimeter per second. And what are we trying to figure out? Well, we're trying to figure out how fast the height of the water is changing with respect to time. We know that the height right now is 2 centimeters, but what we want to figure out is the rate at which the height is changing with respect to time. If we can figure out this, then we have essentially answered the question."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "And what are we trying to figure out? Well, we're trying to figure out how fast the height of the water is changing with respect to time. We know that the height right now is 2 centimeters, but what we want to figure out is the rate at which the height is changing with respect to time. If we can figure out this, then we have essentially answered the question. So one way that we can do this is that we can come up with a relationship between the volume at any moment in time and the height at any moment in time, and then maybe take the derivative of that relationship, possibly using the chain rule, to come up with a relationship between the rate at which the volume is changing and the rate at which the height is changing. So let's try to do it step by step. So first of all, can we come up with a relationship between the volume and the height at any given moment?"}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "If we can figure out this, then we have essentially answered the question. So one way that we can do this is that we can come up with a relationship between the volume at any moment in time and the height at any moment in time, and then maybe take the derivative of that relationship, possibly using the chain rule, to come up with a relationship between the rate at which the volume is changing and the rate at which the height is changing. So let's try to do it step by step. So first of all, can we come up with a relationship between the volume and the height at any given moment? Well, we have also been given the formula for the volume of a cone right over here. The volume of a cone is 1 third times the area of the base of the cone times the height. And we won't prove it here, although we could prove it later on, especially when we start doing solids of revolutions in integral calculus."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "So first of all, can we come up with a relationship between the volume and the height at any given moment? Well, we have also been given the formula for the volume of a cone right over here. The volume of a cone is 1 third times the area of the base of the cone times the height. And we won't prove it here, although we could prove it later on, especially when we start doing solids of revolutions in integral calculus. But we'll just take it on faith right now that this is how we can figure out the volume of a cone. So given this, can we figure out an expression that relates volume to the height of the cone? Well, we could say that volume, and I'll do it in this blue color, the volume of water is what we really care about."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "And we won't prove it here, although we could prove it later on, especially when we start doing solids of revolutions in integral calculus. But we'll just take it on faith right now that this is how we can figure out the volume of a cone. So given this, can we figure out an expression that relates volume to the height of the cone? Well, we could say that volume, and I'll do it in this blue color, the volume of water is what we really care about. The volume of water is going to be equal to 1 third times the area of the surface of the water, area of water surface times our height of the water, so times h. So how can we figure out the area of the water surface, preferably in terms of h? Well, we see right over here the diameter across the top of the cone is 4 centimeters, and the height of the whole cup is 4 centimeters. And so that ratio is going to be true at any depth of water."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we could say that volume, and I'll do it in this blue color, the volume of water is what we really care about. The volume of water is going to be equal to 1 third times the area of the surface of the water, area of water surface times our height of the water, so times h. So how can we figure out the area of the water surface, preferably in terms of h? Well, we see right over here the diameter across the top of the cone is 4 centimeters, and the height of the whole cup is 4 centimeters. And so that ratio is going to be true at any depth of water. It's always going to have the same ratio between the diameter across the top and the height, because these are lines right over here. So at any given point, the ratio between this and this is going to be the same. So at any given point, the diameter across the surface of the water, if the depth is h, the diameter across the surface of the water is also going to be h. And so from that, we can figure out what the radius is going to be."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "And so that ratio is going to be true at any depth of water. It's always going to have the same ratio between the diameter across the top and the height, because these are lines right over here. So at any given point, the ratio between this and this is going to be the same. So at any given point, the diameter across the surface of the water, if the depth is h, the diameter across the surface of the water is also going to be h. And so from that, we can figure out what the radius is going to be. The radius is going to be h over 2. And so the area of the water surface is going to be pi r squared, pi times the radius squared, h over 2 squared. That's the area of the surface of the water."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "So at any given point, the diameter across the surface of the water, if the depth is h, the diameter across the surface of the water is also going to be h. And so from that, we can figure out what the radius is going to be. The radius is going to be h over 2. And so the area of the water surface is going to be pi r squared, pi times the radius squared, h over 2 squared. That's the area of the surface of the water. And of course, we still have the 1 third out here, and we're still multiplying by this h over here. So let me see if I can simplify this. So this gives us 1 third times pi h squared over 4 times another h, which is equal to, we have pi h to the third power over 12."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "That's the area of the surface of the water. And of course, we still have the 1 third out here, and we're still multiplying by this h over here. So let me see if I can simplify this. So this gives us 1 third times pi h squared over 4 times another h, which is equal to, we have pi h to the third power over 12. So that is our volume. Now what we want to do is relate how fast the volume is changing with respect to time and how fast the height is changing with respect to time. So we care with respect to time."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "So this gives us 1 third times pi h squared over 4 times another h, which is equal to, we have pi h to the third power over 12. So that is our volume. Now what we want to do is relate how fast the volume is changing with respect to time and how fast the height is changing with respect to time. So we care with respect to time. Since we care so much about what's happening with respect to time, let's take the derivative of both sides of this equation with respect to time. And just so I have enough space to do that, let me move this over to the right a little bit. So I just move this over to the right."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "So we care with respect to time. Since we care so much about what's happening with respect to time, let's take the derivative of both sides of this equation with respect to time. And just so I have enough space to do that, let me move this over to the right a little bit. So I just move this over to the right. And so now we can take the derivative with respect to time of both sides of this business. So the derivative with respect to time of our volume and the derivative with respect to time of this business. Well, the derivative with respect to time of our volume, we can just rewrite that as dv dt, this thing right over here."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "So I just move this over to the right. And so now we can take the derivative with respect to time of both sides of this business. So the derivative with respect to time of our volume and the derivative with respect to time of this business. Well, the derivative with respect to time of our volume, we can just rewrite that as dv dt, this thing right over here. This is dv dt. And this is going to be equal to, well, we could take the constants out of this. This is going to be equal to pi over 12 times the derivative with respect to t of h to the third power."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the derivative with respect to time of our volume, we can just rewrite that as dv dt, this thing right over here. This is dv dt. And this is going to be equal to, well, we could take the constants out of this. This is going to be equal to pi over 12 times the derivative with respect to t of h to the third power. And just so that the next few things I do will appear a little bit clearer, we're assuming that height is a function of time. In fact, it's definitely a function of time. As time goes on, the height will change because we're pouring more and more water here."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "This is going to be equal to pi over 12 times the derivative with respect to t of h to the third power. And just so that the next few things I do will appear a little bit clearer, we're assuming that height is a function of time. In fact, it's definitely a function of time. As time goes on, the height will change because we're pouring more and more water here. So instead of just writing h to the third power, which I could write over here, let me write h of t to the third power, just to make it clear that this is a function of t. h of t to the third power. Now, what is the derivative with respect to t of h of t to the third power? Now, you might be getting a tingling feeling that the chain rule might be applicable here."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "As time goes on, the height will change because we're pouring more and more water here. So instead of just writing h to the third power, which I could write over here, let me write h of t to the third power, just to make it clear that this is a function of t. h of t to the third power. Now, what is the derivative with respect to t of h of t to the third power? Now, you might be getting a tingling feeling that the chain rule might be applicable here. So let's think about the chain rule. The chain rule tells us, let me rewrite everything else, dv with respect to t is going to be equal to pi over 12 times the derivative of this with respect to t. If we want to take the derivative of this with respect to t, we have something to the third power. So we want to take the derivative of something to the third power with respect to something."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "Now, you might be getting a tingling feeling that the chain rule might be applicable here. So let's think about the chain rule. The chain rule tells us, let me rewrite everything else, dv with respect to t is going to be equal to pi over 12 times the derivative of this with respect to t. If we want to take the derivative of this with respect to t, we have something to the third power. So we want to take the derivative of something to the third power with respect to something. So that's going to be, let me write this in a different color, maybe an orange. So that's going to be 3 times our something squared times the derivative of that something with respect to t times dh, let me, I've already used that pink, times dh dt. Let's just be very clear."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "So we want to take the derivative of something to the third power with respect to something. So that's going to be, let me write this in a different color, maybe an orange. So that's going to be 3 times our something squared times the derivative of that something with respect to t times dh, let me, I've already used that pink, times dh dt. Let's just be very clear. This orange term right over here, and I'm just using the chain rule. This is the derivative of h of t to the third power with respect to h of t. And then we're going to multiply that times the derivative of h of t with respect to t. And then that gives us the derivative of this entire thing, h of t to the third power, with respect to t. This will give us the derivative of h of t to the third power with respect to t, which is exactly what we want to do when we apply this operator. How fast is this changing?"}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "Let's just be very clear. This orange term right over here, and I'm just using the chain rule. This is the derivative of h of t to the third power with respect to h of t. And then we're going to multiply that times the derivative of h of t with respect to t. And then that gives us the derivative of this entire thing, h of t to the third power, with respect to t. This will give us the derivative of h of t to the third power with respect to t, which is exactly what we want to do when we apply this operator. How fast is this changing? How is this changing with respect to time? So we can just rewrite this, just so it gets a little bit cleaner. Let me rewrite everything I've done."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "How fast is this changing? How is this changing with respect to time? So we can just rewrite this, just so it gets a little bit cleaner. Let me rewrite everything I've done. So we've got dv, the rate at which our volume is changing with respect to time. The rate at which our volume is changing with respect to time is equal to pi over 12 times 3h of t squared. Or I could just write that as 3h squared."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "Let me rewrite everything I've done. So we've got dv, the rate at which our volume is changing with respect to time. The rate at which our volume is changing with respect to time is equal to pi over 12 times 3h of t squared. Or I could just write that as 3h squared. Times the rate at which the height is changing with respect to time. Times dh dt. And you might be a little confused."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "Or I could just write that as 3h squared. Times the rate at which the height is changing with respect to time. Times dh dt. And you might be a little confused. You might have been tempted to take the derivative over here with respect to h. But remember, we're thinking about how things are changing with respect to time. So we're assuming we did express volume as a function of height, but we're saying that height itself is a function of time. So we're taking the derivative of everything with respect to time."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "And you might be a little confused. You might have been tempted to take the derivative over here with respect to h. But remember, we're thinking about how things are changing with respect to time. So we're assuming we did express volume as a function of height, but we're saying that height itself is a function of time. So we're taking the derivative of everything with respect to time. So that's why the chain rule came into play when we were taking the derivative of h or the derivative of h of t, because we're assuming that h is a function of time. Now, what does this thing right over here get us? Well, we're telling us at the exact moment that we set up this problem, we know what dv dt is."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "So we're taking the derivative of everything with respect to time. So that's why the chain rule came into play when we were taking the derivative of h or the derivative of h of t, because we're assuming that h is a function of time. Now, what does this thing right over here get us? Well, we're telling us at the exact moment that we set up this problem, we know what dv dt is. We know that it's 1 centimeter cubed per second. We know that this right over here is 1 centimeter cubed per second. We know what our height is right at this moment."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we're telling us at the exact moment that we set up this problem, we know what dv dt is. We know that it's 1 centimeter cubed per second. We know that this right over here is 1 centimeter cubed per second. We know what our height is right at this moment. We were told it is 2 centimeters. Our height right over here, we know it is 2 centimeters. So the only unknown we have over here is a rate at which our height is changing with respect to time, which is exactly what we needed to figure out in the first place."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "We know what our height is right at this moment. We were told it is 2 centimeters. Our height right over here, we know it is 2 centimeters. So the only unknown we have over here is a rate at which our height is changing with respect to time, which is exactly what we needed to figure out in the first place. So we just have to solve for that. So we get 1 cubic centimeter. Let me make it clear."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "So the only unknown we have over here is a rate at which our height is changing with respect to time, which is exactly what we needed to figure out in the first place. So we just have to solve for that. So we get 1 cubic centimeter. Let me make it clear. We get 1 cubic centimeter per second. I won't write the units to save some space. Is equal to pi over 2."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "Let me make it clear. We get 1 cubic centimeter per second. I won't write the units to save some space. Is equal to pi over 2. And I'll write this in a neutral color. Actually, let me write it in the same color. Is equal to pi over 2 times 3 times h squared."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "Is equal to pi over 2. And I'll write this in a neutral color. Actually, let me write it in the same color. Is equal to pi over 2 times 3 times h squared. h is 2, so you're going to get 4 squared centimeters. We kept the units. So 3 times 4."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "Is equal to pi over 2 times 3 times h squared. h is 2, so you're going to get 4 squared centimeters. We kept the units. So 3 times 4. Let me be careful. That wasn't pi over 2. That was pi over 12."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "So 3 times 4. Let me be careful. That wasn't pi over 2. That was pi over 12. This is a pi over 12 right over here. Pi over 12. So you get pi over 12 times 3 times 2 squared times dh dt."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "That was pi over 12. This is a pi over 12 right over here. Pi over 12. So you get pi over 12 times 3 times 2 squared times dh dt. All of this is equal to 1. So now I'll switch to a neutral color. We get 1 is equal to 3 times 4 is 12."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "So you get pi over 12 times 3 times 2 squared times dh dt. All of this is equal to 1. So now I'll switch to a neutral color. We get 1 is equal to 3 times 4 is 12. Cancels out with that 12. We get 1 is equal to pi times dh dt. To solve for dh dt, divide both sides by pi."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "We get 1 is equal to 3 times 4 is 12. Cancels out with that 12. We get 1 is equal to pi times dh dt. To solve for dh dt, divide both sides by pi. And we get our drum roll now. The rate at which our height is changing with respect to time, as we're putting 1 cubic centimeter of water per second in, and right when our height is 2 centimeters high, the rate at which this height is changing with respect to time is 1 over pi. And I haven't done the dimensional analysis, but this is going to be in centimeters per second."}, {"video_title": "Related rates water pouring into a cone AP Calculus AB Khan Academy.mp3", "Sentence": "To solve for dh dt, divide both sides by pi. And we get our drum roll now. The rate at which our height is changing with respect to time, as we're putting 1 cubic centimeter of water per second in, and right when our height is 2 centimeters high, the rate at which this height is changing with respect to time is 1 over pi. And I haven't done the dimensional analysis, but this is going to be in centimeters per second. You could work through the dimensional analysis if you like by putting the dimensions right over here. But there you have it. That's how fast our height is going to be changing at exactly that moment."}, {"video_title": "Definite integral involving natural log AP Calculus AB Khan Academy.mp3", "Sentence": "I have this rational expression, but if we just rewrite this, it might jump out at you how this could be a little bit simpler. So this is equal to the integral from two to four of six over x to the third power plus x squared over x to the third power dx. I just separated this numerator out. I just divided each of those terms by x to the third power. And this I could rewrite. This is equal to the integral from two to four of six x to the negative three power. That's that first term there."}, {"video_title": "Definite integral involving natural log AP Calculus AB Khan Academy.mp3", "Sentence": "I just divided each of those terms by x to the third power. And this I could rewrite. This is equal to the integral from two to four of six x to the negative three power. That's that first term there. And x squared divided by x to the third, well that is going to be one over x. So plus one over x dx. Now this is going to be equal to, let's take the antiderivative of the different parts, and we're going to evaluate that at four, and we're going to evaluate that at two."}, {"video_title": "Definite integral involving natural log AP Calculus AB Khan Academy.mp3", "Sentence": "That's that first term there. And x squared divided by x to the third, well that is going to be one over x. So plus one over x dx. Now this is going to be equal to, let's take the antiderivative of the different parts, and we're going to evaluate that at four, and we're going to evaluate that at two. And we're going to find the difference between this expression, the antiderivative evaluated at four and at two. Now what is the antiderivative of six x to the negative three? Well here once again we can just use, we could use as a power rule for taking the antiderivative, or it's the reverse of the derivative power rule."}, {"video_title": "Definite integral involving natural log AP Calculus AB Khan Academy.mp3", "Sentence": "Now this is going to be equal to, let's take the antiderivative of the different parts, and we're going to evaluate that at four, and we're going to evaluate that at two. And we're going to find the difference between this expression, the antiderivative evaluated at four and at two. Now what is the antiderivative of six x to the negative three? Well here once again we can just use, we could use as a power rule for taking the antiderivative, or it's the reverse of the derivative power rule. We know that if we're taking the integral of x to the n dx, the antiderivative of that is going to be x to the n plus one over n plus one. And if we were just taking an indefinite integral, there would be some plus c. The reason why we don't put the plus c's here is when you evaluated both bounds of integration, the c would cancel out regardless of what it is. So we don't really think about the c much when we're taking indefinite integrals."}, {"video_title": "Definite integral involving natural log AP Calculus AB Khan Academy.mp3", "Sentence": "Well here once again we can just use, we could use as a power rule for taking the antiderivative, or it's the reverse of the derivative power rule. We know that if we're taking the integral of x to the n dx, the antiderivative of that is going to be x to the n plus one over n plus one. And if we were just taking an indefinite integral, there would be some plus c. The reason why we don't put the plus c's here is when you evaluated both bounds of integration, the c would cancel out regardless of what it is. So we don't really think about the c much when we're taking indefinite integrals. But let's apply that to six x to the negative third power. So it's going to be, we're going to take x to the negative three plus one, so it's x to the negative two, and so we're going to divide by negative two as well. And of course we had that six out front from the get-go, so that's the antiderivative six x to the negative three power."}, {"video_title": "Definite integral involving natural log AP Calculus AB Khan Academy.mp3", "Sentence": "So we don't really think about the c much when we're taking indefinite integrals. But let's apply that to six x to the negative third power. So it's going to be, we're going to take x to the negative three plus one, so it's x to the negative two, and so we're going to divide by negative two as well. And of course we had that six out front from the get-go, so that's the antiderivative six x to the negative three power. And what's the antiderivative one over x? You might be tempted to use this same idea right over here. You might be tempted to say, all right, well the antiderivative of x to the negative one, which is the same thing as one over x, would be equal to x to the negative one plus one over negative one plus one."}, {"video_title": "Definite integral involving natural log AP Calculus AB Khan Academy.mp3", "Sentence": "And of course we had that six out front from the get-go, so that's the antiderivative six x to the negative three power. And what's the antiderivative one over x? You might be tempted to use this same idea right over here. You might be tempted to say, all right, well the antiderivative of x to the negative one, which is the same thing as one over x, would be equal to x to the negative one plus one over negative one plus one. But what is negative one plus one? It is zero. So this doesn't fit this property right over here."}, {"video_title": "Definite integral involving natural log AP Calculus AB Khan Academy.mp3", "Sentence": "You might be tempted to say, all right, well the antiderivative of x to the negative one, which is the same thing as one over x, would be equal to x to the negative one plus one over negative one plus one. But what is negative one plus one? It is zero. So this doesn't fit this property right over here. But lucky for us, there is another property. And we went the other way when we were first taking derivatives of natural log functions. The antiderivative of one over x, or x to the negative one, is equal to, sometimes you'll see it written as natural log of x plus c. And sometimes, and I actually prefer this one because you could actually evaluate it for negative values, is to say the absolute value, the natural log of the absolute value of x."}, {"video_title": "Definite integral involving natural log AP Calculus AB Khan Academy.mp3", "Sentence": "So this doesn't fit this property right over here. But lucky for us, there is another property. And we went the other way when we were first taking derivatives of natural log functions. The antiderivative of one over x, or x to the negative one, is equal to, sometimes you'll see it written as natural log of x plus c. And sometimes, and I actually prefer this one because you could actually evaluate it for negative values, is to say the absolute value, the natural log of the absolute value of x. And this is useful because this is defined for negative values, not just positive values. The natural log of x is only defined for positive values of x. But when you take the absolute value, now it could be negative or positive values of x."}, {"video_title": "Definite integral involving natural log AP Calculus AB Khan Academy.mp3", "Sentence": "The antiderivative of one over x, or x to the negative one, is equal to, sometimes you'll see it written as natural log of x plus c. And sometimes, and I actually prefer this one because you could actually evaluate it for negative values, is to say the absolute value, the natural log of the absolute value of x. And this is useful because this is defined for negative values, not just positive values. The natural log of x is only defined for positive values of x. But when you take the absolute value, now it could be negative or positive values of x. And it works, the derivative of this is indeed one over x. Now it's not so relevant here because our bounds of integration are both positive. But if both of our bounds of integration were negative, you could still do this by just reminding yourself that this is the natural log of absolute value of x."}, {"video_title": "Definite integral involving natural log AP Calculus AB Khan Academy.mp3", "Sentence": "But when you take the absolute value, now it could be negative or positive values of x. And it works, the derivative of this is indeed one over x. Now it's not so relevant here because our bounds of integration are both positive. But if both of our bounds of integration were negative, you could still do this by just reminding yourself that this is the natural log of absolute value of x. So this, we could say, is plus the natural log of the absolute value of x. It's not a bad habit to do it, and if everything's positive, well, the absolute value of x is equal to x. And so what is this going to be equal to?"}, {"video_title": "Definite integral involving natural log AP Calculus AB Khan Academy.mp3", "Sentence": "But if both of our bounds of integration were negative, you could still do this by just reminding yourself that this is the natural log of absolute value of x. So this, we could say, is plus the natural log of the absolute value of x. It's not a bad habit to do it, and if everything's positive, well, the absolute value of x is equal to x. And so what is this going to be equal to? This is equal to, let's evaluate everything at four. And actually, before I even evaluate at four, what's six divided by negative two? That's negative three."}, {"video_title": "Definite integral involving natural log AP Calculus AB Khan Academy.mp3", "Sentence": "And so what is this going to be equal to? This is equal to, let's evaluate everything at four. And actually, before I even evaluate at four, what's six divided by negative two? That's negative three. So if we evaluate it at four, it's going to be negative three over four squared. Four to the negative two is one over four squared. And then plus the natural log of the, we could say the absolute value of four, but the absolute value of four is just four."}, {"video_title": "Definite integral involving natural log AP Calculus AB Khan Academy.mp3", "Sentence": "That's negative three. So if we evaluate it at four, it's going to be negative three over four squared. Four to the negative two is one over four squared. And then plus the natural log of the, we could say the absolute value of four, but the absolute value of four is just four. So the natural log of four. And from that, we're going to subtract everything evaluated at two. So let's do that."}, {"video_title": "Definite integral involving natural log AP Calculus AB Khan Academy.mp3", "Sentence": "And then plus the natural log of the, we could say the absolute value of four, but the absolute value of four is just four. So the natural log of four. And from that, we're going to subtract everything evaluated at two. So let's do that. So if we evaluated at two, it's going to be negative three over two squared. So two to the negative two is one over two squared. Over two squared plus the natural log of, the absolute value of positive two is once again, is just two."}, {"video_title": "Definite integral involving natural log AP Calculus AB Khan Academy.mp3", "Sentence": "So let's do that. So if we evaluated at two, it's going to be negative three over two squared. So two to the negative two is one over two squared. Over two squared plus the natural log of, the absolute value of positive two is once again, is just two. And so what does this give us? So let's try to simplify it a little bit. So this is negative 3 16ths."}, {"video_title": "Definite integral involving natural log AP Calculus AB Khan Academy.mp3", "Sentence": "Over two squared plus the natural log of, the absolute value of positive two is once again, is just two. And so what does this give us? So let's try to simplify it a little bit. So this is negative 3 16ths. Let me do that same color. So this is going to be equal to negative three, sorry, not negative 3 16ths, gotta be very careful. Oh, sorry, yes."}, {"video_title": "Definite integral involving natural log AP Calculus AB Khan Academy.mp3", "Sentence": "So this is negative 3 16ths. Let me do that same color. So this is going to be equal to negative three, sorry, not negative 3 16ths, gotta be very careful. Oh, sorry, yes. Sorry, it is negative 3 16ths. For some reason my brain started thinking four to the third power. Negative 3 16ths plus natural log of four."}, {"video_title": "Definite integral involving natural log AP Calculus AB Khan Academy.mp3", "Sentence": "Oh, sorry, yes. Sorry, it is negative 3 16ths. For some reason my brain started thinking four to the third power. Negative 3 16ths plus natural log of four. And then this right over here is negative 3 4ths. Negative 3 4ths, do that same color. This right over here is negative 3 4ths."}, {"video_title": "Definite integral involving natural log AP Calculus AB Khan Academy.mp3", "Sentence": "Negative 3 16ths plus natural log of four. And then this right over here is negative 3 4ths. Negative 3 4ths, do that same color. This right over here is negative 3 4ths. We have this negative sign out front that we're going to have to distribute. So the negative of negative 3 4ths is plus 3 4ths. Plus 3 4ths."}, {"video_title": "Definite integral involving natural log AP Calculus AB Khan Academy.mp3", "Sentence": "This right over here is negative 3 4ths. We have this negative sign out front that we're going to have to distribute. So the negative of negative 3 4ths is plus 3 4ths. Plus 3 4ths. And then we're going to subtract, remember we're distributing this negative sign, the natural log, the natural log of two. And what is this equal to? All right, so this is going to be equal to, and I'm now going to switch to a neutral color."}, {"video_title": "Definite integral involving natural log AP Calculus AB Khan Academy.mp3", "Sentence": "Plus 3 4ths. And then we're going to subtract, remember we're distributing this negative sign, the natural log, the natural log of two. And what is this equal to? All right, so this is going to be equal to, and I'm now going to switch to a neutral color. So let's add these two terms that don't involve the natural log. And let's see, if we have a common denominator, three over four is the same thing. That is the same thing as we multiply the numerator and denominator by four."}, {"video_title": "Definite integral involving natural log AP Calculus AB Khan Academy.mp3", "Sentence": "All right, so this is going to be equal to, and I'm now going to switch to a neutral color. So let's add these two terms that don't involve the natural log. And let's see, if we have a common denominator, three over four is the same thing. That is the same thing as we multiply the numerator and denominator by four. That is 12 over 16. And so you have negative 3 16ths, negative 3 16ths plus 12 16ths will give you 9 16ths. 9 16ths."}, {"video_title": "Definite integral involving natural log AP Calculus AB Khan Academy.mp3", "Sentence": "That is the same thing as we multiply the numerator and denominator by four. That is 12 over 16. And so you have negative 3 16ths, negative 3 16ths plus 12 16ths will give you 9 16ths. 9 16ths. And then we're going to have the ones that do involve the natural log. Natural log of four minus the natural log of two. So we could write this plus the natural log of four minus the natural log of two."}, {"video_title": "Definite integral involving natural log AP Calculus AB Khan Academy.mp3", "Sentence": "9 16ths. And then we're going to have the ones that do involve the natural log. Natural log of four minus the natural log of two. So we could write this plus the natural log of four minus the natural log of two. And you might remember from your logarithm properties that this over here, this is the same thing as the natural log of four divided by two. This comes straight out of your logarithm properties. And so this is going to be the natural log of two."}, {"video_title": "Definite integral involving natural log AP Calculus AB Khan Academy.mp3", "Sentence": "So we could write this plus the natural log of four minus the natural log of two. And you might remember from your logarithm properties that this over here, this is the same thing as the natural log of four divided by two. This comes straight out of your logarithm properties. And so this is going to be the natural log of two. Natural log of two. So we deserve a little bit of a drum roll now. This is all going to be equal to, this is going to be equal to the natural log, sorry, nine over 16 plus the natural log of two."}, {"video_title": "Squeeze theorem or sandwich theorem Limits Differential Calculus Khan Academy.mp3", "Sentence": "We're now going to think about one of my most favorite theorems in mathematics, and that's the Squeeze Theorem. And one of the reasons that it's one of my most favorite theorems in mathematics is that it has the word squeeze in it, a word that you don't see showing up in a lot of mathematics, but it is appropriately named. This is oftentimes also called the Sandwich Theorem, which is also an appropriate name, as we'll see in a second. And since it can be called the Sandwich Theorem, let's first just think about an analogy to kind of get the intuition behind the Squeeze or the Sandwich Theorem. Let's say that there are three people. Let's say that there is Imran. Imran."}, {"video_title": "Squeeze theorem or sandwich theorem Limits Differential Calculus Khan Academy.mp3", "Sentence": "And since it can be called the Sandwich Theorem, let's first just think about an analogy to kind of get the intuition behind the Squeeze or the Sandwich Theorem. Let's say that there are three people. Let's say that there is Imran. Imran. Let's say there's Diya. And let's say there is Sal. And let's say that Imran, on any given day, he always has the fewest amount of calories, and Sal, on any given day, always has the most number of calories."}, {"video_title": "Squeeze theorem or sandwich theorem Limits Differential Calculus Khan Academy.mp3", "Sentence": "Imran. Let's say there's Diya. And let's say there is Sal. And let's say that Imran, on any given day, he always has the fewest amount of calories, and Sal, on any given day, always has the most number of calories. So on a given day, we can always say Diya eats at least as much as Imran. And then we can say Sal eats at least as much, not just to repeat those words, as Diya. And so we could set up a little inequality here."}, {"video_title": "Squeeze theorem or sandwich theorem Limits Differential Calculus Khan Academy.mp3", "Sentence": "And let's say that Imran, on any given day, he always has the fewest amount of calories, and Sal, on any given day, always has the most number of calories. So on a given day, we can always say Diya eats at least as much as Imran. And then we can say Sal eats at least as much, not just to repeat those words, as Diya. And so we could set up a little inequality here. On a given day, we could write that Imran's calories on a given day are going to be less than or equal to Diya's calories on that same day, which is going to be less than or equal to Sal's calories on that same day. Now let's say that it's Tuesday. Let's say on Tuesday, you find out that Imran ate 1,500 calories."}, {"video_title": "Squeeze theorem or sandwich theorem Limits Differential Calculus Khan Academy.mp3", "Sentence": "And so we could set up a little inequality here. On a given day, we could write that Imran's calories on a given day are going to be less than or equal to Diya's calories on that same day, which is going to be less than or equal to Sal's calories on that same day. Now let's say that it's Tuesday. Let's say on Tuesday, you find out that Imran ate 1,500 calories. And on that same day, Sal also ate 1,500 calories. So based on this, how many calories must Diya have eaten that day? Well, she always eats at least as many as Imran's."}, {"video_title": "Squeeze theorem or sandwich theorem Limits Differential Calculus Khan Academy.mp3", "Sentence": "Let's say on Tuesday, you find out that Imran ate 1,500 calories. And on that same day, Sal also ate 1,500 calories. So based on this, how many calories must Diya have eaten that day? Well, she always eats at least as many as Imran's. So she ate 1,500 calories or more. But she always has less than or equal to the number of calories Sal eats. So it must be less than or equal to 1,500."}, {"video_title": "Squeeze theorem or sandwich theorem Limits Differential Calculus Khan Academy.mp3", "Sentence": "Well, she always eats at least as many as Imran's. So she ate 1,500 calories or more. But she always has less than or equal to the number of calories Sal eats. So it must be less than or equal to 1,500. Well, there's only one number that is greater than or equal to 1,500 and less than or equal to 1,500, and that is 1,500 calories. So Diya must have eaten 1,500 calories. This is common sense."}, {"video_title": "Squeeze theorem or sandwich theorem Limits Differential Calculus Khan Academy.mp3", "Sentence": "So it must be less than or equal to 1,500. Well, there's only one number that is greater than or equal to 1,500 and less than or equal to 1,500, and that is 1,500 calories. So Diya must have eaten 1,500 calories. This is common sense. So this Diya must have had 1,500 calories. And the squeeze theorem is essentially the mathematical version of this for functions. And you could even view this is Imran's calories as a function of the day, Sal's calories as a function of the day, and Diya's calories as a function of the day is always going to be in between those."}, {"video_title": "Squeeze theorem or sandwich theorem Limits Differential Calculus Khan Academy.mp3", "Sentence": "This is common sense. So this Diya must have had 1,500 calories. And the squeeze theorem is essentially the mathematical version of this for functions. And you could even view this is Imran's calories as a function of the day, Sal's calories as a function of the day, and Diya's calories as a function of the day is always going to be in between those. So now let's make this a little bit more mathematical. So let me clear this out so we can have some space to do some math in. So let's say that we have the same analogy."}, {"video_title": "Squeeze theorem or sandwich theorem Limits Differential Calculus Khan Academy.mp3", "Sentence": "And you could even view this is Imran's calories as a function of the day, Sal's calories as a function of the day, and Diya's calories as a function of the day is always going to be in between those. So now let's make this a little bit more mathematical. So let me clear this out so we can have some space to do some math in. So let's say that we have the same analogy. So let's say that we have three functions. Let's say f of x over some interval is always less than or equal to g of x over that same interval, which is always less than or equal to h of x over that same interval. So let me depict this graphically."}, {"video_title": "Squeeze theorem or sandwich theorem Limits Differential Calculus Khan Academy.mp3", "Sentence": "So let's say that we have the same analogy. So let's say that we have three functions. Let's say f of x over some interval is always less than or equal to g of x over that same interval, which is always less than or equal to h of x over that same interval. So let me depict this graphically. So let's depict it graphically right over here. So that is my y-axis. This is my x-axis."}, {"video_title": "Squeeze theorem or sandwich theorem Limits Differential Calculus Khan Academy.mp3", "Sentence": "So let me depict this graphically. So let's depict it graphically right over here. So that is my y-axis. This is my x-axis. And I'll just depict some interval in the x-axis right over here. So let's say h of x looks something like that. Let me make it more interesting."}, {"video_title": "Squeeze theorem or sandwich theorem Limits Differential Calculus Khan Academy.mp3", "Sentence": "This is my x-axis. And I'll just depict some interval in the x-axis right over here. So let's say h of x looks something like that. Let me make it more interesting. Whoops. This is the x-axis. So let's say h of x looks something like this."}, {"video_title": "Squeeze theorem or sandwich theorem Limits Differential Calculus Khan Academy.mp3", "Sentence": "Let me make it more interesting. Whoops. This is the x-axis. So let's say h of x looks something like this. So that's my h of x. Let's say f of x looks something like this. Maybe it does some interesting things."}, {"video_title": "Squeeze theorem or sandwich theorem Limits Differential Calculus Khan Academy.mp3", "Sentence": "So let's say h of x looks something like this. So that's my h of x. Let's say f of x looks something like this. Maybe it does some interesting things. And then it comes in, and then it goes up like this. So f of x looks something like that. And then g of x, for any x value, g of x is always in between these two."}, {"video_title": "Squeeze theorem or sandwich theorem Limits Differential Calculus Khan Academy.mp3", "Sentence": "Maybe it does some interesting things. And then it comes in, and then it goes up like this. So f of x looks something like that. And then g of x, for any x value, g of x is always in between these two. So g of x is always in between this. And I think you see where the squeeze is happening and where the sandwich is happening. So this looks like if h of x and f of x were bendy pieces of bread, g of x would be the meat of the bread."}, {"video_title": "Squeeze theorem or sandwich theorem Limits Differential Calculus Khan Academy.mp3", "Sentence": "And then g of x, for any x value, g of x is always in between these two. So g of x is always in between this. And I think you see where the squeeze is happening and where the sandwich is happening. So this looks like if h of x and f of x were bendy pieces of bread, g of x would be the meat of the bread. So it would look something like this. Now let's say that we know this is analogous to saying on a particular day, Sal and Imran ate the same amount. Let's say for a particular x value, the limit as f and h approach that x value, they approach the same limit."}, {"video_title": "Squeeze theorem or sandwich theorem Limits Differential Calculus Khan Academy.mp3", "Sentence": "So this looks like if h of x and f of x were bendy pieces of bread, g of x would be the meat of the bread. So it would look something like this. Now let's say that we know this is analogous to saying on a particular day, Sal and Imran ate the same amount. Let's say for a particular x value, the limit as f and h approach that x value, they approach the same limit. So let's take this x value right over here. Let's say the x value is c right over there. And let's say that the limit of f of x as x approaches c is equal to L. And let's say that the limit as x approaches c of h of x is also equal to L. So notice, as x approaches c, h of x approaches L. As x approaches c from either side, f of x approaches L. So these limits have to be defined."}, {"video_title": "Squeeze theorem or sandwich theorem Limits Differential Calculus Khan Academy.mp3", "Sentence": "Let's say for a particular x value, the limit as f and h approach that x value, they approach the same limit. So let's take this x value right over here. Let's say the x value is c right over there. And let's say that the limit of f of x as x approaches c is equal to L. And let's say that the limit as x approaches c of h of x is also equal to L. So notice, as x approaches c, h of x approaches L. As x approaches c from either side, f of x approaches L. So these limits have to be defined. Actually, the functions don't have to be defined at x approaches c. Just over this interval, they have to be defined as we approach it. But over this interval, this has to be true. And if these limits right over here are defined, because we know that g of x is always sandwiched in between these two functions, therefore, on that day, or for that x value, I should get out of that food eating analogy, this tells us, if all of this is true over this interval, this tells us that the limit as x approaches c of g of x must also be equal to L. And once again, this is common sense."}, {"video_title": "Squeeze theorem or sandwich theorem Limits Differential Calculus Khan Academy.mp3", "Sentence": "And let's say that the limit of f of x as x approaches c is equal to L. And let's say that the limit as x approaches c of h of x is also equal to L. So notice, as x approaches c, h of x approaches L. As x approaches c from either side, f of x approaches L. So these limits have to be defined. Actually, the functions don't have to be defined at x approaches c. Just over this interval, they have to be defined as we approach it. But over this interval, this has to be true. And if these limits right over here are defined, because we know that g of x is always sandwiched in between these two functions, therefore, on that day, or for that x value, I should get out of that food eating analogy, this tells us, if all of this is true over this interval, this tells us that the limit as x approaches c of g of x must also be equal to L. And once again, this is common sense. f of x is approaching L. h of x is approaching L. g of x is sandwiched in between it. So it also has to be approaching L. And you might say, well, this is common sense. Why is this useful?"}, {"video_title": "Integration by parts definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "What we're gonna do in this video is try to evaluate the definite integral from zero to pi of x cosine of x dx. Like always, pause this video and see if you can evaluate it yourself. Well, when you immediately look at this, it's not obvious how you just straight up take the antiderivative here and then evaluate that at pi and then subtract from that and evaluate it at zero. So we're probably going to have to use a slightly more sophisticated technique. And in general, if you see a product of functions right over here, and if one of these functions is fairly straightforward to take the antiderivative of without making it more complicated, like cosine of x, and another of the functions, like x, if you were to take its derivative, it gets simpler. In this case, it would just become one. It's a pretty good sign that we should be using integration by parts."}, {"video_title": "Integration by parts definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So we're probably going to have to use a slightly more sophisticated technique. And in general, if you see a product of functions right over here, and if one of these functions is fairly straightforward to take the antiderivative of without making it more complicated, like cosine of x, and another of the functions, like x, if you were to take its derivative, it gets simpler. In this case, it would just become one. It's a pretty good sign that we should be using integration by parts. So let's just remind ourselves about integration by parts. So integration by parts, I'll do it right over here. If I have the integral, and I'll just write this as an indefinite integral, but here we want to take the indefinite integral and then evaluate it at pi and evaluate it at zero."}, {"video_title": "Integration by parts definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "It's a pretty good sign that we should be using integration by parts. So let's just remind ourselves about integration by parts. So integration by parts, I'll do it right over here. If I have the integral, and I'll just write this as an indefinite integral, but here we want to take the indefinite integral and then evaluate it at pi and evaluate it at zero. So if I have f of x times g prime of x dx, this is going to be equal to, and in other videos we prove this. It really just comes straight out of the product rule that you learned in differential calculus. This is going to be equal to f of x times g of x minus, you then swap these around, minus f prime of x g of x dx."}, {"video_title": "Integration by parts definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "If I have the integral, and I'll just write this as an indefinite integral, but here we want to take the indefinite integral and then evaluate it at pi and evaluate it at zero. So if I have f of x times g prime of x dx, this is going to be equal to, and in other videos we prove this. It really just comes straight out of the product rule that you learned in differential calculus. This is going to be equal to f of x times g of x minus, you then swap these around, minus f prime of x g of x dx. And just to reiterate what I said before, you want to find an f of x that when I take its derivative, it simplifies it. So simplify. And you want to find a g prime of x that when I take its antiderivative, so when I take its antiderivative, it doesn't get more complicated."}, {"video_title": "Integration by parts definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "This is going to be equal to f of x times g of x minus, you then swap these around, minus f prime of x g of x dx. And just to reiterate what I said before, you want to find an f of x that when I take its derivative, it simplifies it. So simplify. And you want to find a g prime of x that when I take its antiderivative, so when I take its antiderivative, it doesn't get more complicated. So not more complicated. Because if the f of x gets simplified when I take its derivative, and the g prime of x does not get more complicated when I take its antiderivative, then this expression will maybe be easier to find the antiderivative of. So let's do that over here."}, {"video_title": "Integration by parts definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "And you want to find a g prime of x that when I take its antiderivative, so when I take its antiderivative, it doesn't get more complicated. So not more complicated. Because if the f of x gets simplified when I take its derivative, and the g prime of x does not get more complicated when I take its antiderivative, then this expression will maybe be easier to find the antiderivative of. So let's do that over here. Between x and cosine of x, which one gets more simple when I take its derivative? Well, the derivative of x is just one. So I'm gonna make that my f of x."}, {"video_title": "Integration by parts definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So let's do that over here. Between x and cosine of x, which one gets more simple when I take its derivative? Well, the derivative of x is just one. So I'm gonna make that my f of x. So I could write that over here. So my f of x, I will say is x, in which case f prime of x, f prime of x is going to be equal to one. And then what would my g prime of x be?"}, {"video_title": "Integration by parts definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So I'm gonna make that my f of x. So I could write that over here. So my f of x, I will say is x, in which case f prime of x, f prime of x is going to be equal to one. And then what would my g prime of x be? Well, my g prime of x, cosine of x, if I take its antiderivative, it doesn't get more complicated. The antiderivative of cosine of x is sine of x. So let me make that my g prime of x."}, {"video_title": "Integration by parts definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "And then what would my g prime of x be? Well, my g prime of x, cosine of x, if I take its antiderivative, it doesn't get more complicated. The antiderivative of cosine of x is sine of x. So let me make that my g prime of x. So g prime of x is equal to cosine of x, in which case g of x, the antiderivative of cosine of x, well, it's just sine of x. Or the other way to think about it, the derivative of sine of x is cosine of x. Now you could think about plus c's and all of that, but remember, this is going to be a definite integral, so all of those arbitrary constants are going to get canceled out."}, {"video_title": "Integration by parts definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So let me make that my g prime of x. So g prime of x is equal to cosine of x, in which case g of x, the antiderivative of cosine of x, well, it's just sine of x. Or the other way to think about it, the derivative of sine of x is cosine of x. Now you could think about plus c's and all of that, but remember, this is going to be a definite integral, so all of those arbitrary constants are going to get canceled out. So now let's think through this. Let's just apply the integration by parts here. In this particular case, all of this is going to be equal to, so I'm saying that is equal to this."}, {"video_title": "Integration by parts definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "Now you could think about plus c's and all of that, but remember, this is going to be a definite integral, so all of those arbitrary constants are going to get canceled out. So now let's think through this. Let's just apply the integration by parts here. In this particular case, all of this is going to be equal to, so I'm saying that is equal to this. I'm gonna skip down here. It's going to be equal to f of x times g of x. So that is f of x is x, g of x is sine of x. F of x times g of x minus the integral of f prime of x. F prime of x is just one."}, {"video_title": "Integration by parts definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "In this particular case, all of this is going to be equal to, so I'm saying that is equal to this. I'm gonna skip down here. It's going to be equal to f of x times g of x. So that is f of x is x, g of x is sine of x. F of x times g of x minus the integral of f prime of x. F prime of x is just one. We could write it like that. One times g of x. G of x is sine of x. So we could write it like this, but one times sine of x, well, we could just rewrite that as sine of x."}, {"video_title": "Integration by parts definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So that is f of x is x, g of x is sine of x. F of x times g of x minus the integral of f prime of x. F prime of x is just one. We could write it like that. One times g of x. G of x is sine of x. So we could write it like this, but one times sine of x, well, we could just rewrite that as sine of x. It'll make it a little bit simpler. Sine of x dx, dx. And then remember, this is a definite integral, so we are going to want to evaluate this whole thing at pi and at zero, and then take the difference between the two."}, {"video_title": "Integration by parts definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So we could write it like this, but one times sine of x, well, we could just rewrite that as sine of x. It'll make it a little bit simpler. Sine of x dx, dx. And then remember, this is a definite integral, so we are going to want to evaluate this whole thing at pi and at zero, and then take the difference between the two. So what is the indefinite integral of sine of x dx? Well, if I were to say the antiderivative of it. We know that the derivative of cosine is negative sine of x."}, {"video_title": "Integration by parts definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "And then remember, this is a definite integral, so we are going to want to evaluate this whole thing at pi and at zero, and then take the difference between the two. So what is the indefinite integral of sine of x dx? Well, if I were to say the antiderivative of it. We know that the derivative of cosine is negative sine of x. And so in fact, what we want, we could bring this negative sine into the integral, so we could say plus the integral of negative sine of x. Now this, clearly the antiderivative here is cosine of x. So this thing is going to be cosine of x."}, {"video_title": "Integration by parts definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "We know that the derivative of cosine is negative sine of x. And so in fact, what we want, we could bring this negative sine into the integral, so we could say plus the integral of negative sine of x. Now this, clearly the antiderivative here is cosine of x. So this thing is going to be cosine of x. And now we just have to evaluate it at the endpoints. So let's first evaluate this whole thing at pi. So this is going to be equal to pi sine of pi, pi sine of pi, plus cosine of pi."}, {"video_title": "Integration by parts definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So this thing is going to be cosine of x. And now we just have to evaluate it at the endpoints. So let's first evaluate this whole thing at pi. So this is going to be equal to pi sine of pi, pi sine of pi, plus cosine of pi. And then from that, I'm going to subtract this whole thing evaluated at zero. So let me do zero in a different color. At zero."}, {"video_title": "Integration by parts definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So this is going to be equal to pi sine of pi, pi sine of pi, plus cosine of pi. And then from that, I'm going to subtract this whole thing evaluated at zero. So let me do zero in a different color. At zero. So it's going to be zero times sine of zero plus cosine of zero. Zero. So let's see."}, {"video_title": "Integration by parts definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "At zero. So it's going to be zero times sine of zero plus cosine of zero. Zero. So let's see. Sine of pi is just zero. So this is just going to cancel out. Cosine of pi, that is negative one."}, {"video_title": "Integration by parts definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So let's see. Sine of pi is just zero. So this is just going to cancel out. Cosine of pi, that is negative one. And then this is zero. And then cosine of zero, that is one. So you have negative one minus one."}, {"video_title": "Integration by parts definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "Cosine of pi, that is negative one. And then this is zero. And then cosine of zero, that is one. So you have negative one minus one. So this all gets us to negative two. And we are done. Using integration by parts, we were able to evaluate this definite integral."}, {"video_title": "Worked example alternating series Series AP Calculus BC Khan Academy.mp3", "Sentence": "This negative one to the n plus one, as n goes from one to two to three, this is just going to alternate between positive one, negative one, positive one, negative one. So we're gonna have alternating signs. So that might be a little bit of a clue of what's going on. And actually, let's just write it out. This is going to be, see, when n equals one, this is going to be to the second power. So it's gonna be positive one. So it's gonna be p over six."}, {"video_title": "Worked example alternating series Series AP Calculus BC Khan Academy.mp3", "Sentence": "And actually, let's just write it out. This is going to be, see, when n equals one, this is going to be to the second power. So it's gonna be positive one. So it's gonna be p over six. And when n is two, this is going to be to the third power. So it's gonna be minus p over six squared, then plus p over six to the third power. And I could even write to the first power right over here."}, {"video_title": "Worked example alternating series Series AP Calculus BC Khan Academy.mp3", "Sentence": "So it's gonna be p over six. And when n is two, this is going to be to the third power. So it's gonna be minus p over six squared, then plus p over six to the third power. And I could even write to the first power right over here. Then minus p over six to the fourth power. And we're gonna just keep going plus, minus, on and on and on and on forever. So this is clearly, this is a classic alternating series right over here."}, {"video_title": "Worked example alternating series Series AP Calculus BC Khan Academy.mp3", "Sentence": "And I could even write to the first power right over here. Then minus p over six to the fourth power. And we're gonna just keep going plus, minus, on and on and on and on forever. So this is clearly, this is a classic alternating series right over here. And so we can actually apply our alternating series test. And our alternating series test tells us that if this part of our expression, the part that is not alternating in sign, I guess you could say, if this part of the expression is monotonically decreasing, monotonically, monotonically decreasing, which is just a fancy way of saying that each successive term is less than the term before it. And if we also know that the limit of this as n approaches infinity, that also has to be equal to zero."}, {"video_title": "Worked example alternating series Series AP Calculus BC Khan Academy.mp3", "Sentence": "So this is clearly, this is a classic alternating series right over here. And so we can actually apply our alternating series test. And our alternating series test tells us that if this part of our expression, the part that is not alternating in sign, I guess you could say, if this part of the expression is monotonically decreasing, monotonically, monotonically decreasing, which is just a fancy way of saying that each successive term is less than the term before it. And if we also know that the limit of this as n approaches infinity, that also has to be equal to zero. So the limit as n approaches infinity of p over six to the nth power also has to be equal to zero. So under what conditions is that going to be true? Well, to meet either one of those conditions, p over six has to be less than one."}, {"video_title": "Worked example alternating series Series AP Calculus BC Khan Academy.mp3", "Sentence": "And if we also know that the limit of this as n approaches infinity, that also has to be equal to zero. So the limit as n approaches infinity of p over six to the nth power also has to be equal to zero. So under what conditions is that going to be true? Well, to meet either one of those conditions, p over six has to be less than one. If p over six was equal to one, if for example p was six, well then we wouldn't be monotonically decreasing. Every term here would just be one. It would be one to the one, one squared, on and on and on."}, {"video_title": "Worked example alternating series Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, to meet either one of those conditions, p over six has to be less than one. If p over six was equal to one, if for example p was six, well then we wouldn't be monotonically decreasing. Every term here would just be one. It would be one to the one, one squared, on and on and on. And if p is greater than six, well then every time we multiply by p over six again, we would get a larger number over and over again. And the limit for sure would not be equal to zero. So we could say p over six needs to be less than one."}, {"video_title": "Worked example alternating series Series AP Calculus BC Khan Academy.mp3", "Sentence": "It would be one to the one, one squared, on and on and on. And if p is greater than six, well then every time we multiply by p over six again, we would get a larger number over and over again. And the limit for sure would not be equal to zero. So we could say p over six needs to be less than one. And so multiply both sides by six and you get p needs to be less than six. And they told us, for what are all the positive, positive values of p? So we also know that p has to be greater than zero."}, {"video_title": "Worked example alternating series Series AP Calculus BC Khan Academy.mp3", "Sentence": "So we could say p over six needs to be less than one. And so multiply both sides by six and you get p needs to be less than six. And they told us, for what are all the positive, positive values of p? So we also know that p has to be greater than zero. So p is greater than zero and less than six, which is that choice right over here. Once again, we're not gonna say less than or equal to six because if p was equal to six, this term is gonna be one to the n. And so we're just gonna have, this would be one, this would be one. It'd be one minus one plus one, on and on and on and on forever."}, {"video_title": "Integrating power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "And like always, if you feel inspired, and I encourage you to feel inspired, pause the video and see if you can work through this on your own, or at any time while I'm working through it, pause it and try to keep on going. All right, well, let's just rewrite this a little bit. This is going to be the same thing as the integral from zero to one. F of x is this series, so I could write the sum from n equals one to infinity of n plus one over four to the n plus one times x to the n. And now what I'm about to do might be the thing that might be new to some of you, but this is essentially, we're taking a definite integral of a sum of terms. And that's the same thing as taking the sum of a bunch of definite integrals. Let me make that clear. So if I had a, let's say this is a definite integral zero to one, and let's say I had a bunch of terms here."}, {"video_title": "Integrating power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "F of x is this series, so I could write the sum from n equals one to infinity of n plus one over four to the n plus one times x to the n. And now what I'm about to do might be the thing that might be new to some of you, but this is essentially, we're taking a definite integral of a sum of terms. And that's the same thing as taking the sum of a bunch of definite integrals. Let me make that clear. So if I had a, let's say this is a definite integral zero to one, and let's say I had a bunch of terms here. I could even call them functions. Let's say it was g of x plus h of x, and I just kept going on and on and on, dx. Well, this is the same thing as the sum of the integrals."}, {"video_title": "Integrating power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "So if I had a, let's say this is a definite integral zero to one, and let's say I had a bunch of terms here. I could even call them functions. Let's say it was g of x plus h of x, and I just kept going on and on and on, dx. Well, this is the same thing as the sum of the integrals. This is the integral from zero to one of g of x. g of x dx plus the integral from zero to one, h of x dx plus, and we go on and on and on forever, however many of these terms are. This comes straight out of our integration properties. So we can do the exact same thing here, although we'll just do it with the sigma notation."}, {"video_title": "Integrating power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, this is the same thing as the sum of the integrals. This is the integral from zero to one of g of x. g of x dx plus the integral from zero to one, h of x dx plus, and we go on and on and on forever, however many of these terms are. This comes straight out of our integration properties. So we can do the exact same thing here, although we'll just do it with the sigma notation. This is going to be equal to, this is going to be equal to the sum from n equals one to infinity of the integral, the definite integral of each of these terms. So I'm gonna write it like this. So of the integral from zero to one of n plus one over four to the n plus oneth power times x to the n, and then it is dx."}, {"video_title": "Integrating power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "So we can do the exact same thing here, although we'll just do it with the sigma notation. This is going to be equal to, this is going to be equal to the sum from n equals one to infinity of the integral, the definite integral of each of these terms. So I'm gonna write it like this. So of the integral from zero to one of n plus one over four to the n plus oneth power times x to the n, and then it is dx. So once again, now we're taking the sum of each of these terms. So let's evaluate, let's evaluate this business right over here. So that is going to, I'll just keep writing it out."}, {"video_title": "Integrating power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "So of the integral from zero to one of n plus one over four to the n plus oneth power times x to the n, and then it is dx. So once again, now we're taking the sum of each of these terms. So let's evaluate, let's evaluate this business right over here. So that is going to, I'll just keep writing it out. This is going to be equal to the sum from n equals one to infinity, and then the stuff that I just underlined in orange, this is going to be, let's see, we take the antiderivative here. We are going to get to x to the n plus one, and then we're gonna divide by n plus one. So we have this original n plus one over four to the n plus one, and that's just a constant when we think in terms of x for any one of these terms."}, {"video_title": "Integrating power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "So that is going to, I'll just keep writing it out. This is going to be equal to the sum from n equals one to infinity, and then the stuff that I just underlined in orange, this is going to be, let's see, we take the antiderivative here. We are going to get to x to the n plus one, and then we're gonna divide by n plus one. So we have this original n plus one over four to the n plus one, and that's just a constant when we think in terms of x for any one of these terms. And then here we'd wanna increment the exponent and then divide by that incremented exponent. This just comes out of, I often call it the inverse, or the anti-power rule, or reversing the power rule. So it's x to the n plus one over n plus one."}, {"video_title": "Integrating power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "So we have this original n plus one over four to the n plus one, and that's just a constant when we think in terms of x for any one of these terms. And then here we'd wanna increment the exponent and then divide by that incremented exponent. This just comes out of, I often call it the inverse, or the anti-power rule, or reversing the power rule. So it's x to the n plus one over n plus one. I just took the antiderivative, and we're gonna go from zero to one for each of these terms. And before we do that, we can simplify. We have an n plus one, we have an n plus one."}, {"video_title": "Integrating power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "So it's x to the n plus one over n plus one. I just took the antiderivative, and we're gonna go from zero to one for each of these terms. And before we do that, we can simplify. We have an n plus one, we have an n plus one. And so we can rewrite all of this. This is going to be the same thing. We're gonna take the sum from n equals one to infinity."}, {"video_title": "Integrating power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "We have an n plus one, we have an n plus one. And so we can rewrite all of this. This is going to be the same thing. We're gonna take the sum from n equals one to infinity. And this is going to be what we have in here when x is equal to one, it is one, we could write one to the n plus one over four to the n plus one. Actually, why don't I write it that way? One to the n plus one over four to the n plus one minus zero to the n plus one over four to the n plus one."}, {"video_title": "Integrating power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "We're gonna take the sum from n equals one to infinity. And this is going to be what we have in here when x is equal to one, it is one, we could write one to the n plus one over four to the n plus one. Actually, why don't I write it that way? One to the n plus one over four to the n plus one minus zero to the n plus one over four to the n plus one. So we're not gonna even have to write that. I could write zero to the n plus one over four to the n plus one, but this is clearly just zero. And then this, and this is starting to get nice and simple now."}, {"video_title": "Integrating power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "One to the n plus one over four to the n plus one minus zero to the n plus one over four to the n plus one. So we're not gonna even have to write that. I could write zero to the n plus one over four to the n plus one, but this is clearly just zero. And then this, and this is starting to get nice and simple now. This is going to be the same thing. This is equal to the sum from n equals one to infinity. We almost are gonna get to our drum roll of 1 1\u2074 to the n plus one."}, {"video_title": "Integrating power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "And then this, and this is starting to get nice and simple now. This is going to be the same thing. This is equal to the sum from n equals one to infinity. We almost are gonna get to our drum roll of 1 1\u2074 to the n plus one. Now you might immediately recognize this. This is an infinite geometric series. What is the first term here?"}, {"video_title": "Integrating power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "We almost are gonna get to our drum roll of 1 1\u2074 to the n plus one. Now you might immediately recognize this. This is an infinite geometric series. What is the first term here? Well, the first term, first, first term is, well, when n is equal to, when n is equal to one, the first term here is 1 1\u2074 to the second power. Did I do that right? Yeah, when n is equal to one, it's going to be, so this is going to be 1\u2074 to the second power, which is equal to one over 16."}, {"video_title": "Integrating power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "What is the first term here? Well, the first term, first, first term is, well, when n is equal to, when n is equal to one, the first term here is 1 1\u2074 to the second power. Did I do that right? Yeah, when n is equal to one, it's going to be, so this is going to be 1\u2074 to the second power, which is equal to one over 16. So that's our first term. And then our common ratio, common, common ratio here. Well, that's gonna be, well, we're gonna, we're just gonna keep multiplying by 1\u2074."}, {"video_title": "Integrating power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "Yeah, when n is equal to one, it's going to be, so this is going to be 1\u2074 to the second power, which is equal to one over 16. So that's our first term. And then our common ratio, common, common ratio here. Well, that's gonna be, well, we're gonna, we're just gonna keep multiplying by 1\u2074. So our common ratio here is 1\u2074. And so for an infinite geometric series, this is, since our common ratio, well, is less than, or its absolute value is less than one, we know that this is going to converge, and it's gonna converge to the value, our first term, one over 16, divided by one minus the common ratio, one minus 1\u2074. So this is 3\u2074, so it's equal to one over 16 times four over three, so one, four."}, {"video_title": "Integrating power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, that's gonna be, well, we're gonna, we're just gonna keep multiplying by 1\u2074. So our common ratio here is 1\u2074. And so for an infinite geometric series, this is, since our common ratio, well, is less than, or its absolute value is less than one, we know that this is going to converge, and it's gonna converge to the value, our first term, one over 16, divided by one minus the common ratio, one minus 1\u2074. So this is 3\u2074, so it's equal to one over 16 times four over three, so one, four. This is going to be equal to 1\u204412. And we're done. And this seemed really daunting at first, but we just have to realize, okay, an integral of a sum, even an infinite sum, well, that's gonna be the sum of these infinite integrals."}, {"video_title": "Integrating power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "So this is 3\u2074, so it's equal to one over 16 times four over three, so one, four. This is going to be equal to 1\u204412. And we're done. And this seemed really daunting at first, but we just have to realize, okay, an integral of a sum, even an infinite sum, well, that's gonna be the sum of these infinite integrals. We take the antiderivative of these infinite integrals, which we were able to do, which is kind of a cool thing, one of the powers of symbolic mathematics. And then we realize, oh, we just have an infinite geometric series, which we know how to find the sum of. And we're done."}, {"video_title": "Basic derivative rules find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Here is her work. And on the right-hand side, it says, Hannah tried to find the derivative of negative three plus eight x using basic differentiation rules. Here is her work. And these are two different examples from differentiation rules exercise on Khan Academy. And I thought I would just do them side by side because we can kind of think about what each of these people are doing correct or incorrect. So these are similar expressions. We have a constant, and then we have a first-degree term, a constant, and then a first-degree term."}, {"video_title": "Basic derivative rules find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And these are two different examples from differentiation rules exercise on Khan Academy. And I thought I would just do them side by side because we can kind of think about what each of these people are doing correct or incorrect. So these are similar expressions. We have a constant, and then we have a first-degree term, a constant, and then a first-degree term. So they're gonna take the derivative. So let's see, step one for Avery. She took, she's separately taking the derivative of seven and separately taking the derivative of five x."}, {"video_title": "Basic derivative rules find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "We have a constant, and then we have a first-degree term, a constant, and then a first-degree term. So they're gonna take the derivative. So let's see, step one for Avery. She took, she's separately taking the derivative of seven and separately taking the derivative of five x. So this is, my spider sense is already going off here. Because what happened to this negative right over here? So it would have made sense for her to do the derivative of seven, and she could have said minus the derivative of five x."}, {"video_title": "Basic derivative rules find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "She took, she's separately taking the derivative of seven and separately taking the derivative of five x. So this is, my spider sense is already going off here. Because what happened to this negative right over here? So it would have made sense for her to do the derivative of seven, and she could have said minus the derivative of five x. That's one possibility that she could have done. The derivative of a difference is equal to the difference of the derivatives. We've seen that property."}, {"video_title": "Basic derivative rules find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So it would have made sense for her to do the derivative of seven, and she could have said minus the derivative of five x. That's one possibility that she could have done. The derivative of a difference is equal to the difference of the derivatives. We've seen that property. Or she could have said the derivative, she could have said this is equal to the derivative of seven plus the derivative with respect to x of negative five x. These two things would have been equivalent to this one. But for this one, she somehow forgot to include the negative."}, {"video_title": "Basic derivative rules find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "We've seen that property. Or she could have said the derivative, she could have said this is equal to the derivative of seven plus the derivative with respect to x of negative five x. These two things would have been equivalent to this one. But for this one, she somehow forgot to include the negative. So I think she had a problem right at step one. Now if you just follow her logic after step one, let's see if she makes any more mistakes. So she takes the derivative of a constant."}, {"video_title": "Basic derivative rules find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "But for this one, she somehow forgot to include the negative. So I think she had a problem right at step one. Now if you just follow her logic after step one, let's see if she makes any more mistakes. So she takes the derivative of a constant. So a constant isn't going to change with respect to x. So that makes sense that that derivative is zero. And so we still have the derivative of five x."}, {"video_title": "Basic derivative rules find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So she takes the derivative of a constant. So a constant isn't going to change with respect to x. So that makes sense that that derivative is zero. And so we still have the derivative of five x. And remember, it should have been negative five x or minus the derivative of five x. And let's see what she does here. So that zero disappears, and now she takes the constant out."}, {"video_title": "Basic derivative rules find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And so we still have the derivative of five x. And remember, it should have been negative five x or minus the derivative of five x. And let's see what she does here. So that zero disappears, and now she takes the constant out. And that's true. The derivative of a constant times something is equal to the constant times the derivative of that something. And then she finds that the derivative with respect to x of x is one."}, {"video_title": "Basic derivative rules find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So that zero disappears, and now she takes the constant out. And that's true. The derivative of a constant times something is equal to the constant times the derivative of that something. And then she finds that the derivative with respect to x of x is one. And that's true. So the slope, if you had the graph of y equals x, the slope there is one. Or what's the rate of change at which x changes with respect to x?"}, {"video_title": "Basic derivative rules find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And then she finds that the derivative with respect to x of x is one. And that's true. So the slope, if you had the graph of y equals x, the slope there is one. Or what's the rate of change at which x changes with respect to x? Well, it's going to be one for one. So the slope here is one. So this is going to be five times one, which is equal to five."}, {"video_title": "Basic derivative rules find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Or what's the rate of change at which x changes with respect to x? Well, it's going to be one for one. So the slope here is one. So this is going to be five times one, which is equal to five. And at the end, they just say, what step did Avery make a mistake? So she clearly made a mistake at step one. This right over here should have been a negative."}, {"video_title": "Basic derivative rules find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be five times one, which is equal to five. And at the end, they just say, what step did Avery make a mistake? So she clearly made a mistake at step one. This right over here should have been a negative. If that's a negative, then that would have been a negative. Then this would have been a negative. Then that would have been a negative."}, {"video_title": "Basic derivative rules find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "This right over here should have been a negative. If that's a negative, then that would have been a negative. Then this would have been a negative. Then that would have been a negative. And then her final answer should have been, should have been a negative five. Now let's go back to Hannah to see if she made any mistakes and where. So she's differentiating a similar expression."}, {"video_title": "Basic derivative rules find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Then that would have been a negative. And then her final answer should have been, should have been a negative five. Now let's go back to Hannah to see if she made any mistakes and where. So she's differentiating a similar expression. So first she takes the derivative of the constant plus the derivative of the first degree term. Derivative of a constant is zero. That looks good."}, {"video_title": "Basic derivative rules find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So she's differentiating a similar expression. So first she takes the derivative of the constant plus the derivative of the first degree term. Derivative of a constant is zero. That looks good. So you get the zero. And then you have the derivative of the first degree term. That's what she's trying to figure out."}, {"video_title": "Basic derivative rules find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "That looks good. So you get the zero. And then you have the derivative of the first degree term. That's what she's trying to figure out. And then, let's see. She's taking, let's see. So this seems off."}, {"video_title": "Basic derivative rules find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "That's what she's trying to figure out. And then, let's see. She's taking, let's see. So this seems off. She is assuming that the derivative of a product is equal to the product of the derivatives. That is not the case. And especially, and it's, if you have a constant here, there's actually a much simpler way of thinking about it, frankly, the way that Avery thought about it."}, {"video_title": "Basic derivative rules find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So this seems off. She is assuming that the derivative of a product is equal to the product of the derivatives. That is not the case. And especially, and it's, if you have a constant here, there's actually a much simpler way of thinking about it, frankly, the way that Avery thought about it. Avery had made a mistake at step one. But this is actually going to be equal to the derivative of a constant times an expression is equal to the same thing as the constant times the derivative of the expression. So this would have been the correct way to go."}, {"video_title": "Basic derivative rules find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And especially, and it's, if you have a constant here, there's actually a much simpler way of thinking about it, frankly, the way that Avery thought about it. Avery had made a mistake at step one. But this is actually going to be equal to the derivative of a constant times an expression is equal to the same thing as the constant times the derivative of the expression. So this would have been the correct way to go. And then the derivative of x with respect to x, well, that's just going to be one. So this should have all simplified to eight. What she did is, she is assumed, she tried to take the derivative of eight and multiply that times the derivative of x."}, {"video_title": "Basic derivative rules find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So this would have been the correct way to go. And then the derivative of x with respect to x, well, that's just going to be one. So this should have all simplified to eight. What she did is, she is assumed, she tried to take the derivative of eight and multiply that times the derivative of x. That is not the way it works. In the future, you will learn something called the product rule. But you won't even have to apply that here because one of these components, I guess you could say, is a constant."}, {"video_title": "Basic derivative rules find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "What she did is, she is assumed, she tried to take the derivative of eight and multiply that times the derivative of x. That is not the way it works. In the future, you will learn something called the product rule. But you won't even have to apply that here because one of these components, I guess you could say, is a constant. So this is the wrong step. This is where Hannah makes a mistake. And you can see, instead of getting a final answer of eight, she is getting a final answer of, she assumes, well, the derivative of eight is zero times the derivative of x is one, zero times one, and she gets zero, which is not the right answer."}, {"video_title": "Radical functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And at first you might say, alright, how do I take the derivative of a fourth root of something? It looks like I have a composite function. I'm taking the fourth root of another expression here, and you'd be right. And if you're dealing with a composite function, the chain rule should be front of mind. But first, let's just make this fourth root a little bit more tractable for us, and just realize that this fourth root is really nothing but a fractional exponent. So this is the same thing as the derivative with respect to x of x to the third plus four x squared plus seven to the 1 4th power. To the 1 4th power."}, {"video_title": "Radical functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And if you're dealing with a composite function, the chain rule should be front of mind. But first, let's just make this fourth root a little bit more tractable for us, and just realize that this fourth root is really nothing but a fractional exponent. So this is the same thing as the derivative with respect to x of x to the third plus four x squared plus seven to the 1 4th power. To the 1 4th power. Now, how do we take the derivative of this? Well, we can view this, as I said a few seconds ago, we can view this as a composite function. What do we do first with our x?"}, {"video_title": "Radical functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "To the 1 4th power. Now, how do we take the derivative of this? Well, we can view this, as I said a few seconds ago, we can view this as a composite function. What do we do first with our x? Well, we do all of this business. And we could call this u of x. And then, whatever we get for u of x, we raise that to the 4th power."}, {"video_title": "Radical functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "What do we do first with our x? Well, we do all of this business. And we could call this u of x. And then, whatever we get for u of x, we raise that to the 4th power. So the way that we would take the derivative, we would take the derivative of this, you could view it as the outer function with respect to u of x, and then multiply that times the derivative of u with respect to x. So let's do that. So what this is going to be, this is going to be equal to, so we're gonna take our outside function, which I'm highlighting in green now, so where I take something to the 1 4th, I'm gonna take the derivative of that with respect to the inside, with respect to u of x."}, {"video_title": "Radical functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And then, whatever we get for u of x, we raise that to the 4th power. So the way that we would take the derivative, we would take the derivative of this, you could view it as the outer function with respect to u of x, and then multiply that times the derivative of u with respect to x. So let's do that. So what this is going to be, this is going to be equal to, so we're gonna take our outside function, which I'm highlighting in green now, so where I take something to the 1 4th, I'm gonna take the derivative of that with respect to the inside, with respect to u of x. Well, I'm just gonna use the power rule here. I'm just gonna bring that 1 4th out front. So it's going to be 1 4th times whatever I'm taking the derivative with respect to, to the 1 4th minus one power."}, {"video_title": "Radical functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So what this is going to be, this is going to be equal to, so we're gonna take our outside function, which I'm highlighting in green now, so where I take something to the 1 4th, I'm gonna take the derivative of that with respect to the inside, with respect to u of x. Well, I'm just gonna use the power rule here. I'm just gonna bring that 1 4th out front. So it's going to be 1 4th times whatever I'm taking the derivative with respect to, to the 1 4th minus one power. Look, all I did is use the power rule here. I didn't have an x here. Now I'm taking the derivative with respect to u of x, with respect to this polynomial expression here."}, {"video_title": "Radical functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So it's going to be 1 4th times whatever I'm taking the derivative with respect to, to the 1 4th minus one power. Look, all I did is use the power rule here. I didn't have an x here. Now I'm taking the derivative with respect to u of x, with respect to this polynomial expression here. So I could just throw the u of x in here if I like. Actually, let me just do that. So this is going to be x to the 3rd plus four x squared plus seven."}, {"video_title": "Radical functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Now I'm taking the derivative with respect to u of x, with respect to this polynomial expression here. So I could just throw the u of x in here if I like. Actually, let me just do that. So this is going to be x to the 3rd plus four x squared plus seven. And then I wanna multiply that, and this is the chain rule. I took the derivative of the outside with respect to the inside, and I'm gonna multiply that times the derivative of the inside. So what's the derivative of u of x?"}, {"video_title": "Radical functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be x to the 3rd plus four x squared plus seven. And then I wanna multiply that, and this is the chain rule. I took the derivative of the outside with respect to the inside, and I'm gonna multiply that times the derivative of the inside. So what's the derivative of u of x? u prime of x. Let's see, we're just gonna use the power rule a bunch of times. It's going to be three x squared plus two times four is eight."}, {"video_title": "Radical functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So what's the derivative of u of x? u prime of x. Let's see, we're just gonna use the power rule a bunch of times. It's going to be three x squared plus two times four is eight. X to the two minus one is just one power, first power. So that's just, I can just write that as eight x. And then the derivative with respect to x of seven, well, derivative with respect to x of a constant is just gonna be zero."}, {"video_title": "Radical functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "It's going to be three x squared plus two times four is eight. X to the two minus one is just one power, first power. So that's just, I can just write that as eight x. And then the derivative with respect to x of seven, well, derivative with respect to x of a constant is just gonna be zero. So that's u prime of x. So then I'm just gonna multiply by u prime of x, which is three x squared plus, three x squared plus eight x. And so, well, I can clean this up a little bit."}, {"video_title": "Radical functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And then the derivative with respect to x of seven, well, derivative with respect to x of a constant is just gonna be zero. So that's u prime of x. So then I'm just gonna multiply by u prime of x, which is three x squared plus, three x squared plus eight x. And so, well, I can clean this up a little bit. So this would be equal to, this would be equal to, actually, let me just rewrite that exponent there. So this 1 4th minus one, I can rewrite it, 1 4th minus one is negative 3 4ths. Negative 3 4ths."}, {"video_title": "Radical functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And so, well, I can clean this up a little bit. So this would be equal to, this would be equal to, actually, let me just rewrite that exponent there. So this 1 4th minus one, I can rewrite it, 1 4th minus one is negative 3 4ths. Negative 3 4ths. Negative 3 4ths power. And you could manipulate this in different ways if you like, but the key is to just recognize that this is an application of the chain rule. Derivative of the outside, well, actually, the first thing to realize is the fourth root is the same thing as taking something to the 1 4th power, basic exponent property."}, {"video_title": "Radical functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Negative 3 4ths. Negative 3 4ths power. And you could manipulate this in different ways if you like, but the key is to just recognize that this is an application of the chain rule. Derivative of the outside, well, actually, the first thing to realize is the fourth root is the same thing as taking something to the 1 4th power, basic exponent property. And then realize that, okay, I have a composite function here. So I can take the derivative of the outside with respect to the inside, that's what we did here, times the derivative of the inside with respect to x. And so if someone were to tell you, if someone were to say, all right, f of x, f of x is equal to the fourth root of x to the third plus four x squared plus seven."}, {"video_title": "Radical functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Derivative of the outside, well, actually, the first thing to realize is the fourth root is the same thing as taking something to the 1 4th power, basic exponent property. And then realize that, okay, I have a composite function here. So I can take the derivative of the outside with respect to the inside, that's what we did here, times the derivative of the inside with respect to x. And so if someone were to tell you, if someone were to say, all right, f of x, f of x is equal to the fourth root of x to the third plus four x squared plus seven. And then they said, well, what is f prime of, I don't know, negative three? Well, you would evaluate this at negative three. So let me just do that."}, {"video_title": "Radical functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And so if someone were to tell you, if someone were to say, all right, f of x, f of x is equal to the fourth root of x to the third plus four x squared plus seven. And then they said, well, what is f prime of, I don't know, negative three? Well, you would evaluate this at negative three. So let me just do that. So it's 1 4th times, see, you have negative 27. I hope this works out reasonably well, plus 36, plus 36, plus seven to the negative 3 4ths. What does this result to?"}, {"video_title": "Radical functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So let me just do that. So it's 1 4th times, see, you have negative 27. I hope this works out reasonably well, plus 36, plus 36, plus seven to the negative 3 4ths. What does this result to? This is going to be equal to, this right over here is 16, right? Negative 27 plus seven is negative 20 plus 36. So this is 16."}, {"video_title": "Radical functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "What does this result to? This is going to be equal to, this right over here is 16, right? Negative 27 plus seven is negative 20 plus 36. So this is 16. I think this is gonna work out nicely. And then times, three times negative, so three times nine, which is 27 minus 24. So this is going to be right over here."}, {"video_title": "Radical functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So this is 16. I think this is gonna work out nicely. And then times, three times negative, so three times nine, which is 27 minus 24. So this is going to be right over here. That is going to be three. Now what is 16 to the negative 3 4ths? So let me, this is 1 4th times, so 16 to the 1 4th is two, and then you raise that to the, let me, actually, I don't wanna skip steps here."}, {"video_title": "Radical functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be right over here. That is going to be three. Now what is 16 to the negative 3 4ths? So let me, this is 1 4th times, so 16 to the 1 4th is two, and then you raise that to the, let me, actually, I don't wanna skip steps here. But this is, at this point, we are dealing with algebra, or maybe even pre-algebra. So this is going to be times, times 16 to the 1 4th, and then we're gonna raise that to the negative three, times that three out front. So we could put that three there."}, {"video_title": "Radical functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So let me, this is 1 4th times, so 16 to the 1 4th is two, and then you raise that to the, let me, actually, I don't wanna skip steps here. But this is, at this point, we are dealing with algebra, or maybe even pre-algebra. So this is going to be times, times 16 to the 1 4th, and then we're gonna raise that to the negative three, times that three out front. So we could put that three there. 16 to the 1 4th is two. Two to the 3rd is eight. So two to the negative 3rd power is 1 8th."}, {"video_title": "Radical functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So we could put that three there. 16 to the 1 4th is two. Two to the 3rd is eight. So two to the negative 3rd power is 1 8th. So that is 1 8th. So we have 3 4ths times 1 8th, which is equal to three over 32. Three 30 seconds."}, {"video_title": "Derivatives of sec(x) and csc(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And what I want to do in this video is to keep going and find the derivatives of secant of x and cosecant of x. So let's start with secant of x. The derivative with respect to x of secant of x. Well, secant of x is the same thing as, so we're going to find the derivative with respect to x of, secant of x is the same thing as one over, one over the cosine of x. And that's just the definition of secant. And you could, there's multiple ways you could do this. When you learn the chain rule, that actually might be a more natural thing to use to evaluate the derivative here."}, {"video_title": "Derivatives of sec(x) and csc(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, secant of x is the same thing as, so we're going to find the derivative with respect to x of, secant of x is the same thing as one over, one over the cosine of x. And that's just the definition of secant. And you could, there's multiple ways you could do this. When you learn the chain rule, that actually might be a more natural thing to use to evaluate the derivative here. But we know the quotient rule, so we will apply the quotient rule here. And it's no coincidence that you get to the same answer. The quotient rule actually can be derived based on the chain rule and the product rule."}, {"video_title": "Derivatives of sec(x) and csc(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "When you learn the chain rule, that actually might be a more natural thing to use to evaluate the derivative here. But we know the quotient rule, so we will apply the quotient rule here. And it's no coincidence that you get to the same answer. The quotient rule actually can be derived based on the chain rule and the product rule. But I won't keep going into that. Let's just apply the quotient rule right over here. So this derivative is going to be equal to, it's going to be equal to the derivative of the top."}, {"video_title": "Derivatives of sec(x) and csc(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "The quotient rule actually can be derived based on the chain rule and the product rule. But I won't keep going into that. Let's just apply the quotient rule right over here. So this derivative is going to be equal to, it's going to be equal to the derivative of the top. Well, what's the derivative of one with respect to x? Well, that's just zero times the function on the bottom. So times cosine of x, cosine of x, minus, minus the function on the top, well, that's just one, times the derivative on the bottom."}, {"video_title": "Derivatives of sec(x) and csc(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So this derivative is going to be equal to, it's going to be equal to the derivative of the top. Well, what's the derivative of one with respect to x? Well, that's just zero times the function on the bottom. So times cosine of x, cosine of x, minus, minus the function on the top, well, that's just one, times the derivative on the bottom. Well, the derivative of the bottom is, derivative of cosine of x is negative sine of x. So we could put the sine of x there, but it's negative sine of x, so you have a minus and there would be a negative, so we could just make that a positive. And then all of that over the function on the bottom squared."}, {"video_title": "Derivatives of sec(x) and csc(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So times cosine of x, cosine of x, minus, minus the function on the top, well, that's just one, times the derivative on the bottom. Well, the derivative of the bottom is, derivative of cosine of x is negative sine of x. So we could put the sine of x there, but it's negative sine of x, so you have a minus and there would be a negative, so we could just make that a positive. And then all of that over the function on the bottom squared. So cosine of x squared. And so zero times cosine of x, that is just zero. And so all we are left with is sine of x over cosine of x squared."}, {"video_title": "Derivatives of sec(x) and csc(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And then all of that over the function on the bottom squared. So cosine of x squared. And so zero times cosine of x, that is just zero. And so all we are left with is sine of x over cosine of x squared. And there's multiple ways that you could rewrite this if you like. You could say that this is the same thing as sine of x over cosine of x times one over cosine of x. And of course, this is tangent of x times secant of x. Secant of x."}, {"video_title": "Derivatives of sec(x) and csc(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And so all we are left with is sine of x over cosine of x squared. And there's multiple ways that you could rewrite this if you like. You could say that this is the same thing as sine of x over cosine of x times one over cosine of x. And of course, this is tangent of x times secant of x. Secant of x. So you could say derivative of secant x is sine of x over cosine squared of x, or it is tangent of x times secant of x. So now let's do cosecant. So the derivative with respect to x of cosecant of x, well, that's the same thing as the derivative with respect to x of one over sine of x. Cosecant is one over sine of x. I remember that, because you think it's cosecant, maybe it's the reciprocal of cosine, but it's not, it's the opposite of what you would expect."}, {"video_title": "Derivatives of sec(x) and csc(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And of course, this is tangent of x times secant of x. Secant of x. So you could say derivative of secant x is sine of x over cosine squared of x, or it is tangent of x times secant of x. So now let's do cosecant. So the derivative with respect to x of cosecant of x, well, that's the same thing as the derivative with respect to x of one over sine of x. Cosecant is one over sine of x. I remember that, because you think it's cosecant, maybe it's the reciprocal of cosine, but it's not, it's the opposite of what you would expect. Cosine's reciprocal isn't cosecant, it is secant. Once again, opposite of what you would expect. That starts with an S, this starts with a C. That starts with a C, that starts with an S. Just the way that it happened to be defined."}, {"video_title": "Derivatives of sec(x) and csc(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So the derivative with respect to x of cosecant of x, well, that's the same thing as the derivative with respect to x of one over sine of x. Cosecant is one over sine of x. I remember that, because you think it's cosecant, maybe it's the reciprocal of cosine, but it's not, it's the opposite of what you would expect. Cosine's reciprocal isn't cosecant, it is secant. Once again, opposite of what you would expect. That starts with an S, this starts with a C. That starts with a C, that starts with an S. Just the way that it happened to be defined. But anyway, let's just evaluate this. Once again, we'll do the quotient rule, but you could also do this using the chain rule. So it's going to be the derivative of the expression on top, which is zero, times the expression on the bottom, which is sine of x, sine of x, minus the expression on top, which is just one, times the derivative of the expression on the bottom, which is cosine of x, all of that over the expression on the bottom squared."}, {"video_title": "Derivatives of sec(x) and csc(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "That starts with an S, this starts with a C. That starts with a C, that starts with an S. Just the way that it happened to be defined. But anyway, let's just evaluate this. Once again, we'll do the quotient rule, but you could also do this using the chain rule. So it's going to be the derivative of the expression on top, which is zero, times the expression on the bottom, which is sine of x, sine of x, minus the expression on top, which is just one, times the derivative of the expression on the bottom, which is cosine of x, all of that over the expression on the bottom squared. Sine squared of x, that's zero, so we get negative cosine of x over sine, over sine squared of x. So that's one way to think about it. Or, if you like, you could view this, the same thing we did over here, this is the same thing as negative cosine of x over sine of x, times one over sine of x."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "I've drawn an arbitrary function here. And what we're going to try to do is approximate this arbitrary function. We don't know what it is. Using a polynomial, we'll keep adding terms to that polynomial. But to do this, we're going to assume that we can evaluate the function at 0, that it gives us some value, and that we can keep taking the derivative of the function and evaluating the first, the second, and the third derivative, so on and so forth, at 0 as well. So we're assuming that we know what f of 0 is. We're assuming that we know what f prime of 0 is."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Using a polynomial, we'll keep adding terms to that polynomial. But to do this, we're going to assume that we can evaluate the function at 0, that it gives us some value, and that we can keep taking the derivative of the function and evaluating the first, the second, and the third derivative, so on and so forth, at 0 as well. So we're assuming that we know what f of 0 is. We're assuming that we know what f prime of 0 is. We're assuming that we know the second derivative at 0. We're assuming that we know the third derivative at 0. So maybe I'll write it third derivative."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "We're assuming that we know what f prime of 0 is. We're assuming that we know the second derivative at 0. We're assuming that we know the third derivative at 0. So maybe I'll write it third derivative. I'll just write f prime prime at 0, so forth, and so on. So let's think about how we can approximate this using polynomials of ever-increasing length. So we could have a polynomial of just one term."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So maybe I'll write it third derivative. I'll just write f prime prime at 0, so forth, and so on. So let's think about how we can approximate this using polynomials of ever-increasing length. So we could have a polynomial of just one term. And it would just be a constant term. So this would be a polynomial of degree 0. And if we have a constant term, we at least might want to make that constant polynomial, it really is just a constant function, equal the function at f of 0."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So we could have a polynomial of just one term. And it would just be a constant term. So this would be a polynomial of degree 0. And if we have a constant term, we at least might want to make that constant polynomial, it really is just a constant function, equal the function at f of 0. So at first, maybe we just want p of 0, where p is the polynomial that we're going to construct. We want p of 0 to be equal to f of 0. So if we want to do that using a polynomial of only one term, of only one constant term, we can just set p of x is equal to f of 0."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And if we have a constant term, we at least might want to make that constant polynomial, it really is just a constant function, equal the function at f of 0. So at first, maybe we just want p of 0, where p is the polynomial that we're going to construct. We want p of 0 to be equal to f of 0. So if we want to do that using a polynomial of only one term, of only one constant term, we can just set p of x is equal to f of 0. So if I were to graph it, it would look like this. It would just be a horizontal line at f of 0. And you could say, Sal, that's a horrible approximation."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So if we want to do that using a polynomial of only one term, of only one constant term, we can just set p of x is equal to f of 0. So if I were to graph it, it would look like this. It would just be a horizontal line at f of 0. And you could say, Sal, that's a horrible approximation. It only approximates the function at this point. Looks like we got lucky at a couple of other points. But it's really bad everywhere else."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And you could say, Sal, that's a horrible approximation. It only approximates the function at this point. Looks like we got lucky at a couple of other points. But it's really bad everywhere else. And I would tell you, well, try to do any better using a horizontal line. At least we got it right at f of 0. So this is about as good as we can do with just a constant."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "But it's really bad everywhere else. And I would tell you, well, try to do any better using a horizontal line. At least we got it right at f of 0. So this is about as good as we can do with just a constant. And even though, I just want to remind you, this might not look like a constant, but we're assuming that given the function, we can evaluate it at 0, and that'll just give us a number. So whatever number that was, we would put it right over here, and we'd say p of x is equal to that number. And it would just be a horizontal line right there at f of 0."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So this is about as good as we can do with just a constant. And even though, I just want to remind you, this might not look like a constant, but we're assuming that given the function, we can evaluate it at 0, and that'll just give us a number. So whatever number that was, we would put it right over here, and we'd say p of x is equal to that number. And it would just be a horizontal line right there at f of 0. But that, obviously, is not so great. So let's add some more constraints. Beyond the fact that we want p of 0 to be equal to f of 0, let's say that we also want p prime at 0 to be the same thing as f prime at 0."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And it would just be a horizontal line right there at f of 0. But that, obviously, is not so great. So let's add some more constraints. Beyond the fact that we want p of 0 to be equal to f of 0, let's say that we also want p prime at 0 to be the same thing as f prime at 0. Let me do this in a new color. So we also want, in the new color, we also want p prime. We want the first derivative of our polynomial, when evaluated at 0, to be the same thing as the first derivative of the function when evaluated at 0."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Beyond the fact that we want p of 0 to be equal to f of 0, let's say that we also want p prime at 0 to be the same thing as f prime at 0. Let me do this in a new color. So we also want, in the new color, we also want p prime. We want the first derivative of our polynomial, when evaluated at 0, to be the same thing as the first derivative of the function when evaluated at 0. And we don't want to lose this right over here. So what if we set p of x as being equal to f of 0. So we're taking our old p of x, but now we're going to add another term so that the derivatives match up."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "We want the first derivative of our polynomial, when evaluated at 0, to be the same thing as the first derivative of the function when evaluated at 0. And we don't want to lose this right over here. So what if we set p of x as being equal to f of 0. So we're taking our old p of x, but now we're going to add another term so that the derivatives match up. Plus f prime of 0 times x. So let's think about this a little bit. If we use this as our new polynomial, what happens?"}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So we're taking our old p of x, but now we're going to add another term so that the derivatives match up. Plus f prime of 0 times x. So let's think about this a little bit. If we use this as our new polynomial, what happens? What is p of 0? p of 0 is going to be equal to, you're going to have f of 0 plus whatever this f prime of 0 is times 0. If you put a 0 in for x, this term's just going to be 0."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "If we use this as our new polynomial, what happens? What is p of 0? p of 0 is going to be equal to, you're going to have f of 0 plus whatever this f prime of 0 is times 0. If you put a 0 in for x, this term's just going to be 0. So you're going to be left with p of 0 is equal to f of 0. That's cool. That's just as good as our first version."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "If you put a 0 in for x, this term's just going to be 0. So you're going to be left with p of 0 is equal to f of 0. That's cool. That's just as good as our first version. Now what's the derivative over here? So the derivative is p prime of x is equal to, take the derivative of this. This is just a constant."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "That's just as good as our first version. Now what's the derivative over here? So the derivative is p prime of x is equal to, take the derivative of this. This is just a constant. So its derivative is 0. The derivative of a coefficient times x is just going to be the coefficient. So it's going to be f prime of 0."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "This is just a constant. So its derivative is 0. The derivative of a coefficient times x is just going to be the coefficient. So it's going to be f prime of 0. So if you evaluate it at 0, so p prime of 0, or the derivative of our polynomial evaluated at 0. I know it's a little weird because we're doing p prime of x of f of 0 and all of this. But just remember what's the variable, what's the constant, and hopefully it'll make sense."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So it's going to be f prime of 0. So if you evaluate it at 0, so p prime of 0, or the derivative of our polynomial evaluated at 0. I know it's a little weird because we're doing p prime of x of f of 0 and all of this. But just remember what's the variable, what's the constant, and hopefully it'll make sense. So this is just obviously going to be f prime of 0. Its derivative is a constant value. This is a constant value right here."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "But just remember what's the variable, what's the constant, and hopefully it'll make sense. So this is just obviously going to be f prime of 0. Its derivative is a constant value. This is a constant value right here. We're assuming that we could take the derivative of our function and evaluate that thing at 0 to give a constant value. So if p prime of x is equal to this constant value, obviously p prime of x evaluated at 0 is going to be that value. But what's cool about this right here, this polynomial that has a 0 degree term and a first degree term, is now this polynomial is equal to our function at x is equal to 0."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "This is a constant value right here. We're assuming that we could take the derivative of our function and evaluate that thing at 0 to give a constant value. So if p prime of x is equal to this constant value, obviously p prime of x evaluated at 0 is going to be that value. But what's cool about this right here, this polynomial that has a 0 degree term and a first degree term, is now this polynomial is equal to our function at x is equal to 0. And it also has the same first derivative. It also has the same slope at x is equal to 0. So this thing will look, this new polynomial with two terms, getting a little bit better."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "But what's cool about this right here, this polynomial that has a 0 degree term and a first degree term, is now this polynomial is equal to our function at x is equal to 0. And it also has the same first derivative. It also has the same slope at x is equal to 0. So this thing will look, this new polynomial with two terms, getting a little bit better. It will look something like that. It will essentially have, it'll look like a tangent line at f of 0, at x is equal to 0. So we're doing better, but still not a super good approximation."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So this thing will look, this new polynomial with two terms, getting a little bit better. It will look something like that. It will essentially have, it'll look like a tangent line at f of 0, at x is equal to 0. So we're doing better, but still not a super good approximation. It kind of is going in the same general direction as our function around 0. But maybe we can do better by making sure that they have the same second derivative. And to try to have the same second derivative while still having the same first derivative and the same value at 0, let's try to do something interesting."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So we're doing better, but still not a super good approximation. It kind of is going in the same general direction as our function around 0. But maybe we can do better by making sure that they have the same second derivative. And to try to have the same second derivative while still having the same first derivative and the same value at 0, let's try to do something interesting. Let's define p of x. So let's make it clear. This was our first try."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And to try to have the same second derivative while still having the same first derivative and the same value at 0, let's try to do something interesting. Let's define p of x. So let's make it clear. This was our first try. This is our second try right over here. And I'm about to embark on our third try. So in our third try, my goal is that the value of my polynomial is the same as the value of the function at 0."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "This was our first try. This is our second try right over here. And I'm about to embark on our third try. So in our third try, my goal is that the value of my polynomial is the same as the value of the function at 0. They have the same derivative at 0. And they also have the same second derivative at 0. So let's define my polynomial to be equal to, so I'm going to do the first two terms of these guys right over here."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So in our third try, my goal is that the value of my polynomial is the same as the value of the function at 0. They have the same derivative at 0. And they also have the same second derivative at 0. So let's define my polynomial to be equal to, so I'm going to do the first two terms of these guys right over here. So it's going to be f of 0 plus f prime of 0 times x. So exactly what we did here. But now let me add another term."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's define my polynomial to be equal to, so I'm going to do the first two terms of these guys right over here. So it's going to be f of 0 plus f prime of 0 times x. So exactly what we did here. But now let me add another term. I'll do the other term in a new color. Plus, and I'm going to put a 1 half out here, and hopefully it might make sense why I'm about to do this, plus 1 half times f, the second derivative of our function, evaluated at 0x squared. And when we evaluate the derivative of this, I think you'll see why this 1 half is there."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "But now let me add another term. I'll do the other term in a new color. Plus, and I'm going to put a 1 half out here, and hopefully it might make sense why I'm about to do this, plus 1 half times f, the second derivative of our function, evaluated at 0x squared. And when we evaluate the derivative of this, I think you'll see why this 1 half is there. Because now let's evaluate this and its derivatives at 0. So if we evaluate p of 0, p of 0 is going to be equal to what? Well, you have this constant term."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And when we evaluate the derivative of this, I think you'll see why this 1 half is there. Because now let's evaluate this and its derivatives at 0. So if we evaluate p of 0, p of 0 is going to be equal to what? Well, you have this constant term. If you evaluate at 0, this x and this x squared are both going to be 0. So those terms are going to go away. So p of 0 is still equal to f of 0."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, you have this constant term. If you evaluate at 0, this x and this x squared are both going to be 0. So those terms are going to go away. So p of 0 is still equal to f of 0. If you take the derivative of p of x, so let me take the derivative right here. I'll do it in yellow. So the derivative of my new p of x is going to be equal to, so this term's going to go away."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So p of 0 is still equal to f of 0. If you take the derivative of p of x, so let me take the derivative right here. I'll do it in yellow. So the derivative of my new p of x is going to be equal to, so this term's going to go away. It's a constant term. It's going to be equal to f prime of 0. That's the coefficient on this."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So the derivative of my new p of x is going to be equal to, so this term's going to go away. It's a constant term. It's going to be equal to f prime of 0. That's the coefficient on this. Plus, this is the power rule right here, 2 times 1 half is just 1 plus f prime prime of 0 times x. Take the 2, multiply it times 1 half, and decrement that 2 right there. And I think you now have a sense of why we put the 1 half there."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "That's the coefficient on this. Plus, this is the power rule right here, 2 times 1 half is just 1 plus f prime prime of 0 times x. Take the 2, multiply it times 1 half, and decrement that 2 right there. And I think you now have a sense of why we put the 1 half there. It's kind of counter, it's making it so that we don't end up with a 2 coefficient out front. Now, what is p prime of 0? So let me write it right here."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And I think you now have a sense of why we put the 1 half there. It's kind of counter, it's making it so that we don't end up with a 2 coefficient out front. Now, what is p prime of 0? So let me write it right here. p prime of 0 is what? Well, this term right here is just going to be 0. So you're left with this constant value right over here, so it's going to be f prime of 0."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let me write it right here. p prime of 0 is what? Well, this term right here is just going to be 0. So you're left with this constant value right over here, so it's going to be f prime of 0. So, so far, our third generation polynomial has all the properties of the first two. And let's see how it does on its third derivative. I should say the second derivative."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So you're left with this constant value right over here, so it's going to be f prime of 0. So, so far, our third generation polynomial has all the properties of the first two. And let's see how it does on its third derivative. I should say the second derivative. So p prime prime of x is equal to, this is a constant, so its derivative is 0. So you just take the coefficient on the second term, is equal to f prime prime of 0. So what's the second derivative of p evaluated at 0?"}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "I should say the second derivative. So p prime prime of x is equal to, this is a constant, so its derivative is 0. So you just take the coefficient on the second term, is equal to f prime prime of 0. So what's the second derivative of p evaluated at 0? Well, it's just going to be this constant value. It's going to be f prime prime of 0. So notice, by adding this term, now not only is our polynomial value the same thing as our function value at 0, its derivative at 0 is the same thing as the derivative of the function at 0."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So what's the second derivative of p evaluated at 0? Well, it's just going to be this constant value. It's going to be f prime prime of 0. So notice, by adding this term, now not only is our polynomial value the same thing as our function value at 0, its derivative at 0 is the same thing as the derivative of the function at 0. And its second derivative at 0 is the same thing as the second derivative of the function at 0. So we're getting pretty good at this. And you might guess that there's a pattern here."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So notice, by adding this term, now not only is our polynomial value the same thing as our function value at 0, its derivative at 0 is the same thing as the derivative of the function at 0. And its second derivative at 0 is the same thing as the second derivative of the function at 0. So we're getting pretty good at this. And you might guess that there's a pattern here. Every term we add, it'll allow us to set up the situation so that the nth derivative of our approximation at 0 will be the same thing as the nth derivative of our function at 0. So in general, if we wanted to keep doing this, if we had a lot of time on our hands and we wanted to just keep adding terms to our polynomial, we could, and let me do this in a new color, maybe I'll do it in a color I already used, we could make our polynomial approximation. So the first term, the constant term, will just be f of 0."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And you might guess that there's a pattern here. Every term we add, it'll allow us to set up the situation so that the nth derivative of our approximation at 0 will be the same thing as the nth derivative of our function at 0. So in general, if we wanted to keep doing this, if we had a lot of time on our hands and we wanted to just keep adding terms to our polynomial, we could, and let me do this in a new color, maybe I'll do it in a color I already used, we could make our polynomial approximation. So the first term, the constant term, will just be f of 0. Then the next term will be f prime of 0 times x. Then the next term will be f prime prime of 0 times 1 half times x squared. I just rewrote that in a slightly different order."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So the first term, the constant term, will just be f of 0. Then the next term will be f prime of 0 times x. Then the next term will be f prime prime of 0 times 1 half times x squared. I just rewrote that in a slightly different order. Then the next term, if we want to make their third derivatives the same at 0, would be f prime prime prime of 0, the third derivative of the function at 0, times 1 half times 1 third. So 1 over 2 times 3 times x to the third. And we can keep going."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "I just rewrote that in a slightly different order. Then the next term, if we want to make their third derivatives the same at 0, would be f prime prime prime of 0, the third derivative of the function at 0, times 1 half times 1 third. So 1 over 2 times 3 times x to the third. And we can keep going. Maybe you'll start to see a pattern here. Plus, if we want to make their fourth derivatives at 0 coincide, it would be the fourth derivative of the function. I could put a 4 up there, but this is really emphasizing it."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And we can keep going. Maybe you'll start to see a pattern here. Plus, if we want to make their fourth derivatives at 0 coincide, it would be the fourth derivative of the function. I could put a 4 up there, but this is really emphasizing it. It's the fourth derivative at 0 times 1 over, and I'll change the order, instead of writing it in increasing order, I'll write it as 4 times 3 times 2 times x to the fourth. And you can verify it for yourself. If we just had this only, and if you were to take the fourth derivative of this, evaluate it at 0, it'll be the same thing as the fourth derivative of the function evaluated at 0."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "I could put a 4 up there, but this is really emphasizing it. It's the fourth derivative at 0 times 1 over, and I'll change the order, instead of writing it in increasing order, I'll write it as 4 times 3 times 2 times x to the fourth. And you can verify it for yourself. If we just had this only, and if you were to take the fourth derivative of this, evaluate it at 0, it'll be the same thing as the fourth derivative of the function evaluated at 0. And in general, you can keep adding terms where the nth term will look like this. The nth derivative of your function, evaluated at 0, times x to the n over n factorial. Notice, this is the same thing as 4 factorial."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "If we just had this only, and if you were to take the fourth derivative of this, evaluate it at 0, it'll be the same thing as the fourth derivative of the function evaluated at 0. And in general, you can keep adding terms where the nth term will look like this. The nth derivative of your function, evaluated at 0, times x to the n over n factorial. Notice, this is the same thing as 4 factorial. 4 factorial is equal to 4 times 3 times 2 times 1. You don't have to write the 1 there, but you could put it there. This right here is the same thing as 3 factorial."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Notice, this is the same thing as 4 factorial. 4 factorial is equal to 4 times 3 times 2 times 1. You don't have to write the 1 there, but you could put it there. This right here is the same thing as 3 factorial. 3 times 2 times 1. I didn't put the 1 there. This right here is the same thing as 2 factorial."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "This right here is the same thing as 3 factorial. 3 times 2 times 1. I didn't put the 1 there. This right here is the same thing as 2 factorial. 2 times 1. This is the same thing. We didn't write anything, but you could divide this by 1 factorial, which is the same thing as 1."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "This right here is the same thing as 2 factorial. 2 times 1. This is the same thing. We didn't write anything, but you could divide this by 1 factorial, which is the same thing as 1. And you could divide this by 0 factorial, which also happens to be 1. We won't have to study it too much over here. But this general series that I've kind of set up right here is called the Maclaurin series."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "We didn't write anything, but you could divide this by 1 factorial, which is the same thing as 1. And you could divide this by 0 factorial, which also happens to be 1. We won't have to study it too much over here. But this general series that I've kind of set up right here is called the Maclaurin series. And you can approximate a polynomial, and we'll see at least some pretty powerful results later on. But what happens, and I don't have the computing power in my brain to draw the graph properly, is that when only the functions equal, you get that horizontal line. When you make the function equal at 0 and their first derivative is equal at 0, then you have something that looks like the tangent line."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "But this general series that I've kind of set up right here is called the Maclaurin series. And you can approximate a polynomial, and we'll see at least some pretty powerful results later on. But what happens, and I don't have the computing power in my brain to draw the graph properly, is that when only the functions equal, you get that horizontal line. When you make the function equal at 0 and their first derivative is equal at 0, then you have something that looks like the tangent line. When you add another degree, it might approximate the polynomial something like this. When you add another degree, it might look something like that. And as you keep adding more and more degrees, or when you keep adding more and more terms, it gets closer and closer around, especially as you get close to x is equal to 0, but in theory, if you add an infinite number of terms, you shouldn't be able to do it."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "When you make the function equal at 0 and their first derivative is equal at 0, then you have something that looks like the tangent line. When you add another degree, it might approximate the polynomial something like this. When you add another degree, it might look something like that. And as you keep adding more and more degrees, or when you keep adding more and more terms, it gets closer and closer around, especially as you get close to x is equal to 0, but in theory, if you add an infinite number of terms, you shouldn't be able to do it. I haven't proven this to you, so that's why I'm saying it's in theory. I haven't proved it yet to you. But if you add an infinite number of terms, all of the derivatives should be the same, and then the functions should pretty much look like each other."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And as you keep adding more and more degrees, or when you keep adding more and more terms, it gets closer and closer around, especially as you get close to x is equal to 0, but in theory, if you add an infinite number of terms, you shouldn't be able to do it. I haven't proven this to you, so that's why I'm saying it's in theory. I haven't proved it yet to you. But if you add an infinite number of terms, all of the derivatives should be the same, and then the functions should pretty much look like each other. In the next video, I'll do this with some actual functions, just so it makes a little bit more sense. And just so you know, the Maclaurin series is a special case of the Taylor series, because we're centering it at 0. And when you're doing a Taylor series, you could pick any center point, but we'll focus on the Maclaurin right now."}, {"video_title": "Intermediate value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "I'm not going to prove it here, but I think the conceptual underpinning here should be straightforward. So the theorem tells us that suppose f is a function continuous at every point of the interval, the closed interval, so we're including a and b, so it's continuous at every point of the interval a, b. And so let me just draw a couple of examples of what f could look like just based on these first lines. So suppose f is a function continuous at every point of the interval a, b. So let me draw some axes here. So that's my y-axis, and this is my x-axis. So one situation, if this is a, and this is b, f is continuous at every point of the closed interval a and b, so that means it's got to be, for sure, defined at every point as well as being continuous."}, {"video_title": "Intermediate value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "So suppose f is a function continuous at every point of the interval a, b. So let me draw some axes here. So that's my y-axis, and this is my x-axis. So one situation, if this is a, and this is b, f is continuous at every point of the closed interval a and b, so that means it's got to be, for sure, defined at every point as well as being continuous. To be continuous, you have to be defined at every point, and the limit of the function as you approach that point should be equal to the value of the function of that point. And so the function is definitely going to be defined at f of a, so it's definitely going to have an f of a right over here, this right over here is f of a. Maybe f of b is higher, although we can look at different cases."}, {"video_title": "Intermediate value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "So one situation, if this is a, and this is b, f is continuous at every point of the closed interval a and b, so that means it's got to be, for sure, defined at every point as well as being continuous. To be continuous, you have to be defined at every point, and the limit of the function as you approach that point should be equal to the value of the function of that point. And so the function is definitely going to be defined at f of a, so it's definitely going to have an f of a right over here, this right over here is f of a. Maybe f of b is higher, although we can look at different cases. So that would be our f of b. And they tell us it is a continuous function. It is a continuous function."}, {"video_title": "Intermediate value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "Maybe f of b is higher, although we can look at different cases. So that would be our f of b. And they tell us it is a continuous function. It is a continuous function. So if you're trying to imagine continuous functions, one way to think about it is, if we're continuous over an interval, we take the value of the function at one point of the interval, and if it's continuous, we need to be able to get to the value of the function at the other point of the interval without picking up our pencil. So I can do all sorts of things, and it still has to be a function, so I can't do something like that, but as long as I don't pick up my pencil, this is a continuous function. So there you go."}, {"video_title": "Intermediate value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "It is a continuous function. So if you're trying to imagine continuous functions, one way to think about it is, if we're continuous over an interval, we take the value of the function at one point of the interval, and if it's continuous, we need to be able to get to the value of the function at the other point of the interval without picking up our pencil. So I can do all sorts of things, and it still has to be a function, so I can't do something like that, but as long as I don't pick up my pencil, this is a continuous function. So there you go. If somehow the graph, I had to pick up my pencil, if I had to do something like this. Oops, I gotta pick up my pencil and do something like that. Well, it's not continuous anymore."}, {"video_title": "Intermediate value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "So there you go. If somehow the graph, I had to pick up my pencil, if I had to do something like this. Oops, I gotta pick up my pencil and do something like that. Well, it's not continuous anymore. If I had to do something like this, and, oops, pick up my pencil, not continuous anymore. If I had to do something like, whoop, okay pick up my pencil, go down here, not continuous anymore. So this is what a continuous function, a function that is continuous over the closed interval A, B looks like."}, {"video_title": "Intermediate value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "Well, it's not continuous anymore. If I had to do something like this, and, oops, pick up my pencil, not continuous anymore. If I had to do something like, whoop, okay pick up my pencil, go down here, not continuous anymore. So this is what a continuous function, a function that is continuous over the closed interval A, B looks like. I can draw some other examples. In fact, let me do that. So let me draw one, maybe where F of B is less than F of A."}, {"video_title": "Intermediate value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "So this is what a continuous function, a function that is continuous over the closed interval A, B looks like. I can draw some other examples. In fact, let me do that. So let me draw one, maybe where F of B is less than F of A. So that's my Y axis, and this is my X axis. And once again, A and B don't both have to be positive. They could both be negative."}, {"video_title": "Intermediate value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "So let me draw one, maybe where F of B is less than F of A. So that's my Y axis, and this is my X axis. And once again, A and B don't both have to be positive. They could both be negative. One could be, A could be negative, B could be positive, and maybe in this situation, and F of A and F of B, it could also be positive or negative. But let's take a situation where this is F of A. So that right over there is F of A."}, {"video_title": "Intermediate value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "They could both be negative. One could be, A could be negative, B could be positive, and maybe in this situation, and F of A and F of B, it could also be positive or negative. But let's take a situation where this is F of A. So that right over there is F of A. This right over here is F of B. F of B. And once again, we're saying F is a continuous function. So I should be able to go from F of A to F of B. F of B, draw a function without having to pick up my pencil."}, {"video_title": "Intermediate value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "So that right over there is F of A. This right over here is F of B. F of B. And once again, we're saying F is a continuous function. So I should be able to go from F of A to F of B. F of B, draw a function without having to pick up my pencil. So it could do something like this. Actually, I don't want to make it go vertical. It could go like this, and then go down, and then do something like that."}, {"video_title": "Intermediate value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "So I should be able to go from F of A to F of B. F of B, draw a function without having to pick up my pencil. So it could do something like this. Actually, I don't want to make it go vertical. It could go like this, and then go down, and then do something like that. So these are both cases, and I could draw an infinite number of cases, where F is a function continuous at every point of the interval, the closed interval, from A to B. Now, given that, there's two ways to state the conclusion for the Intermediate Value Theorem. You'll see it written in one of these ways, or something close to one of these ways, and that's why I included both of these."}, {"video_title": "Intermediate value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "It could go like this, and then go down, and then do something like that. So these are both cases, and I could draw an infinite number of cases, where F is a function continuous at every point of the interval, the closed interval, from A to B. Now, given that, there's two ways to state the conclusion for the Intermediate Value Theorem. You'll see it written in one of these ways, or something close to one of these ways, and that's why I included both of these. So one way to say it is, well, if this first statement is true, then F will take on every value between F of A and F of B over the interval. And you see in both of these cases, every interval, sorry, every value between F of A and F of B, so every value here, is being taken on at some point. You can pick some value."}, {"video_title": "Intermediate value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "You'll see it written in one of these ways, or something close to one of these ways, and that's why I included both of these. So one way to say it is, well, if this first statement is true, then F will take on every value between F of A and F of B over the interval. And you see in both of these cases, every interval, sorry, every value between F of A and F of B, so every value here, is being taken on at some point. You can pick some value. You could pick some value, an arbitrary value, L right over here. Well, look, L happened right over there. If you pick L, well, L happened right over there, and actually, it also happened there, and it also happened there."}, {"video_title": "Intermediate value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "You can pick some value. You could pick some value, an arbitrary value, L right over here. Well, look, L happened right over there. If you pick L, well, L happened right over there, and actually, it also happened there, and it also happened there. And this second bullet point describes the Intermediate Value Theorem more that way. For any L between the values of F of A and F of B, there exists a number C in the closed interval from A to B, for which F of C equals L. So there exists at least one C. So in this case, that would be our C. Over here, there's potential, there's multiple candidates for C. That could be a candidate for C. That could be a C. So we could say there exists at least one number. At least one number."}, {"video_title": "Intermediate value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "If you pick L, well, L happened right over there, and actually, it also happened there, and it also happened there. And this second bullet point describes the Intermediate Value Theorem more that way. For any L between the values of F of A and F of B, there exists a number C in the closed interval from A to B, for which F of C equals L. So there exists at least one C. So in this case, that would be our C. Over here, there's potential, there's multiple candidates for C. That could be a candidate for C. That could be a C. So we could say there exists at least one number. At least one number. I'll throw that in there. At least one number C in the interval for which this is true. And something that might amuse you for a few minutes is try to draw a function where this first statement is true but somehow the second statement is not true."}, {"video_title": "Intermediate value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "At least one number. I'll throw that in there. At least one number C in the interval for which this is true. And something that might amuse you for a few minutes is try to draw a function where this first statement is true but somehow the second statement is not true. So you say, okay, well, let's assume that there's an L where there isn't a C in the interval. Well, let me try to do that. I'll draw it big so that we can really see how obvious that we have to take on all of the values between F of A and F of B is."}, {"video_title": "Intermediate value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "And something that might amuse you for a few minutes is try to draw a function where this first statement is true but somehow the second statement is not true. So you say, okay, well, let's assume that there's an L where there isn't a C in the interval. Well, let me try to do that. I'll draw it big so that we can really see how obvious that we have to take on all of the values between F of A and F of B is. So let me draw a big axis this time. So that's my Y axis. And that is my X axis."}, {"video_title": "Intermediate value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "I'll draw it big so that we can really see how obvious that we have to take on all of the values between F of A and F of B is. So let me draw a big axis this time. So that's my Y axis. And that is my X axis. And I'll just do the case where, just for simplicity, that is A and that is B. And let's say that this is F of A. So that is F of A."}, {"video_title": "Intermediate value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "And that is my X axis. And I'll just do the case where, just for simplicity, that is A and that is B. And let's say that this is F of A. So that is F of A. And let's say that this is F of B. The dotted line. All right, F of B."}, {"video_title": "Intermediate value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "So that is F of A. And let's say that this is F of B. The dotted line. All right, F of B. And we assume that we have a continuous function here. So the graph, I could draw it from F of A to F of B, from this point to this point without picking up my pencil. From this coordinate, A comma F of A, to this coordinate, B comma F of B without picking up my pencil."}, {"video_title": "Intermediate value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "All right, F of B. And we assume that we have a continuous function here. So the graph, I could draw it from F of A to F of B, from this point to this point without picking up my pencil. From this coordinate, A comma F of A, to this coordinate, B comma F of B without picking up my pencil. Well, let's assume that there is some L that we don't take on. Let's say there's some value L right over here. And we never take on this value."}, {"video_title": "Intermediate value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "From this coordinate, A comma F of A, to this coordinate, B comma F of B without picking up my pencil. Well, let's assume that there is some L that we don't take on. Let's say there's some value L right over here. And we never take on this value. This continuous function never takes on this value as we go from X equaling A to X equal B. Let's see if I can draw that. Let's see if I can get from here to here without ever essentially crossing this dotted line."}, {"video_title": "Intermediate value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "And we never take on this value. This continuous function never takes on this value as we go from X equaling A to X equal B. Let's see if I can draw that. Let's see if I can get from here to here without ever essentially crossing this dotted line. Well, let's see, I could, ooh, maybe I'll avoid it a little bit. I'll avoid, but gee, how am I going to get there? Well, without picking up my pencil."}, {"video_title": "Intermediate value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "Let's see if I can get from here to here without ever essentially crossing this dotted line. Well, let's see, I could, ooh, maybe I'll avoid it a little bit. I'll avoid, but gee, how am I going to get there? Well, without picking up my pencil. Well, I really need to cross that line. All right, well, there you go. I found we took on the value L and it happened at C, which is in that closed interval."}, {"video_title": "Intermediate value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "Well, without picking up my pencil. Well, I really need to cross that line. All right, well, there you go. I found we took on the value L and it happened at C, which is in that closed interval. So once again, I'm not giving you a proof here, but hopefully you have a good intuition that the Intermediate Value Theorem is kind of common sense. The key is you're dealing with a continuous function. If you were to make its graph, if you were to draw it between the coordinates A comma F of A and B comma F of B, and you don't pick up your pencil, which would be true of a continuous function, well, it's going to take on every value between F of A and F of B."}, {"video_title": "Quotient rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "In a future video, we can prove it using the product rule, and we'll see it has some similarities to the product rule. But here, we'll learn about what it is and how and where to actually apply it. So, for example, if I have some function, f of x, and it can be expressed as the quotient of two expressions. So let's say u of x of x over v of x, then the quotient rule tells us that f prime of x is going to be equal to, and this is going to look a little bit complicated, but once we apply it, you'll hopefully get a little bit more comfortable with it. It's going to be equal to the derivative of the numerator function, u prime of x, times the denominator function, v of x, minus the numerator function, u of x, do that in that blue color, u of x times the derivative of the denominator function, times v prime of x. And this already looks very similar to the product rule. If this was u of x times v of x, then this is what we would get when we took the derivative if this was a plus sign, but this is here a minus sign."}, {"video_title": "Quotient rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say u of x of x over v of x, then the quotient rule tells us that f prime of x is going to be equal to, and this is going to look a little bit complicated, but once we apply it, you'll hopefully get a little bit more comfortable with it. It's going to be equal to the derivative of the numerator function, u prime of x, times the denominator function, v of x, minus the numerator function, u of x, do that in that blue color, u of x times the derivative of the denominator function, times v prime of x. And this already looks very similar to the product rule. If this was u of x times v of x, then this is what we would get when we took the derivative if this was a plus sign, but this is here a minus sign. But we're not done yet. We would then divide by the denominator function squared, v of x squared. So let's actually apply this idea."}, {"video_title": "Quotient rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "If this was u of x times v of x, then this is what we would get when we took the derivative if this was a plus sign, but this is here a minus sign. But we're not done yet. We would then divide by the denominator function squared, v of x squared. So let's actually apply this idea. So let's say that we have f of x is equal to x squared over cosine of x. Well, what could be our u of x and what could be our v of x? Well, our u of x could be our x squared, so that is u of x and u prime of x would be equal to two x."}, {"video_title": "Quotient rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So let's actually apply this idea. So let's say that we have f of x is equal to x squared over cosine of x. Well, what could be our u of x and what could be our v of x? Well, our u of x could be our x squared, so that is u of x and u prime of x would be equal to two x. And then this could be our v of x. So this is v of x and v prime of x. The derivative of cosine of x with respect to x is equal to negative sine of x."}, {"video_title": "Quotient rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, our u of x could be our x squared, so that is u of x and u prime of x would be equal to two x. And then this could be our v of x. So this is v of x and v prime of x. The derivative of cosine of x with respect to x is equal to negative sine of x. And then we just apply this. So based on that, f prime of x is going to be equal to the derivative of the numerator function, that's two x right over here, that's that there. So it's gonna be two x times the denominator function."}, {"video_title": "Quotient rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "The derivative of cosine of x with respect to x is equal to negative sine of x. And then we just apply this. So based on that, f prime of x is going to be equal to the derivative of the numerator function, that's two x right over here, that's that there. So it's gonna be two x times the denominator function. V of x is just cosine of x times cosine of x minus the numerator function, which is just x squared, x squared, times the derivative of the denominator function. The derivative of cosine of x is negative sine x. So negative sine of x."}, {"video_title": "Quotient rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So it's gonna be two x times the denominator function. V of x is just cosine of x times cosine of x minus the numerator function, which is just x squared, x squared, times the derivative of the denominator function. The derivative of cosine of x is negative sine x. So negative sine of x. Negative sine of x. All of that over, all of that over, the denominator function squared. So that's cosine of x and I'm going to square it."}, {"video_title": "Quotient rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So negative sine of x. Negative sine of x. All of that over, all of that over, the denominator function squared. So that's cosine of x and I'm going to square it. I could write it, of course, like this. Actually, let me write it like that just to make it a little bit clearer. And at this point, we just have to simplify."}, {"video_title": "Quotient rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So that's cosine of x and I'm going to square it. I could write it, of course, like this. Actually, let me write it like that just to make it a little bit clearer. And at this point, we just have to simplify. This is going to be equal to, let's see, we're gonna get two x times cosine of x, x cosine of x, negative times a negative is a positive, plus x squared, x squared times sine of x, sine of x, all of that over, cosine of x squared, which I could write like this as well. And we're done. You could try to simplify it."}, {"video_title": "Quotient rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And at this point, we just have to simplify. This is going to be equal to, let's see, we're gonna get two x times cosine of x, x cosine of x, negative times a negative is a positive, plus x squared, x squared times sine of x, sine of x, all of that over, cosine of x squared, which I could write like this as well. And we're done. You could try to simplify it. In fact, there's no obvious ways to simplify this any further. Now what you'll see in the future, you might already know something called the chain rule or you might learn in the future, but you could also do the quotient rule using the product and the chain rules, which you might learn in the future. But if you don't know the chain rule yet, this is fairly useful."}, {"video_title": "Analyzing functions for discontinuities (continuous example) AP Calculus AB Khan Academy.mp3", "Sentence": "And once again, this three is right at the interface between these two clauses or these two cases. We go to this first case when x is between zero and three, when it's greater than zero and less than three, and then at three, we hit this case. So in order to find the limit, we want to find the limit from the left-hand side, which will have us dealing with this situation, because if we're less than three, we're in this clause, and we also want to find the limit from the right-hand side, which would put us in this clause right over here. And then if both of those limits exist and if they are the same, then that is going to be the limit of this. So let's do that. So let me first go from the left-hand side. So the limit as x approaches three from values less than three, so we're gonna approach from the left, of g of x."}, {"video_title": "Analyzing functions for discontinuities (continuous example) AP Calculus AB Khan Academy.mp3", "Sentence": "And then if both of those limits exist and if they are the same, then that is going to be the limit of this. So let's do that. So let me first go from the left-hand side. So the limit as x approaches three from values less than three, so we're gonna approach from the left, of g of x. Well, this is equivalent to saying this is the limit as x approaches three from the negative side. When x is less than three, which is what's happening here, we're approaching three from the left, we're in this clause right over here. So we're gonna be operating right over there."}, {"video_title": "Analyzing functions for discontinuities (continuous example) AP Calculus AB Khan Academy.mp3", "Sentence": "So the limit as x approaches three from values less than three, so we're gonna approach from the left, of g of x. Well, this is equivalent to saying this is the limit as x approaches three from the negative side. When x is less than three, which is what's happening here, we're approaching three from the left, we're in this clause right over here. So we're gonna be operating right over there. That is what g of x is when we are less than three. So log of three x. And since this function right over here is defined and continuous over the interval we care about, it's defined and continuous for all x's greater than zero, well, we can just substitute three in here to see what it would be approaching."}, {"video_title": "Analyzing functions for discontinuities (continuous example) AP Calculus AB Khan Academy.mp3", "Sentence": "So we're gonna be operating right over there. That is what g of x is when we are less than three. So log of three x. And since this function right over here is defined and continuous over the interval we care about, it's defined and continuous for all x's greater than zero, well, we can just substitute three in here to see what it would be approaching. So this would be, this would be equal log of, log of three times three, or logarithm of nine. And once again, when people just write log here without writing the base, it's implied that we're dealing, that it is 10 right over here. So this is log base 10."}, {"video_title": "Analyzing functions for discontinuities (continuous example) AP Calculus AB Khan Academy.mp3", "Sentence": "And since this function right over here is defined and continuous over the interval we care about, it's defined and continuous for all x's greater than zero, well, we can just substitute three in here to see what it would be approaching. So this would be, this would be equal log of, log of three times three, or logarithm of nine. And once again, when people just write log here without writing the base, it's implied that we're dealing, that it is 10 right over here. So this is log base 10. That's just a good thing to know that sometimes gets missed a little bit. All right, now let's think about the other case. Let's think about the situation where we are approaching three from the right-hand side, from values greater than three."}, {"video_title": "Analyzing functions for discontinuities (continuous example) AP Calculus AB Khan Academy.mp3", "Sentence": "So this is log base 10. That's just a good thing to know that sometimes gets missed a little bit. All right, now let's think about the other case. Let's think about the situation where we are approaching three from the right-hand side, from values greater than three. Well, we are now going to be in this scenario right over there. So this is going to be equal to the limit as x approaches three from the positive direction, from the right-hand side, of, well, g of x is in this clause when we are greater than three. So four minus x times log of nine."}, {"video_title": "Analyzing functions for discontinuities (continuous example) AP Calculus AB Khan Academy.mp3", "Sentence": "Let's think about the situation where we are approaching three from the right-hand side, from values greater than three. Well, we are now going to be in this scenario right over there. So this is going to be equal to the limit as x approaches three from the positive direction, from the right-hand side, of, well, g of x is in this clause when we are greater than three. So four minus x times log of nine. Four times log of nine. And this looks like some type of a logarithm expression at first until you realize that log of nine is just a constant. Log base 10 of nine, it's gonna be some number close to one."}, {"video_title": "Analyzing functions for discontinuities (continuous example) AP Calculus AB Khan Academy.mp3", "Sentence": "So four minus x times log of nine. Four times log of nine. And this looks like some type of a logarithm expression at first until you realize that log of nine is just a constant. Log base 10 of nine, it's gonna be some number close to one. This is just, this expression would actually define a line. For x greater than or equal to three, g of x is just a line, even though it looks a little bit complicated. And so this is actually defined for all real numbers."}, {"video_title": "Analyzing functions for discontinuities (continuous example) AP Calculus AB Khan Academy.mp3", "Sentence": "Log base 10 of nine, it's gonna be some number close to one. This is just, this expression would actually define a line. For x greater than or equal to three, g of x is just a line, even though it looks a little bit complicated. And so this is actually defined for all real numbers. And so, and it's also, it's continuous for any x that you put into it. So to find this limit, to think about what is this expression approaching as we approach three from the positive direction, well, we can just evaluate it at three. So it's going to be four minus three times log of nine."}, {"video_title": "Analyzing functions for discontinuities (continuous example) AP Calculus AB Khan Academy.mp3", "Sentence": "And so this is actually defined for all real numbers. And so, and it's also, it's continuous for any x that you put into it. So to find this limit, to think about what is this expression approaching as we approach three from the positive direction, well, we can just evaluate it at three. So it's going to be four minus three times log of nine. Well, that's just one. So that's equal to log base 10 of nine. So the limit from the left equals the limit from the right."}, {"video_title": "Analyzing functions for discontinuities (continuous example) AP Calculus AB Khan Academy.mp3", "Sentence": "So it's going to be four minus three times log of nine. Well, that's just one. So that's equal to log base 10 of nine. So the limit from the left equals the limit from the right. They're both log nine. So the answer here is log, log of nine. Log, log of nine, and we are done."}, {"video_title": "Volume with cross sections semicircle AP Calculus AB Khan Academy.mp3", "Sentence": "So this region right over here is the base of a three-dimensional figure. And what we know about this three-dimensional figure is if we were to take cross sections that are perpendicular to the x-axis, so cross sections, I guess we could say, that are parallel to the y-axis. So let's say we were to take a cross section like that. We know that this will be a semicircle. So if we take a slightly different view of the same three-dimensional figure, we would see something like this. I've kind of laid the coordinate plane down flat and I'm looking at it from above. So this cross section, if we're looking at it at an angle, and if the figure were transparent, it would be this cross section right over here."}, {"video_title": "Volume with cross sections semicircle AP Calculus AB Khan Academy.mp3", "Sentence": "We know that this will be a semicircle. So if we take a slightly different view of the same three-dimensional figure, we would see something like this. I've kind of laid the coordinate plane down flat and I'm looking at it from above. So this cross section, if we're looking at it at an angle, and if the figure were transparent, it would be this cross section right over here. It would be that cross section right over here, which is a semicircle. If we were to take this cross section right over here along the y-axis, that would be this cross section, this larger semicircle. So given that, given what I've just told you, I encourage you to pause the video and see if you can figure out the volume of this thing that I've, I could shade it in a little bit."}, {"video_title": "Volume with cross sections semicircle AP Calculus AB Khan Academy.mp3", "Sentence": "So this cross section, if we're looking at it at an angle, and if the figure were transparent, it would be this cross section right over here. It would be that cross section right over here, which is a semicircle. If we were to take this cross section right over here along the y-axis, that would be this cross section, this larger semicircle. So given that, given what I've just told you, I encourage you to pause the video and see if you can figure out the volume of this thing that I've, I could shade it in a little bit. The volume of this thing that I've just started, this three-dimensional figure that we are attempting to visualize. So I'm assuming you've had a go at it. And one way to think about it is, well, let's just think about each of these."}, {"video_title": "Volume with cross sections semicircle AP Calculus AB Khan Academy.mp3", "Sentence": "So given that, given what I've just told you, I encourage you to pause the video and see if you can figure out the volume of this thing that I've, I could shade it in a little bit. The volume of this thing that I've just started, this three-dimensional figure that we are attempting to visualize. So I'm assuming you've had a go at it. And one way to think about it is, well, let's just think about each of these. Let's split up the figure into a bunch of disks. Into a bunch of disks. If you figure out the volume of each of those disks, if you figure out the volume of each of those disks and sum them up, then that would be a pretty good approximation for the volume of the whole thing."}, {"video_title": "Volume with cross sections semicircle AP Calculus AB Khan Academy.mp3", "Sentence": "And one way to think about it is, well, let's just think about each of these. Let's split up the figure into a bunch of disks. Into a bunch of disks. If you figure out the volume of each of those disks, if you figure out the volume of each of those disks and sum them up, then that would be a pretty good approximation for the volume of the whole thing. And then if you took the limit as you get an infinite number of disks that are infinitely thin, then you're going to get the exact volume. So let's just say, let's just take the approximating case first. So let's say that right over here at x, what is going to be, what is going to be the diameter, what is going to be the diameter of the disk at x?"}, {"video_title": "Volume with cross sections semicircle AP Calculus AB Khan Academy.mp3", "Sentence": "If you figure out the volume of each of those disks, if you figure out the volume of each of those disks and sum them up, then that would be a pretty good approximation for the volume of the whole thing. And then if you took the limit as you get an infinite number of disks that are infinitely thin, then you're going to get the exact volume. So let's just say, let's just take the approximating case first. So let's say that right over here at x, what is going to be, what is going to be the diameter, what is going to be the diameter of the disk at x? Well, to think about that, we just have to re-express x plus y is equal to one. That's the same thing as saying that the function that f of x, or y is equal to f of x, is equal to one minus x. And so the diameter of this circle right over here, let me make it clear, the diameter of this circle is going to be this height."}, {"video_title": "Volume with cross sections semicircle AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say that right over here at x, what is going to be, what is going to be the diameter, what is going to be the diameter of the disk at x? Well, to think about that, we just have to re-express x plus y is equal to one. That's the same thing as saying that the function that f of x, or y is equal to f of x, is equal to one minus x. And so the diameter of this circle right over here, let me make it clear, the diameter of this circle is going to be this height. It's going to be the difference between one minus x and the x-axis, or between one minus x and x equals zero. And so this is just going to be, as a function of x, the diameter is going to be one minus x. Now, if you want to find the surface area for a circle, or if we want to find the area, I guess you could say, of a circle, we know that area is pi r squared."}, {"video_title": "Volume with cross sections semicircle AP Calculus AB Khan Academy.mp3", "Sentence": "And so the diameter of this circle right over here, let me make it clear, the diameter of this circle is going to be this height. It's going to be the difference between one minus x and the x-axis, or between one minus x and x equals zero. And so this is just going to be, as a function of x, the diameter is going to be one minus x. Now, if you want to find the surface area for a circle, or if we want to find the area, I guess you could say, of a circle, we know that area is pi r squared. Now, for our semicircle, you're going to divide by two. So what's the radius going to be? So let me zoom in a little bit."}, {"video_title": "Volume with cross sections semicircle AP Calculus AB Khan Academy.mp3", "Sentence": "Now, if you want to find the surface area for a circle, or if we want to find the area, I guess you could say, of a circle, we know that area is pi r squared. Now, for our semicircle, you're going to divide by two. So what's the radius going to be? So let me zoom in a little bit. So what is the radius of one of these semicircles going to be? So the radius, I could draw it like this. My best attempt to draw one of these things."}, {"video_title": "Volume with cross sections semicircle AP Calculus AB Khan Academy.mp3", "Sentence": "So let me zoom in a little bit. So what is the radius of one of these semicircles going to be? So the radius, I could draw it like this. My best attempt to draw one of these things. So it's going to look something like this. I'm trying to draw it at an angle. So the radius, what's going to be half the diameter?"}, {"video_title": "Volume with cross sections semicircle AP Calculus AB Khan Academy.mp3", "Sentence": "My best attempt to draw one of these things. So it's going to look something like this. I'm trying to draw it at an angle. So the radius, what's going to be half the diameter? The diameter is one minus x. The radius is going to be one minus x over two. That distance is one minus x over two."}, {"video_title": "Volume with cross sections semicircle AP Calculus AB Khan Academy.mp3", "Sentence": "So the radius, what's going to be half the diameter? The diameter is one minus x. The radius is going to be one minus x over two. That distance is one minus x over two. That distance is one minus x over two. That distance is one minus x over two. So that is one minus x over two is equal to the radius."}, {"video_title": "Volume with cross sections semicircle AP Calculus AB Khan Academy.mp3", "Sentence": "That distance is one minus x over two. That distance is one minus x over two. That distance is one minus x over two. So that is one minus x over two is equal to the radius. And so what would be the area of, what would be the area of this side right over here? Well, if it was a full circle, it would be pi r squared, but it's a semicircle, so it's pi r squared over two. So let me write this."}, {"video_title": "Volume with cross sections semicircle AP Calculus AB Khan Academy.mp3", "Sentence": "So that is one minus x over two is equal to the radius. And so what would be the area of, what would be the area of this side right over here? Well, if it was a full circle, it would be pi r squared, but it's a semicircle, so it's pi r squared over two. So let me write this. So the area is equal to pi r squared, r squared over two, because it's a semicircle. So in terms of x, I guess I could say our area as a function of x right over there. Let me write it that way."}, {"video_title": "Volume with cross sections semicircle AP Calculus AB Khan Academy.mp3", "Sentence": "So let me write this. So the area is equal to pi r squared, r squared over two, because it's a semicircle. So in terms of x, I guess I could say our area as a function of x right over there. Let me write it that way. Our area as a function of x is going to be pi over two. Let me just write that. It's going to be pi over two times one minus x over two squared."}, {"video_title": "Volume with cross sections semicircle AP Calculus AB Khan Academy.mp3", "Sentence": "Let me write it that way. Our area as a function of x is going to be pi over two. Let me just write that. It's going to be pi over two times one minus x over two squared. So it's one minus x over two, over two squared. And so if I wanted a volume of just this disc right over here, I'd then multiply that area times the depth. I'd just multiply the area times this depth here."}, {"video_title": "Volume with cross sections semicircle AP Calculus AB Khan Academy.mp3", "Sentence": "It's going to be pi over two times one minus x over two squared. So it's one minus x over two, over two squared. And so if I wanted a volume of just this disc right over here, I'd then multiply that area times the depth. I'd just multiply the area times this depth here. And we could call that dx, or we could call that delta x. So we could call the depth there delta x. Now, that's the, actually let me be clear."}, {"video_title": "Volume with cross sections semicircle AP Calculus AB Khan Academy.mp3", "Sentence": "I'd just multiply the area times this depth here. And we could call that dx, or we could call that delta x. So we could call the depth there delta x. Now, that's the, actually let me be clear. That's the area, the volume, the volume of one of those shells is going to be equal to, and I'll just write this in one color. It's going to be pi over two times one minus x over two squared times the depth, the area times the depth. So this is volume of one of these half discs."}, {"video_title": "Volume with cross sections semicircle AP Calculus AB Khan Academy.mp3", "Sentence": "Now, that's the, actually let me be clear. That's the area, the volume, the volume of one of those shells is going to be equal to, and I'll just write this in one color. It's going to be pi over two times one minus x over two squared times the depth, the area times the depth. So this is volume of one of these half discs. Half discs. So the volume of the whole thing we can approximate as the sum of these, or we could take the limit as our delta x's get much, much, much, much smaller, and we have an infinite number of these things, which is essentially, if we're taking that limit, we're gonna take the definite integral. So we could take the definite integral, so the volume that we care about, the volume of this figure is going to be equal to the definite integral from x equals zero to x equals one."}, {"video_title": "Volume with cross sections semicircle AP Calculus AB Khan Academy.mp3", "Sentence": "So this is volume of one of these half discs. Half discs. So the volume of the whole thing we can approximate as the sum of these, or we could take the limit as our delta x's get much, much, much, much smaller, and we have an infinite number of these things, which is essentially, if we're taking that limit, we're gonna take the definite integral. So we could take the definite integral, so the volume that we care about, the volume of this figure is going to be equal to the definite integral from x equals zero to x equals one. That's where we intersect the x-axis. X equals zero to x equals one of, we're integrating an infinite number of these things that are infinitely, I guess, thin. So it's going to be, let me write it."}, {"video_title": "Volume with cross sections semicircle AP Calculus AB Khan Academy.mp3", "Sentence": "So we could take the definite integral, so the volume that we care about, the volume of this figure is going to be equal to the definite integral from x equals zero to x equals one. That's where we intersect the x-axis. X equals zero to x equals one of, we're integrating an infinite number of these things that are infinitely, I guess, thin. So it's going to be, let me write it. It's gonna be pi over two times, I'll just write it, what's one minus x squared? Let me just expand it out for fun right over here. So that's going to be, that's the same thing as x minus one squared, so it's gonna be x squared minus two x plus one, and then two squared is four, over four, and instead of delta x, I'm gonna write dx now."}, {"video_title": "Volume with cross sections semicircle AP Calculus AB Khan Academy.mp3", "Sentence": "So it's going to be, let me write it. It's gonna be pi over two times, I'll just write it, what's one minus x squared? Let me just expand it out for fun right over here. So that's going to be, that's the same thing as x minus one squared, so it's gonna be x squared minus two x plus one, and then two squared is four, over four, and instead of delta x, I'm gonna write dx now. I'm gonna write dx, because I'm taking the limit as these become infinitely small, and I have an infinite, and I'm summing an infinite number of them. So the volume is just going to be, we just have to evaluate this definite integral, and if you feel so inspired, if you feel inspired at the moment, feel free to pause and try to evaluate this definite integral. So let's just take some of these constants out."}, {"video_title": "Worked example separable differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "What we're going to do in this video is get some practice finding general solutions to separable differential equations so let's say that I had the differential equation dy dx, the derivative of y with respect to x is equal to e to the x over y. See if you can find the general solution to this differential equation. I'm giving you a huge hint. It is a separable differential equation. All right, so when we're dealing with a separable differential equation, what we want to do is get the y's and the dy's on one side and then the x's and the dx's on the other side and we really treat these differentials kind of like variables which is a little hand wavy with the mathematics but that's what we will do. So let's see, if we multiply both sides times y, so we're gonna multiply both sides times y, what are we going to get? We're gonna get y times the derivative of y with respect to x is equal to e to the x and now we can multiply both sides by the differential dx, multiply both of them by dx, those cancel out and we are left with y times dy is equal to e to the x dx and now we can take the integral of both sides so let us do that."}, {"video_title": "Worked example separable differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "It is a separable differential equation. All right, so when we're dealing with a separable differential equation, what we want to do is get the y's and the dy's on one side and then the x's and the dx's on the other side and we really treat these differentials kind of like variables which is a little hand wavy with the mathematics but that's what we will do. So let's see, if we multiply both sides times y, so we're gonna multiply both sides times y, what are we going to get? We're gonna get y times the derivative of y with respect to x is equal to e to the x and now we can multiply both sides by the differential dx, multiply both of them by dx, those cancel out and we are left with y times dy is equal to e to the x dx and now we can take the integral of both sides so let us do that. So what is the integral of y dy? Well, here we would just use the reverse power rule. We would increment the exponent so it's y to the first but so now when we take the antiderivative, it would be y squared and then we divide by that incremented exponent is equal to, well, the exciting thing about e to the x is its antiderivative is, and its derivative is e to the x is equal to e to the x plus, is equal to e to the x plus c and so we can leave it like this if we like."}, {"video_title": "Worked example separable differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "We're gonna get y times the derivative of y with respect to x is equal to e to the x and now we can multiply both sides by the differential dx, multiply both of them by dx, those cancel out and we are left with y times dy is equal to e to the x dx and now we can take the integral of both sides so let us do that. So what is the integral of y dy? Well, here we would just use the reverse power rule. We would increment the exponent so it's y to the first but so now when we take the antiderivative, it would be y squared and then we divide by that incremented exponent is equal to, well, the exciting thing about e to the x is its antiderivative is, and its derivative is e to the x is equal to e to the x plus, is equal to e to the x plus c and so we can leave it like this if we like. In fact, this right over here, this isn't an explicit function. Y here isn't an explicit function of x. You could actually say y is equal to the plus or minus square root of two times all of this business but this would be a pretty general relationship which would satisfy this separable differential equation."}, {"video_title": "Worked example separable differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "We would increment the exponent so it's y to the first but so now when we take the antiderivative, it would be y squared and then we divide by that incremented exponent is equal to, well, the exciting thing about e to the x is its antiderivative is, and its derivative is e to the x is equal to e to the x plus, is equal to e to the x plus c and so we can leave it like this if we like. In fact, this right over here, this isn't an explicit function. Y here isn't an explicit function of x. You could actually say y is equal to the plus or minus square root of two times all of this business but this would be a pretty general relationship which would satisfy this separable differential equation. Let's do another example. So let's say that we have the derivative of y with respect to x is equal to, let's say it's equal to y squared times sine of x. Pause the video and see if you can find the general solution here."}, {"video_title": "Worked example separable differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "You could actually say y is equal to the plus or minus square root of two times all of this business but this would be a pretty general relationship which would satisfy this separable differential equation. Let's do another example. So let's say that we have the derivative of y with respect to x is equal to, let's say it's equal to y squared times sine of x. Pause the video and see if you can find the general solution here. So once again, we want to separate our y's and our x's. So let's see, we can multiply both sides times y to the negative two power. Y to the negative two, y to the negative two."}, {"video_title": "Worked example separable differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "Pause the video and see if you can find the general solution here. So once again, we want to separate our y's and our x's. So let's see, we can multiply both sides times y to the negative two power. Y to the negative two, y to the negative two. These become one and then we can also multiply both sides times dx. So if we multiply dx here, those cancel out and then we multiply dx here. And so we're left with y to the negative two power times dy is equal to sine of x dx and now we just can integrate both sides."}, {"video_title": "Worked example separable differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "Y to the negative two, y to the negative two. These become one and then we can also multiply both sides times dx. So if we multiply dx here, those cancel out and then we multiply dx here. And so we're left with y to the negative two power times dy is equal to sine of x dx and now we just can integrate both sides. Now what is the antiderivative of y to the negative two? Well once again, we use the reverse power rule. We increment the exponent, so it's gonna be y to the negative one and then we divide by that newly incremented exponent."}, {"video_title": "Worked example separable differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "And so we're left with y to the negative two power times dy is equal to sine of x dx and now we just can integrate both sides. Now what is the antiderivative of y to the negative two? Well once again, we use the reverse power rule. We increment the exponent, so it's gonna be y to the negative one and then we divide by that newly incremented exponent. So we divide by negative one. Well that would just make this thing negative. That is going to be equal to, so what's the antiderivative of sine of x?"}, {"video_title": "Worked example separable differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "We increment the exponent, so it's gonna be y to the negative one and then we divide by that newly incremented exponent. So we divide by negative one. Well that would just make this thing negative. That is going to be equal to, so what's the antiderivative of sine of x? Well it is, you might recognize it if I put a negative there and a negative there. The antiderivative of negative sine of x, well that's cosine of x. So this whole thing is gonna be negative cosine of x or another way to write this, I could multiply both sides times the negative one and so these would both become positive and so I could write one over y is equal to cosine of x and actually let me write it this way, plus c. Don't wanna forget my plus c's."}, {"video_title": "Worked example separable differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "That is going to be equal to, so what's the antiderivative of sine of x? Well it is, you might recognize it if I put a negative there and a negative there. The antiderivative of negative sine of x, well that's cosine of x. So this whole thing is gonna be negative cosine of x or another way to write this, I could multiply both sides times the negative one and so these would both become positive and so I could write one over y is equal to cosine of x and actually let me write it this way, plus c. Don't wanna forget my plus c's. Plus c or I can take the reciprocal of both sides. If I wanna solve explicitly for y, I could get y is equal to one over cosine of x plus c as our general solution and we're done. That was strangely fun."}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "But instead of rotating around the x-axis this time, I want to rotate around the y-axis. And instead of going between zero and some point, I'm going to go between y is equal to one and y is equal to four. So what I'm going to do is I'm going to take this graph right over here, I'm going to take this curve, instead of rotating it around the x-axis like we did in the last few videos, I'm going to rotate it around the y-axis. So I'm going to rotate it around just like that. So what's the shape that we would get? Let me see if we can visualize this. So the base is going to look something like that."}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So I'm going to rotate it around just like that. So what's the shape that we would get? Let me see if we can visualize this. So the base is going to look something like that. We could see through it. It's going to look something like that. And then this up here, the top of it, would look something like that."}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So the base is going to look something like that. We could see through it. It's going to look something like that. And then this up here, the top of it, would look something like that. And we care about the stuff in between. So we care about this part right over here, not the very bottom of it. And let me shade it in a little bit."}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "And then this up here, the top of it, would look something like that. And we care about the stuff in between. So we care about this part right over here, not the very bottom of it. And let me shade it in a little bit. So it would look something like that. So let me draw it separately just so we can visualize it. So I'll draw it at different angles."}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "And let me shade it in a little bit. So it would look something like that. So let me draw it separately just so we can visualize it. So I'll draw it at different angles. So if I were to draw it with the y-axis kind of coming out the back, it would look something like this. It looks something like it's a little bit smaller like that. And then it gets cut off right over here like that."}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So I'll draw it at different angles. So if I were to draw it with the y-axis kind of coming out the back, it would look something like this. It looks something like it's a little bit smaller like that. And then it gets cut off right over here like that. So it looks, I don't know what shape you could call it. But I think hopefully you're conceptualizing this. Let me do that same yellow color."}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "And then it gets cut off right over here like that. So it looks, I don't know what shape you could call it. But I think hopefully you're conceptualizing this. Let me do that same yellow color. The visualization here is probably the hardest part. But as we can see, it's not too bad. So it looks something like this."}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Let me do that same yellow color. The visualization here is probably the hardest part. But as we can see, it's not too bad. So it looks something like this. It looks like maybe a truffle, an upside down truffle. So this right here, let me draw the y-axis just to show how we're oriented. So the y-axis is popping out in this example like that."}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So it looks something like this. It looks like maybe a truffle, an upside down truffle. So this right here, let me draw the y-axis just to show how we're oriented. So the y-axis is popping out in this example like that. Then it goes down over here. And then the x-axis is going like this. So I just tilted this over."}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So the y-axis is popping out in this example like that. Then it goes down over here. And then the x-axis is going like this. So I just tilted this over. I tilted it over a little bit to be able to view it at a different angle. This top right over here is this top right over there. So that gives you an idea of what it looks like."}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So I just tilted this over. I tilted it over a little bit to be able to view it at a different angle. This top right over here is this top right over there. So that gives you an idea of what it looks like. But we still haven't thought about how do we actually find the volume of this thing. Well, what we can do instead of creating disks where the depth is in little dx's, what if we created disks where the depth is in dy's? So let's think about that a little bit."}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So that gives you an idea of what it looks like. But we still haven't thought about how do we actually find the volume of this thing. Well, what we can do instead of creating disks where the depth is in little dx's, what if we created disks where the depth is in dy's? So let's think about that a little bit. So let's think about constructing a disk at a certain y value. So let's think about a certain y value. We're going to construct a disk right over there that has the same radius of the shape at that point."}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So let's think about that a little bit. So let's think about constructing a disk at a certain y value. So let's think about a certain y value. We're going to construct a disk right over there that has the same radius of the shape at that point. So that's our disk right over here. And then it has a depth. Instead of saying it has a depth of dx, let's say it has a depth of dy."}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "We're going to construct a disk right over there that has the same radius of the shape at that point. So that's our disk right over here. And then it has a depth. Instead of saying it has a depth of dx, let's say it has a depth of dy. So this depth right over here is dy. So what is the volume of this disk in terms of y? And as you can imagine, we're going to do this definite integral with respect to y."}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Instead of saying it has a depth of dx, let's say it has a depth of dy. So this depth right over here is dy. So what is the volume of this disk in terms of y? And as you can imagine, we're going to do this definite integral with respect to y. So what's the volume of this thing? Well, like we did in the last video, we have to figure out the area of the top of each of these disks, or I guess you could say the face of this coin. Well, to find that area, it's pi r squared."}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "And as you can imagine, we're going to do this definite integral with respect to y. So what's the volume of this thing? Well, like we did in the last video, we have to figure out the area of the top of each of these disks, or I guess you could say the face of this coin. Well, to find that area, it's pi r squared. If we can figure out this radius right over here, we know the area. So what's that radius? So to think about that radius in terms of y, we just have to solve this equation explicitly in terms of y."}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Well, to find that area, it's pi r squared. If we can figure out this radius right over here, we know the area. So what's that radius? So to think about that radius in terms of y, we just have to solve this equation explicitly in terms of y. So instead of saying it's y is equal to x squared, we can take the principal root of both sides, and we could say that the square root of y is equal to x. And this right over here is only defined for non-negative y's, but that's OK, because we are in the positive x-axis right over here. So we can also call this function right over here x is equal to the square root of y."}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So to think about that radius in terms of y, we just have to solve this equation explicitly in terms of y. So instead of saying it's y is equal to x squared, we can take the principal root of both sides, and we could say that the square root of y is equal to x. And this right over here is only defined for non-negative y's, but that's OK, because we are in the positive x-axis right over here. So we can also call this function right over here x is equal to the square root of y. And we're essentially looking at this side of it. We're not looking at this stuff right over here. So we're only looking at this side right over here."}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So we can also call this function right over here x is equal to the square root of y. And we're essentially looking at this side of it. We're not looking at this stuff right over here. So we're only looking at this side right over here. We've now expressed this graph, this curve, as x as a function of y. So if we do it that way, what's our radius right over here? Well, our radius right over here is going to be f of y."}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So we're only looking at this side right over here. We've now expressed this graph, this curve, as x as a function of y. So if we do it that way, what's our radius right over here? Well, our radius right over here is going to be f of y. It's going to be the square root of y. So it's going to be a function of y. I don't want to confuse you if you thought this was f of x and I'm saying this is f of y. It's going to be a function of y."}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Well, our radius right over here is going to be f of y. It's going to be the square root of y. So it's going to be a function of y. I don't want to confuse you if you thought this was f of x and I'm saying this is f of y. It's going to be a function of y. We could call it g of y. It's going to be the square root of y. So area is equal to pi r squared, which means that the area of this thing is going to be pi times our radius squared."}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "It's going to be a function of y. We could call it g of y. It's going to be the square root of y. So area is equal to pi r squared, which means that the area of this thing is going to be pi times our radius squared. Our radius is square root of y. So this thing is going to be equal to pi squared of y squared is just pi times y. Now, if we want the volume, we just have to multiply the area of the surface times the depth, times dy."}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So area is equal to pi r squared, which means that the area of this thing is going to be pi times our radius squared. Our radius is square root of y. So this thing is going to be equal to pi squared of y squared is just pi times y. Now, if we want the volume, we just have to multiply the area of the surface times the depth, times dy. So the volume of each of those disks is going to be pi y times dy. This gives you the volume of a disk. Now, if we want the volume of this entire thing, we just have to sum all of these disks for all the y values between y is equal to 1 and y is equal to 4."}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Now, if we want the volume, we just have to multiply the area of the surface times the depth, times dy. So the volume of each of those disks is going to be pi y times dy. This gives you the volume of a disk. Now, if we want the volume of this entire thing, we just have to sum all of these disks for all the y values between y is equal to 1 and y is equal to 4. So let's do that. So we just take the definite integral from y is equal to 1 to y equals 4. And just as a reminder, definite integral is a very special type of sum."}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Now, if we want the volume of this entire thing, we just have to sum all of these disks for all the y values between y is equal to 1 and y is equal to 4. So let's do that. So we just take the definite integral from y is equal to 1 to y equals 4. And just as a reminder, definite integral is a very special type of sum. We're summing up all of these things, but we're taking the limit of that sum as these dy's get shorter, or gets, I guess, squatter and squatter, smaller and smaller. And we have a larger and larger number of these disks. Really, as these become infinitely small and we have an infinite number of disks, so that our sum doesn't just approximate the volume."}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "And just as a reminder, definite integral is a very special type of sum. We're summing up all of these things, but we're taking the limit of that sum as these dy's get shorter, or gets, I guess, squatter and squatter, smaller and smaller. And we have a larger and larger number of these disks. Really, as these become infinitely small and we have an infinite number of disks, so that our sum doesn't just approximate the volume. It actually is the volume at the limit. So to figure out the volume of this entire thing, we just have to evaluate this definite integral in terms of y. And so how do we do that?"}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Really, as these become infinitely small and we have an infinite number of disks, so that our sum doesn't just approximate the volume. It actually is the volume at the limit. So to figure out the volume of this entire thing, we just have to evaluate this definite integral in terms of y. And so how do we do that? What's going to be equal to? Well, we could take the pi outside. It's going to be pi times the antiderivative of y, which is just y squared over 2, evaluated from 1 to 4, which is equal to pi times, well, you evaluate it at 4, you get 16 over 2."}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "And so how do we do that? What's going to be equal to? Well, we could take the pi outside. It's going to be pi times the antiderivative of y, which is just y squared over 2, evaluated from 1 to 4, which is equal to pi times, well, you evaluate it at 4, you get 16 over 2. Let me just write it out like this. 4 squared over 2 minus 1 squared over 2, which is equal to pi times 16 over 2 is 8 minus 1 half. And so we could do this as 16 halves minus 1 half, which is equal to 15 halves."}, {"video_title": "Disc method around y-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "It's going to be pi times the antiderivative of y, which is just y squared over 2, evaluated from 1 to 4, which is equal to pi times, well, you evaluate it at 4, you get 16 over 2. Let me just write it out like this. 4 squared over 2 minus 1 squared over 2, which is equal to pi times 16 over 2 is 8 minus 1 half. And so we could do this as 16 halves minus 1 half, which is equal to 15 halves. So this is equal to 15 halves times pi. Or another way of thinking of it is 7 and 1 half times pi. But this is a little bit clearer."}, {"video_title": "Worked example slope field from equation AP Calculus AB Khan Academy.mp3", "Sentence": "which slope field is generated by the differential equation the derivative of y with respect to x is equal to x minus y. And like always, pause this video and see if you can figure it out on your own. Well, the easiest way to think about a slope field, if I needed to plot this slope field by hand, I would sample a bunch of x and y points and then I would figure out what the derivative would have to be at that point. And so what we can do here, since they've already drawn some candidate slope fields for us, is figure out what we think the slope field should be at some point and see which of these diagrams, these graphs or these slope fields, actually show that. So let's, let me make a little table here. So I'm gonna have, I'm gonna have x, y, and then the derivative of y with respect to x. And we can do it at a bunch of values, so let's think about it."}, {"video_title": "Worked example slope field from equation AP Calculus AB Khan Academy.mp3", "Sentence": "And so what we can do here, since they've already drawn some candidate slope fields for us, is figure out what we think the slope field should be at some point and see which of these diagrams, these graphs or these slope fields, actually show that. So let's, let me make a little table here. So I'm gonna have, I'm gonna have x, y, and then the derivative of y with respect to x. And we can do it at a bunch of values, so let's think about it. Let's think about when, we're at this point right over here, when x is two and y is two. When x is two and y is two, the derivative of y with respect to x is going to be two minus two. It's going to be equal to zero."}, {"video_title": "Worked example slope field from equation AP Calculus AB Khan Academy.mp3", "Sentence": "And we can do it at a bunch of values, so let's think about it. Let's think about when, we're at this point right over here, when x is two and y is two. When x is two and y is two, the derivative of y with respect to x is going to be two minus two. It's going to be equal to zero. And just with that, let's see, here, this slope on this slope field does not look like it's zero. This looks like it's negative one. So already I could rule this one out."}, {"video_title": "Worked example slope field from equation AP Calculus AB Khan Academy.mp3", "Sentence": "It's going to be equal to zero. And just with that, let's see, here, this slope on this slope field does not look like it's zero. This looks like it's negative one. So already I could rule this one out. This slope right over here looks like it's positive one, so I'll rule that out. It's definitely not zero. This slope also looks like positive one, so I can rule that one out."}, {"video_title": "Worked example slope field from equation AP Calculus AB Khan Academy.mp3", "Sentence": "So already I could rule this one out. This slope right over here looks like it's positive one, so I'll rule that out. It's definitely not zero. This slope also looks like positive one, so I can rule that one out. This slope at two comma two actually does look like zero, so I'm liking this one right over here. This slope at two comma two looks larger than one, so I could rule that out. So it was that straightforward to deduce that this choice right over here is, if any of these are going to be the accurate slope field, it's this one."}, {"video_title": "Worked example slope field from equation AP Calculus AB Khan Academy.mp3", "Sentence": "This slope also looks like positive one, so I can rule that one out. This slope at two comma two actually does look like zero, so I'm liking this one right over here. This slope at two comma two looks larger than one, so I could rule that out. So it was that straightforward to deduce that this choice right over here is, if any of these are going to be the accurate slope field, it's this one. But just for kicks, we could keep going to verify that this is indeed the slope field. So let's think about what happens when x is equal to, well, one, whenever x is equal to y, you're going to get the derivative equaling zero, and you see that here. When you're at four, four, derivative equals zero."}, {"video_title": "Worked example slope field from equation AP Calculus AB Khan Academy.mp3", "Sentence": "So it was that straightforward to deduce that this choice right over here is, if any of these are going to be the accurate slope field, it's this one. But just for kicks, we could keep going to verify that this is indeed the slope field. So let's think about what happens when x is equal to, well, one, whenever x is equal to y, you're going to get the derivative equaling zero, and you see that here. When you're at four, four, derivative equals zero. When it's six, six, derivative equals zero. At negative two, negative two, derivative equals zero. So that feels good that this is the right slope field."}, {"video_title": "Worked example slope field from equation AP Calculus AB Khan Academy.mp3", "Sentence": "When you're at four, four, derivative equals zero. When it's six, six, derivative equals zero. At negative two, negative two, derivative equals zero. So that feels good that this is the right slope field. And then we could pick other arbitrary points. Let's say when x is four, y is two, then the derivative here should be four minus two, which is going to be two. So when x is four, y is two, we do indeed see that the slope field is indicating a slope that looks like two right over here."}, {"video_title": "Worked example slope field from equation AP Calculus AB Khan Academy.mp3", "Sentence": "So that feels good that this is the right slope field. And then we could pick other arbitrary points. Let's say when x is four, y is two, then the derivative here should be four minus two, which is going to be two. So when x is four, y is two, we do indeed see that the slope field is indicating a slope that looks like two right over here. And if it was the other way around, when x is, when x is, let's say, x is negative four, and y is negative two. So negative four, negative two. Well, negative four minus negative two is going to be negative two."}, {"video_title": "Worked example slope field from equation AP Calculus AB Khan Academy.mp3", "Sentence": "So when x is four, y is two, we do indeed see that the slope field is indicating a slope that looks like two right over here. And if it was the other way around, when x is, when x is, let's say, x is negative four, and y is negative two. So negative four, negative two. Well, negative four minus negative two is going to be negative two. And you can see that right over here. Negative four, negative two. You can see the slope right over here."}, {"video_title": "Worked example slope field from equation AP Calculus AB Khan Academy.mp3", "Sentence": "Well, negative four minus negative two is going to be negative two. And you can see that right over here. Negative four, negative two. You can see the slope right over here. It's a little harder to see. Looks like negative two. So once again, using even just this first two comma two coordinates, we were able to deduce that this was the choice."}, {"video_title": "Interpreting direction of motion from position-time graph AP Calculus AB Khan Academy.mp3", "Sentence": "The following graph gives the object's position relative to its starting point over time. For each point on the graph, is the object moving forward, backward, or neither? So pause this video and try to figure that out. All right, so we see we have position in meters versus time. So for example, this point right over here tells us that after one second, we are four meters ahead of our starting point. Or for example, this point right over here says that after four seconds, we are almost, it seems, almost four meters behind our starting point. So let's look at each of these points and think about whether we're moving forward, backward, or neither."}, {"video_title": "Interpreting direction of motion from position-time graph AP Calculus AB Khan Academy.mp3", "Sentence": "All right, so we see we have position in meters versus time. So for example, this point right over here tells us that after one second, we are four meters ahead of our starting point. Or for example, this point right over here says that after four seconds, we are almost, it seems, almost four meters behind our starting point. So let's look at each of these points and think about whether we're moving forward, backward, or neither. So at this point right over here, at that moment, we're about two and a half meters in front of our starting point. We're at a positive position of two and a half meters. But as time goes on, we are moving backwards closer and closer to the starting point."}, {"video_title": "Interpreting direction of motion from position-time graph AP Calculus AB Khan Academy.mp3", "Sentence": "So let's look at each of these points and think about whether we're moving forward, backward, or neither. So at this point right over here, at that moment, we're about two and a half meters in front of our starting point. We're at a positive position of two and a half meters. But as time goes on, we are moving backwards closer and closer to the starting point. So this is, we are moving backward. One way to think about it, at this time, we're at two and a half meters. If you go forward about half a second, we are then back at our starting point."}, {"video_title": "Interpreting direction of motion from position-time graph AP Calculus AB Khan Academy.mp3", "Sentence": "But as time goes on, we are moving backwards closer and closer to the starting point. So this is, we are moving backward. One way to think about it, at this time, we're at two and a half meters. If you go forward about half a second, we are then back at our starting point. So we have to go backwards. And if we look at this point right over here, it looks like we were going backwards this entire time while our curve is downward sloping. But at this point right over here, when we are about, it looks like, five meters behind our starting point, we start going forward again."}, {"video_title": "Interpreting direction of motion from position-time graph AP Calculus AB Khan Academy.mp3", "Sentence": "If you go forward about half a second, we are then back at our starting point. So we have to go backwards. And if we look at this point right over here, it looks like we were going backwards this entire time while our curve is downward sloping. But at this point right over here, when we are about, it looks like, five meters behind our starting point, we start going forward again. But right at that moment, we are going neither forward nor backwards. It's right at that moment where we just finished going backwards and we're about to go forward. And one way to think about it is, what would be the slope of the tangent line at that point?"}, {"video_title": "Interpreting direction of motion from position-time graph AP Calculus AB Khan Academy.mp3", "Sentence": "But at this point right over here, when we are about, it looks like, five meters behind our starting point, we start going forward again. But right at that moment, we are going neither forward nor backwards. It's right at that moment where we just finished going backwards and we're about to go forward. And one way to think about it is, what would be the slope of the tangent line at that point? And the slope of the tangent line at that point would be horizontal. And so this is neither. So we can use that same technique to think about this point."}, {"video_title": "Interpreting direction of motion from position-time graph AP Calculus AB Khan Academy.mp3", "Sentence": "And one way to think about it is, what would be the slope of the tangent line at that point? And the slope of the tangent line at that point would be horizontal. And so this is neither. So we can use that same technique to think about this point. The slope is positive. And we see that, all right, right at that moment, it looks like we are at the starting point. But if you fast forward even a few, even a fraction of a second, we are now in front of our starting point."}, {"video_title": "Interpreting direction of motion from position-time graph AP Calculus AB Khan Academy.mp3", "Sentence": "So we can use that same technique to think about this point. The slope is positive. And we see that, all right, right at that moment, it looks like we are at the starting point. But if you fast forward even a few, even a fraction of a second, we are now in front of our starting point. So we are moving forward. We are moving forward right over here. And at this point, we are at our starting point."}, {"video_title": "Interpreting direction of motion from position-time graph AP Calculus AB Khan Academy.mp3", "Sentence": "But if you fast forward even a few, even a fraction of a second, we are now in front of our starting point. So we are moving forward. We are moving forward right over here. And at this point, we are at our starting point. But if you think about what's going to happen a moment later, a moment later, we're going to be a little bit behind our starting point. And so here we are moving backward. And we're done."}, {"video_title": "Worked example Riemann sums in summation notation AP Calculus AB Khan Academy.mp3", "Sentence": "That's this curve right over here. And then we have these rectangles that are trying to approximate the area under the curve, the area under the function f between x equals 0 and x equals 8. And the way that this diagram, or the way that we're attempting to do it, is by splitting it into four rectangles. And so we could call this rectangle one, this is rectangle two, rectangle three, and rectangle four. And each of their heights, let's see, the interval looks like they each have a width of two, so they're equally spaced. So we go from zero to eight, and we split into four sections, so each has a width of two. So they're each going to be two wide."}, {"video_title": "Worked example Riemann sums in summation notation AP Calculus AB Khan Academy.mp3", "Sentence": "And so we could call this rectangle one, this is rectangle two, rectangle three, and rectangle four. And each of their heights, let's see, the interval looks like they each have a width of two, so they're equally spaced. So we go from zero to eight, and we split into four sections, so each has a width of two. So they're each going to be two wide. So that's two, that's two, that's two, that's two. And their height seems to be based on, their height seems to be based on the midpoint. So between the start, between the left side and the right side of the rectangle, you take the value of the function at at the middle value right over here."}, {"video_title": "Worked example Riemann sums in summation notation AP Calculus AB Khan Academy.mp3", "Sentence": "So they're each going to be two wide. So that's two, that's two, that's two, that's two. And their height seems to be based on, their height seems to be based on the midpoint. So between the start, between the left side and the right side of the rectangle, you take the value of the function at at the middle value right over here. So for example, this height right over here looks like f of one. This height right over here looks like f of three. This height of this rectangle is f of five."}, {"video_title": "Worked example Riemann sums in summation notation AP Calculus AB Khan Academy.mp3", "Sentence": "So between the start, between the left side and the right side of the rectangle, you take the value of the function at at the middle value right over here. So for example, this height right over here looks like f of one. This height right over here looks like f of three. This height of this rectangle is f of five. This height right over here is f of seven. So given the way that this has been constructed, and we want to take the sum of the areas of these rectangles as an approximation under, as the area under this curve, how would we write that as sigma notation? And I'll get us started, and then I encourage you to pause the video and try to finish it."}, {"video_title": "Worked example Riemann sums in summation notation AP Calculus AB Khan Academy.mp3", "Sentence": "This height of this rectangle is f of five. This height right over here is f of seven. So given the way that this has been constructed, and we want to take the sum of the areas of these rectangles as an approximation under, as the area under this curve, how would we write that as sigma notation? And I'll get us started, and then I encourage you to pause the video and try to finish it. So the sum of these, the sum of these rectangles, we could say it's the sum of, so we'll have n equals one to four, because we have four rectangles. And I encourage you to, and I encourage you to, to finish this up. Once, actually just write it in terms of the function, use function notation."}, {"video_title": "Worked example Riemann sums in summation notation AP Calculus AB Khan Academy.mp3", "Sentence": "And I'll get us started, and then I encourage you to pause the video and try to finish it. So the sum of these, the sum of these rectangles, we could say it's the sum of, so we'll have n equals one to four, because we have four rectangles. And I encourage you to, and I encourage you to, to finish this up. Once, actually just write it in terms of the function, use function notation. You don't have to write it out as one plus point oh one times something squared. So I'm assuming you've had a go at it. So for each of these, so for the first, so for the first rectangle over here, we're going to multiply two times the height."}, {"video_title": "Worked example Riemann sums in summation notation AP Calculus AB Khan Academy.mp3", "Sentence": "Once, actually just write it in terms of the function, use function notation. You don't have to write it out as one plus point oh one times something squared. So I'm assuming you've had a go at it. So for each of these, so for the first, so for the first rectangle over here, we're going to multiply two times the height. So the height right over here is one, and this is the first rectangle, so you might be tempted to say times f of n. But then that breaks down as we go into the second rectangle. The second rectangle, the two still applies. This two is the width of the rectangle."}, {"video_title": "Worked example Riemann sums in summation notation AP Calculus AB Khan Academy.mp3", "Sentence": "So for each of these, so for the first, so for the first rectangle over here, we're going to multiply two times the height. So the height right over here is one, and this is the first rectangle, so you might be tempted to say times f of n. But then that breaks down as we go into the second rectangle. The second rectangle, the two still applies. This two is the width of the rectangle. But now we want to multiply it times f of three, not f of two. So this f of n isn't going to, this f of n isn't going to, isn't going to pass muster. And so let's see, let's see how we want to think about it."}, {"video_title": "Worked example Riemann sums in summation notation AP Calculus AB Khan Academy.mp3", "Sentence": "This two is the width of the rectangle. But now we want to multiply it times f of three, not f of two. So this f of n isn't going to, this f of n isn't going to, isn't going to pass muster. And so let's see, let's see how we want to think about it. So when n is, so when n is one, two, three, four, we're gonna take f of, so f of n, or I should say f of n, we're gonna take f of something. So here, this first one, we're gonna take f of one. Then here, for the second rectangle, we're taking f of three."}, {"video_title": "Worked example Riemann sums in summation notation AP Calculus AB Khan Academy.mp3", "Sentence": "And so let's see, let's see how we want to think about it. So when n is, so when n is one, two, three, four, we're gonna take f of, so f of n, or I should say f of n, we're gonna take f of something. So here, this first one, we're gonna take f of one. Then here, for the second rectangle, we're taking f of three. f of three for the height. For the third rectangle, we're taking f of five. f of five."}, {"video_title": "Worked example Riemann sums in summation notation AP Calculus AB Khan Academy.mp3", "Sentence": "Then here, for the second rectangle, we're taking f of three. f of three for the height. For the third rectangle, we're taking f of five. f of five. And then, so for the fourth rectangle, we're taking f of seven. f of seven. So what's the relationship over here?"}, {"video_title": "Worked example Riemann sums in summation notation AP Calculus AB Khan Academy.mp3", "Sentence": "f of five. And then, so for the fourth rectangle, we're taking f of seven. f of seven. So what's the relationship over here? Let's see, it looks like if you multiply by two and subtract one, so two times one minus one is one, two times two minus one is three. Two times three minus five, two times three minus one is five. Two times four minus one is seven."}, {"video_title": "Worked example Riemann sums in summation notation AP Calculus AB Khan Academy.mp3", "Sentence": "So what's the relationship over here? Let's see, it looks like if you multiply by two and subtract one, so two times one minus one is one, two times two minus one is three. Two times three minus five, two times three minus one is five. Two times four minus one is seven. So this is two n minus one. So the area of each of these rectangles, the base is two, and the height is f of two n minus, f of two n minus one. So that's, that hopefully makes a little bit clearer, kind of mapping between the sigma notation and what we're actually trying to do."}, {"video_title": "Worked example Riemann sums in summation notation AP Calculus AB Khan Academy.mp3", "Sentence": "Two times four minus one is seven. So this is two n minus one. So the area of each of these rectangles, the base is two, and the height is f of two n minus, f of two n minus one. So that's, that hopefully makes a little bit clearer, kind of mapping between the sigma notation and what we're actually trying to do. And now let's just, just for fun, let's actually try to evaluate this thing. What is this thing going to evaluate to? Well, this is going to evaluate to two times f of, when n is equal to one, this is one, f of one, plus two times, when n is two, this is going to be f of two times two minus one is three, f of three."}, {"video_title": "Worked example Riemann sums in summation notation AP Calculus AB Khan Academy.mp3", "Sentence": "So that's, that hopefully makes a little bit clearer, kind of mapping between the sigma notation and what we're actually trying to do. And now let's just, just for fun, let's actually try to evaluate this thing. What is this thing going to evaluate to? Well, this is going to evaluate to two times f of, when n is equal to one, this is one, f of one, plus two times, when n is two, this is going to be f of two times two minus one is three, f of three. When n is three, this is going to be two times f of five. When n is four, this is going to be two times, two times f of seven. Four times two minus one is seven."}, {"video_title": "Worked example Riemann sums in summation notation AP Calculus AB Khan Academy.mp3", "Sentence": "Well, this is going to evaluate to two times f of, when n is equal to one, this is one, f of one, plus two times, when n is two, this is going to be f of two times two minus one is three, f of three. When n is three, this is going to be two times f of five. When n is four, this is going to be two times, two times f of seven. Four times two minus one is seven. f of seven. And so that is going to be, we're going to have to evaluate a bunch of these things over here. So let me actually, let me clear this out so I have a little bit more real estate."}, {"video_title": "Worked example Riemann sums in summation notation AP Calculus AB Khan Academy.mp3", "Sentence": "Four times two minus one is seven. f of seven. And so that is going to be, we're going to have to evaluate a bunch of these things over here. So let me actually, let me clear this out so I have a little bit more real estate. I'm feeling this might get a little bit messy now. So this is going to be, actually we could factor out a two. So this is going to be equal to two times, f of one is one plus 0.1 times one squared."}, {"video_title": "Worked example Riemann sums in summation notation AP Calculus AB Khan Academy.mp3", "Sentence": "So let me actually, let me clear this out so I have a little bit more real estate. I'm feeling this might get a little bit messy now. So this is going to be, actually we could factor out a two. So this is going to be equal to two times, f of one is one plus 0.1 times one squared. So it's one plus 0.1, so it's one point, let me color code a little bit so we can keep track of things. So this right over here is one point one. So one plus 0.1 is 1.1."}, {"video_title": "Worked example Riemann sums in summation notation AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be equal to two times, f of one is one plus 0.1 times one squared. So it's one plus 0.1, so it's one point, let me color code a little bit so we can keep track of things. So this right over here is one point one. So one plus 0.1 is 1.1. This right over here, f of three, so it's one plus 0.1 times three squared, nine. So one plus 0.9. So one plus, so it's 1.9."}, {"video_title": "Worked example Riemann sums in summation notation AP Calculus AB Khan Academy.mp3", "Sentence": "So one plus 0.1 is 1.1. This right over here, f of three, so it's one plus 0.1 times three squared, nine. So one plus 0.9. So one plus, so it's 1.9. And then let's see, this one right over here, f of five, this is going to be one plus, see five squared is 25 times 0.1 is 2.5. So one plus 2.5 is going to be 3.5. And then finally f of seven, f of seven is going to be one plus 0.1 times seven squared."}, {"video_title": "Worked example Riemann sums in summation notation AP Calculus AB Khan Academy.mp3", "Sentence": "So one plus, so it's 1.9. And then let's see, this one right over here, f of five, this is going to be one plus, see five squared is 25 times 0.1 is 2.5. So one plus 2.5 is going to be 3.5. And then finally f of seven, f of seven is going to be one plus 0.1 times seven squared. So this is 49 times 0.1. 49 times 0.1 is 4.9 plus one, so plus 5.9. And so what is this going to be equal to?"}, {"video_title": "Worked example Riemann sums in summation notation AP Calculus AB Khan Academy.mp3", "Sentence": "And then finally f of seven, f of seven is going to be one plus 0.1 times seven squared. So this is 49 times 0.1. 49 times 0.1 is 4.9 plus one, so plus 5.9. And so what is this going to be equal to? So let's see, 1.1 plus 1.9, these two are going to sum up to be equal to three. And then these two are going to sum up to be, let's see, if we add the five, we get to 8.5, and then we add the 0.9, we get to 9.4. So plus 9.4, did I do that right?"}, {"video_title": "Worked example Riemann sums in summation notation AP Calculus AB Khan Academy.mp3", "Sentence": "And so what is this going to be equal to? So let's see, 1.1 plus 1.9, these two are going to sum up to be equal to three. And then these two are going to sum up to be, let's see, if we add the five, we get to 8.5, and then we add the 0.9, we get to 9.4. So plus 9.4, did I do that right? Three plus five is 8.5 plus 0.9 is 1.4, yep. And so this is going to be, so once again we have the two times it all, so this is going to be equal to two times 12.4, which is equal to 24.8, which is our approximation. Once again, this is just an approximation using these rectangles of the area under the curve between x equals zero and x equals eight."}, {"video_title": "Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "And what we'll see in this video is the solution to a differential equation isn't a value or a set of values. It's a function or a set of functions. But before we go about actually trying to solve this or figure out all of the solutions, let's test whether certain equations, certain functions are solutions to this differential equation. So for example, if I have y is equal to four x, is this a solution to this differential equation? Pause the video and see if you can figure it out. Well, to see if this is a solution, what we have to do is figure out the derivative of y with respect to x, and see is that truly equal to four times y over x. And I'm gonna try to express everything in terms of x to see if I really have an equality there."}, {"video_title": "Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "So for example, if I have y is equal to four x, is this a solution to this differential equation? Pause the video and see if you can figure it out. Well, to see if this is a solution, what we have to do is figure out the derivative of y with respect to x, and see is that truly equal to four times y over x. And I'm gonna try to express everything in terms of x to see if I really have an equality there. So first, let's figure out the derivative of y with respect to x. Well, that's just going to be equal to four. We've seen that many times before."}, {"video_title": "Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "And I'm gonna try to express everything in terms of x to see if I really have an equality there. So first, let's figure out the derivative of y with respect to x. Well, that's just going to be equal to four. We've seen that many times before. And so what we need to test is, is four, the derivative of y with respect to x, equal to four times, I could write y, but instead of y, let's write four x. I'm gonna put everything in terms of x. So y is equal to four x. So instead of four y, I could write four times four x."}, {"video_title": "Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "We've seen that many times before. And so what we need to test is, is four, the derivative of y with respect to x, equal to four times, I could write y, but instead of y, let's write four x. I'm gonna put everything in terms of x. So y is equal to four x. So instead of four y, I could write four times four x. All of that over x. Is this true? Well, that x cancels with that, and I'm gonna get four is equal to 16, which it clearly is not."}, {"video_title": "Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "So instead of four y, I could write four times four x. All of that over x. Is this true? Well, that x cancels with that, and I'm gonna get four is equal to 16, which it clearly is not. And so this is not a solution. Not a solution to our differential equation. Let's look at another equation."}, {"video_title": "Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "Well, that x cancels with that, and I'm gonna get four is equal to 16, which it clearly is not. And so this is not a solution. Not a solution to our differential equation. Let's look at another equation. What about y is equal to x to the fourth power? Pause this video and see if this is a solution to our original differential equation. Well, we're going to do the same thing."}, {"video_title": "Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "Let's look at another equation. What about y is equal to x to the fourth power? Pause this video and see if this is a solution to our original differential equation. Well, we're going to do the same thing. What's the derivative of y with respect to x? This is equal to, just using the power rule, four x to the third power. And so what we have to test is, is four x to the third power, that's the derivative of y with respect to x, equal to four times y."}, {"video_title": "Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we're going to do the same thing. What's the derivative of y with respect to x? This is equal to, just using the power rule, four x to the third power. And so what we have to test is, is four x to the third power, that's the derivative of y with respect to x, equal to four times y. Instead of writing a y, I'm gonna write it all in terms of x. So is that equal to four times x to the fourth, because x to the fourth is the same thing as y, divided by x. And so let's see, x to the fourth divided by x, that is going to be x to the third."}, {"video_title": "Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "And so what we have to test is, is four x to the third power, that's the derivative of y with respect to x, equal to four times y. Instead of writing a y, I'm gonna write it all in terms of x. So is that equal to four times x to the fourth, because x to the fourth is the same thing as y, divided by x. And so let's see, x to the fourth divided by x, that is going to be x to the third. And so you will indeed get four x to the third is equal to four x to the third. So check, this is a solution. So is a solution."}, {"video_title": "Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "And so let's see, x to the fourth divided by x, that is going to be x to the third. And so you will indeed get four x to the third is equal to four x to the third. So check, this is a solution. So is a solution. It's not necessarily the only solution, but it is a solution to that differential equation. Let's look at another differential equation. Let's say that I had, and I'm gonna write it with different notation, f prime of x is equal to f of x minus x."}, {"video_title": "Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "So is a solution. It's not necessarily the only solution, but it is a solution to that differential equation. Let's look at another differential equation. Let's say that I had, and I'm gonna write it with different notation, f prime of x is equal to f of x minus x. And the first function that I wanna test, let's say I have f of x is equal to two x. Is this a solution to this differential equation? Pause the video again and see if you can figure it out."}, {"video_title": "Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say that I had, and I'm gonna write it with different notation, f prime of x is equal to f of x minus x. And the first function that I wanna test, let's say I have f of x is equal to two x. Is this a solution to this differential equation? Pause the video again and see if you can figure it out. Well, to figure that out, you have to say, well, what is f prime of x? F prime of x is just going to be equal to two. And then test the equality."}, {"video_title": "Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "Pause the video again and see if you can figure it out. Well, to figure that out, you have to say, well, what is f prime of x? F prime of x is just going to be equal to two. And then test the equality. Is two, is f prime of x equal to f of x, which is two x, minus x, minus x. And so let's see, we are going to get two is equal to x. So you might be tempted to say, oh, hey, I just solved for x or something like that."}, {"video_title": "Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "And then test the equality. Is two, is f prime of x equal to f of x, which is two x, minus x, minus x. And so let's see, we are going to get two is equal to x. So you might be tempted to say, oh, hey, I just solved for x or something like that. But this would tell you that this is not a solution because this needs to be true for any x that is in the domain of this function. And so this is, I'll just put an x there, or I'll put an incorrect there to say not, not a, not a solution. Just to be clear again, this needs, in order for a function to be a solution of this differential equation, it needs to work for any x that you can put into the function."}, {"video_title": "Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "So you might be tempted to say, oh, hey, I just solved for x or something like that. But this would tell you that this is not a solution because this needs to be true for any x that is in the domain of this function. And so this is, I'll just put an x there, or I'll put an incorrect there to say not, not a, not a solution. Just to be clear again, this needs, in order for a function to be a solution of this differential equation, it needs to work for any x that you can put into the function. Let's look at another one. Let's say that we have f of x is equal to x plus one. Pause the video and see, is this a solution to our differential equation?"}, {"video_title": "Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "Just to be clear again, this needs, in order for a function to be a solution of this differential equation, it needs to work for any x that you can put into the function. Let's look at another one. Let's say that we have f of x is equal to x plus one. Pause the video and see, is this a solution to our differential equation? Well, same drill. f prime of x is going to be equal to one. And so we have to see, is f prime of x, which is equal to one, is it equal to f of x, which is x plus one, x plus one, minus x?"}, {"video_title": "Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "Pause the video and see, is this a solution to our differential equation? Well, same drill. f prime of x is going to be equal to one. And so we have to see, is f prime of x, which is equal to one, is it equal to f of x, which is x plus one, x plus one, minus x? And so here, you see no matter what x is, this equation is going to be true. So this is a solution, is a solution. Let's do a few more of these."}, {"video_title": "Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "And so we have to see, is f prime of x, which is equal to one, is it equal to f of x, which is x plus one, x plus one, minus x? And so here, you see no matter what x is, this equation is going to be true. So this is a solution, is a solution. Let's do a few more of these. Let me scroll down a little bit so I have a little bit more, a little bit more space, but make sure we see our original differential equation. Let's test whether, I'm gonna do it in a red color. Let's test whether f of x equals e to the x plus x plus one is a solution to this differential equation."}, {"video_title": "Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "Let's do a few more of these. Let me scroll down a little bit so I have a little bit more, a little bit more space, but make sure we see our original differential equation. Let's test whether, I'm gonna do it in a red color. Let's test whether f of x equals e to the x plus x plus one is a solution to this differential equation. Pause the video again and see if you can figure it out. All right, well, let's figure out the derivative here. f prime of x is going to be equal to, derivative of e to the x with respect to x is e to the x, which I always find amazing, and so, and then plus one, and the derivative of this with respect to x is just zero."}, {"video_title": "Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "Let's test whether f of x equals e to the x plus x plus one is a solution to this differential equation. Pause the video again and see if you can figure it out. All right, well, let's figure out the derivative here. f prime of x is going to be equal to, derivative of e to the x with respect to x is e to the x, which I always find amazing, and so, and then plus one, and the derivative of this with respect to x is just zero. And then let's substitute this into our original differential equation. So f prime of x is e to the x plus one. Is that equal to f of x, which is e to the x plus x plus one minus x, minus x?"}, {"video_title": "Verifying solutions to differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "f prime of x is going to be equal to, derivative of e to the x with respect to x is e to the x, which I always find amazing, and so, and then plus one, and the derivative of this with respect to x is just zero. And then let's substitute this into our original differential equation. So f prime of x is e to the x plus one. Is that equal to f of x, which is e to the x plus x plus one minus x, minus x? And if that x cancels out with that x, it is indeed. They are indeed equal. So this is also a solution."}, {"video_title": "Volume with cross sections squares and rectangles (no graph) AP Calculus AB Khan Academy.mp3", "Sentence": "The base of a solid is the region enclosed by the graphs of y is equal to negative x squared plus six x minus one, and y is equal to four. Cross sections of the solid perpendicular to the x-axis are rectangles whose height is x. Express the volume of the solid with a definite integral. So pause this video and see if you can have a go at that. All right, now what's interesting about this is they've just given us the equations for the graphs, but we haven't visualized them yet. And we need to visualize them, or at least I like to visualize them, so I can think about this region that they're talking about. So maybe a first thing to do is think about, well, where do these two lines intersect?"}, {"video_title": "Volume with cross sections squares and rectangles (no graph) AP Calculus AB Khan Academy.mp3", "Sentence": "So pause this video and see if you can have a go at that. All right, now what's interesting about this is they've just given us the equations for the graphs, but we haven't visualized them yet. And we need to visualize them, or at least I like to visualize them, so I can think about this region that they're talking about. So maybe a first thing to do is think about, well, where do these two lines intersect? So when do we have the same y value? Or another way to think about it is when does this thing equal four? So if we set them equal to each other, we have negative x squared plus six x minus one is equal to four."}, {"video_title": "Volume with cross sections squares and rectangles (no graph) AP Calculus AB Khan Academy.mp3", "Sentence": "So maybe a first thing to do is think about, well, where do these two lines intersect? So when do we have the same y value? Or another way to think about it is when does this thing equal four? So if we set them equal to each other, we have negative x squared plus six x minus one is equal to four. This will give us the x values where these two lines intersect. And so we will get, if we wanna solve for x, we can subtract four from both sides. Negative x squared plus six x minus five is equal to zero."}, {"video_title": "Volume with cross sections squares and rectangles (no graph) AP Calculus AB Khan Academy.mp3", "Sentence": "So if we set them equal to each other, we have negative x squared plus six x minus one is equal to four. This will give us the x values where these two lines intersect. And so we will get, if we wanna solve for x, we can subtract four from both sides. Negative x squared plus six x minus five is equal to zero. We can multiply both sides by negative one. We will get x squared minus six x plus five is equal to zero and then this is pretty straightforward to factor. One times five is five, or actually I say negative one times negative five is five, and negative one plus negative five is negative six."}, {"video_title": "Volume with cross sections squares and rectangles (no graph) AP Calculus AB Khan Academy.mp3", "Sentence": "Negative x squared plus six x minus five is equal to zero. We can multiply both sides by negative one. We will get x squared minus six x plus five is equal to zero and then this is pretty straightforward to factor. One times five is five, or actually I say negative one times negative five is five, and negative one plus negative five is negative six. So it's going to be x minus one times x minus five is equal to zero. And so these intersect when x is equal to one or x is equal to five. Since we have a negative out front of the second degree term right over here, we know it's going to be a downward opening parabola."}, {"video_title": "Volume with cross sections squares and rectangles (no graph) AP Calculus AB Khan Academy.mp3", "Sentence": "One times five is five, or actually I say negative one times negative five is five, and negative one plus negative five is negative six. So it's going to be x minus one times x minus five is equal to zero. And so these intersect when x is equal to one or x is equal to five. Since we have a negative out front of the second degree term right over here, we know it's going to be a downward opening parabola. And we know that we intersect y equals four when x is equal to one and x equals five. And so the vertex must be right in between them. So the vertex is going to be at x equals three."}, {"video_title": "Volume with cross sections squares and rectangles (no graph) AP Calculus AB Khan Academy.mp3", "Sentence": "Since we have a negative out front of the second degree term right over here, we know it's going to be a downward opening parabola. And we know that we intersect y equals four when x is equal to one and x equals five. And so the vertex must be right in between them. So the vertex is going to be at x equals three. So let's actually visualize this a little bit. So it's going to look something like this. I'll draw it with some perspective because we're gonna have to think about a three-dimensional shape."}, {"video_title": "Volume with cross sections squares and rectangles (no graph) AP Calculus AB Khan Academy.mp3", "Sentence": "So the vertex is going to be at x equals three. So let's actually visualize this a little bit. So it's going to look something like this. I'll draw it with some perspective because we're gonna have to think about a three-dimensional shape. So that's our y-axis. This is our x-axis. And let me draw some y values."}, {"video_title": "Volume with cross sections squares and rectangles (no graph) AP Calculus AB Khan Academy.mp3", "Sentence": "I'll draw it with some perspective because we're gonna have to think about a three-dimensional shape. So that's our y-axis. This is our x-axis. And let me draw some y values. So one, two, three, four, five, six, seven, eight. This is probably sufficient. Now we have y is equal to four, which is going to look something like this."}, {"video_title": "Volume with cross sections squares and rectangles (no graph) AP Calculus AB Khan Academy.mp3", "Sentence": "And let me draw some y values. So one, two, three, four, five, six, seven, eight. This is probably sufficient. Now we have y is equal to four, which is going to look something like this. So that is y is equal to four. And then we have y is equal to negative x squared plus six x minus one, which we know intersects y equals four at x equals one or x equals five. So let's see, one, two, three, four, five."}, {"video_title": "Volume with cross sections squares and rectangles (no graph) AP Calculus AB Khan Academy.mp3", "Sentence": "Now we have y is equal to four, which is going to look something like this. So that is y is equal to four. And then we have y is equal to negative x squared plus six x minus one, which we know intersects y equals four at x equals one or x equals five. So let's see, one, two, three, four, five. So x equals one. So we have that point right over there. One comma four, and then we have five comma four."}, {"video_title": "Volume with cross sections squares and rectangles (no graph) AP Calculus AB Khan Academy.mp3", "Sentence": "So let's see, one, two, three, four, five. So x equals one. So we have that point right over there. One comma four, and then we have five comma four. And then we know the vertex is when x is equal to three. So it might look something like this. We could substitute three back in here."}, {"video_title": "Volume with cross sections squares and rectangles (no graph) AP Calculus AB Khan Academy.mp3", "Sentence": "One comma four, and then we have five comma four. And then we know the vertex is when x is equal to three. So it might look something like this. We could substitute three back in here. So let's see, y will equal to negative nine, three squared, plus 18 minus one. And so what is that going to be? That's going to be y is going to be equal to eight."}, {"video_title": "Volume with cross sections squares and rectangles (no graph) AP Calculus AB Khan Academy.mp3", "Sentence": "We could substitute three back in here. So let's see, y will equal to negative nine, three squared, plus 18 minus one. And so what is that going to be? That's going to be y is going to be equal to eight. So we have the point three comma eight. So this is five, six, seven, eight. Yep, right about there."}, {"video_title": "Volume with cross sections squares and rectangles (no graph) AP Calculus AB Khan Academy.mp3", "Sentence": "That's going to be y is going to be equal to eight. So we have the point three comma eight. So this is five, six, seven, eight. Yep, right about there. And so we are dealing with a situation, we are dealing with a situation that looks something like this. This is the region in question. So that's going to be the base of our solid."}, {"video_title": "Volume with cross sections squares and rectangles (no graph) AP Calculus AB Khan Academy.mp3", "Sentence": "Yep, right about there. And so we are dealing with a situation, we are dealing with a situation that looks something like this. This is the region in question. So that's going to be the base of our solid. And they say cross sections of the solid perpendicular to the x-axis. So let me draw one of those cross sections. So this is a cross section perpendicular to the x-axis."}, {"video_title": "Volume with cross sections squares and rectangles (no graph) AP Calculus AB Khan Academy.mp3", "Sentence": "So that's going to be the base of our solid. And they say cross sections of the solid perpendicular to the x-axis. So let me draw one of those cross sections. So this is a cross section perpendicular to the x-axis. Our rectangles whose height is x. So this is going to have height x right over here. Height x."}, {"video_title": "Volume with cross sections squares and rectangles (no graph) AP Calculus AB Khan Academy.mp3", "Sentence": "So this is a cross section perpendicular to the x-axis. Our rectangles whose height is x. So this is going to have height x right over here. Height x. Now what is this, the width, I guess we could say, of this rectangle? Well, it's going to be the difference between these two functions. It's going to be this upper function minus this lower function."}, {"video_title": "Volume with cross sections squares and rectangles (no graph) AP Calculus AB Khan Academy.mp3", "Sentence": "Height x. Now what is this, the width, I guess we could say, of this rectangle? Well, it's going to be the difference between these two functions. It's going to be this upper function minus this lower function. So it's going to be, that right over there is going to be negative x squared plus six x minus one. And then minus four, minus the lower function. So that could be simplified as negative x squared plus six x minus five."}, {"video_title": "Volume with cross sections squares and rectangles (no graph) AP Calculus AB Khan Academy.mp3", "Sentence": "It's going to be this upper function minus this lower function. So it's going to be, that right over there is going to be negative x squared plus six x minus one. And then minus four, minus the lower function. So that could be simplified as negative x squared plus six x minus five. And so if we wanna figure out the volume of this little section right over here, we'd multiply x times this. And then we would multiply that times an infinitesimally small depth, a dx. And then we can just integrate from x equals one to x equals five."}, {"video_title": "Volume with cross sections squares and rectangles (no graph) AP Calculus AB Khan Academy.mp3", "Sentence": "So that could be simplified as negative x squared plus six x minus five. And so if we wanna figure out the volume of this little section right over here, we'd multiply x times this. And then we would multiply that times an infinitesimally small depth, a dx. And then we can just integrate from x equals one to x equals five. So let's do that. The volume of just this little slice over here is going to be the base, which is negative x squared plus six x minus five times the height, times x, times the depth, times dx. And then what we wanna do is we wanna sum up all of these."}, {"video_title": "Volume with cross sections squares and rectangles (no graph) AP Calculus AB Khan Academy.mp3", "Sentence": "And then we can just integrate from x equals one to x equals five. So let's do that. The volume of just this little slice over here is going to be the base, which is negative x squared plus six x minus five times the height, times x, times the depth, times dx. And then what we wanna do is we wanna sum up all of these. And so you could imagine right over here, you would have, or like right over here, you would have a cross section that looks like this. X is now much larger. The height is x."}, {"video_title": "Volume with cross sections squares and rectangles (no graph) AP Calculus AB Khan Academy.mp3", "Sentence": "And then what we wanna do is we wanna sum up all of these. And so you could imagine right over here, you would have, or like right over here, you would have a cross section that looks like this. X is now much larger. The height is x. So now it looks something like this. So I'm just drawing two cross sections just so you get the idea. So these are the, this is any one cross section for a given x."}, {"video_title": "Volume with cross sections squares and rectangles (no graph) AP Calculus AB Khan Academy.mp3", "Sentence": "The height is x. So now it looks something like this. So I'm just drawing two cross sections just so you get the idea. So these are the, this is any one cross section for a given x. But now we wanna integrate. Our x is going from x equals one to x equals five. X equals one to x equals five."}, {"video_title": "Volume with cross sections squares and rectangles (no graph) AP Calculus AB Khan Academy.mp3", "Sentence": "So these are the, this is any one cross section for a given x. But now we wanna integrate. Our x is going from x equals one to x equals five. X equals one to x equals five. And there you have it. We have expressed the volume of that solid as a definite integral. And it's worth noting that this definite integral, if you distribute this x, if you multiply it by all of these terms, it's very solvable."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say that we've got some function f that is continuous on the interval a to b. So let's try to see if we can visualize that. So this is my y-axis. That's my y-axis. This right over here. I'm going to make it my t-axis. We'll use x a little bit later."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "That's my y-axis. This right over here. I'm going to make it my t-axis. We'll use x a little bit later. So I'll call this my t-axis. And then let's say that this right over here is the graph of y is equal to f of t. y is equal to f of t. And we're saying it's continuous on the interval from a to b. So this is t is equal to a."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "We'll use x a little bit later. So I'll call this my t-axis. And then let's say that this right over here is the graph of y is equal to f of t. y is equal to f of t. And we're saying it's continuous on the interval from a to b. So this is t is equal to a. This is t is equal to b. So we're saying that it is continuous. It is continuous over this whole interval."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "So this is t is equal to a. This is t is equal to b. So we're saying that it is continuous. It is continuous over this whole interval. Now, for fun, let's define a function capital F of x. And I will do it in blue. Let's define capital F of x as equal to the definite integral from a, from a as a lower bound, to x of f of t, of f of t, dt, where x is in this interval."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "It is continuous over this whole interval. Now, for fun, let's define a function capital F of x. And I will do it in blue. Let's define capital F of x as equal to the definite integral from a, from a as a lower bound, to x of f of t, of f of t, dt, where x is in this interval. Where a is less than or equal to x is less than or equal to b. Or that's just another way of saying that x is in this interval right over here. Now, when you see this, you might say, oh, you know, the definite integral, this has to do with differentiation and antiderivatives and all that."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "Let's define capital F of x as equal to the definite integral from a, from a as a lower bound, to x of f of t, of f of t, dt, where x is in this interval. Where a is less than or equal to x is less than or equal to b. Or that's just another way of saying that x is in this interval right over here. Now, when you see this, you might say, oh, you know, the definite integral, this has to do with differentiation and antiderivatives and all that. But we don't know that yet. All we know right now is that this is the area under the curve F between a and x. So between a and let's say this right over here, this right over here is x."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "Now, when you see this, you might say, oh, you know, the definite integral, this has to do with differentiation and antiderivatives and all that. But we don't know that yet. All we know right now is that this is the area under the curve F between a and x. So between a and let's say this right over here, this right over here is x. So f of x is just this area right over here. That's all we know about it. We don't know it has anything to do with antiderivatives just yet."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "So between a and let's say this right over here, this right over here is x. So f of x is just this area right over here. That's all we know about it. We don't know it has anything to do with antiderivatives just yet. That's what we're going to try to prove in this video. So just for fun, let's take the derivative of f. And we're going to do it just using the definition of derivatives and see what we get when we take the derivative using the definition of derivatives. So we would get the derivative f prime of x."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "We don't know it has anything to do with antiderivatives just yet. That's what we're going to try to prove in this video. So just for fun, let's take the derivative of f. And we're going to do it just using the definition of derivatives and see what we get when we take the derivative using the definition of derivatives. So we would get the derivative f prime of x. Well, this definition of derivatives, it's a limit as delta x approaches 0 of capital F of x plus delta x minus f of x, all of that over delta x. This is just the definition of the derivative. Now, what is this equal to?"}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "So we would get the derivative f prime of x. Well, this definition of derivatives, it's a limit as delta x approaches 0 of capital F of x plus delta x minus f of x, all of that over delta x. This is just the definition of the derivative. Now, what is this equal to? Well, let me rewrite it using these integrals right up here. This is going to be equal to the limit as delta x approaches 0 of what's f of x plus delta x? Well, put x in right over here."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "Now, what is this equal to? Well, let me rewrite it using these integrals right up here. This is going to be equal to the limit as delta x approaches 0 of what's f of x plus delta x? Well, put x in right over here. You're going to get the definite integral from a to x plus delta x of f of t dt. And then from that, you are going to subtract this business, f of x, which we've already written as the definite integral from a to x of f of t dt. And then all of that is over delta x."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "Well, put x in right over here. You're going to get the definite integral from a to x plus delta x of f of t dt. And then from that, you are going to subtract this business, f of x, which we've already written as the definite integral from a to x of f of t dt. And then all of that is over delta x. Now, what does this represent? Remember, we don't know anything about definite integrals or somehow dealing with something with an antiderivative and all that. We just know this is another way of saying the area under the curve f between a and x plus delta x."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "And then all of that is over delta x. Now, what does this represent? Remember, we don't know anything about definite integrals or somehow dealing with something with an antiderivative and all that. We just know this is another way of saying the area under the curve f between a and x plus delta x. So it's the area under the curve f between a and x plus delta x. So it's this entire area right over here. So that's this part."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "We just know this is another way of saying the area under the curve f between a and x plus delta x. So it's the area under the curve f between a and x plus delta x. So it's this entire area right over here. So that's this part. We already know what this blue stuff is. This blue stuff, that same shade of blue, so this blue stuff right over here, this is equal to all of this business. We've already shaded this in."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "So that's this part. We already know what this blue stuff is. This blue stuff, that same shade of blue, so this blue stuff right over here, this is equal to all of this business. We've already shaded this in. It's equal to all of this business right over here. So if you were to take all of this green area, which is from a to x plus delta x, and subtract out this blue area, which is exactly what we're doing in the numerator, what are you left with? Well, you're going to be left with what color have I not used yet?"}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "We've already shaded this in. It's equal to all of this business right over here. So if you were to take all of this green area, which is from a to x plus delta x, and subtract out this blue area, which is exactly what we're doing in the numerator, what are you left with? Well, you're going to be left with what color have I not used yet? Maybe I will use this pink color. Well, no, I already used that. I'll use this purple color."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "Well, you're going to be left with what color have I not used yet? Maybe I will use this pink color. Well, no, I already used that. I'll use this purple color. You're going to be left with this area right over here. So what's another way of writing that? Well, another way of writing this area right over here is the definite integral between x and x plus delta x of f of t dt."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "I'll use this purple color. You're going to be left with this area right over here. So what's another way of writing that? Well, another way of writing this area right over here is the definite integral between x and x plus delta x of f of t dt. So we can rewrite this entire expression, the derivative of capital F of x. This is capital F prime of x. We can rewrite it now as being equal to the limit as delta x approaches 0."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "Well, another way of writing this area right over here is the definite integral between x and x plus delta x of f of t dt. So we can rewrite this entire expression, the derivative of capital F of x. This is capital F prime of x. We can rewrite it now as being equal to the limit as delta x approaches 0. This I can write as 1 over delta x times the numerator. We already figured out the numerator. The green area minus the blue area is just the purple area."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "We can rewrite it now as being equal to the limit as delta x approaches 0. This I can write as 1 over delta x times the numerator. We already figured out the numerator. The green area minus the blue area is just the purple area. And another way of denoting that area is this expression right over here. So 1 over delta x times the definite integral from x to x plus delta x of f of t dt. Now, this expression is interesting."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "The green area minus the blue area is just the purple area. And another way of denoting that area is this expression right over here. So 1 over delta x times the definite integral from x to x plus delta x of f of t dt. Now, this expression is interesting. This might look familiar from the mean value theorem of definite integrals. The mean value theorem of definite integrals tells us, the mean value theorem of definite integrals tells us, there exists a c in the interval. So I could see where, I'll write it this way, where a is less than or equal to c, which is less than, or actually, let me make it clear, the interval that we now care about is between x and x plus delta x."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "Now, this expression is interesting. This might look familiar from the mean value theorem of definite integrals. The mean value theorem of definite integrals tells us, the mean value theorem of definite integrals tells us, there exists a c in the interval. So I could see where, I'll write it this way, where a is less than or equal to c, which is less than, or actually, let me make it clear, the interval that we now care about is between x and x plus delta x. Where x is less than or equal to c, which is less than or equal to x plus delta x, such that the function evaluated at c, so let me draw this c. So there's a c someplace over here. So if I were to take the function evaluated at the c, that's f of c right over here. So if I were to take the function evaluated at the c, which would essentially be the height of this line, and I multiply it times the base, this interval, if I multiply it times the interval, and this interval is just delta x, x plus delta x minus x is just delta x."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "So I could see where, I'll write it this way, where a is less than or equal to c, which is less than, or actually, let me make it clear, the interval that we now care about is between x and x plus delta x. Where x is less than or equal to c, which is less than or equal to x plus delta x, such that the function evaluated at c, so let me draw this c. So there's a c someplace over here. So if I were to take the function evaluated at the c, that's f of c right over here. So if I were to take the function evaluated at the c, which would essentially be the height of this line, and I multiply it times the base, this interval, if I multiply it times the interval, and this interval is just delta x, x plus delta x minus x is just delta x. So if we just multiply the height times the base, that this is going to be equal to the area under the curve, which is the definite integral from x to x plus delta x, x to x plus delta x of f of t dt. This is what the mean value theorem of integrals tells us. If f is a continuous function, there exists a c in this interval between our two endpoints, where the function evaluated at the c is essentially, you can view it as the mean height."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "So if I were to take the function evaluated at the c, which would essentially be the height of this line, and I multiply it times the base, this interval, if I multiply it times the interval, and this interval is just delta x, x plus delta x minus x is just delta x. So if we just multiply the height times the base, that this is going to be equal to the area under the curve, which is the definite integral from x to x plus delta x, x to x plus delta x of f of t dt. This is what the mean value theorem of integrals tells us. If f is a continuous function, there exists a c in this interval between our two endpoints, where the function evaluated at the c is essentially, you can view it as the mean height. And if you take that mean value of the function and you multiply it times the base, you're going to get the area of the curve. Or another way of rewriting this, you could say that f of c, there exists a c in that interval where f of c is equal to 1 over delta x. I'm just dividing both sides by delta x, times the definite integral from x to x plus delta x of f of t dt. And this is often viewed as the mean value of the function over the interval."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "If f is a continuous function, there exists a c in this interval between our two endpoints, where the function evaluated at the c is essentially, you can view it as the mean height. And if you take that mean value of the function and you multiply it times the base, you're going to get the area of the curve. Or another way of rewriting this, you could say that f of c, there exists a c in that interval where f of c is equal to 1 over delta x. I'm just dividing both sides by delta x, times the definite integral from x to x plus delta x of f of t dt. And this is often viewed as the mean value of the function over the interval. Why is that? Well, this part right over here gives you the area, and then you divide the area by the base, and you get the mean height. Or another way you could say it is, if you were to take the height right over here, multiply it times the base, you get a rectangle that has the exact same area as the area under the curve."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "And this is often viewed as the mean value of the function over the interval. Why is that? Well, this part right over here gives you the area, and then you divide the area by the base, and you get the mean height. Or another way you could say it is, if you were to take the height right over here, multiply it times the base, you get a rectangle that has the exact same area as the area under the curve. Well, this is useful, because this is exactly what we got as the derivative of f prime of x. So there must exist a c such that f of c is equal to this stuff, or we could say that the limit, and let me rewrite all of this down a new color. So there exists a c in the interval x to x plus delta x, where f prime of x, which we know is equal to this, we can now say is now equal to the limit as delta x approaches 0."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "Or another way you could say it is, if you were to take the height right over here, multiply it times the base, you get a rectangle that has the exact same area as the area under the curve. Well, this is useful, because this is exactly what we got as the derivative of f prime of x. So there must exist a c such that f of c is equal to this stuff, or we could say that the limit, and let me rewrite all of this down a new color. So there exists a c in the interval x to x plus delta x, where f prime of x, which we know is equal to this, we can now say is now equal to the limit as delta x approaches 0. And instead of writing this, we know that there's some c that's equal to all of this business, of f of c. Now we're in the home stretch. We just have to figure out what the limit as delta x approaches 0 of f of c is. And the main realization is this part right over here."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "So there exists a c in the interval x to x plus delta x, where f prime of x, which we know is equal to this, we can now say is now equal to the limit as delta x approaches 0. And instead of writing this, we know that there's some c that's equal to all of this business, of f of c. Now we're in the home stretch. We just have to figure out what the limit as delta x approaches 0 of f of c is. And the main realization is this part right over here. We know that c is always sandwiched in between x and x plus delta x. And intuitively, you could tell that, look, as delta x approaches 0, as this green line right over here moves more and more to the left, as it approaches this blue line, the c has to be in between. And so the c is going to approach x."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "And the main realization is this part right over here. We know that c is always sandwiched in between x and x plus delta x. And intuitively, you could tell that, look, as delta x approaches 0, as this green line right over here moves more and more to the left, as it approaches this blue line, the c has to be in between. And so the c is going to approach x. So we know intuitively that c approaches x as delta x approaches 0. Or another way of saying it is that f of c is going to approach f of x as delta x approaches 0. And so intuitively, we could say that this is going to be equal to f of x."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "And so the c is going to approach x. So we know intuitively that c approaches x as delta x approaches 0. Or another way of saying it is that f of c is going to approach f of x as delta x approaches 0. And so intuitively, we could say that this is going to be equal to f of x. Now you might say, OK, that's intuitively, but we're kind of working on a little bit of a proof here, Sal. Let me know for sure that x is going to approach c. Don't just do this little thing where you drew this diagram and it makes sense that c is going to have to get closer and closer to x. And if you want that, you could just resort to the squeeze theorem."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "And so intuitively, we could say that this is going to be equal to f of x. Now you might say, OK, that's intuitively, but we're kind of working on a little bit of a proof here, Sal. Let me know for sure that x is going to approach c. Don't just do this little thing where you drew this diagram and it makes sense that c is going to have to get closer and closer to x. And if you want that, you could just resort to the squeeze theorem. And to resort to the squeeze theorem, you just have to view c as a function of delta x. And it really is. Depending on your delta x, c is going to be further to the left or to the right, possibly."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "And if you want that, you could just resort to the squeeze theorem. And to resort to the squeeze theorem, you just have to view c as a function of delta x. And it really is. Depending on your delta x, c is going to be further to the left or to the right, possibly. And so I can just rewrite this expression as x is less than or equal to c as a function of delta x, which is less than or equal to x plus delta x. So now you see that c is always sandwiched between x and x plus delta x. Well, what's the limit of x as delta x approaches zero?"}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "Depending on your delta x, c is going to be further to the left or to the right, possibly. And so I can just rewrite this expression as x is less than or equal to c as a function of delta x, which is less than or equal to x plus delta x. So now you see that c is always sandwiched between x and x plus delta x. Well, what's the limit of x as delta x approaches zero? Well, x isn't dependent on delta x in any way. So this is just going to be equal to x. What's the limit of x plus delta x as delta x approaches zero?"}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "Well, what's the limit of x as delta x approaches zero? Well, x isn't dependent on delta x in any way. So this is just going to be equal to x. What's the limit of x plus delta x as delta x approaches zero? Well, as delta x approaches zero, this is just going to be equal to x. So if this approaches x, as delta x approaches zero, and it's less than this function, and if this approaches x as delta x approaches zero and it's always greater than this, then we know from the squeeze theorem or the sandwich theorem that the limit as delta x approaches 0 of c as a function of delta x is going to be equal to x as well. It has to approach the same thing that that and that is."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "What's the limit of x plus delta x as delta x approaches zero? Well, as delta x approaches zero, this is just going to be equal to x. So if this approaches x, as delta x approaches zero, and it's less than this function, and if this approaches x as delta x approaches zero and it's always greater than this, then we know from the squeeze theorem or the sandwich theorem that the limit as delta x approaches 0 of c as a function of delta x is going to be equal to x as well. It has to approach the same thing that that and that is. It's sandwiched in between. And so that's a slight, we resort to the sandwich theorem. It's a little bit more rigorous to get to this exact result."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "It has to approach the same thing that that and that is. It's sandwiched in between. And so that's a slight, we resort to the sandwich theorem. It's a little bit more rigorous to get to this exact result. As delta x approaches 0, c approaches x. If c is approaching x, then f of c is going to approach f of x. And then we essentially have our proof."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "It's a little bit more rigorous to get to this exact result. As delta x approaches 0, c approaches x. If c is approaching x, then f of c is going to approach f of x. And then we essentially have our proof. f is a continuous function. We defined f in this way, capital F in this way. And we were able to use just the definition of the derivative to figure out that the derivative of capital F of x is equal to f of x."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "And then we essentially have our proof. f is a continuous function. We defined f in this way, capital F in this way. And we were able to use just the definition of the derivative to figure out that the derivative of capital F of x is equal to f of x. And once again, why is this a big deal? Well, it tells you that if you have any continuous function f, and that's what we assumed. We assumed that f is continuous over the interval."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "And we were able to use just the definition of the derivative to figure out that the derivative of capital F of x is equal to f of x. And once again, why is this a big deal? Well, it tells you that if you have any continuous function f, and that's what we assumed. We assumed that f is continuous over the interval. There exists some function. There exists a function. You can just define the function this way as the area under the curve between some end point or the beginning of the interval and some x."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "We assumed that f is continuous over the interval. There exists some function. There exists a function. You can just define the function this way as the area under the curve between some end point or the beginning of the interval and some x. If you define a function in that way, the derivative of this function is going to be equal to your continuous function. Or another way of saying it is that you always have an antiderivative, that any continuous function has an antiderivative. And so it's a couple of cool things."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "You can just define the function this way as the area under the curve between some end point or the beginning of the interval and some x. If you define a function in that way, the derivative of this function is going to be equal to your continuous function. Or another way of saying it is that you always have an antiderivative, that any continuous function has an antiderivative. And so it's a couple of cool things. Any continuous function has an antiderivative. It's going to be that capital F of x. And this is why it's called the fundamental theorem of calculus."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "And so it's a couple of cool things. Any continuous function has an antiderivative. It's going to be that capital F of x. And this is why it's called the fundamental theorem of calculus. It ties together these two ideas. And you have differential calculus. You have the idea of a derivative."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "And this is why it's called the fundamental theorem of calculus. It ties together these two ideas. And you have differential calculus. You have the idea of a derivative. And then in integral calculus, you have the idea of an integral. Before this proof, all we viewed an integral as is the area under the curve. It was just literally a notation to say the area under the curve."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "You have the idea of a derivative. And then in integral calculus, you have the idea of an integral. Before this proof, all we viewed an integral as is the area under the curve. It was just literally a notation to say the area under the curve. But now we've been able to make a connection, that there's a connection between the integral and the derivative, or a connection between the integral and the antiderivative in particular. So it connects all of calculus together in a very, very, very powerful. And we're so used to it now."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "It was just literally a notation to say the area under the curve. But now we've been able to make a connection, that there's a connection between the integral and the derivative, or a connection between the integral and the antiderivative in particular. So it connects all of calculus together in a very, very, very powerful. And we're so used to it now. And now we can say almost a somewhat obvious way. But it wasn't obvious. Remember, we always think of integrals as somehow doing an antiderivative."}, {"video_title": "Proof of fundamental theorem of calculus AP Calculus AB Khan Academy.mp3", "Sentence": "And we're so used to it now. And now we can say almost a somewhat obvious way. But it wasn't obvious. Remember, we always think of integrals as somehow doing an antiderivative. But it wasn't clear. If you just viewed an integral as only an area, you would have to go through this process and say, wow, no. It's connected."}, {"video_title": "Derivatives of tan(x) and cot(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "We know that the derivative with respect to x of sine of x is equal to cosine of x. We know that the derivative with respect to x of cosine of x is equal to negative sine of x. And so what we want to do in this video is find the derivatives of the other basic trig functions. So in particular, we know, let's figure out what the derivative with respect to x, let's first do tangent of x. Tangent of x. Well, this is the same thing as trying to find the derivative with respect to x of, well, tangent of x is just sine of x. Sine of x over cosine of x. And since it can be expressed as the quotient of two functions, we can apply the quotient rule here to evaluate this or to figure out what this is going to be."}, {"video_title": "Derivatives of tan(x) and cot(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So in particular, we know, let's figure out what the derivative with respect to x, let's first do tangent of x. Tangent of x. Well, this is the same thing as trying to find the derivative with respect to x of, well, tangent of x is just sine of x. Sine of x over cosine of x. And since it can be expressed as the quotient of two functions, we can apply the quotient rule here to evaluate this or to figure out what this is going to be. The quotient rule tells us that this is going to be the derivative of the top function, which we know is cosine of x, times the bottom function, which is cosine of x, so times cosine of x, minus the top function, which is sine of x, the top function, which is sine of x, sine of x, times the derivative of the bottom function, so the derivative of cosine of x is negative sine of x. So I could put the sine of x there, but where the negative can just cancel that out, and it's going to be over, over the bottom function squared. So cosine squared of x."}, {"video_title": "Derivatives of tan(x) and cot(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And since it can be expressed as the quotient of two functions, we can apply the quotient rule here to evaluate this or to figure out what this is going to be. The quotient rule tells us that this is going to be the derivative of the top function, which we know is cosine of x, times the bottom function, which is cosine of x, so times cosine of x, minus the top function, which is sine of x, the top function, which is sine of x, sine of x, times the derivative of the bottom function, so the derivative of cosine of x is negative sine of x. So I could put the sine of x there, but where the negative can just cancel that out, and it's going to be over, over the bottom function squared. So cosine squared of x. Now what is this? Well, what we have here, this is just cosine squared of x. This is just sine squared of x."}, {"video_title": "Derivatives of tan(x) and cot(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So cosine squared of x. Now what is this? Well, what we have here, this is just cosine squared of x. This is just sine squared of x. And we know from the Pythagorean identity, and this really just comes out of the unit circle definition, that cosine squared of x plus sine squared of x, well that's going to be equal to one for any x. So all of this is equal to one. And so we end up with one over cosine squared x, which is the same thing as secant of x squared."}, {"video_title": "Derivatives of tan(x) and cot(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "This is just sine squared of x. And we know from the Pythagorean identity, and this really just comes out of the unit circle definition, that cosine squared of x plus sine squared of x, well that's going to be equal to one for any x. So all of this is equal to one. And so we end up with one over cosine squared x, which is the same thing as secant of x squared. One over cosine of x is secant, so this is just secant of x squared. So that was pretty straightforward. Now let's just do the inverse of the, or you could say the reciprocal, I should say, of the tangent function, which is the cotangent."}, {"video_title": "Derivatives of tan(x) and cot(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And so we end up with one over cosine squared x, which is the same thing as secant of x squared. One over cosine of x is secant, so this is just secant of x squared. So that was pretty straightforward. Now let's just do the inverse of the, or you could say the reciprocal, I should say, of the tangent function, which is the cotangent. So that was fun, so let's do that. D dx of cotangent, not cosine, of cotangent of x, well, same idea. That's the derivative with respect to x."}, {"video_title": "Derivatives of tan(x) and cot(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's just do the inverse of the, or you could say the reciprocal, I should say, of the tangent function, which is the cotangent. So that was fun, so let's do that. D dx of cotangent, not cosine, of cotangent of x, well, same idea. That's the derivative with respect to x. And this time, let me make some sufficiently large brackets. So now this is cosine of x over sine of x. Over sine of x."}, {"video_title": "Derivatives of tan(x) and cot(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "That's the derivative with respect to x. And this time, let me make some sufficiently large brackets. So now this is cosine of x over sine of x. Over sine of x. But once again, we can use the quotient rule here. So this is going to be the derivative of the top function, which is negative, we're doing that magenta color, that is negative sine of x times the bottom function. So times sine of x, sine of x, minus, minus the top function, cosine of x, cosine of x, times the derivative of the bottom function, which is just going to be another cosine of x, and then all of that over the bottom function squared."}, {"video_title": "Derivatives of tan(x) and cot(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Over sine of x. But once again, we can use the quotient rule here. So this is going to be the derivative of the top function, which is negative, we're doing that magenta color, that is negative sine of x times the bottom function. So times sine of x, sine of x, minus, minus the top function, cosine of x, cosine of x, times the derivative of the bottom function, which is just going to be another cosine of x, and then all of that over the bottom function squared. So sine of x squared. Now what does this simplify to? Up here, let's see, this is sine squared of x, although we have a negative there, minus cosine squared of x, but we could factor out the negative, and this would be negative sine squared of x plus cosine squared of x."}, {"video_title": "Derivatives of tan(x) and cot(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So times sine of x, sine of x, minus, minus the top function, cosine of x, cosine of x, times the derivative of the bottom function, which is just going to be another cosine of x, and then all of that over the bottom function squared. So sine of x squared. Now what does this simplify to? Up here, let's see, this is sine squared of x, although we have a negative there, minus cosine squared of x, but we could factor out the negative, and this would be negative sine squared of x plus cosine squared of x. Well, this is just one by the Pythagorean identity, and so this is negative one over sine squared x. Negative one over sine squared x, and that is the same thing as negative cosecant squared of, I'm running out of space, of x. There you go."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "In this case, it will be with respect to our parameter t. So let me draw some new stuff right here. So let's say I have the vector-valued function R of t, and this is no different than what I did in the last video, x of t times unit vector i plus y of t times the unit vector j. If we're dealing in three dimensions, we'd add a z of t times k, but let's keep things relatively simple. And let's say that this describes a curve, and let's say the curve we're dealing with between t is between a and b, and this curve will look something like... let me do my best effort to draw the curve. Let's draw some random curve here. So let's say the curve looks something like that. This is when t is equal to a, so it's going to go in this direction."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "And let's say that this describes a curve, and let's say the curve we're dealing with between t is between a and b, and this curve will look something like... let me do my best effort to draw the curve. Let's draw some random curve here. So let's say the curve looks something like that. This is when t is equal to a, so it's going to go in this direction. This is when t is equal to b right here. This is t is equal to a, so this right here would be x of a. This right here is y of a."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "This is when t is equal to a, so it's going to go in this direction. This is when t is equal to b right here. This is t is equal to a, so this right here would be x of a. This right here is y of a. And similarly, this up here, this is x of b, and this over here is y of b. Now, we saw in the last video that the endpoints of these position vectors are what's describing this curve. So r of a we saw in the last video."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "This right here is y of a. And similarly, this up here, this is x of b, and this over here is y of b. Now, we saw in the last video that the endpoints of these position vectors are what's describing this curve. So r of a we saw in the last video. It describes that point right there. I don't want to review that too much. But what I want to do is think about what is the difference between two points."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "So r of a we saw in the last video. It describes that point right there. I don't want to review that too much. But what I want to do is think about what is the difference between two points. So let's say that we take some random point here. Let's say some random t here. Let's call that r of t. Actually, I'm going to do a different point just because I want to make it a little bit clearer."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "But what I want to do is think about what is the difference between two points. So let's say that we take some random point here. Let's say some random t here. Let's call that r of t. Actually, I'm going to do a different point just because I want to make it a little bit clearer. So let's say that that right there is r of some t, some particular t right there. That is r of t. It's going to be a plus something. So that's some particular t. And let's say that we want to figure out, and let's say we increase t by a little bit, by h. So let's say that r of t plus h, well, if we view the parameter t as time, we can kind of view we've moved forward in time by some amount."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "Let's call that r of t. Actually, I'm going to do a different point just because I want to make it a little bit clearer. So let's say that that right there is r of some t, some particular t right there. That is r of t. It's going to be a plus something. So that's some particular t. And let's say that we want to figure out, and let's say we increase t by a little bit, by h. So let's say that r of t plus h, well, if we view the parameter t as time, we can kind of view we've moved forward in time by some amount. So our little particle has moved a little bit. And let's say that we're over here. So that is that right there in yellow is r of t plus h. Just a slightly larger value for h. Now, one question we might ask ourselves is how quickly is r changing with respect to t?"}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "So that's some particular t. And let's say that we want to figure out, and let's say we increase t by a little bit, by h. So let's say that r of t plus h, well, if we view the parameter t as time, we can kind of view we've moved forward in time by some amount. So our little particle has moved a little bit. And let's say that we're over here. So that is that right there in yellow is r of t plus h. Just a slightly larger value for h. Now, one question we might ask ourselves is how quickly is r changing with respect to t? So the first thing we might want to say is what's the difference between these two? If I were to take, and I want to visualize it, if I were to take r, the position vector that we get by evaluating r at t plus h, and from that I were to subtract r of t, what do we get? Well, you might want to review some of your vector algebra, but we're essentially just going to get this vector."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "So that is that right there in yellow is r of t plus h. Just a slightly larger value for h. Now, one question we might ask ourselves is how quickly is r changing with respect to t? So the first thing we might want to say is what's the difference between these two? If I were to take, and I want to visualize it, if I were to take r, the position vector that we get by evaluating r at t plus h, and from that I were to subtract r of t, what do we get? Well, you might want to review some of your vector algebra, but we're essentially just going to get this vector. We're going to let me do it in a nice vibrant color. We're going to get this vector right there that I'm doing in magenta. So that magenta vector right there is the vector r of t plus h minus r of t. And it should make sense because when you add vectors, you go heads to tails."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "Well, you might want to review some of your vector algebra, but we're essentially just going to get this vector. We're going to let me do it in a nice vibrant color. We're going to get this vector right there that I'm doing in magenta. So that magenta vector right there is the vector r of t plus h minus r of t. And it should make sense because when you add vectors, you go heads to tails. You could alternatively write this as r of t plus this character right here, plus r of t plus h minus r of t. When you add two vectors, you're adding, let me make it very clear, I'm adding this vector to this vector right here. You put the tail of the second vector at the head of the first. So this is the first vector, and then I put the tail of the second there, and then the sum of those two, as we predicted, should be equal to this last one."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "So that magenta vector right there is the vector r of t plus h minus r of t. And it should make sense because when you add vectors, you go heads to tails. You could alternatively write this as r of t plus this character right here, plus r of t plus h minus r of t. When you add two vectors, you're adding, let me make it very clear, I'm adding this vector to this vector right here. You put the tail of the second vector at the head of the first. So this is the first vector, and then I put the tail of the second there, and then the sum of those two, as we predicted, should be equal to this last one. It should be equal to r of t plus h. And we see that is the case, and even algebraically, you would see that obviously this guy and that guy are going to cancel out. So hopefully that satisfies you. And I want to be clear, this all of a sudden, this isn't a position vector."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "So this is the first vector, and then I put the tail of the second there, and then the sum of those two, as we predicted, should be equal to this last one. It should be equal to r of t plus h. And we see that is the case, and even algebraically, you would see that obviously this guy and that guy are going to cancel out. So hopefully that satisfies you. And I want to be clear, this all of a sudden, this isn't a position vector. We're not saying that, hey, let's nail this guy's tail at the origin and use this guy to describe a unique position. Now all of a sudden, it's just kind of a pure vector. It's describing just a change between two other position vectors."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "And I want to be clear, this all of a sudden, this isn't a position vector. We're not saying that, hey, let's nail this guy's tail at the origin and use this guy to describe a unique position. Now all of a sudden, it's just kind of a pure vector. It's describing just a change between two other position vectors. So this guy is right out here, but this vector literally describes the change. But let's say we care, and how would this look algebraically if we were to expand it like that? So this is going to be equal to, what's r of t plus h?"}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "It's describing just a change between two other position vectors. So this guy is right out here, but this vector literally describes the change. But let's say we care, and how would this look algebraically if we were to expand it like that? So this is going to be equal to, what's r of t plus h? That's the same thing as x of t plus h times the unit vector i plus y of t plus h times the unit vector j. That's just that piece. That piece right there is that piece."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "So this is going to be equal to, what's r of t plus h? That's the same thing as x of t plus h times the unit vector i plus y of t plus h times the unit vector j. That's just that piece. That piece right there is that piece. Minus, so this piece, so minus, I'll do it in the second line. I could have done it out here, but I'm running out of space. Minus x of t, right?"}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "That piece right there is that piece. Minus, so this piece, so minus, I'll do it in the second line. I could have done it out here, but I'm running out of space. Minus x of t, right? R of t is just x of t times i plus, but I'll just distribute the minus sign. So it's minus y of t times j. Actually, let me write it this way, plus this."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "Minus x of t, right? R of t is just x of t times i plus, but I'll just distribute the minus sign. So it's minus y of t times j. Actually, let me write it this way, plus this. So you realize that this is really just this guy right here. I'm just evaluating at t. So you have x of t and y of t, and then later we can distribute. If you distribute this minus sign, you get a minus x of t and a minus y of t. And in vector addition, you might need a little review on this if you haven't seen it in a while, you can just add the corresponding components."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "Actually, let me write it this way, plus this. So you realize that this is really just this guy right here. I'm just evaluating at t. So you have x of t and y of t, and then later we can distribute. If you distribute this minus sign, you get a minus x of t and a minus y of t. And in vector addition, you might need a little review on this if you haven't seen it in a while, you can just add the corresponding components. You can add the x components and you can add the y components. So this is going to be equal to, let me rewrite it over here, because I think I'm going to need some space later on. So let me rewrite it over here."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "If you distribute this minus sign, you get a minus x of t and a minus y of t. And in vector addition, you might need a little review on this if you haven't seen it in a while, you can just add the corresponding components. You can add the x components and you can add the y components. So this is going to be equal to, let me rewrite it over here, because I think I'm going to need some space later on. So let me rewrite it over here. So I have R of t plus h minus R of t is equal to, now I'm just going to group the x and the y components. This is equal to the x components added together, but this is a negative, so we're going to subtract this guy from that guy. So we have x of t plus h minus x of t, and then all of that times our unit vector in the x direction."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "So let me rewrite it over here. So I have R of t plus h minus R of t is equal to, now I'm just going to group the x and the y components. This is equal to the x components added together, but this is a negative, so we're going to subtract this guy from that guy. So we have x of t plus h minus x of t, and then all of that times our unit vector in the x direction. And then we'll have plus y of t plus h minus y of t times the unit vector in the j direction. I'm just rearranging things right now. So this will tell us what is our change between any two R's for a given change in distance."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "So we have x of t plus h minus x of t, and then all of that times our unit vector in the x direction. And then we'll have plus y of t plus h minus y of t times the unit vector in the j direction. I'm just rearranging things right now. So this will tell us what is our change between any two R's for a given change in distance. And our change in distance here is h between any two position vectors. Now, what I set out at the beginning of this video, I said, well, I want to figure out the change, and we're going to think about the instantaneous change with respect to t. So I want to see, well, how much did this change over a period of h? Instead of writing h, we could have written delta t. It would have been the same thing."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "So this will tell us what is our change between any two R's for a given change in distance. And our change in distance here is h between any two position vectors. Now, what I set out at the beginning of this video, I said, well, I want to figure out the change, and we're going to think about the instantaneous change with respect to t. So I want to see, well, how much did this change over a period of h? Instead of writing h, we could have written delta t. It would have been the same thing. So I want to divide this by h. So I want to say, look, my vector has changed this much, but I want to say it's over a period of h. And this is analogous to when we do slope. We say rise over run over delta y or change in y over change in x. This is kind of the change in our function per change in x."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "Instead of writing h, we could have written delta t. It would have been the same thing. So I want to divide this by h. So I want to say, look, my vector has changed this much, but I want to say it's over a period of h. And this is analogous to when we do slope. We say rise over run over delta y or change in y over change in x. This is kind of the change in our function per change in x. Let's just divide everything, or I shouldn't say change in x, per change in t. So here our change in t is h. The difference between t plus h and t is just going to be h. And so we're going to divide everything by h. When you multiply a vector by some scalar or divide it by some scalar, you're just taking each of its components and multiplying or dividing by that scalar. And we get that right there. So this, for any finite difference right here, h, this will tell us how much our vector changes per h. But if we want to find the instantaneous change, just like what we did when we first learned differential calculus, we said, okay, this is kind of analogous to a slope."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "This is kind of the change in our function per change in x. Let's just divide everything, or I shouldn't say change in x, per change in t. So here our change in t is h. The difference between t plus h and t is just going to be h. And so we're going to divide everything by h. When you multiply a vector by some scalar or divide it by some scalar, you're just taking each of its components and multiplying or dividing by that scalar. And we get that right there. So this, for any finite difference right here, h, this will tell us how much our vector changes per h. But if we want to find the instantaneous change, just like what we did when we first learned differential calculus, we said, okay, this is kind of analogous to a slope. This would be good. This would work out well for us if the path under question looked something like this, if it was a linear path. If our path looked something like this, we could just calculate this and we'll essentially have the average change in our position vectors."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "So this, for any finite difference right here, h, this will tell us how much our vector changes per h. But if we want to find the instantaneous change, just like what we did when we first learned differential calculus, we said, okay, this is kind of analogous to a slope. This would be good. This would work out well for us if the path under question looked something like this, if it was a linear path. If our path looked something like this, we could just calculate this and we'll essentially have the average change in our position vectors. So you can imagine two position vectors. That's one of them. Well, actually, they'd all be parallel."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "If our path looked something like this, we could just calculate this and we'll essentially have the average change in our position vectors. So you can imagine two position vectors. That's one of them. Well, actually, they'd all be parallel. Well, they don't always. Well, the position vectors, they don't have to be parallel, but I won't. They don't have to be parallel."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "Well, actually, they'd all be parallel. Well, they don't always. Well, the position vectors, they don't have to be parallel, but I won't. They don't have to be parallel. They could be like that. And then this would just describe the change between these two per h, or how quickly are the position vectors changing per our change in our parameter. The h you could also consider as kind of a delta t. Sometimes people find the h simpler."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "They don't have to be parallel. They could be like that. And then this would just describe the change between these two per h, or how quickly are the position vectors changing per our change in our parameter. The h you could also consider as kind of a delta t. Sometimes people find the h simpler. Sometimes they find the delta t. But anyway, I'm concerned with the instantaneous. We're dealing with curves. We're dealing with calculus."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "The h you could also consider as kind of a delta t. Sometimes people find the h simpler. Sometimes they find the delta t. But anyway, I'm concerned with the instantaneous. We're dealing with curves. We're dealing with calculus. This would have been okay if we were just in an algebraic linear world. So what do we do? Well, maybe we can just take the limit as h approaches zero."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "We're dealing with calculus. This would have been okay if we were just in an algebraic linear world. So what do we do? Well, maybe we can just take the limit as h approaches zero. So let me scroll this over. So let's just take the limit. Let me do this in a nice, vibrant color."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "Well, maybe we can just take the limit as h approaches zero. So let me scroll this over. So let's just take the limit. Let me do this in a nice, vibrant color. Let's take the limit as h approaches zero of both sides of this. So here, too, I'm going to take the limit as h approaches zero. And here, too, I'm going to take the limit as h approaches zero."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "Let me do this in a nice, vibrant color. Let's take the limit as h approaches zero of both sides of this. So here, too, I'm going to take the limit as h approaches zero. And here, too, I'm going to take the limit as h approaches zero. So I just want to say, well, what happens? How much do I change per a change in my parameter t? But what's kind of the instantaneous change?"}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "And here, too, I'm going to take the limit as h approaches zero. So I just want to say, well, what happens? How much do I change per a change in my parameter t? But what's kind of the instantaneous change? As the difference gets smaller and smaller and smaller. This is exactly what we first learned when we learned about instantaneous slope or instantaneous velocity or slope of a tangent line. Well, this thing looks a little bit undefined to me right now."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "But what's kind of the instantaneous change? As the difference gets smaller and smaller and smaller. This is exactly what we first learned when we learned about instantaneous slope or instantaneous velocity or slope of a tangent line. Well, this thing looks a little bit undefined to me right now. We haven't defined limits for vector-valued functions. We haven't defined derivatives for vector-valued functions. But lucky for us, all of this stuff here looks pretty familiar."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "Well, this thing looks a little bit undefined to me right now. We haven't defined limits for vector-valued functions. We haven't defined derivatives for vector-valued functions. But lucky for us, all of this stuff here looks pretty familiar. This is actually the definition of our derivative. And these are scalar-valued functions right here. They're multiplied by vectors in order for us to get vector-valued functions."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "But lucky for us, all of this stuff here looks pretty familiar. This is actually the definition of our derivative. And these are scalar-valued functions right here. They're multiplied by vectors in order for us to get vector-valued functions. But this right here, by definition, this is the derivative. This is x prime of t. Or this is dx dt. This right here is y prime of t. Or we could write that as dy dt."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "They're multiplied by vectors in order for us to get vector-valued functions. But this right here, by definition, this is the derivative. This is x prime of t. Or this is dx dt. This right here is y prime of t. Or we could write that as dy dt. So all of a sudden, we can define, we can say, and I'm being a little hand-wavy here, but I want to give you the intuition more than anything. We can say that the derivative, we can call this expression right here as the derivative of my vector-valued function r with respect to t. Or we could call it dr dt. Notice I keep the vector signs there."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "This right here is y prime of t. Or we could write that as dy dt. So all of a sudden, we can define, we can say, and I'm being a little hand-wavy here, but I want to give you the intuition more than anything. We can say that the derivative, we can call this expression right here as the derivative of my vector-valued function r with respect to t. Or we could call it dr dt. Notice I keep the vector signs there. This is its derivative, and all it's going to be equal to, all it's going to be equal to, r prime of t, is going to be equal to, well, this is just the derivative of x with respect to t, is equal to x prime of t times the x unit vector, the horizontal unit vector, plus y prime of t times the y unit vector, times j, the unit vector in the horizontal direction. That's a pretty nice and simple outcome. But the harder thing, maybe, is to kind of visualize what it represents."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "Notice I keep the vector signs there. This is its derivative, and all it's going to be equal to, all it's going to be equal to, r prime of t, is going to be equal to, well, this is just the derivative of x with respect to t, is equal to x prime of t times the x unit vector, the horizontal unit vector, plus y prime of t times the y unit vector, times j, the unit vector in the horizontal direction. That's a pretty nice and simple outcome. But the harder thing, maybe, is to kind of visualize what it represents. So if we think about what happens, let me draw a big graph, just to get the visualization going in a healthy way. So let's say my curve looks something like this. That's my curve."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "But the harder thing, maybe, is to kind of visualize what it represents. So if we think about what happens, let me draw a big graph, just to get the visualization going in a healthy way. So let's say my curve looks something like this. That's my curve. Let's say that this is, we want to figure out the instantaneous change at this point right here, so that is r of t. And then if we take r of t plus h, we saw this already, t plus h might be something like right there. So this is r of t plus h. Now, right now, the difference between these two, and this is just the numerator when you take the difference, or how fast we're changing from this vector to that vector, in terms of t, and it's hard to visualize here. And I'm going to do a whole video so we can think about the magnitude here."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "That's my curve. Let's say that this is, we want to figure out the instantaneous change at this point right here, so that is r of t. And then if we take r of t plus h, we saw this already, t plus h might be something like right there. So this is r of t plus h. Now, right now, the difference between these two, and this is just the numerator when you take the difference, or how fast we're changing from this vector to that vector, in terms of t, and it's hard to visualize here. And I'm going to do a whole video so we can think about the magnitude here. That might be some vector. Well, the difference between these two is just going to be that. But then when you divide it by h, it's going to be a larger vector."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "And I'm going to do a whole video so we can think about the magnitude here. That might be some vector. Well, the difference between these two is just going to be that. But then when you divide it by h, it's going to be a larger vector. If we assume h is a small number, let's say h is less than 1, we're going to get a larger vector. But this is kind of the average change over this time. But as h gets smaller and smaller and smaller, this r prime of t, its direction is going to be tangential to the curve."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "But then when you divide it by h, it's going to be a larger vector. If we assume h is a small number, let's say h is less than 1, we're going to get a larger vector. But this is kind of the average change over this time. But as h gets smaller and smaller and smaller, this r prime of t, its direction is going to be tangential to the curve. And I think you can visualize that. As these two guys get closer and closer and closer, the dr's get smaller, so the change, the dr, the difference between the two, the delta r's get smaller and smaller. You can imagine if h was even smaller, if it was right here."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "But as h gets smaller and smaller and smaller, this r prime of t, its direction is going to be tangential to the curve. And I think you can visualize that. As these two guys get closer and closer and closer, the dr's get smaller, so the change, the dr, the difference between the two, the delta r's get smaller and smaller. You can imagine if h was even smaller, if it was right here. All of a sudden the difference between those two vectors is getting smaller. And it's getting more and more tangential to the curve. But then we're also dividing by a smaller h. So the actual derivative, as the limit as h approaches 0, it might be, maybe it's even a bigger number there."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "You can imagine if h was even smaller, if it was right here. All of a sudden the difference between those two vectors is getting smaller. And it's getting more and more tangential to the curve. But then we're also dividing by a smaller h. So the actual derivative, as the limit as h approaches 0, it might be, maybe it's even a bigger number there. And actually, the magnitude of this vector, it's a little hard to visualize, it's going to be dependent on a parametrization for the curve. It's not dependent on the shape of the curve. The direction of this vector is dependent on the shape of this curve."}, {"video_title": "Derivative of a position vector valued function Multivariable Calculus Khan Academy.mp3", "Sentence": "But then we're also dividing by a smaller h. So the actual derivative, as the limit as h approaches 0, it might be, maybe it's even a bigger number there. And actually, the magnitude of this vector, it's a little hard to visualize, it's going to be dependent on a parametrization for the curve. It's not dependent on the shape of the curve. The direction of this vector is dependent on the shape of this curve. So the direction, this will be tangent to the curve. Or you could imagine that this vector is on the tangent line to the curve. The magnitude of it is a little bit harder to understand."}, {"video_title": "Solving the logistic differential equation part 2 Khan Academy.mp3", "Sentence": "So we left off in part one getting pretty close to finding our N of T that satisfies the logistic differential equation where its initial condition is between zero and K, and now we just have to really just do some algebra to finish things up. So we left with this, that for our N of T, this must be true. Now we could use a little bit of logarithm properties to rewrite this left-hand side as the logarithm of, have the logarithm of something minus the logarithm of something else, that's going to be the first something, the logarithm of the first something divided by the second something. One minus N over K, and this is of course going to be equal to all this business that we have, it's going to be equal to, actually let me just write it, it's going to be equal to, equal to R times T plus C, plus C, and now what we could do, this is the same thing as saying that E to the RT plus C is going to be equal to this thing, right over here. The natural log of this is equal to the exponent that I have to raise E to to get to this right over here, so I can just write that. We could just write, or another way of thinking about it is we could take E to, if this is equal to that, we could take E to this power on the left-hand side and E to this power on this right-hand side, and they should be equal. So E to this power is just going to be what's inside the parentheses, it's just going to be N, it's just going to be N over one minus, one minus N over K, and I'll do that in green color so you can just keep track of where things came from, is equal to E to this business, is equal to E to the RT, I'll do the T in white, RT plus C. Now, this I could rewrite if I want to."}, {"video_title": "Solving the logistic differential equation part 2 Khan Academy.mp3", "Sentence": "One minus N over K, and this is of course going to be equal to all this business that we have, it's going to be equal to, actually let me just write it, it's going to be equal to, equal to R times T plus C, plus C, and now what we could do, this is the same thing as saying that E to the RT plus C is going to be equal to this thing, right over here. The natural log of this is equal to the exponent that I have to raise E to to get to this right over here, so I can just write that. We could just write, or another way of thinking about it is we could take E to, if this is equal to that, we could take E to this power on the left-hand side and E to this power on this right-hand side, and they should be equal. So E to this power is just going to be what's inside the parentheses, it's just going to be N, it's just going to be N over one minus, one minus N over K, and I'll do that in green color so you can just keep track of where things came from, is equal to E to this business, is equal to E to the RT, I'll do the T in white, RT plus C. Now, this I could rewrite if I want to. If I have E to something plus something else, I could rewrite this as to the RT times E to the C, and just to simplify things, this is just going to be another constant here, and if this is C, I could call it C1, but I'm just going to call that a constant. So I could just say that that's going to be equal to some constant times E to the RT, and now we just need to solve for N, and once again, if at any point while we're working on this, you get inspired, feel free to solve for N. So let's see, one way to solve for N, let's see, if we could take the reciprocal of both sides of this, we're going to get one, let me draw a line here just so you know that we're, so let me draw a line. So if we take the reciprocal of both sides of this, we're going to get one over N, one minus N over K over light green N, is equal to, is equal to, and so let's, you know, we could say it's equal to, it's equal to one over C times E to the negative RT, but one over C, that's just going to be another constant."}, {"video_title": "Solving the logistic differential equation part 2 Khan Academy.mp3", "Sentence": "So E to this power is just going to be what's inside the parentheses, it's just going to be N, it's just going to be N over one minus, one minus N over K, and I'll do that in green color so you can just keep track of where things came from, is equal to E to this business, is equal to E to the RT, I'll do the T in white, RT plus C. Now, this I could rewrite if I want to. If I have E to something plus something else, I could rewrite this as to the RT times E to the C, and just to simplify things, this is just going to be another constant here, and if this is C, I could call it C1, but I'm just going to call that a constant. So I could just say that that's going to be equal to some constant times E to the RT, and now we just need to solve for N, and once again, if at any point while we're working on this, you get inspired, feel free to solve for N. So let's see, one way to solve for N, let's see, if we could take the reciprocal of both sides of this, we're going to get one, let me draw a line here just so you know that we're, so let me draw a line. So if we take the reciprocal of both sides of this, we're going to get one over N, one minus N over K over light green N, is equal to, is equal to, and so let's, you know, we could say it's equal to, it's equal to one over C times E to the negative RT, but one over C, that's just going to be another constant. So I could write one over C here, when I take the reciprocal, but once again, this is just going to be another constant, so I'm going to be a little bit hand-wavy. Well, okay, we're going to get another constant here. I could have called, you know, this C1, this C2, I could call this C3 if I want to make it clear that these are not going to be the same number."}, {"video_title": "Solving the logistic differential equation part 2 Khan Academy.mp3", "Sentence": "So if we take the reciprocal of both sides of this, we're going to get one over N, one minus N over K over light green N, is equal to, is equal to, and so let's, you know, we could say it's equal to, it's equal to one over C times E to the negative RT, but one over C, that's just going to be another constant. So I could write one over C here, when I take the reciprocal, but once again, this is just going to be another constant, so I'm going to be a little bit hand-wavy. Well, okay, we're going to get another constant here. I could have called, you know, this C1, this C2, I could call this C3 if I want to make it clear that these are not going to be the same number. This is E to this power, this is the reciprocal of that. Actually, maybe I'll do that just to make it a little bit, a little bit clearer. That would have been C1 right over there."}, {"video_title": "Solving the logistic differential equation part 2 Khan Academy.mp3", "Sentence": "I could have called, you know, this C1, this C2, I could call this C3 if I want to make it clear that these are not going to be the same number. This is E to this power, this is the reciprocal of that. Actually, maybe I'll do that just to make it a little bit, a little bit clearer. That would have been C1 right over there. And so this is going to be, sorry, the reciprocal of this is C3, and E to the negative, the reciprocal of E to the RT is E to the negative RT, E to the negative, E to the negative RE to the negative RT. And let's see, if we divide the numerator and the denominator by N, or if we divide, one way to think about it, we divide this term by N, we're going to get, so this term by N is going to be 1 over N, and then this term by N is just going to be minus 1 over K, so this is just going to be minus 1 over K is equal to this, is equal to this business. So copy and paste."}, {"video_title": "Solving the logistic differential equation part 2 Khan Academy.mp3", "Sentence": "That would have been C1 right over there. And so this is going to be, sorry, the reciprocal of this is C3, and E to the negative, the reciprocal of E to the RT is E to the negative RT, E to the negative, E to the negative RE to the negative RT. And let's see, if we divide the numerator and the denominator by N, or if we divide, one way to think about it, we divide this term by N, we're going to get, so this term by N is going to be 1 over N, and then this term by N is just going to be minus 1 over K, so this is just going to be minus 1 over K is equal to this, is equal to this business. So copy and paste. It's going to be equal to that. This is good algebra practice here. Now let's see, we could take this 1 over K, add it to both sides, so let's do that."}, {"video_title": "Solving the logistic differential equation part 2 Khan Academy.mp3", "Sentence": "So copy and paste. It's going to be equal to that. This is good algebra practice here. Now let's see, we could take this 1 over K, add it to both sides, so let's do that. So let me just cut, so let me cut and paste it. I'm going to add it to both sides, so this should be a plus 1 over K. So plus 1 over K. And now to solve for N, I just take the reciprocal of both sides. So I'm going to get N, and I'll write it in kind of the function notation, N of T, actually let me make my T in white since I've been taking the trouble all this time of rewriting this in white, is equal to 1 over, is going to be equal to 1 over all of this business, is going to be equal to 1 over all of this business."}, {"video_title": "Solving the logistic differential equation part 2 Khan Academy.mp3", "Sentence": "Now let's see, we could take this 1 over K, add it to both sides, so let's do that. So let me just cut, so let me cut and paste it. I'm going to add it to both sides, so this should be a plus 1 over K. So plus 1 over K. And now to solve for N, I just take the reciprocal of both sides. So I'm going to get N, and I'll write it in kind of the function notation, N of T, actually let me make my T in white since I've been taking the trouble all this time of rewriting this in white, is equal to 1 over, is going to be equal to 1 over all of this business, is going to be equal to 1 over all of this business. So copy and paste is going to be equal to that. And that by itself, that by itself is already interesting. That by itself is already interesting."}, {"video_title": "Solving the logistic differential equation part 2 Khan Academy.mp3", "Sentence": "So I'm going to get N, and I'll write it in kind of the function notation, N of T, actually let me make my T in white since I've been taking the trouble all this time of rewriting this in white, is equal to 1 over, is going to be equal to 1 over all of this business, is going to be equal to 1 over all of this business. So copy and paste is going to be equal to that. And that by itself, that by itself is already interesting. That by itself is already interesting. So I could write it like this, and if I want, if I don't like, let's see, well yeah, I could just, if I don't like having this K, you know, kind of a fraction in a fraction, I could rewrite it as, actually maybe I'll do it over here, N of T is equal to, I'll just multiply the numerator and the denominator by, I'll just multiply it, actually let me just leave it like that for now. And what I'm going to do, what I'm now going to do is I'm going to say, well look, you know we're assuming that N of 0 is N sub naught. So we're assuming that N of 0, N of 0 is equal to N sub naught."}, {"video_title": "Solving the logistic differential equation part 2 Khan Academy.mp3", "Sentence": "That by itself is already interesting. So I could write it like this, and if I want, if I don't like, let's see, well yeah, I could just, if I don't like having this K, you know, kind of a fraction in a fraction, I could rewrite it as, actually maybe I'll do it over here, N of T is equal to, I'll just multiply the numerator and the denominator by, I'll just multiply it, actually let me just leave it like that for now. And what I'm going to do, what I'm now going to do is I'm going to say, well look, you know we're assuming that N of 0 is N sub naught. So we're assuming that N of 0, N of 0 is equal to N sub naught. So let's write this thing, let's solve for the constant, let's figure out what this could be if we know what our initial condition is. So N of 0, N of 0 is going to be equal to, it's going to be equal to 1, 1 over, when T is 0, this is just going to be equal to 1, this is just going to be our constant, C3 plus 1 over K, plus 1 over the maximum population that our environment can handle, and that's going to be equal to N naught, and now we can solve for our constant. So we get, actually I'm probably going to need a lot of real estate for this, I can take the reciprocal of both sides again, so this is something that we're doing a lot of, C3 plus 1 over K is equal to 1 over N naught, just took the reciprocal of both sides, and so we get our constant, C3 is equal to 1 over N naught, minus 1 over K, and so we can rewrite our solution, which we'll call the logistic function, we get, this is fun now, N of T is equal to 1 over, our constant is this, so it's going to be, let me copy and paste this, so copy and paste, it's going to be that, that's our constant, so it's going to be that, times E to the negative RT, to the negative RT, plus 1 over K, and if we don't like having all of these denominators, all of these fractions in the denominator, why don't we multiply everything times, the numerator and the denominator by N naught K, so I'm going to multiply the numerator times N naught K, and I'm going to multiply the denominator by N naught K, and so what do we get?"}, {"video_title": "Solving the logistic differential equation part 2 Khan Academy.mp3", "Sentence": "So we're assuming that N of 0, N of 0 is equal to N sub naught. So let's write this thing, let's solve for the constant, let's figure out what this could be if we know what our initial condition is. So N of 0, N of 0 is going to be equal to, it's going to be equal to 1, 1 over, when T is 0, this is just going to be equal to 1, this is just going to be our constant, C3 plus 1 over K, plus 1 over the maximum population that our environment can handle, and that's going to be equal to N naught, and now we can solve for our constant. So we get, actually I'm probably going to need a lot of real estate for this, I can take the reciprocal of both sides again, so this is something that we're doing a lot of, C3 plus 1 over K is equal to 1 over N naught, just took the reciprocal of both sides, and so we get our constant, C3 is equal to 1 over N naught, minus 1 over K, and so we can rewrite our solution, which we'll call the logistic function, we get, this is fun now, N of T is equal to 1 over, our constant is this, so it's going to be, let me copy and paste this, so copy and paste, it's going to be that, that's our constant, so it's going to be that, times E to the negative RT, to the negative RT, plus 1 over K, and if we don't like having all of these denominators, all of these fractions in the denominator, why don't we multiply everything times, the numerator and the denominator by N naught K, so I'm going to multiply the numerator times N naught K, and I'm going to multiply the denominator by N naught K, and so what do we get? This is all going to be equal to, in the numerator, I have N naught times K, and in the denominator, I am going to have, let's see, if I multiply this term right over here, times N naught K, I'm going to have K, if I multiply this term times N naught K, I'm going to have N naught, so it's going to be minus N naught, minus N naught, times E to the negative RT, times E to the negative RT, negative RT, and then if I multiply this times N naught K, I'm going to get N naught, so plus N naught, and there you have it, we have found a solution for the logistic differential equation. We will call this the logistic function, and in future videos, we will explore it more, and we will see that it actually does, if you were to plot this, I encourage you to do so, either on the internet, you could try Wolfram-Alpha, or if you're on your graphing calculator, you will see that it has the exact properties that we want it to have. It starts at N naught, it starts to increase at increasing rates, but then it starts to slow down as we reach our maximum population, and so that is actually a very neat function."}, {"video_title": "Partial sums term value from partial sum Series AP Calculus BC Khan Academy.mp3", "Sentence": "And they want us to figure out what is the actual seventh term. And like always, pause this video and see if you can figure it out on your own before we work through it together. All right, so one way to think about it is a sub seven, let's think about how that relates to different sums. So if we have a sub one plus a sub two, I'll just go all the way, a sub three plus a sub four plus a sub five plus a sub six plus a sub seven. So if I were to sum all of these things together, that, this entire sum, that would be s sub seven. And if I wanted to figure out a sub seven, well, I could subtract from that, I could subtract out the sum of the first six terms. So I could sum, I could subtract out s sub, I could subtract out s sub six."}, {"video_title": "Partial sums term value from partial sum Series AP Calculus BC Khan Academy.mp3", "Sentence": "So if we have a sub one plus a sub two, I'll just go all the way, a sub three plus a sub four plus a sub five plus a sub six plus a sub seven. So if I were to sum all of these things together, that, this entire sum, that would be s sub seven. And if I wanted to figure out a sub seven, well, I could subtract from that, I could subtract out the sum of the first six terms. So I could sum, I could subtract out s sub, I could subtract out s sub six. So once again, what am I doing here? What is my strategy? I know the formula for the sum of the first n terms."}, {"video_title": "Partial sums term value from partial sum Series AP Calculus BC Khan Academy.mp3", "Sentence": "So I could sum, I could subtract out s sub, I could subtract out s sub six. So once again, what am I doing here? What is my strategy? I know the formula for the sum of the first n terms. I can use that to say, okay, I can figure out the sum of the first seven terms, that's gonna be the sum of all of these, and then I can use that same formula to figure out the sum of the first six terms. And the difference between the two, well, that's going to be our a sub seven. So another way of saying what I just said is that a sub seven is going to be the sum of the first seven terms minus the sum of the first six terms."}, {"video_title": "Partial sums term value from partial sum Series AP Calculus BC Khan Academy.mp3", "Sentence": "I know the formula for the sum of the first n terms. I can use that to say, okay, I can figure out the sum of the first seven terms, that's gonna be the sum of all of these, and then I can use that same formula to figure out the sum of the first six terms. And the difference between the two, well, that's going to be our a sub seven. So another way of saying what I just said is that a sub seven is going to be the sum of the first seven terms minus the sum of the first six terms. Sum of the first six terms. And if you were doing this problem on your own, you wouldn't have to write it out this way. I just wrote it out this way, hopefully making this statement a little bit more intuitive."}, {"video_title": "Partial sums term value from partial sum Series AP Calculus BC Khan Academy.mp3", "Sentence": "So another way of saying what I just said is that a sub seven is going to be the sum of the first seven terms minus the sum of the first six terms. Sum of the first six terms. And if you were doing this problem on your own, you wouldn't have to write it out this way. I just wrote it out this way, hopefully making this statement a little bit more intuitive. Well, what is this going to be? Well, s sub seven, the sum of the first seven terms, we just, wherever we see an n, we replace it with a seven. So it's going to be seven squared plus one over seven plus one, and from that, we are going to subtract s sub six, the sum of the first six terms."}, {"video_title": "Partial sums term value from partial sum Series AP Calculus BC Khan Academy.mp3", "Sentence": "I just wrote it out this way, hopefully making this statement a little bit more intuitive. Well, what is this going to be? Well, s sub seven, the sum of the first seven terms, we just, wherever we see an n, we replace it with a seven. So it's going to be seven squared plus one over seven plus one, and from that, we are going to subtract s sub six, the sum of the first six terms. Well, that's going to be six squared plus one over six plus one, over six plus one. And from here, we just have to do a little bit of arithmetic. So this is going to be, let's see, seven squared plus one, this is 49 plus one, so that is 50, over eight."}, {"video_title": "Partial sums term value from partial sum Series AP Calculus BC Khan Academy.mp3", "Sentence": "So it's going to be seven squared plus one over seven plus one, and from that, we are going to subtract s sub six, the sum of the first six terms. Well, that's going to be six squared plus one over six plus one, over six plus one. And from here, we just have to do a little bit of arithmetic. So this is going to be, let's see, seven squared plus one, this is 49 plus one, so that is 50, over eight. And this is six squared plus one, that is 36 plus one, that's 37, over seven. So let's see, we want to find a common denominator between eight and seven. That would be 56."}, {"video_title": "Partial sums term value from partial sum Series AP Calculus BC Khan Academy.mp3", "Sentence": "So this is going to be, let's see, seven squared plus one, this is 49 plus one, so that is 50, over eight. And this is six squared plus one, that is 36 plus one, that's 37, over seven. So let's see, we want to find a common denominator between eight and seven. That would be 56. So this is going to be something over 56, something over 56, minus something else over 56, minus something else over 56. Now, to go from eight to 56, I multiply by seven, so I need to multiply the numerator by seven as well. 50 times seven is 350."}, {"video_title": "Partial sums term value from partial sum Series AP Calculus BC Khan Academy.mp3", "Sentence": "That would be 56. So this is going to be something over 56, something over 56, minus something else over 56, minus something else over 56. Now, to go from eight to 56, I multiply by seven, so I need to multiply the numerator by seven as well. 50 times seven is 350. And then this second fraction, I multiply the denominator by eight to get to 56. So I have to multiply 37 times eight. And see, 37 times eight is going to be 240 plus 56, so that is 296."}, {"video_title": "Partial sums term value from partial sum Series AP Calculus BC Khan Academy.mp3", "Sentence": "50 times seven is 350. And then this second fraction, I multiply the denominator by eight to get to 56. So I have to multiply 37 times eight. And see, 37 times eight is going to be 240 plus 56, so that is 296. 296. And so this is going to be equal to, so I have a denominator of 56. 350 minus 296 is 54."}, {"video_title": "Partial sums term value from partial sum Series AP Calculus BC Khan Academy.mp3", "Sentence": "And see, 37 times eight is going to be 240 plus 56, so that is 296. 296. And so this is going to be equal to, so I have a denominator of 56. 350 minus 296 is 54. So it's 54, 56. And if we wanted to reduce this a little bit, before we rewrite it maybe in a simpler form, we're not really making it, we're rewriting the same value. This would be, let's see if we can write it as 27 over 28, 27 28s."}, {"video_title": "Partial sums term value from partial sum Series AP Calculus BC Khan Academy.mp3", "Sentence": "350 minus 296 is 54. So it's 54, 56. And if we wanted to reduce this a little bit, before we rewrite it maybe in a simpler form, we're not really making it, we're rewriting the same value. This would be, let's see if we can write it as 27 over 28, 27 28s. And let's see, is that about, yep, that's about as simplified as we can get. But there you go, that's what a sub seven is. It's 27 28s, the difference between the sum of the first seven terms and the sum of the first six terms."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "And so in this video, let's actually evaluate this integral. So the first thing that we could do is maybe factor out this pi. So this is going to be equal to pi times the definite integral from 0 to 1. And then let's square this stuff that we have right here in green. So 2 squared is going to be 4. And then we're going to have 2 times the product of both of these terms. So 2 times negative 2y squared times 2 is going to be negative 4y squared."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "And then let's square this stuff that we have right here in green. So 2 squared is going to be 4. And then we're going to have 2 times the product of both of these terms. So 2 times negative 2y squared times 2 is going to be negative 4y squared. And then negative y squared squared is plus y to the fourth. And then from that, we are going to subtract this thing squared. We are going to subtract this business square, which is going to be 4 minus 4 square roots of y plus the square root of y squared is just going to be y."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "So 2 times negative 2y squared times 2 is going to be negative 4y squared. And then negative y squared squared is plus y to the fourth. And then from that, we are going to subtract this thing squared. We are going to subtract this business square, which is going to be 4 minus 4 square roots of y plus the square root of y squared is just going to be y. And then all of that dy. Let me write that in that same color. All of that dy."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "We are going to subtract this business square, which is going to be 4 minus 4 square roots of y plus the square root of y squared is just going to be y. And then all of that dy. Let me write that in that same color. All of that dy. And so this is going to be equal to pi times the definite integral from 0 to 1. And let's see if we can simplify this. We have a positive 4 here, but then when you distribute this negative, you're going to have a negative 4."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "All of that dy. And so this is going to be equal to pi times the definite integral from 0 to 1. And let's see if we can simplify this. We have a positive 4 here, but then when you distribute this negative, you're going to have a negative 4. So that cancels with that. And let's see. The highest degree term here is going to be our y to the fourth."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "We have a positive 4 here, but then when you distribute this negative, you're going to have a negative 4. So that cancels with that. And let's see. The highest degree term here is going to be our y to the fourth. So we have a y to the fourth. I'll write it in that same color. y to the fourth."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "The highest degree term here is going to be our y to the fourth. So we have a y to the fourth. I'll write it in that same color. y to the fourth. And so the next highest degree term right here is this negative 4y squared. So then you have negative. Let me do that same color."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "y to the fourth. And so the next highest degree term right here is this negative 4y squared. So then you have negative. Let me do that same color. We have negative 4y squared. That's that one right over there. And then we have this y, but remember we have this negative out front."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "Let me do that same color. We have negative 4y squared. That's that one right over there. And then we have this y, but remember we have this negative out front. So it's a negative y. So this one right over here is a negative y. And then we have a negative times a negative, which is going to give us a positive 4 square roots of y."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "And then we have this y, but remember we have this negative out front. So it's a negative y. So this one right over here is a negative y. And then we have a negative times a negative, which is going to give us a positive 4 square roots of y. So this is going to end up being a positive 4 square roots of y. And actually, just to make it clear, when we take the antiderivative, I'm going to write that as 4y to the 1 half power. And we're going to multiply all of that stuff by dy."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "And then we have a negative times a negative, which is going to give us a positive 4 square roots of y. So this is going to end up being a positive 4 square roots of y. And actually, just to make it clear, when we take the antiderivative, I'm going to write that as 4y to the 1 half power. And we're going to multiply all of that stuff by dy. Now we're ready to take the antiderivative. So it's going to be equal to pi times the antiderivative of y to the fourth is y to the fifth over 5. Antiderivative of negative 4y squared is negative 4 thirds y to the third power."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "And we're going to multiply all of that stuff by dy. Now we're ready to take the antiderivative. So it's going to be equal to pi times the antiderivative of y to the fourth is y to the fifth over 5. Antiderivative of negative 4y squared is negative 4 thirds y to the third power. Antiderivative of the negative y is negative y squared over 2. And then the antiderivative of 4y to the 1 half, let's see, we're going to increment. It's going to be y to the 3 halves."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "Antiderivative of negative 4y squared is negative 4 thirds y to the third power. Antiderivative of the negative y is negative y squared over 2. And then the antiderivative of 4y to the 1 half, let's see, we're going to increment. It's going to be y to the 3 halves. Multiply by 2 thirds. We're going to get 8 thirds plus 8 thirds y to the 3 halves. And let's see."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "It's going to be y to the 3 halves. Multiply by 2 thirds. We're going to get 8 thirds plus 8 thirds y to the 3 halves. And let's see. Yep, that all works out. And we're going to evaluate this at 1 and at 0. And lucky for us, when you evaluate at 0, this whole thing turns out to be 0."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "And let's see. Yep, that all works out. And we're going to evaluate this at 1 and at 0. And lucky for us, when you evaluate at 0, this whole thing turns out to be 0. So this is all going to be equal to pi times evaluating all this business at 1. So that's going to be 1 fifth minus 4 thirds. I'll do that in green color, minus 4 thirds minus 1 half."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "And lucky for us, when you evaluate at 0, this whole thing turns out to be 0. So this is all going to be equal to pi times evaluating all this business at 1. So that's going to be 1 fifth minus 4 thirds. I'll do that in green color, minus 4 thirds minus 1 half. So whenever you evaluate at 1, it's just going to be. So plus 8 thirds. Plus 8 thirds."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "I'll do that in green color, minus 4 thirds minus 1 half. So whenever you evaluate at 1, it's just going to be. So plus 8 thirds. Plus 8 thirds. Plus 8 thirds. And let's see. What's the least common multiple over here?"}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "Plus 8 thirds. Plus 8 thirds. And let's see. What's the least common multiple over here? Let's see. 5, a 3, and a 2. Looks like we're going to have a denominator of 30."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "What's the least common multiple over here? Let's see. 5, a 3, and a 2. Looks like we're going to have a denominator of 30. So we could rewrite this as equal to pi. And we could put everything over a denominator of 30. 1 fifth is 6 over 30."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "Looks like we're going to have a denominator of 30. So we could rewrite this as equal to pi. And we could put everything over a denominator of 30. 1 fifth is 6 over 30. 4 thirds is 40 over 30. So this is minus 40. That's the different shade of green."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "1 fifth is 6 over 30. 4 thirds is 40 over 30. So this is minus 40. That's the different shade of green. Well, actually, let me make that other shade of green. So this is minus 40 over 30. Negative 1 half, that's minus 15 over 30."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "That's the different shade of green. Well, actually, let me make that other shade of green. So this is minus 40 over 30. Negative 1 half, that's minus 15 over 30. And then finally, 8 thirds is the same thing as 80 over 30. So that's plus 80. So this simplifies to."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "Negative 1 half, that's minus 15 over 30. And then finally, 8 thirds is the same thing as 80 over 30. So that's plus 80. So this simplifies to. So let's see. We have 86 minus 55. Actually, let me make sure I'm doing the math right over here."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "So this simplifies to. So let's see. We have 86 minus 55. Actually, let me make sure I'm doing the math right over here. So 80 minus 40 gets us 40. Plus 6 is 46. Minus 15 is 31."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "Actually, let me make sure I'm doing the math right over here. So 80 minus 40 gets us 40. Plus 6 is 46. Minus 15 is 31. So this is equal to 31 pi over 30. I have a suspicion that I might have done something shady in this last part right over here. So this is going to be."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "Minus 15 is 31. So this is equal to 31 pi over 30. I have a suspicion that I might have done something shady in this last part right over here. So this is going to be. Let's see. 86 negative 51 plus 80. I think that seems right."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be. Let's see. 86 negative 51 plus 80. I think that seems right. I'm going to do it one more time. Let's see. 80 minus 40 is 40."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "I think that seems right. I'm going to do it one more time. Let's see. 80 minus 40 is 40. 46 minus 10 is 36. Minus another 5 is 31. So yes, we get 31 pi over 30 for our volume."}, {"video_title": "Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "But all it is, it's describing the rate of change of a vertical variable with respect to a horizontal variable. So for example, here I have our classic y-axis in the vertical direction and x-axis in the horizontal direction. And if I wanted to figure out the slope of this line, I could pick two points, say that point and that point. I could say, okay, from this point to this point, what is my change in x? Well, my change in x would be this distance right over here. Change in x, the Greek letter delta, this triangle here, it's just shorthand for change. So change in x."}, {"video_title": "Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "I could say, okay, from this point to this point, what is my change in x? Well, my change in x would be this distance right over here. Change in x, the Greek letter delta, this triangle here, it's just shorthand for change. So change in x. And I could also calculate the change in y. So this point going up to that point, our change in y would be this right over here, our change in y. And then we would define slope, or we have defined slope, as change in y over change in x."}, {"video_title": "Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So change in x. And I could also calculate the change in y. So this point going up to that point, our change in y would be this right over here, our change in y. And then we would define slope, or we have defined slope, as change in y over change in x. So slope is equal to the rate of change of our vertical variable over the rate of change of our horizontal variable. It's sometimes described as rise over run. And for any line, it's associated with a slope because it has a constant rate of change."}, {"video_title": "Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And then we would define slope, or we have defined slope, as change in y over change in x. So slope is equal to the rate of change of our vertical variable over the rate of change of our horizontal variable. It's sometimes described as rise over run. And for any line, it's associated with a slope because it has a constant rate of change. If you took any two points on this line, no matter how far apart or no matter how close together, anywhere they sit on the line, if you were to do this calculation, you would get the same slope. That's what makes it a line. But what's fascinating about calculus is we're going to build the tools so that we can think about the rate of change not just of a line, which we've called slope in the past, we can think about the rate of change, the instantaneous rate of change of a curve, of something whose rate of change is possibly constantly changing."}, {"video_title": "Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And for any line, it's associated with a slope because it has a constant rate of change. If you took any two points on this line, no matter how far apart or no matter how close together, anywhere they sit on the line, if you were to do this calculation, you would get the same slope. That's what makes it a line. But what's fascinating about calculus is we're going to build the tools so that we can think about the rate of change not just of a line, which we've called slope in the past, we can think about the rate of change, the instantaneous rate of change of a curve, of something whose rate of change is possibly constantly changing. So for example, here's a curve where the rate of change of y with respect to x is constantly changing. Even if we wanted to use our traditional tools, if we said, okay, we can calculate the average rate of change, let's say between this point and this point, well, what would it be? Well, the average rate of change between this point and this point would be the slope of the line that connects them."}, {"video_title": "Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "But what's fascinating about calculus is we're going to build the tools so that we can think about the rate of change not just of a line, which we've called slope in the past, we can think about the rate of change, the instantaneous rate of change of a curve, of something whose rate of change is possibly constantly changing. So for example, here's a curve where the rate of change of y with respect to x is constantly changing. Even if we wanted to use our traditional tools, if we said, okay, we can calculate the average rate of change, let's say between this point and this point, well, what would it be? Well, the average rate of change between this point and this point would be the slope of the line that connects them. So it'd be the slope of this line, of the secant line. But if we pick two different points, if we pick this point and this point, the average rate of change between those points all of a sudden looks quite different. It looks like it has a higher slope."}, {"video_title": "Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the average rate of change between this point and this point would be the slope of the line that connects them. So it'd be the slope of this line, of the secant line. But if we pick two different points, if we pick this point and this point, the average rate of change between those points all of a sudden looks quite different. It looks like it has a higher slope. So even when we take the slopes between two points on the line, the secant lines, you can see that those slopes are changing. But what if we wanted to ask ourselves an even more interesting question? What is the instantaneous rate of change at a point?"}, {"video_title": "Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "It looks like it has a higher slope. So even when we take the slopes between two points on the line, the secant lines, you can see that those slopes are changing. But what if we wanted to ask ourselves an even more interesting question? What is the instantaneous rate of change at a point? So for example, how fast is y changing with respect to x exactly at that point, exactly when x is equal to that value? Let's call it x one. Well, one way you could think about it is what if we could draw a tangent line to this point, a line that just touches the graph right over there?"}, {"video_title": "Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "What is the instantaneous rate of change at a point? So for example, how fast is y changing with respect to x exactly at that point, exactly when x is equal to that value? Let's call it x one. Well, one way you could think about it is what if we could draw a tangent line to this point, a line that just touches the graph right over there? And we can calculate the slope of that line. Well, that should be the rate of change at that point, the instantaneous rate of change. So in this case, the tangent line might look something like that."}, {"video_title": "Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Well, one way you could think about it is what if we could draw a tangent line to this point, a line that just touches the graph right over there? And we can calculate the slope of that line. Well, that should be the rate of change at that point, the instantaneous rate of change. So in this case, the tangent line might look something like that. If we know the slope of this, well, then we could say that that's the instantaneous rate of change at that point. Why do I say instantaneous rate of change? Well, think about the video on the sprinters, the Usain Bolt example."}, {"video_title": "Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So in this case, the tangent line might look something like that. If we know the slope of this, well, then we could say that that's the instantaneous rate of change at that point. Why do I say instantaneous rate of change? Well, think about the video on the sprinters, the Usain Bolt example. If we wanted to figure out the speed of Usain Bolt at a given instant, well, maybe this describes his position with respect to time. If y was position and x is time, usually you would see t as time, but let's say x is time. So then if we're talking about right at this time, we're talking about the instantaneous rate."}, {"video_title": "Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Well, think about the video on the sprinters, the Usain Bolt example. If we wanted to figure out the speed of Usain Bolt at a given instant, well, maybe this describes his position with respect to time. If y was position and x is time, usually you would see t as time, but let's say x is time. So then if we're talking about right at this time, we're talking about the instantaneous rate. And this idea is a central idea of differential calculus, and it's known as a derivative. The slope of the tangent line, which you could also view as the instantaneous rate of change, I'm putting exclamation mark because it's so conceptually important here. So how can we denote a derivative?"}, {"video_title": "Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So then if we're talking about right at this time, we're talking about the instantaneous rate. And this idea is a central idea of differential calculus, and it's known as a derivative. The slope of the tangent line, which you could also view as the instantaneous rate of change, I'm putting exclamation mark because it's so conceptually important here. So how can we denote a derivative? One way is known as Leibniz's notation, and Leibniz is one of the fathers of calculus along with Isaac Newton. And his notation, you would denote the slope of the tangent line as equaling dy over dx. Now why do I like this notation?"}, {"video_title": "Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So how can we denote a derivative? One way is known as Leibniz's notation, and Leibniz is one of the fathers of calculus along with Isaac Newton. And his notation, you would denote the slope of the tangent line as equaling dy over dx. Now why do I like this notation? Because it really comes from this idea of a slope, which is change in y over change in x. As you'll see in future videos, one way to think about the slope of the tangent line is, well, let's calculate the slope of secant lines. Let's say between that point and that point."}, {"video_title": "Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Now why do I like this notation? Because it really comes from this idea of a slope, which is change in y over change in x. As you'll see in future videos, one way to think about the slope of the tangent line is, well, let's calculate the slope of secant lines. Let's say between that point and that point. But then let's get even closer. Let's say that point and that point, and then let's get even closer, and that point and that point, and then let's get even closer. And let's see what happens as the change in x approaches zero."}, {"video_title": "Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say between that point and that point. But then let's get even closer. Let's say that point and that point, and then let's get even closer, and that point and that point, and then let's get even closer. And let's see what happens as the change in x approaches zero. And so using these d's instead of deltas, this was Leibniz's way of saying, hey, what happens if my changes in, say, x become close to zero? So this idea, this is known as sometimes differential notation, Leibniz's notation, is instead of just change in y over change in x, super small changes in y for a super small change in x, especially as the change in x approaches zero. And as you will see, that is how we will calculate the derivative."}, {"video_title": "Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And let's see what happens as the change in x approaches zero. And so using these d's instead of deltas, this was Leibniz's way of saying, hey, what happens if my changes in, say, x become close to zero? So this idea, this is known as sometimes differential notation, Leibniz's notation, is instead of just change in y over change in x, super small changes in y for a super small change in x, especially as the change in x approaches zero. And as you will see, that is how we will calculate the derivative. Now there's other notations. If this curve is described as y is equal to f of x, the slope of the tangent line at that point could be denoted as equaling f prime of x one. So this notation, it takes a little bit of time getting used to, the Lagrange notation."}, {"video_title": "Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And as you will see, that is how we will calculate the derivative. Now there's other notations. If this curve is described as y is equal to f of x, the slope of the tangent line at that point could be denoted as equaling f prime of x one. So this notation, it takes a little bit of time getting used to, the Lagrange notation. It's saying f prime is representing the derivative. It's telling us the slope of the tangent line for a given point. So if you input an x into this function, into f, you're getting the corresponding y value."}, {"video_title": "Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So this notation, it takes a little bit of time getting used to, the Lagrange notation. It's saying f prime is representing the derivative. It's telling us the slope of the tangent line for a given point. So if you input an x into this function, into f, you're getting the corresponding y value. If you input an x into f prime, you're getting the slope of the tangent line at that point. Now another notation that you'll see less likely in a calculus class, but you might see in a physics class, is the notation y with a dot over it. So you could write this as y with a dot over it, which also denotes the derivative."}, {"video_title": "Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So if you input an x into this function, into f, you're getting the corresponding y value. If you input an x into f prime, you're getting the slope of the tangent line at that point. Now another notation that you'll see less likely in a calculus class, but you might see in a physics class, is the notation y with a dot over it. So you could write this as y with a dot over it, which also denotes the derivative. You might also see y prime. This would be more common in a math class. Now as we march forward in our calculus adventure, we will build the tools to actually calculate these things."}, {"video_title": "Derivative as a concept Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So you could write this as y with a dot over it, which also denotes the derivative. You might also see y prime. This would be more common in a math class. Now as we march forward in our calculus adventure, we will build the tools to actually calculate these things. And if you're already familiar with limits, they will be very useful, as you can imagine, because we're really going to be taking the limit of our change in y over change in x as our change in x approaches zero. And we're not just going to be able to figure it out for a point, we're going to be able to figure out general equations that describe the derivative for any given point. So be very, very excited."}, {"video_title": "Worked example Area between two polar graphs AP Calculus BC Khan Academy.mp3", "Sentence": "So I encourage you to pause the video and give it a go. Alright, so I assume you've tried. And what's interesting here is we're clearly bounded by two different polar graphs. And it looks like they intersect right over here. If we eyeball it, it looks like they're intersecting at when theta is equal to pi over four. And we can verify that. Cosine of pi over four is the same thing as sine of pi over four."}, {"video_title": "Worked example Area between two polar graphs AP Calculus BC Khan Academy.mp3", "Sentence": "And it looks like they intersect right over here. If we eyeball it, it looks like they're intersecting at when theta is equal to pi over four. And we can verify that. Cosine of pi over four is the same thing as sine of pi over four. So it is indeed the case that these two things are going to be equal to each other. They're going to be, their point of intersection happens at theta is equal to pi over four. And if that wasn't as obvious, you would set these two equal to each other and figure out that theta is where this actually happened."}, {"video_title": "Worked example Area between two polar graphs AP Calculus BC Khan Academy.mp3", "Sentence": "Cosine of pi over four is the same thing as sine of pi over four. So it is indeed the case that these two things are going to be equal to each other. They're going to be, their point of intersection happens at theta is equal to pi over four. And if that wasn't as obvious, you would set these two equal to each other and figure out that theta is where this actually happened. But here it jumps out at you a little bit more. So this is theta is equal to pi over four. And so the key is to realize is that for theta, theta being between zero and pi over four, we're bounded by the red circle."}, {"video_title": "Worked example Area between two polar graphs AP Calculus BC Khan Academy.mp3", "Sentence": "And if that wasn't as obvious, you would set these two equal to each other and figure out that theta is where this actually happened. But here it jumps out at you a little bit more. So this is theta is equal to pi over four. And so the key is to realize is that for theta, theta being between zero and pi over four, we're bounded by the red circle. We're bounded by three, r is equal to three sine theta. And then as we go from pi over four to pi over two, we're bounded by the black circle. We're bounded by r is equal to three cosine theta."}, {"video_title": "Worked example Area between two polar graphs AP Calculus BC Khan Academy.mp3", "Sentence": "And so the key is to realize is that for theta, theta being between zero and pi over four, we're bounded by the red circle. We're bounded by three, r is equal to three sine theta. And then as we go from pi over four to pi over two, we're bounded by the black circle. We're bounded by r is equal to three cosine theta. So we could just break up our area into those two regions. So this first area right over here, we already know is going to be one half times the definite integral from zero to pi over four, zero to pi over four, of, what are we bounded by? Three sine of theta, three sine of theta."}, {"video_title": "Worked example Area between two polar graphs AP Calculus BC Khan Academy.mp3", "Sentence": "We're bounded by r is equal to three cosine theta. So we could just break up our area into those two regions. So this first area right over here, we already know is going to be one half times the definite integral from zero to pi over four, zero to pi over four, of, what are we bounded by? Three sine of theta, three sine of theta. And we're gonna square that thing, d theta. That's the orange region. And then this, I guess you could say this blue region right over here is going to be one half times the definite integral."}, {"video_title": "Worked example Area between two polar graphs AP Calculus BC Khan Academy.mp3", "Sentence": "Three sine of theta, three sine of theta. And we're gonna square that thing, d theta. That's the orange region. And then this, I guess you could say this blue region right over here is going to be one half times the definite integral. And now we're gonna go from pi over four to pi over two, to pi over two of three cosine theta squared, d theta. That's this region right over here. Now one thing that might jump out at you is that they're gonna be the same area."}, {"video_title": "Worked example Area between two polar graphs AP Calculus BC Khan Academy.mp3", "Sentence": "And then this, I guess you could say this blue region right over here is going to be one half times the definite integral. And now we're gonna go from pi over four to pi over two, to pi over two of three cosine theta squared, d theta. That's this region right over here. Now one thing that might jump out at you is that they're gonna be the same area. These two circles, they are, I guess you could say, symmetric around this line. Theta is equal to pi over four. So these are going to be the same area."}, {"video_title": "Worked example Area between two polar graphs AP Calculus BC Khan Academy.mp3", "Sentence": "Now one thing that might jump out at you is that they're gonna be the same area. These two circles, they are, I guess you could say, symmetric around this line. Theta is equal to pi over four. So these are going to be the same area. So one thing that we could do is just solve for one of these and then double it, and we will get the total region that we care about. So the total area, and you can verify that for yourself if you like, but I'm just gonna say that the total area, I'm just gonna double this right over here. If I just double this, just the orange expression, I'm gonna get the definite integral from zero to pi over four of, let's see, let me just, of nine, three squared is nine, sine squared theta, d theta."}, {"video_title": "Worked example Area between two polar graphs AP Calculus BC Khan Academy.mp3", "Sentence": "So these are going to be the same area. So one thing that we could do is just solve for one of these and then double it, and we will get the total region that we care about. So the total area, and you can verify that for yourself if you like, but I'm just gonna say that the total area, I'm just gonna double this right over here. If I just double this, just the orange expression, I'm gonna get the definite integral from zero to pi over four of, let's see, let me just, of nine, three squared is nine, sine squared theta, d theta. And you could evaluate this by hand, you could evaluate this calculator. Let's evaluate this analytically. So sine squared theta is the same thing as 1 1\u20442 times one minus, 1 1\u20442 times one minus cosine of two theta."}, {"video_title": "Worked example Area between two polar graphs AP Calculus BC Khan Academy.mp3", "Sentence": "If I just double this, just the orange expression, I'm gonna get the definite integral from zero to pi over four of, let's see, let me just, of nine, three squared is nine, sine squared theta, d theta. And you could evaluate this by hand, you could evaluate this calculator. Let's evaluate this analytically. So sine squared theta is the same thing as 1 1\u20442 times one minus, 1 1\u20442 times one minus cosine of two theta. That's a trigonometric identity that we've seen a lot in trigonometry class. Actually, let me just write it up here. So sine squared theta is equal to 1 1\u20442 times one minus cosine of two theta."}, {"video_title": "Worked example Area between two polar graphs AP Calculus BC Khan Academy.mp3", "Sentence": "So sine squared theta is the same thing as 1 1\u20442 times one minus, 1 1\u20442 times one minus cosine of two theta. That's a trigonometric identity that we've seen a lot in trigonometry class. Actually, let me just write it up here. So sine squared theta is equal to 1 1\u20442 times one minus cosine of two theta. So if we replace this with this, it's going to be equal to, this is going to be equal to, let's take the 1 1\u20442 out, so we're gonna get 9 1\u20442 times the definite integral from zero to pi over four of one minus cosine two theta, d theta. And so this is going to be equal to 9 1\u20442, antiderivative of one is theta, and let's see, cosine two theta, it's going to be negative sine of two theta. Did I do that?"}, {"video_title": "Worked example Area between two polar graphs AP Calculus BC Khan Academy.mp3", "Sentence": "So sine squared theta is equal to 1 1\u20442 times one minus cosine of two theta. So if we replace this with this, it's going to be equal to, this is going to be equal to, let's take the 1 1\u20442 out, so we're gonna get 9 1\u20442 times the definite integral from zero to pi over four of one minus cosine two theta, d theta. And so this is going to be equal to 9 1\u20442, antiderivative of one is theta, and let's see, cosine two theta, it's going to be negative sine of two theta. Did I do that? Negative sine of two theta over two, over two. Actually, let me just notice this. Negative 1 1\u20442 sine of two theta."}, {"video_title": "Worked example Area between two polar graphs AP Calculus BC Khan Academy.mp3", "Sentence": "Did I do that? Negative sine of two theta over two, over two. Actually, let me just notice this. Negative 1 1\u20442 sine of two theta. And you could do u substitution and do it like this, but this you might be able to do in your head, and you can verify. The derivative of sine of two theta is going to be two cosine of two theta, and then you multiply it times the negative 1\u20442, you just get a negative one right over there. And so then we are going to evaluate this at pi over four and at zero."}, {"video_title": "Worked example Area between two polar graphs AP Calculus BC Khan Academy.mp3", "Sentence": "Negative 1 1\u20442 sine of two theta. And you could do u substitution and do it like this, but this you might be able to do in your head, and you can verify. The derivative of sine of two theta is going to be two cosine of two theta, and then you multiply it times the negative 1\u20442, you just get a negative one right over there. And so then we are going to evaluate this at pi over four and at zero. So if you evaluate it at, well, luckily, if you evaluate this thing at zero, this whole thing is going to be zero, so we really just have to evaluate it at pi over four. So this is going to be equal to 9 1\u20442 times pi over four minus 1 1\u20442 sine of two times pi over four is pi over two. Sine of pi over two, sine of pi over two, we already know is one."}, {"video_title": "Worked example Area between two polar graphs AP Calculus BC Khan Academy.mp3", "Sentence": "And so then we are going to evaluate this at pi over four and at zero. So if you evaluate it at, well, luckily, if you evaluate this thing at zero, this whole thing is going to be zero, so we really just have to evaluate it at pi over four. So this is going to be equal to 9 1\u20442 times pi over four minus 1 1\u20442 sine of two times pi over four is pi over two. Sine of pi over two, sine of pi over two, we already know is one. So it is really pi over four, so this right over here is just going to be equal to one. So this is going to be 9 1\u20442 times, we could say pi over four minus 1 1\u20442, or we could say pi over four minus two over four. So we could write it like that, or we could multiply everything out."}, {"video_title": "Differentiating rational functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say that y is equal to five minus three x over x squared plus three x. And we want to figure out what is the derivative of y with respect to x. Now, it might immediately jump out at you that look, look, y is being defined as a rational expression here, as the quotient of two different expressions. Or you could even view this as two different functions. You could view this one up here as u of x. So you could say this is the same thing. This is the same thing as u of x over, you could view the one in the denominator as v of x."}, {"video_title": "Differentiating rational functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Or you could even view this as two different functions. You could view this one up here as u of x. So you could say this is the same thing. This is the same thing as u of x over, you could view the one in the denominator as v of x. So that one right there is v of x. And so if you're taking the derivative of something that can be expressed in this way as the quotient of two different functions, well then you could use the quotient rule. And I'll give you my little aside, like I always do, the quotient rule, if you ever forget it, it can be derived from the product rule."}, {"video_title": "Differentiating rational functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "This is the same thing as u of x over, you could view the one in the denominator as v of x. So that one right there is v of x. And so if you're taking the derivative of something that can be expressed in this way as the quotient of two different functions, well then you could use the quotient rule. And I'll give you my little aside, like I always do, the quotient rule, if you ever forget it, it can be derived from the product rule. And we have videos there, because the product rule's a little bit easier to remember. But what I can do is just say, look, dy dx, if y is just u of x over v of x, I'm just gonna restate the quotient rule. This is going to be, this is going to be the derivative of the function in the numerator."}, {"video_title": "Differentiating rational functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And I'll give you my little aside, like I always do, the quotient rule, if you ever forget it, it can be derived from the product rule. And we have videos there, because the product rule's a little bit easier to remember. But what I can do is just say, look, dy dx, if y is just u of x over v of x, I'm just gonna restate the quotient rule. This is going to be, this is going to be the derivative of the function in the numerator. So d dx of u of x times the function in the denominator, times v of x, minus, minus, I'll do the, minus the function in the numerator, u of x, times the derivative of the function in the denominator, times d dx, v of x, and we're almost there, and then over, over, the function in the denominator squared. The function in the denominator squared. So this might look messy, but all we have to do now is think about, well, what is the derivative of u of x?"}, {"video_title": "Differentiating rational functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "This is going to be, this is going to be the derivative of the function in the numerator. So d dx of u of x times the function in the denominator, times v of x, minus, minus, I'll do the, minus the function in the numerator, u of x, times the derivative of the function in the denominator, times d dx, v of x, and we're almost there, and then over, over, the function in the denominator squared. The function in the denominator squared. So this might look messy, but all we have to do now is think about, well, what is the derivative of u of x? What is the derivative of v of x? And we should just be able to substitute those things back into this expression we just wrote down. So let's do that."}, {"video_title": "Differentiating rational functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So this might look messy, but all we have to do now is think about, well, what is the derivative of u of x? What is the derivative of v of x? And we should just be able to substitute those things back into this expression we just wrote down. So let's do that. So the derivative with respect to x of u of x, of u of x, is equal to, let's see, five minus three x, the derivative of five is zero. The derivative of negative three x, well, that's just gonna be negative three. That's just negative three."}, {"video_title": "Differentiating rational functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So let's do that. So the derivative with respect to x of u of x, of u of x, is equal to, let's see, five minus three x, the derivative of five is zero. The derivative of negative three x, well, that's just gonna be negative three. That's just negative three. If any of that looks completely unfamiliar to you, I encourage you to review the derivative properties and maybe the power rule. And now let's think about what is the derivative with respect to x, derivative with respect to x, of v of x, of v of x. Well, derivative of x squared, we just bring that exponent out front."}, {"video_title": "Differentiating rational functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "That's just negative three. If any of that looks completely unfamiliar to you, I encourage you to review the derivative properties and maybe the power rule. And now let's think about what is the derivative with respect to x, derivative with respect to x, of v of x, of v of x. Well, derivative of x squared, we just bring that exponent out front. It's gonna be two times x to the two minus one, or two x to the first power, or just two x. And then the derivative of three x is just three. So two x plus three."}, {"video_title": "Differentiating rational functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, derivative of x squared, we just bring that exponent out front. It's gonna be two times x to the two minus one, or two x to the first power, or just two x. And then the derivative of three x is just three. So two x plus three. And now we know everything we need to substitute back in here. The derivative of u with respect to x, this right over here is just negative three. V of x, this we know is x squared plus three x."}, {"video_title": "Differentiating rational functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So two x plus three. And now we know everything we need to substitute back in here. The derivative of u with respect to x, this right over here is just negative three. V of x, this we know is x squared plus three x. We know that this right over here is v of x. And then u of x we know is five minus three x. Five minus three x."}, {"video_title": "Differentiating rational functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "V of x, this we know is x squared plus three x. We know that this right over here is v of x. And then u of x we know is five minus three x. Five minus three x. The derivative of v with respect to x we know is two x plus three. Two x plus three. And then finally, v of x we know is x squared plus three x."}, {"video_title": "Differentiating rational functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Five minus three x. The derivative of v with respect to x we know is two x plus three. Two x plus three. And then finally, v of x we know is x squared plus three x. So this is x squared plus three x. And so what do we get? Well, we are going to get, and it's gonna look a little bit hairy, it's going to be equal to negative, I'll focus, so first we have, first we have this business up here."}, {"video_title": "Differentiating rational functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And then finally, v of x we know is x squared plus three x. So this is x squared plus three x. And so what do we get? Well, we are going to get, and it's gonna look a little bit hairy, it's going to be equal to negative, I'll focus, so first we have, first we have this business up here. So it's negative three times x squared plus three x. So I'm just gonna distribute the negative three. So it's negative three x squared minus nine x."}, {"video_title": "Differentiating rational functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we are going to get, and it's gonna look a little bit hairy, it's going to be equal to negative, I'll focus, so first we have, first we have this business up here. So it's negative three times x squared plus three x. So I'm just gonna distribute the negative three. So it's negative three x squared minus nine x. And then from that we are going to subtract the product of these two expressions. And so let's see, what is that going to be? Well, we have a five times two x, which is 10 x."}, {"video_title": "Differentiating rational functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So it's negative three x squared minus nine x. And then from that we are going to subtract the product of these two expressions. And so let's see, what is that going to be? Well, we have a five times two x, which is 10 x. A five times three, which is 15. We have a negative three x times two x. So that is going to be negative six x squared."}, {"video_title": "Differentiating rational functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we have a five times two x, which is 10 x. A five times three, which is 15. We have a negative three x times two x. So that is going to be negative six x squared. Minus six x squared. And then a negative three x times three. So negative nine x."}, {"video_title": "Differentiating rational functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So that is going to be negative six x squared. Minus six x squared. And then a negative three x times three. So negative nine x. And let's see, we can simplify that a little bit. 10 x minus nine x, well that's just going to leave us with x. So if we, 10 x minus nine x is just going to be x."}, {"video_title": "Differentiating rational functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So negative nine x. And let's see, we can simplify that a little bit. 10 x minus nine x, well that's just going to leave us with x. So if we, 10 x minus nine x is just going to be x. And then in our denominator, we're almost there. In our denominator, we could just write that as x plus three x, x squared plus three x squared. Or if we want, we could expand it out."}, {"video_title": "Differentiating rational functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So if we, 10 x minus nine x is just going to be x. And then in our denominator, we're almost there. In our denominator, we could just write that as x plus three x, x squared plus three x squared. Or if we want, we could expand it out. I'll just leave it like that. x squared plus three x squared. And so if we want to, let's just simplify, or attempt to simplify this a little bit."}, {"video_title": "Differentiating rational functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Or if we want, we could expand it out. I'll just leave it like that. x squared plus three x squared. And so if we want to, let's just simplify, or attempt to simplify this a little bit. It's going to be negative three x squared minus nine x. And then, let's see, you're going to have a negative minus x, minus x. And then minus 15."}, {"video_title": "Differentiating rational functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And so if we want to, let's just simplify, or attempt to simplify this a little bit. It's going to be negative three x squared minus nine x. And then, let's see, you're going to have a negative minus x, minus x. And then minus 15. And then minus negative six x squared. So plus six x squared. All of that, all of that, over x squared plus three x squared."}, {"video_title": "Differentiating rational functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And then minus 15. And then minus negative six x squared. So plus six x squared. All of that, all of that, over x squared plus three x squared. Or x squared plus three x squared. I should say it that way. And let's see, this numerator I can simplify a little bit."}, {"video_title": "Differentiating rational functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "All of that, all of that, over x squared plus three x squared. Or x squared plus three x squared. I should say it that way. And let's see, this numerator I can simplify a little bit. Negative three x squared plus six x squared, that's going to be positive three x squared. And then we have, I'll do it in orange. We have negative nine x minus an x, well that's going to be minus 10 x."}, {"video_title": "Differentiating rational functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And let's see, this numerator I can simplify a little bit. Negative three x squared plus six x squared, that's going to be positive three x squared. And then we have, I'll do it in orange. We have negative nine x minus an x, well that's going to be minus 10 x. Minus 10 x. And then we have minus 15. So minus 15."}, {"video_title": "Differentiating rational functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "We have negative nine x minus an x, well that's going to be minus 10 x. Minus 10 x. And then we have minus 15. So minus 15. And so there you have it. We finally have finished. This is all going to be equal to, this is all going to be equal to three x squared minus 10 x minus 15 over x squared plus three x squared."}, {"video_title": "Worked example range of solution curve from slope field AP Calculus AB Khan Academy.mp3", "Sentence": "So we have a slope field here for a differential equation, and we're saying, okay, if we have a solution where the initial condition is zero comma six, so zero comma six is part of that solution. So, let's see, zero comma six, so this is part of the solution, and we want to know the range of the solution curve. So the solution curve, you can eyeball a little bit by looking at the slope field. So as x, remember, x is gonna be greater than or equal to zero, so it's going to include this point right over here, and as x increases, you can tell from the slope, okay, y is gonna decrease, but it's gonna keep decreasing at a slower and slower rate, and it looks like it's asymptoting towards the line y is equal to four. So it's gonna get really, as x gets larger and larger and larger, it's gonna get infinitely close to y is equal to four, but it's not quite gonna get there. So the range, the y values that this is going to take on, y is going to be greater than four. It's not ever gonna be equal to four, so I'll do, it's going to be greater than four, that's gonna be the bottom end of my range, and at the top end of my range, I will be equal to six."}, {"video_title": "Worked example range of solution curve from slope field AP Calculus AB Khan Academy.mp3", "Sentence": "So as x, remember, x is gonna be greater than or equal to zero, so it's going to include this point right over here, and as x increases, you can tell from the slope, okay, y is gonna decrease, but it's gonna keep decreasing at a slower and slower rate, and it looks like it's asymptoting towards the line y is equal to four. So it's gonna get really, as x gets larger and larger and larger, it's gonna get infinitely close to y is equal to four, but it's not quite gonna get there. So the range, the y values that this is going to take on, y is going to be greater than four. It's not ever gonna be equal to four, so I'll do, it's going to be greater than four, that's gonna be the bottom end of my range, and at the top end of my range, I will be equal to six. Six is the largest value that I am going to take on. Another way I could have written this is four is less than y, is less than or equal to six. Either way, this is a way of describing the range, the y values that the solution will take on for x being greater than or equal to zero."}, {"video_title": "Worked example range of solution curve from slope field AP Calculus AB Khan Academy.mp3", "Sentence": "It's not ever gonna be equal to four, so I'll do, it's going to be greater than four, that's gonna be the bottom end of my range, and at the top end of my range, I will be equal to six. Six is the largest value that I am going to take on. Another way I could have written this is four is less than y, is less than or equal to six. Either way, this is a way of describing the range, the y values that the solution will take on for x being greater than or equal to zero. If they said for all x's, well, then you might have been able to go back this way and keep going, but they're saying the range of the solution curve for x is greater than or equal to zero so we won't consider those values of x less than zero. So there you go. The solution curve would look something like that, and you can see the highest value it takes on is six, and it actually does take on that value because we're including x equaling zero, and then it keeps going down, approaching four, getting very, very close to four, but never quite equaling four."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "So that is our first term. We also know the common ratio of our geometric series and we're gonna call that r. So this is the common ratio. And we also know that it's a finite geometric series. So let me write this. It is finite. So it has a finite number of terms. And let's say that n is equal to the number of terms."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "So let me write this. It is finite. So it has a finite number of terms. And let's say that n is equal to the number of terms. The number of terms. And we're gonna use a notation. We're going to use a notation s sub n to denote the sum."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "And let's say that n is equal to the number of terms. The number of terms. And we're gonna use a notation. We're going to use a notation s sub n to denote the sum. The sum. The sum of first. First."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "We're going to use a notation s sub n to denote the sum. The sum. The sum of first. First. Of first n terms. And the goal of this whole video is using this information, coming up with a general formula for the sum of the first n terms. A formula for evaluating a geometric series."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "First. Of first n terms. And the goal of this whole video is using this information, coming up with a general formula for the sum of the first n terms. A formula for evaluating a geometric series. So let's write out s sub n. Just get a feeling for what it would look like. So s sub n is going to be equal to, well you have your first term here, which is an a. And then what's our second term going to be?"}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "A formula for evaluating a geometric series. So let's write out s sub n. Just get a feeling for what it would look like. So s sub n is going to be equal to, well you have your first term here, which is an a. And then what's our second term going to be? Well it's going to be, this is a geometric series. So it's going to be a times the common ratio. So it's going to be the first term times the common ratio."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "And then what's our second term going to be? Well it's going to be, this is a geometric series. So it's going to be a times the common ratio. So it's going to be the first term times the common ratio. So the first term times r. Now what's the third term going to be? Well it's going to be the second term times our common ratio again. So it's going to be ar times r. Or ar squared."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "So it's going to be the first term times the common ratio. So the first term times r. Now what's the third term going to be? Well it's going to be the second term times our common ratio again. So it's going to be ar times r. Or ar squared. Ar squared. And we could go all the way to our nth term. So we're going to go all the way to the nth term."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "So it's going to be ar times r. Or ar squared. Ar squared. And we could go all the way to our nth term. So we're going to go all the way to the nth term. And you might be tempted to say it's going to be a times r to the nth power. But we have to be careful here. Because notice, our first term is really ar to the zeroth power."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "So we're going to go all the way to the nth term. And you might be tempted to say it's going to be a times r to the nth power. But we have to be careful here. Because notice, our first term is really ar to the zeroth power. Our second term is ar to the first power. Our third term is ar to the second power. So whatever term we're on, the exponent is that term number minus one."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "Because notice, our first term is really ar to the zeroth power. Our second term is ar to the first power. Our third term is ar to the second power. So whatever term we're on, the exponent is that term number minus one. So if we're on the nth term, it's going to be ar to the n minus oneth power. So we want to come up with a nice, clean formula for evaluating this. And we're going to use a little trick to do it."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "So whatever term we're on, the exponent is that term number minus one. So if we're on the nth term, it's going to be ar to the n minus oneth power. So we want to come up with a nice, clean formula for evaluating this. And we're going to use a little trick to do it. To do it, we're going to think about what r times the sum is. And we're going to subtract that out. So we're going to take the r times that sum."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "And we're going to use a little trick to do it. To do it, we're going to think about what r times the sum is. And we're going to subtract that out. So we're going to take the r times that sum. R times the sum of the first nth term. Actually, let's just multiply negative r. Negative r times the sum. And then we can just add these two things."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "So we're going to take the r times that sum. R times the sum of the first nth term. Actually, let's just multiply negative r. Negative r times the sum. And then we can just add these two things. And you'll see that it cleans this thing up nicely. So what is this going to be equal to? This is going to be equal to, well, if you multiply, if we multiply a times negative r, we will get negative ar."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "And then we can just add these two things. And you'll see that it cleans this thing up nicely. So what is this going to be equal to? This is going to be equal to, well, if you multiply, if we multiply a times negative r, we will get negative ar. And I'm just going to write it right underneath this one. So if you multiply this times negative r, I'm just going to multiply every one of these terms by negative r. That's the equivalent of multiplying negative r times the sum. Distributing the negative r. So if I multiply it times this term, a times negative r, that's going to be negative, that's going to be negative ar."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "This is going to be equal to, well, if you multiply, if we multiply a times negative r, we will get negative ar. And I'm just going to write it right underneath this one. So if you multiply this times negative r, I'm just going to multiply every one of these terms by negative r. That's the equivalent of multiplying negative r times the sum. Distributing the negative r. So if I multiply it times this term, a times negative r, that's going to be negative, that's going to be negative ar. And if I multiply ar times negative r, that's going to be negative ar squared. Negative ar squared. You might see where this is going."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "Distributing the negative r. So if I multiply it times this term, a times negative r, that's going to be negative, that's going to be negative ar. And if I multiply ar times negative r, that's going to be negative ar squared. Negative ar squared. You might see where this is going. And just to be clear what's going on, that's that term times negative r. Times negative r. This is that term times negative r. And we would keep going all the way to the term, the term, this, the term before this times negative r. So if I, the term before this times negative r is going to be a, is going to be negative. Actually, let me put subtraction signs. It's going to be negative a times r to the n minus one power."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "You might see where this is going. And just to be clear what's going on, that's that term times negative r. Times negative r. This is that term times negative r. And we would keep going all the way to the term, the term, this, the term before this times negative r. So if I, the term before this times negative r is going to be a, is going to be negative. Actually, let me put subtraction signs. It's going to be negative a times r to the n minus one power. r to the n minus one power. That was the term right before this. That was a times r to the n minus two times negative r is going to give us this."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "It's going to be negative a times r to the n minus one power. r to the n minus one power. That was the term right before this. That was a times r to the n minus two times negative r is going to give us this. So it's going to get us right over there. And then finally we take this last term and you multiply it by negative r, what do you get? You get negative a, negative a, and then times r to the n. r to the n. You multiply this times the negative, you get the negative a."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "That was a times r to the n minus two times negative r is going to give us this. So it's going to get us right over there. And then finally we take this last term and you multiply it by negative r, what do you get? You get negative a, negative a, and then times r to the n. r to the n. You multiply this times the negative, you get the negative a. And then r to the n minus one times r, or times r to the first, well this is going to be r to the n. And now what's interesting here is this we can add up the left side and we can add up the right hand side. So let's do that. So let's do that."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "You get negative a, negative a, and then times r to the n. r to the n. You multiply this times the negative, you get the negative a. And then r to the n minus one times r, or times r to the first, well this is going to be r to the n. And now what's interesting here is this we can add up the left side and we can add up the right hand side. So let's do that. So let's do that. On the left hand side, we get s sub n minus r, minus r, minus r times s sub n, s sub n. And on the right hand side, we have something very cool happening. Notice this a, we still have that. The a sits there."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "So let's do that. On the left hand side, we get s sub n minus r, minus r, minus r times s sub n, s sub n. And on the right hand side, we have something very cool happening. Notice this a, we still have that. The a sits there. But everything else, except for this last thing, is going to cancel out. So these two are gonna cancel out. These two are gonna cancel out."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "The a sits there. But everything else, except for this last thing, is going to cancel out. So these two are gonna cancel out. These two are gonna cancel out. Let me do that a little neater. These two are gonna cancel out. These two are gonna cancel out."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "These two are gonna cancel out. Let me do that a little neater. These two are gonna cancel out. These two are gonna cancel out. And all we're gonna have left with is a negative a r to the n. So it's gonna be a minus a times r to the nth power. And now we can just solve for s sub n and we have our formula, what we were looking for. So let's see, we can factor out an s sub n on the left hand side."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "These two are gonna cancel out. And all we're gonna have left with is a negative a r to the n. So it's gonna be a minus a times r to the nth power. And now we can just solve for s sub n and we have our formula, what we were looking for. So let's see, we can factor out an s sub n on the left hand side. So you get an s sub n, the sum of our first n terms. You factor that out. So it's gonna be times, times one minus r is going to be equal to, and on the right hand side we can actually factor out an a."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "So let's see, we can factor out an s sub n on the left hand side. So you get an s sub n, the sum of our first n terms. You factor that out. So it's gonna be times, times one minus r is going to be equal to, and on the right hand side we can actually factor out an a. So it's gonna be a times one minus r to the n. And so to solve for s sub n, the sum of our first n terms, we deserve a little bit of a drum roll here. S sub n is going to be equal to this divided by one minus r. So it's going to be a times one minus r to the n one minus r to the n over one minus r. And we're done. We have figured out our formula for the sum or for the sum of a finite geometric series."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "So it's gonna be times, times one minus r is going to be equal to, and on the right hand side we can actually factor out an a. So it's gonna be a times one minus r to the n. And so to solve for s sub n, the sum of our first n terms, we deserve a little bit of a drum roll here. S sub n is going to be equal to this divided by one minus r. So it's going to be a times one minus r to the n one minus r to the n over one minus r. And we're done. We have figured out our formula for the sum or for the sum of a finite geometric series. And so in the next few videos or in future videos we will apply this. And I encourage you, whenever you use this formula, it's very important now that you know where it came from that you really keep close track of how many terms you're actually summing up. Sometimes you might have a sigma notation that starts, it might start its index at zero and then goes up to a number, in which case you're gonna have that number plus one term."}, {"video_title": "Finite geometric series formula justification High School Math Khan Academy.mp3", "Sentence": "We have figured out our formula for the sum or for the sum of a finite geometric series. And so in the next few videos or in future videos we will apply this. And I encourage you, whenever you use this formula, it's very important now that you know where it came from that you really keep close track of how many terms you're actually summing up. Sometimes you might have a sigma notation that starts, it might start its index at zero and then goes up to a number, in which case you're gonna have that number plus one term. So you have to be very careful. This is the number of terms. This is the first term here, we define it up here."}, {"video_title": "Sigma notation for sums Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "What I want to do in this video is introduce you to the idea of sigma notation, which will be used extensively through your mathematical career. So let's just say you wanted to find a sum of some terms. And these terms have a pattern. So let's say you want to find a sum of the first 10 numbers. You could say 1 plus 2 plus 3 plus, and you go all the way to plus 9 plus 10. And I clearly could have even written this whole thing out. But you can imagine it becomes a lot harder if you wanted to find the sum of the first 100 numbers."}, {"video_title": "Sigma notation for sums Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "So let's say you want to find a sum of the first 10 numbers. You could say 1 plus 2 plus 3 plus, and you go all the way to plus 9 plus 10. And I clearly could have even written this whole thing out. But you can imagine it becomes a lot harder if you wanted to find the sum of the first 100 numbers. So that would be 1 plus 2 plus 3 plus, and you would go all the way to 99 plus 100. So mathematicians said, well, let's find some notation instead of having to do this dot, dot, dot thing, which you will see sometimes done, so that we can more cleanly express these types of sums. And that's where sigma notation comes from."}, {"video_title": "Sigma notation for sums Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "But you can imagine it becomes a lot harder if you wanted to find the sum of the first 100 numbers. So that would be 1 plus 2 plus 3 plus, and you would go all the way to 99 plus 100. So mathematicians said, well, let's find some notation instead of having to do this dot, dot, dot thing, which you will see sometimes done, so that we can more cleanly express these types of sums. And that's where sigma notation comes from. So this sum up here right over here, this first one, it could be represented as sigma. This is a capital sigma. It's a Greek letter right over here."}, {"video_title": "Sigma notation for sums Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "And that's where sigma notation comes from. So this sum up here right over here, this first one, it could be represented as sigma. This is a capital sigma. It's a Greek letter right over here. And what you do is you define an index. And you could start your index at some value. So let's say your index starts at 1."}, {"video_title": "Sigma notation for sums Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "It's a Greek letter right over here. And what you do is you define an index. And you could start your index at some value. So let's say your index starts at 1. I'll just use i for index. So let's say that i starts at 1, and I'm going to go to 10. So i starts at 1, and it goes to 10."}, {"video_title": "Sigma notation for sums Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "So let's say your index starts at 1. I'll just use i for index. So let's say that i starts at 1, and I'm going to go to 10. So i starts at 1, and it goes to 10. And I'm going to sum up the i's. So how does this translate into this right over here? Well, what you do is you start wherever the index is."}, {"video_title": "Sigma notation for sums Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "So i starts at 1, and it goes to 10. And I'm going to sum up the i's. So how does this translate into this right over here? Well, what you do is you start wherever the index is. The index is at 1. Set i equal to 1. Write the 1 down."}, {"video_title": "Sigma notation for sums Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "Well, what you do is you start wherever the index is. The index is at 1. Set i equal to 1. Write the 1 down. And then you increment the index. And so i will then be equal to 2. i is 2. Put the 2 down."}, {"video_title": "Sigma notation for sums Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "Write the 1 down. And then you increment the index. And so i will then be equal to 2. i is 2. Put the 2 down. And you're summing each of these terms as you go. And you go all the way until i is equal to 10. All the way until i is equal to 10."}, {"video_title": "Sigma notation for sums Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "Put the 2 down. And you're summing each of these terms as you go. And you go all the way until i is equal to 10. All the way until i is equal to 10. So given what I just told you, I encourage you to pause this video and write the sigma notation for this sum right over here. Assuming you've given a go at it, well, this would be the sum. The first term, well, it might be easy to just say, we'll start at i equals 1 again."}, {"video_title": "Sigma notation for sums Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "All the way until i is equal to 10. So given what I just told you, I encourage you to pause this video and write the sigma notation for this sum right over here. Assuming you've given a go at it, well, this would be the sum. The first term, well, it might be easy to just say, we'll start at i equals 1 again. i equals 1. But now we're not going to stop until i equals 100. And we're going to sum up all of the i's."}, {"video_title": "Sigma notation for sums Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "The first term, well, it might be easy to just say, we'll start at i equals 1 again. i equals 1. But now we're not going to stop until i equals 100. And we're going to sum up all of the i's. Let's do another example. Let's imagine the sum. Let's imagine the sum from i equals 0 to 50 of, I don't know, let me say pi i squared."}, {"video_title": "Sigma notation for sums Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "And we're going to sum up all of the i's. Let's do another example. Let's imagine the sum. Let's imagine the sum from i equals 0 to 50 of, I don't know, let me say pi i squared. What would this sum look like? And once again, I encourage you to pause the video and write it out. I kind of expand out the sum."}, {"video_title": "Sigma notation for sums Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "Let's imagine the sum from i equals 0 to 50 of, I don't know, let me say pi i squared. What would this sum look like? And once again, I encourage you to pause the video and write it out. I kind of expand out the sum. Well, let's just go step by step. When i equals 0, this will be pi times 0 squared. And that's clearly 0, but I'll write it out."}, {"video_title": "Sigma notation for sums Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "I kind of expand out the sum. Well, let's just go step by step. When i equals 0, this will be pi times 0 squared. And that's clearly 0, but I'll write it out. Pi times 0 squared. Then we increase our i. And we make sure that we haven't hit this, that our i isn't already this top boundary right over here or this top value."}, {"video_title": "Sigma notation for sums Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "And that's clearly 0, but I'll write it out. Pi times 0 squared. Then we increase our i. And we make sure that we haven't hit this, that our i isn't already this top boundary right over here or this top value. So now we said i equals 1. Pi times 1 squared. So plus pi times 1 squared."}, {"video_title": "Sigma notation for sums Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "And we make sure that we haven't hit this, that our i isn't already this top boundary right over here or this top value. So now we said i equals 1. Pi times 1 squared. So plus pi times 1 squared. Well, is 1 our top value right over here where we stop? No. So we keep going."}, {"video_title": "Sigma notation for sums Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "So plus pi times 1 squared. Well, is 1 our top value right over here where we stop? No. So we keep going. So then we go i equals 2. Pi times 2 squared. So plus pi times 2 squared."}, {"video_title": "Sigma notation for sums Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "So we keep going. So then we go i equals 2. Pi times 2 squared. So plus pi times 2 squared. I think you see the pattern here. And we're just going to keep going all the way until at some point we're going to keep incrementing our i. i is going to be 49. So it's going to be pi times 49 squared."}, {"video_title": "Sigma notation for sums Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "So plus pi times 2 squared. I think you see the pattern here. And we're just going to keep going all the way until at some point we're going to keep incrementing our i. i is going to be 49. So it's going to be pi times 49 squared. And then finally, we increment i. i equals becomes 50. And we're going to have plus pi times 50 squared. And then we say, OK, our i is finally equal to this top boundary."}, {"video_title": "Sigma notation for sums Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "So it's going to be pi times 49 squared. And then finally, we increment i. i equals becomes 50. And we're going to have plus pi times 50 squared. And then we say, OK, our i is finally equal to this top boundary. And now we can stop. And so you can see this notation, this sigma notation for this sum, was a much cleaner way, a much purer way of representing this than having to write out the entire sum. But you'll see people switch back and forth between the two."}, {"video_title": "Function as a geometric series Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so you might immediately go out and say, all right, well, let me evaluate this function at zero, evaluate its derivative at zero, its second derivative at zero, so on and so forth, and then I can use the formula for the Maclaurin series to just expand it out. But very quickly, you will run into roadblocks, because evaluating this at x equals zero is pretty straightforward, evaluating the first derivative is pretty straightforward, but then once you start taking the second and third derivatives, it gets very hairy very fast. You could do a simplification, where you could say, well, let me find the Maclaurin series for f of u is equal to six over one plus u, where u is equal to x to the third. So you find this Maclaurin expansion in terms of u, and then you substitute for x to the third, and actually, that makes it a good bit simpler. So that is another way to approach it. But the simplest way to approach it is to say, hey, you know what, this form right over here, this rational expression, it looks similar. It looks like the sum of a geometric series."}, {"video_title": "Function as a geometric series Series AP Calculus BC Khan Academy.mp3", "Sentence": "So you find this Maclaurin expansion in terms of u, and then you substitute for x to the third, and actually, that makes it a good bit simpler. So that is another way to approach it. But the simplest way to approach it is to say, hey, you know what, this form right over here, this rational expression, it looks similar. It looks like the sum of a geometric series. Let's just remind ourselves what the sum of a geometric series looks like. If I have a plus a times r, so a is my first term, r is my common ratio, plus, I'm gonna multiply times r again, plus a times r squared, plus a times r to the third power, and I keep going on and on and on forever, we know that this is going to be equal to a, our first term, over one minus our common ratio. And this just comes from the sum of a geometric series."}, {"video_title": "Function as a geometric series Series AP Calculus BC Khan Academy.mp3", "Sentence": "It looks like the sum of a geometric series. Let's just remind ourselves what the sum of a geometric series looks like. If I have a plus a times r, so a is my first term, r is my common ratio, plus, I'm gonna multiply times r again, plus a times r squared, plus a times r to the third power, and I keep going on and on and on forever, we know that this is going to be equal to a, our first term, over one minus our common ratio. And this just comes from the sum of a geometric series. And notice that what we have here, our f of x, our definition of f of x, and the sum of a geometric series look very, very similar. If we say that this right over here is a, so a is equal to six, and if negative r is equal to x to the third, or we could say, let me rewrite this, I could write this denominator as one minus negative x to the third. And so now, you could say, okay, well, r could be equal to negative x to the third."}, {"video_title": "Function as a geometric series Series AP Calculus BC Khan Academy.mp3", "Sentence": "And this just comes from the sum of a geometric series. And notice that what we have here, our f of x, our definition of f of x, and the sum of a geometric series look very, very similar. If we say that this right over here is a, so a is equal to six, and if negative r is equal to x to the third, or we could say, let me rewrite this, I could write this denominator as one minus negative x to the third. And so now, you could say, okay, well, r could be equal to negative x to the third. And just like that, we can expand it out. Well, if a is equal to six, and r is equal to negative x to the third, well, then we can just write this out as a geometric series, which is very straightforward. So let's do that."}, {"video_title": "Function as a geometric series Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so now, you could say, okay, well, r could be equal to negative x to the third. And just like that, we can expand it out. Well, if a is equal to six, and r is equal to negative x to the third, well, then we can just write this out as a geometric series, which is very straightforward. So let's do that. And I will do this in, I'll do this in this nice pink color. So the first term would be six, plus six times our common ratio, six times negative x to the third. And so actually, let me just write that as negative six x to the third."}, {"video_title": "Function as a geometric series Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's do that. And I will do this in, I'll do this in this nice pink color. So the first term would be six, plus six times our common ratio, six times negative x to the third. And so actually, let me just write that as negative six x to the third. And then we're gonna multiply by negative x to the third again, so that's going to be, if I multiply this times negative x to the third, that's gonna be positive six times x to the sixth power. And then I'm gonna multiply it by times negative x to the third again, so it's gonna be minus six times x to the ninth power. And I'm gonna go on and on and on."}, {"video_title": "Function as a geometric series Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so actually, let me just write that as negative six x to the third. And then we're gonna multiply by negative x to the third again, so that's going to be, if I multiply this times negative x to the third, that's gonna be positive six times x to the sixth power. And then I'm gonna multiply it by times negative x to the third again, so it's gonna be minus six times x to the ninth power. And I'm gonna go on and on and on. So it's gonna be, and then I can keep going. I multiply times negative x to the third, I will get six x to the 12th power. And we can go on and on and on and on forever."}, {"video_title": "Function as a geometric series Series AP Calculus BC Khan Academy.mp3", "Sentence": "And I'm gonna go on and on and on. So it's gonna be, and then I can keep going. I multiply times negative x to the third, I will get six x to the 12th power. And we can go on and on and on and on forever. And so the key here was, and this is the Maclaurin Series expansion for our f of x, but the key is to not have to go through all of this business and just to recognize that, hey, the way this function was defined is, it looks a lot like the sum of a geometric series. And it can be considered the sum of a geometric series. And we can use that to find the power series expansion for our function."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "So let's start off with some function, some expression, that can be expressed as the product of two functions, or the composition of two functions. So it can be expressed as f of g of x. So it's a function that can be expressed as a composition or expression that can be expressed as a composition of two functions. Let me get that same color. I want the colors to be accurate. And my goal is to take the derivative of this business, the derivative with respect to x. And what the chain rule tells us is that this is going to be equal to the derivative of the outer function with respect to the inner function."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "Let me get that same color. I want the colors to be accurate. And my goal is to take the derivative of this business, the derivative with respect to x. And what the chain rule tells us is that this is going to be equal to the derivative of the outer function with respect to the inner function. And we can write that as f prime of not x, but f prime of g of x, of the inner function, f prime of g of x times the derivative of the inner function with respect to x. Now this might seem all very abstract and mathy. How do you actually apply it?"}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "And what the chain rule tells us is that this is going to be equal to the derivative of the outer function with respect to the inner function. And we can write that as f prime of not x, but f prime of g of x, of the inner function, f prime of g of x times the derivative of the inner function with respect to x. Now this might seem all very abstract and mathy. How do you actually apply it? Well, let's try it with a real example. Let's say we were trying to take the derivative of the square root of 3x squared minus x. So how could we define an f and a g so that this really is the composition of f of x and g of x?"}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "How do you actually apply it? Well, let's try it with a real example. Let's say we were trying to take the derivative of the square root of 3x squared minus x. So how could we define an f and a g so that this really is the composition of f of x and g of x? Well, we could define f of x as being equal to the square root of x. And if we defined g of x as being equal to 3x squared minus x, then what is f of g of x? Well, f of g of x is going to be equal to, and I'm going to try to keep all the colors accurate."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "So how could we define an f and a g so that this really is the composition of f of x and g of x? Well, we could define f of x as being equal to the square root of x. And if we defined g of x as being equal to 3x squared minus x, then what is f of g of x? Well, f of g of x is going to be equal to, and I'm going to try to keep all the colors accurate. Hopefully it'll help with the understanding. f of g of x is equal to, where everywhere you see the x, you replace with the g of x, the principal root of g of x, which is equal to the principal root of, we've defined g of x right over here, 3x squared minus x. So this thing right over here is exactly f of g of x if we define f of x in this way and g of x in this way."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "Well, f of g of x is going to be equal to, and I'm going to try to keep all the colors accurate. Hopefully it'll help with the understanding. f of g of x is equal to, where everywhere you see the x, you replace with the g of x, the principal root of g of x, which is equal to the principal root of, we've defined g of x right over here, 3x squared minus x. So this thing right over here is exactly f of g of x if we define f of x in this way and g of x in this way. Fair enough. So let's apply the chain rule. What is f prime of g of x going to be equal to, the derivative of f with respect to g?"}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "So this thing right over here is exactly f of g of x if we define f of x in this way and g of x in this way. Fair enough. So let's apply the chain rule. What is f prime of g of x going to be equal to, the derivative of f with respect to g? Well, what's f prime of x? f prime of x is equal to, this is the same thing as x to the 1 half power, so we can just apply the power rule. So it's going to be 1 half times x to the, and then we just take 1 away from the exponent, 1 half minus 1 is negative 1 half."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "What is f prime of g of x going to be equal to, the derivative of f with respect to g? Well, what's f prime of x? f prime of x is equal to, this is the same thing as x to the 1 half power, so we can just apply the power rule. So it's going to be 1 half times x to the, and then we just take 1 away from the exponent, 1 half minus 1 is negative 1 half. And so what is f prime of g of x? f prime of g of x. Well, where everywhere in the derivative we saw an x, we can replace it with a g of x."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "So it's going to be 1 half times x to the, and then we just take 1 away from the exponent, 1 half minus 1 is negative 1 half. And so what is f prime of g of x? f prime of g of x. Well, where everywhere in the derivative we saw an x, we can replace it with a g of x. It's going to be 1 half times, instead of an x to the negative 1 half, we can write a g of x. We can write a g of x to the 1 half. And this is just going to be equal to, let me write it right over here, it's going to be equal to 1 half times all of this business to the negative 1 half power."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "Well, where everywhere in the derivative we saw an x, we can replace it with a g of x. It's going to be 1 half times, instead of an x to the negative 1 half, we can write a g of x. We can write a g of x to the 1 half. And this is just going to be equal to, let me write it right over here, it's going to be equal to 1 half times all of this business to the negative 1 half power. So 3x squared minus x, which is exactly what we need to solve right over here. f prime of g of x is equal to this. So this part right over here, I will, let me square it off in green."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "And this is just going to be equal to, let me write it right over here, it's going to be equal to 1 half times all of this business to the negative 1 half power. So 3x squared minus x, which is exactly what we need to solve right over here. f prime of g of x is equal to this. So this part right over here, I will, let me square it off in green. What we're trying to solve right over here, f prime of g of x, we have just figured out, is exactly this thing right over here. It's the derivative of f of the outer function with respect to the inner function. So let me write it."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "So this part right over here, I will, let me square it off in green. What we're trying to solve right over here, f prime of g of x, we have just figured out, is exactly this thing right over here. It's the derivative of f of the outer function with respect to the inner function. So let me write it. It is equal to 1 half times g of x to the negative 1 half times 3x squared minus x. This is exactly this, based on how we've defined f of x and how we've defined g of x. Conceptually, if you're just looking at this, the derivative of the outer thing, you're taking something to the 1 half power."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "So let me write it. It is equal to 1 half times g of x to the negative 1 half times 3x squared minus x. This is exactly this, based on how we've defined f of x and how we've defined g of x. Conceptually, if you're just looking at this, the derivative of the outer thing, you're taking something to the 1 half power. So the derivative of that whole thing with respect to your something is going to be 1 half times that something to the negative 1 half power. That's essentially what we're saying. But now we have to take the derivative of our something with respect to x."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "Conceptually, if you're just looking at this, the derivative of the outer thing, you're taking something to the 1 half power. So the derivative of that whole thing with respect to your something is going to be 1 half times that something to the negative 1 half power. That's essentially what we're saying. But now we have to take the derivative of our something with respect to x. The derivative of our something with respect to x. And that's more straightforward. g prime of x, we just use the power rule for each of these terms, is equal to 6x to the first, or just 6x, minus 1."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "But now we have to take the derivative of our something with respect to x. The derivative of our something with respect to x. And that's more straightforward. g prime of x, we just use the power rule for each of these terms, is equal to 6x to the first, or just 6x, minus 1. So this part right over here is just going to be 6x minus 1. Just to be clear. This right over here is this right over here."}, {"video_title": "Worked example Derivative of \u00c3\u0083\u00c2\u0083(3x_-x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "g prime of x, we just use the power rule for each of these terms, is equal to 6x to the first, or just 6x, minus 1. So this part right over here is just going to be 6x minus 1. Just to be clear. This right over here is this right over here. And we're multiplying. And we're done. We have just applied the power rule."}, {"video_title": "Analyzing motion problems total distance traveled AP Calculus AB Khan Academy.mp3", "Sentence": "Alexei received the following problem. A particle moves in a straight line with velocity v of t is equal to negative t squared plus eight meters per second where t is time in seconds. At t is equal to two, the particle's distance from the starting point was five meters. What is the total distance the particle has traveled between t equals two and t equals six seconds? Which expression should Alexei use to solve the problem? So we don't actually have to figure the actual answer out. We just have to figure out what is the appropriate expression."}, {"video_title": "Analyzing motion problems total distance traveled AP Calculus AB Khan Academy.mp3", "Sentence": "What is the total distance the particle has traveled between t equals two and t equals six seconds? Which expression should Alexei use to solve the problem? So we don't actually have to figure the actual answer out. We just have to figure out what is the appropriate expression. So like always, pause this video and see if you can work through it on your own. So now let's tackle this together. So the key question is what is the total distance the particle has traveled between t equals two and t equals six?"}, {"video_title": "Analyzing motion problems total distance traveled AP Calculus AB Khan Academy.mp3", "Sentence": "We just have to figure out what is the appropriate expression. So like always, pause this video and see if you can work through it on your own. So now let's tackle this together. So the key question is what is the total distance the particle has traveled between t equals two and t equals six? So we just care what happens between those points. We don't care that the particle's distance from the starting point was five meters at t equals two. So this right over here is actually unnecessary information."}, {"video_title": "Analyzing motion problems total distance traveled AP Calculus AB Khan Academy.mp3", "Sentence": "So the key question is what is the total distance the particle has traveled between t equals two and t equals six? So we just care what happens between those points. We don't care that the particle's distance from the starting point was five meters at t equals two. So this right over here is actually unnecessary information. So the first thing that you might want to think about is well maybe distance is just the integral of the velocity function. We see that multiple times. If you want to find the change in a quantity, you just say the starting time and the ending time, and then you integrate the rate function."}, {"video_title": "Analyzing motion problems total distance traveled AP Calculus AB Khan Academy.mp3", "Sentence": "So this right over here is actually unnecessary information. So the first thing that you might want to think about is well maybe distance is just the integral of the velocity function. We see that multiple times. If you want to find the change in a quantity, you just say the starting time and the ending time, and then you integrate the rate function. So wouldn't it just be that? Now we have to be very, very careful. If the question was what is the displacement for the particle between time equals two and times equals six, this would have been the correct answer."}, {"video_title": "Analyzing motion problems total distance traveled AP Calculus AB Khan Academy.mp3", "Sentence": "If you want to find the change in a quantity, you just say the starting time and the ending time, and then you integrate the rate function. So wouldn't it just be that? Now we have to be very, very careful. If the question was what is the displacement for the particle between time equals two and times equals six, this would have been the correct answer. So this would be displacement, displacement from t equals two to t is equal to six. But they're not saying displacement. They're saying total distance the particle has traveled."}, {"video_title": "Analyzing motion problems total distance traveled AP Calculus AB Khan Academy.mp3", "Sentence": "If the question was what is the displacement for the particle between time equals two and times equals six, this would have been the correct answer. So this would be displacement, displacement from t equals two to t is equal to six. But they're not saying displacement. They're saying total distance the particle has traveled. So this is the total path length for the particle. So one way to think about it, this is you would integrate not the velocity function. If you integrate velocity, you get displacement."}, {"video_title": "Analyzing motion problems total distance traveled AP Calculus AB Khan Academy.mp3", "Sentence": "They're saying total distance the particle has traveled. So this is the total path length for the particle. So one way to think about it, this is you would integrate not the velocity function. If you integrate velocity, you get displacement. Instead, you would integrate the speed function. Now what is speed? It is the magnitude of velocity."}, {"video_title": "Analyzing motion problems total distance traveled AP Calculus AB Khan Academy.mp3", "Sentence": "If you integrate velocity, you get displacement. Instead, you would integrate the speed function. Now what is speed? It is the magnitude of velocity. In one dimension, it would just be the absolute value of your velocity function. And so the absolute value of the velocity function, this would give you, integrating the speed, this would give you the distance. Distance from t equals two to t is equal to six."}, {"video_title": "Analyzing motion problems total distance traveled AP Calculus AB Khan Academy.mp3", "Sentence": "It is the magnitude of velocity. In one dimension, it would just be the absolute value of your velocity function. And so the absolute value of the velocity function, this would give you, integrating the speed, this would give you the distance. Distance from t equals two to t is equal to six. And let's see, we have that choice right over here. The displacement one here, this is an interesting distractor, but that is not going to be the choice. This one right over here, v prime of six, that gives you the acceleration."}, {"video_title": "Analyzing motion problems total distance traveled AP Calculus AB Khan Academy.mp3", "Sentence": "Distance from t equals two to t is equal to six. And let's see, we have that choice right over here. The displacement one here, this is an interesting distractor, but that is not going to be the choice. This one right over here, v prime of six, that gives you the acceleration. If you're taking the derivative of the velocity function, the acceleration at six seconds, that's not what we're interested in. And this gives you the absolute difference in velocity between time six and time two. That's not what we were trying to figure out either."}, {"video_title": "Area between curves with multiple boundaries.mp3", "Sentence": "And the way that we can tackle this is by dividing this area into two sections, or dividing this region into two regions, the region on the left and the region on the right, where for this first region, which I'll color even more in yellow, for this first region, over that entire interval in x, and it looks like x is going between 0 and 1, y equals, when x is equal to 1, this function is equal to 1, when x is equal to 1, this function is also equal to 1. So this is the point 1, 1, that's where they intersect. So for this section, this sub-region right over here, y equals square root of x is the upper function the entire time. And then we can set up a different, we can separately tackle figuring out the area of this region, from x is equal to 1 to x is equal to 2, where y equals 2 minus x is the upper function. So let's do it. So let's first think about this first region. Well, that's going to be the definite integral from x is equal to 0 to x is equal to 1, and our upper function is square root of x, so square root of x, and then from that we want to subtract our lower function, square root of x minus x squared over 4 minus 1."}, {"video_title": "Area between curves with multiple boundaries.mp3", "Sentence": "And then we can set up a different, we can separately tackle figuring out the area of this region, from x is equal to 1 to x is equal to 2, where y equals 2 minus x is the upper function. So let's do it. So let's first think about this first region. Well, that's going to be the definite integral from x is equal to 0 to x is equal to 1, and our upper function is square root of x, so square root of x, and then from that we want to subtract our lower function, square root of x minus x squared over 4 minus 1. So minus x squared over 4 minus 1. And then, of course, we have our dx. So this right over here, this is describing the area in yellow, and you can imagine it that this part right over here, the difference between these two functions, is essentially this height, we'll do it in a different color, is essentially this height, this height, and then you multiply it times dx, you get a little rectangle with width dx, and then you do that for each x, each x you get a different rectangle, and then you sum them all up, and you take the limit as your change in x approaches 0, so you get ultra, ultra thin rectangles, and you have an infinite number of them."}, {"video_title": "Area between curves with multiple boundaries.mp3", "Sentence": "Well, that's going to be the definite integral from x is equal to 0 to x is equal to 1, and our upper function is square root of x, so square root of x, and then from that we want to subtract our lower function, square root of x minus x squared over 4 minus 1. So minus x squared over 4 minus 1. And then, of course, we have our dx. So this right over here, this is describing the area in yellow, and you can imagine it that this part right over here, the difference between these two functions, is essentially this height, we'll do it in a different color, is essentially this height, this height, and then you multiply it times dx, you get a little rectangle with width dx, and then you do that for each x, each x you get a different rectangle, and then you sum them all up, and you take the limit as your change in x approaches 0, so you get ultra, ultra thin rectangles, and you have an infinite number of them. And that's our definition, the Riemann definition of what a definite integral is. And so this is the area of the left region, and by the exact same logic, we can figure out the area of the right region. The right region, and then we can just sum the two things together."}, {"video_title": "Area between curves with multiple boundaries.mp3", "Sentence": "So this right over here, this is describing the area in yellow, and you can imagine it that this part right over here, the difference between these two functions, is essentially this height, we'll do it in a different color, is essentially this height, this height, and then you multiply it times dx, you get a little rectangle with width dx, and then you do that for each x, each x you get a different rectangle, and then you sum them all up, and you take the limit as your change in x approaches 0, so you get ultra, ultra thin rectangles, and you have an infinite number of them. And that's our definition, the Riemann definition of what a definite integral is. And so this is the area of the left region, and by the exact same logic, we can figure out the area of the right region. The right region, and then we can just sum the two things together. The right region, we're going from x is equal to 0 to x, sorry, x is equal to 1, x is equal to 2, 1 to 2. The upper function is 2 minus x, and from that we're going to subtract, from that we're going to subtract the lower function, which is x squared over 4 minus 1. x squared over 4 minus 1. And now we just have to evaluate."}, {"video_title": "Area between curves with multiple boundaries.mp3", "Sentence": "The right region, and then we can just sum the two things together. The right region, we're going from x is equal to 0 to x, sorry, x is equal to 1, x is equal to 2, 1 to 2. The upper function is 2 minus x, and from that we're going to subtract, from that we're going to subtract the lower function, which is x squared over 4 minus 1. x squared over 4 minus 1. And now we just have to evaluate. So let's first simplify this right over here. This is equal to the definite integral from 0 to 1 of square root of x minus x squared over 4 plus 1 dx, I'm going to write it all in one color now, plus the definite integral from 1 to 2 of 2 minus x minus x squared over 4, then subtracting a negative 1 is a positive 3, or a positive 1, we could just add it to this 2, and so this 2 just becomes a 3. I said 2 minus negative 1 is 3 dx, and now we just have to take the antiderivative and evaluate it at 1 and 0."}, {"video_title": "Volume with cross sections perpendicular to y-axis AP Calculus AB Khan Academy.mp3", "Sentence": "Let R be the region enclosed by Y is equal to four times the square root of nine minus X and the axes in the first quadrant. And we can see that region R in gray right over here. Region R is the base of a solid. For each Y value, the cross section of the solid taken perpendicular to the Y axis is a rectangle whose base lies in R and whose height is Y. Express the volume of the solid with a definite integral. So pause this video and see if you can do that. All right, now let's do this together."}, {"video_title": "Volume with cross sections perpendicular to y-axis AP Calculus AB Khan Academy.mp3", "Sentence": "For each Y value, the cross section of the solid taken perpendicular to the Y axis is a rectangle whose base lies in R and whose height is Y. Express the volume of the solid with a definite integral. So pause this video and see if you can do that. All right, now let's do this together. And first let's just try to visualize the solid. And I'll try to do it by drawing this with a little bit of perspective. So if that's our Y axis, and then this is our X axis right over here."}, {"video_title": "Volume with cross sections perpendicular to y-axis AP Calculus AB Khan Academy.mp3", "Sentence": "All right, now let's do this together. And first let's just try to visualize the solid. And I'll try to do it by drawing this with a little bit of perspective. So if that's our Y axis, and then this is our X axis right over here. And I can redraw region R. It looks something like this. And now let's just imagine a cross section of our solid. So it says the cross section of the solid taken perpendicular to the Y axis."}, {"video_title": "Volume with cross sections perpendicular to y-axis AP Calculus AB Khan Academy.mp3", "Sentence": "So if that's our Y axis, and then this is our X axis right over here. And I can redraw region R. It looks something like this. And now let's just imagine a cross section of our solid. So it says the cross section of the solid taken perpendicular to the Y axis. So let's pick a Y value right over here. We're gonna go perpendicular to the Y axis. It says whose base lies in R. So the base would look like that."}, {"video_title": "Volume with cross sections perpendicular to y-axis AP Calculus AB Khan Academy.mp3", "Sentence": "So it says the cross section of the solid taken perpendicular to the Y axis. So let's pick a Y value right over here. We're gonna go perpendicular to the Y axis. It says whose base lies in R. So the base would look like that. It would actually be the X value that corresponds to that particular Y value. So I'll just write X right over here. And then the height is Y."}, {"video_title": "Volume with cross sections perpendicular to y-axis AP Calculus AB Khan Academy.mp3", "Sentence": "It says whose base lies in R. So the base would look like that. It would actually be the X value that corresponds to that particular Y value. So I'll just write X right over here. And then the height is Y. So the height is going to be whatever our Y value is. And then if we wanted to calculate the volume of just a little bit, a slice that has an infinitesimal depth, we could think about that infinitesimal depth in terms of Y. So we could say its depth right over here is DY, DY."}, {"video_title": "Volume with cross sections perpendicular to y-axis AP Calculus AB Khan Academy.mp3", "Sentence": "And then the height is Y. So the height is going to be whatever our Y value is. And then if we wanted to calculate the volume of just a little bit, a slice that has an infinitesimal depth, we could think about that infinitesimal depth in terms of Y. So we could say its depth right over here is DY, DY. And we could draw other cross sections. For example, right over here, our Y is much lower. It might look some, so our height will be like that."}, {"video_title": "Volume with cross sections perpendicular to y-axis AP Calculus AB Khan Academy.mp3", "Sentence": "So we could say its depth right over here is DY, DY. And we could draw other cross sections. For example, right over here, our Y is much lower. It might look some, so our height will be like that. But then our base is the corresponding X value that sits on the curve right over that XY pair that would sit on that curve. And so this cross section would look like this. And once again, if we wanted to put, if we wanted to calculate its volume, we could say there's an infinitesimal volume and it would have depth DY."}, {"video_title": "Volume with cross sections perpendicular to y-axis AP Calculus AB Khan Academy.mp3", "Sentence": "It might look some, so our height will be like that. But then our base is the corresponding X value that sits on the curve right over that XY pair that would sit on that curve. And so this cross section would look like this. And once again, if we wanted to put, if we wanted to calculate its volume, we could say there's an infinitesimal volume and it would have depth DY. And so as we've learned many times in integration, what we wanna do is think about the volume of one of these, I guess you could say slices, and then integrate across all of them. Now there's a couple of ways to approach it. You could try to integrate with respect to X or you could integrate with respect to Y. I'm gonna argue it's much easier to integrate with respect to Y here because we already have things in terms of DY."}, {"video_title": "Volume with cross sections perpendicular to y-axis AP Calculus AB Khan Academy.mp3", "Sentence": "And once again, if we wanted to put, if we wanted to calculate its volume, we could say there's an infinitesimal volume and it would have depth DY. And so as we've learned many times in integration, what we wanna do is think about the volume of one of these, I guess you could say slices, and then integrate across all of them. Now there's a couple of ways to approach it. You could try to integrate with respect to X or you could integrate with respect to Y. I'm gonna argue it's much easier to integrate with respect to Y here because we already have things in terms of DY. The volume of this little slice is going to be Y times X times DY. Now if we wanna integrate with respect to Y, we want everything in terms of Y. And so what we need to do is express X in terms of Y."}, {"video_title": "Volume with cross sections perpendicular to y-axis AP Calculus AB Khan Academy.mp3", "Sentence": "You could try to integrate with respect to X or you could integrate with respect to Y. I'm gonna argue it's much easier to integrate with respect to Y here because we already have things in terms of DY. The volume of this little slice is going to be Y times X times DY. Now if we wanna integrate with respect to Y, we want everything in terms of Y. And so what we need to do is express X in terms of Y. So here we just have to solve for X. So one way to do this is, let's see, we can square both sides of, well actually let's divide both sides by four. So you get Y over four is equal to the square root of nine minus X."}, {"video_title": "Volume with cross sections perpendicular to y-axis AP Calculus AB Khan Academy.mp3", "Sentence": "And so what we need to do is express X in terms of Y. So here we just have to solve for X. So one way to do this is, let's see, we can square both sides of, well actually let's divide both sides by four. So you get Y over four is equal to the square root of nine minus X. Now we can square both sides. Y squared over 16 is equal to nine minus X. And then let's see, we could multiply both sides by negative one."}, {"video_title": "Volume with cross sections perpendicular to y-axis AP Calculus AB Khan Academy.mp3", "Sentence": "So you get Y over four is equal to the square root of nine minus X. Now we can square both sides. Y squared over 16 is equal to nine minus X. And then let's see, we could multiply both sides by negative one. So negative Y squared over 16 is equal to X minus nine. And now we could add nine to both sides and we get nine minus Y squared over 16 is equal to X. And so we could substitute that right over there."}, {"video_title": "Volume with cross sections perpendicular to y-axis AP Calculus AB Khan Academy.mp3", "Sentence": "And then let's see, we could multiply both sides by negative one. So negative Y squared over 16 is equal to X minus nine. And now we could add nine to both sides and we get nine minus Y squared over 16 is equal to X. And so we could substitute that right over there. So another way to express the volume of this little slice right over here of infinitesimal depth, DY depth, is going to be Y times nine minus Y squared over 16 DY. Y squared over 16 DY. And if we want to find the volume of the whole figure, that's gonna look something like that, we're just going to integrate from Y equals zero to Y is equal to 12."}, {"video_title": "Volume with cross sections perpendicular to y-axis AP Calculus AB Khan Academy.mp3", "Sentence": "And so we could substitute that right over there. So another way to express the volume of this little slice right over here of infinitesimal depth, DY depth, is going to be Y times nine minus Y squared over 16 DY. Y squared over 16 DY. And if we want to find the volume of the whole figure, that's gonna look something like that, we're just going to integrate from Y equals zero to Y is equal to 12. So integrate from Y is equal to zero to Y is equal to 12. And that's all they asked us to do to express the volume as a definite integral. But this is actually a definite integral that you could solve without a calculator."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "That's my y-axis. That is my x-axis and maybe f of x looks something like that. What I want to do is I want to approximate f of x with a Taylor polynomial centered around x is equal to a. This is the x-axis. This is the y-axis. I want a Taylor polynomial centered around there. We've seen how this works."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "This is the x-axis. This is the y-axis. I want a Taylor polynomial centered around there. We've seen how this works. The Taylor polynomial comes out of the idea that for all of the derivatives up to and including the degree of the polynomial, those derivatives of that polynomial evaluated at a should be equal to the derivatives of our function evaluated at a and that polynomial evaluated at a should also be equal to that function evaluated at a. Our Taylor polynomial approximation would look something like this. I'll call it p of x and sometimes you might see a subscript, a big N there to say it's an nth degree approximation."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "We've seen how this works. The Taylor polynomial comes out of the idea that for all of the derivatives up to and including the degree of the polynomial, those derivatives of that polynomial evaluated at a should be equal to the derivatives of our function evaluated at a and that polynomial evaluated at a should also be equal to that function evaluated at a. Our Taylor polynomial approximation would look something like this. I'll call it p of x and sometimes you might see a subscript, a big N there to say it's an nth degree approximation. Sometimes you'll see something like this. Sometimes you'll see something like N, a to say it's an nth degree approximation centered at a. Actually, I'll write that right now."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "I'll call it p of x and sometimes you might see a subscript, a big N there to say it's an nth degree approximation. Sometimes you'll see something like this. Sometimes you'll see something like N, a to say it's an nth degree approximation centered at a. Actually, I'll write that right now. Maybe we might lose it if we have to keep writing it over and over but you should assume that it is an nth degree polynomial centered at a. It's going to look like this. It is going to be f of a plus f prime of a times x minus a plus f prime prime of a times x minus a squared over either you could write 2 or 2 factorial."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Actually, I'll write that right now. Maybe we might lose it if we have to keep writing it over and over but you should assume that it is an nth degree polynomial centered at a. It's going to look like this. It is going to be f of a plus f prime of a times x minus a plus f prime prime of a times x minus a squared over either you could write 2 or 2 factorial. They're the same value. I'll write 2 factorial. You could write it divided by 1 factorial over here if you like."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "It is going to be f of a plus f prime of a times x minus a plus f prime prime of a times x minus a squared over either you could write 2 or 2 factorial. They're the same value. I'll write 2 factorial. You could write it divided by 1 factorial over here if you like. Then plus you go to the third derivative of f at a times x minus a to the third power. I think you see where this is going over 3 factorial. You keep going."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "You could write it divided by 1 factorial over here if you like. Then plus you go to the third derivative of f at a times x minus a to the third power. I think you see where this is going over 3 factorial. You keep going. I'll go to this line right here all the way to your nth degree term which is the nth derivative of f evaluated at a times x minus a to the n over n factorial. This polynomial right over here, this nth degree polynomial centered at a, it's definitely f of a is going to be the same or p of a is going to be the same thing as f of a. You can verify that because all of these other terms have an x minus a here."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "You keep going. I'll go to this line right here all the way to your nth degree term which is the nth derivative of f evaluated at a times x minus a to the n over n factorial. This polynomial right over here, this nth degree polynomial centered at a, it's definitely f of a is going to be the same or p of a is going to be the same thing as f of a. You can verify that because all of these other terms have an x minus a here. If you put an a in the polynomial, all of these other terms are going to be 0. You'll have p of a is equal to f of a. Let me write that down."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "You can verify that because all of these other terms have an x minus a here. If you put an a in the polynomial, all of these other terms are going to be 0. You'll have p of a is equal to f of a. Let me write that down. p of a is equal to f of a. It might look something like this. It's going to fit the curve better the more of these terms that we actually have."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let me write that down. p of a is equal to f of a. It might look something like this. It's going to fit the curve better the more of these terms that we actually have. It might look something like this. I'll try my best to show what it might look like. What I want to do in this video, this is all review."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "It's going to fit the curve better the more of these terms that we actually have. It might look something like this. I'll try my best to show what it might look like. What I want to do in this video, this is all review. I have this polynomial that's approximating this function. The more terms I have, the higher degree of this polynomial, the better that it will fit this curve, the further that I get away from a. What I want to do in this video is think about if we can bound how good it's fitting this function as we move away from a."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "What I want to do in this video, this is all review. I have this polynomial that's approximating this function. The more terms I have, the higher degree of this polynomial, the better that it will fit this curve, the further that I get away from a. What I want to do in this video is think about if we can bound how good it's fitting this function as we move away from a. What I want to do is define a remainder function. Sometimes I've seen some textbooks call it an error function. I'm going to call this an error."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "What I want to do in this video is think about if we can bound how good it's fitting this function as we move away from a. What I want to do is define a remainder function. Sometimes I've seen some textbooks call it an error function. I'm going to call this an error. Just so you're consistent with all the different notations you might see in a book, some people will call this a remainder function and sometimes they'll write a remainder function for an nth degree polynomial centered at a. Sometimes you'll see this as an error function. The error function is sometimes avoided because it looks like expected value from probability."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "I'm going to call this an error. Just so you're consistent with all the different notations you might see in a book, some people will call this a remainder function and sometimes they'll write a remainder function for an nth degree polynomial centered at a. Sometimes you'll see this as an error function. The error function is sometimes avoided because it looks like expected value from probability. You'll see this often. This is e for error, r for remainder, and sometimes they'll also have the subscripts over there like that. What we'll do is we'll just define this function to be the difference between f of x and our approximation of f of x for any given x."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "The error function is sometimes avoided because it looks like expected value from probability. You'll see this often. This is e for error, r for remainder, and sometimes they'll also have the subscripts over there like that. What we'll do is we'll just define this function to be the difference between f of x and our approximation of f of x for any given x. It's really just going to be f of x minus p of x where this is an nth degree polynomial centered at a. For example, if someone were to ask you or if you wanted to visualize, what are they talking about if they're saying the error of this nth degree polynomial centered at a when we are at x is equal to b? What is this thing equal to or how should you think about this?"}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "What we'll do is we'll just define this function to be the difference between f of x and our approximation of f of x for any given x. It's really just going to be f of x minus p of x where this is an nth degree polynomial centered at a. For example, if someone were to ask you or if you wanted to visualize, what are they talking about if they're saying the error of this nth degree polynomial centered at a when we are at x is equal to b? What is this thing equal to or how should you think about this? If b is right over here, the error of b is going to be f of b minus the polynomial at b. f of b there, the polynomial is right over there, so it'll be this distance right over here. If you measure the error at a, it would actually be 0 because the polynomial and the function are the same there. f of a is equal to p of a, so the error at a is equal to 0."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "What is this thing equal to or how should you think about this? If b is right over here, the error of b is going to be f of b minus the polynomial at b. f of b there, the polynomial is right over there, so it'll be this distance right over here. If you measure the error at a, it would actually be 0 because the polynomial and the function are the same there. f of a is equal to p of a, so the error at a is equal to 0. Let me actually write that down because that's an interesting property and it'll help us bound it eventually. Let me write that. The error function at a, and for the rest of this video, you can assume that I could write a subscript."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "f of a is equal to p of a, so the error at a is equal to 0. Let me actually write that down because that's an interesting property and it'll help us bound it eventually. Let me write that. The error function at a, and for the rest of this video, you can assume that I could write a subscript. This is for the nth degree polynomial centered at a. I'm just going to not write that every time just to save ourselves a little bit of time in writing to keep my hand fresh. The error at a is equal to f of a minus p of a. Once again, I won't write the sub n sub a."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "The error function at a, and for the rest of this video, you can assume that I could write a subscript. This is for the nth degree polynomial centered at a. I'm just going to not write that every time just to save ourselves a little bit of time in writing to keep my hand fresh. The error at a is equal to f of a minus p of a. Once again, I won't write the sub n sub a. You can assume it. This is an nth degree polynomial centered at a. These two things are equal to each other, so this is going to be equal to 0."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Once again, I won't write the sub n sub a. You can assume it. This is an nth degree polynomial centered at a. These two things are equal to each other, so this is going to be equal to 0. We see that right over here. The distance between the two functions is 0 there. Now, let's think about something else."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "These two things are equal to each other, so this is going to be equal to 0. We see that right over here. The distance between the two functions is 0 there. Now, let's think about something else. Let's think about what the derivative of the error function of value at a is. That's going to be the derivative of our function at a minus the first derivative of our polynomial at a. If we assume that this is higher than degree 1, we know that these derivatives are going to be the same at a."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Now, let's think about something else. Let's think about what the derivative of the error function of value at a is. That's going to be the derivative of our function at a minus the first derivative of our polynomial at a. If we assume that this is higher than degree 1, we know that these derivatives are going to be the same at a. You can try to take the first derivative here. If you take the first derivative of this whole mess, and this is actually why Taylor polynomials are so useful, is that up to and including the degree of the polynomial, when you evaluate the derivatives of your polynomial at a, they're going to be the same as the derivatives of the function at a. That's what starts to make it a good approximation."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "If we assume that this is higher than degree 1, we know that these derivatives are going to be the same at a. You can try to take the first derivative here. If you take the first derivative of this whole mess, and this is actually why Taylor polynomials are so useful, is that up to and including the degree of the polynomial, when you evaluate the derivatives of your polynomial at a, they're going to be the same as the derivatives of the function at a. That's what starts to make it a good approximation. If you took a derivative here, this term right here will disappear. It'll go to 0. I'll cross it out for now."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "That's what starts to make it a good approximation. If you took a derivative here, this term right here will disappear. It'll go to 0. I'll cross it out for now. This term right over here will just be f prime of a. Then all of these other terms are going to be left with some type of an x minus a in them. When you evaluate at a, all the terms with an x minus a disappear because you have an a minus a on them."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "I'll cross it out for now. This term right over here will just be f prime of a. Then all of these other terms are going to be left with some type of an x minus a in them. When you evaluate at a, all the terms with an x minus a disappear because you have an a minus a on them. This one already disappeared, and you're literally just left with p prime of a will equal to f prime of a. We've seen that before. Let me write that."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "When you evaluate at a, all the terms with an x minus a disappear because you have an a minus a on them. This one already disappeared, and you're literally just left with p prime of a will equal to f prime of a. We've seen that before. Let me write that. Because we know that p prime of a is equal to f prime of a, when you evaluate the error function, the derivative of the error function at a, that also is going to be equal to 0. This general property right over here is true up to and including n. Let me write this down. We already know that p of a is equal to f of a."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let me write that. Because we know that p prime of a is equal to f prime of a, when you evaluate the error function, the derivative of the error function at a, that also is going to be equal to 0. This general property right over here is true up to and including n. Let me write this down. We already know that p of a is equal to f of a. We already know that p prime of a is equal to f prime of a. This really comes straight out of the definition of the Taylor polynomials. This is going to be true all the way until the nth derivative of our polynomial is going, evaluated at a, not everywhere, just evaluated at a, is going to be equal to the nth derivative of our function evaluated at a."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "We already know that p of a is equal to f of a. We already know that p prime of a is equal to f prime of a. This really comes straight out of the definition of the Taylor polynomials. This is going to be true all the way until the nth derivative of our polynomial is going, evaluated at a, not everywhere, just evaluated at a, is going to be equal to the nth derivative of our function evaluated at a. What that tells us is that we could keep doing this with the error function all the way to the nth derivative of the error function evaluated at a is going to be equal to the nth derivative of f evaluated at a minus the nth derivative of our polynomial evaluated at a. We already said that these are going to be equal to each other up to the nth derivative when we evaluate them at a. These are all going to be equal to 0."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "This is going to be true all the way until the nth derivative of our polynomial is going, evaluated at a, not everywhere, just evaluated at a, is going to be equal to the nth derivative of our function evaluated at a. What that tells us is that we could keep doing this with the error function all the way to the nth derivative of the error function evaluated at a is going to be equal to the nth derivative of f evaluated at a minus the nth derivative of our polynomial evaluated at a. We already said that these are going to be equal to each other up to the nth derivative when we evaluate them at a. These are all going to be equal to 0. This is an interesting property. It's also going to be useful when we start to try to bound this error function. That's the whole point of where I'm going with this video and probably the next video."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "These are all going to be equal to 0. This is an interesting property. It's also going to be useful when we start to try to bound this error function. That's the whole point of where I'm going with this video and probably the next video. We're going to try to bound it so we know how good of an estimate we have, especially as we go further and further from where we are centered, from where our approximation is centered. Now let's think about when we take a derivative beyond that. Let's think about what happens when we take the n plus 1th derivative."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "That's the whole point of where I'm going with this video and probably the next video. We're going to try to bound it so we know how good of an estimate we have, especially as we go further and further from where we are centered, from where our approximation is centered. Now let's think about when we take a derivative beyond that. Let's think about what happens when we take the n plus 1th derivative. What's a good place to write? I have some screen real estate right over here. What happens if I take the n plus 1th derivative of our error function?"}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let's think about what happens when we take the n plus 1th derivative. What's a good place to write? I have some screen real estate right over here. What happens if I take the n plus 1th derivative of our error function? Not even if I'm just evaluating it today. If I just say generally the error function E of X, what's the n plus 1th derivative of it? It's going to be the n plus 1th derivative of our function minus the n plus 1th derivative of our function."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "What happens if I take the n plus 1th derivative of our error function? Not even if I'm just evaluating it today. If I just say generally the error function E of X, what's the n plus 1th derivative of it? It's going to be the n plus 1th derivative of our function minus the n plus 1th derivative of our function. I'm literally just taking the n plus 1th derivative of both sides of this equation right over here. It's literally the n plus 1th derivative of our function minus the n plus 1th derivative of our nth degree polynomial. Once again, I could write an n here, I could write an a here to show it's an nth degree centered at a."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "It's going to be the n plus 1th derivative of our function minus the n plus 1th derivative of our function. I'm literally just taking the n plus 1th derivative of both sides of this equation right over here. It's literally the n plus 1th derivative of our function minus the n plus 1th derivative of our nth degree polynomial. Once again, I could write an n here, I could write an a here to show it's an nth degree centered at a. What is the n plus 1th derivative of an nth degree polynomial? If you want some hints, take the second derivative of Y is equal to X. It's a first degree polynomial."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Once again, I could write an n here, I could write an a here to show it's an nth degree centered at a. What is the n plus 1th derivative of an nth degree polynomial? If you want some hints, take the second derivative of Y is equal to X. It's a first degree polynomial. Take the second derivative, you're going to get zero. Take the third derivative of Y is equal to X squared. The first derivative is 2X, the second derivative is 2, the third derivative is zero."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "It's a first degree polynomial. Take the second derivative, you're going to get zero. Take the third derivative of Y is equal to X squared. The first derivative is 2X, the second derivative is 2, the third derivative is zero. In general, if you take an n plus 1th derivative of an nth degree polynomial, and you could prove it for yourself, you could even prove it generally, but I think it might make a little sense to you, it's going to be equal to zero. This thing right here, this is an n plus 1th derivative of an nth degree polynomial. This is going to be equal to zero."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "The first derivative is 2X, the second derivative is 2, the third derivative is zero. In general, if you take an n plus 1th derivative of an nth degree polynomial, and you could prove it for yourself, you could even prove it generally, but I think it might make a little sense to you, it's going to be equal to zero. This thing right here, this is an n plus 1th derivative of an nth degree polynomial. This is going to be equal to zero. The n plus 1th derivative of our error function, or our remainder function we could call it, is equal to the n plus 1th derivative of our function. What we could do now, and we'll probably have to continue this in the next video, is figure out at least can we bound this? If we are able to bound this, if we're able to figure out an upper bound on its magnitude, so actually what we want to do is we want to bound its overall magnitude."}, {"video_title": "Taylor polynomial remainder (part 1) Series AP Calculus BC Khan Academy.mp3", "Sentence": "This is going to be equal to zero. The n plus 1th derivative of our error function, or our remainder function we could call it, is equal to the n plus 1th derivative of our function. What we could do now, and we'll probably have to continue this in the next video, is figure out at least can we bound this? If we are able to bound this, if we're able to figure out an upper bound on its magnitude, so actually what we want to do is we want to bound its overall magnitude. We want to bound its absolute value. If we can determine that it is less than or equal to some value m, so if we can actually bound it, maybe we can do a little bit of calculus, we could keep integrating it, and maybe we can go back to the original function and bound that in some way, if we do know some type of bound like this over here. I'll take that up in the next video."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And the way I'll do it is by setting up 5 trapezoids of equal width. So this will be the left boundary of the first trapezoid. This will be its right boundary, which will also be the left boundary of the second trapezoid. This will be the right boundary of the second trapezoid. This is the right boundary of the third trapezoid. This will be the right boundary of the fourth trapezoid. And then finally, this will be the route boundary of the fifth trapezoid."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "This will be the right boundary of the second trapezoid. This is the right boundary of the third trapezoid. This will be the right boundary of the fourth trapezoid. And then finally, this will be the route boundary of the fifth trapezoid. And since we're traveling, we're going from 1 to 6, so we're traveling 6 minus 1 in the x direction. And I want to split it into five sections. The width of each trapezoid is just going to be equal to 1."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And then finally, this will be the route boundary of the fifth trapezoid. And since we're traveling, we're going from 1 to 6, so we're traveling 6 minus 1 in the x direction. And I want to split it into five sections. The width of each trapezoid is just going to be equal to 1. And so if we say that the width of a trapezoid is delta x, we just can now say that delta x is equal to 1. So let's set up our trapezoids. So the first trapezoid is going to look like that."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "The width of each trapezoid is just going to be equal to 1. And so if we say that the width of a trapezoid is delta x, we just can now say that delta x is equal to 1. So let's set up our trapezoids. So the first trapezoid is going to look like that. Actually, it's going to be a triangle, not really a trapezoid. Then the second trapezoid is going to look like this. I guess you could say a trapezoid where one of the sides has length 0 turns into a triangle."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So the first trapezoid is going to look like that. Actually, it's going to be a triangle, not really a trapezoid. Then the second trapezoid is going to look like this. I guess you could say a trapezoid where one of the sides has length 0 turns into a triangle. Then the third trapezoid is going to look like this. And then the fourth trapezoid is going to look like that. And then finally, you have the fifth trapezoid."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "I guess you could say a trapezoid where one of the sides has length 0 turns into a triangle. Then the third trapezoid is going to look like this. And then the fourth trapezoid is going to look like that. And then finally, you have the fifth trapezoid. So let's calculate the area of each of these. And then we will have our approximation for the area under the curve. So let's do trapezoid, or I really should say triangle, this first shape, whatever you want to call it."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And then finally, you have the fifth trapezoid. So let's calculate the area of each of these. And then we will have our approximation for the area under the curve. So let's do trapezoid, or I really should say triangle, this first shape, whatever you want to call it. What is the area of that going to be? Well, the area of a trapezoid, and you'll see this will just turn into the area of a triangle, it's the average of the heights of the two sides of the trapezoid, the way we've looked at it. Or you could say the average of the heights of the two parallel sides."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So let's do trapezoid, or I really should say triangle, this first shape, whatever you want to call it. What is the area of that going to be? Well, the area of a trapezoid, and you'll see this will just turn into the area of a triangle, it's the average of the heights of the two sides of the trapezoid, the way we've looked at it. Or you could say the average of the heights of the two parallel sides. I guess that's the best way to say it. So f of 1, that's the height here, plus f of 2, all of that over 2. And then we're going to multiply it times our delta x."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Or you could say the average of the heights of the two parallel sides. I guess that's the best way to say it. So f of 1, that's the height here, plus f of 2, all of that over 2. And then we're going to multiply it times our delta x. Actually, let me do that in that same red color to show you that this is the area of that first trapezoid. And as you see right over here, if you look at it, the f of 1 is just going to be 0. So you're going to have f of 2 times, so it's going to be that this height times this base times 1 half, which is just the area of a triangle."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And then we're going to multiply it times our delta x. Actually, let me do that in that same red color to show you that this is the area of that first trapezoid. And as you see right over here, if you look at it, the f of 1 is just going to be 0. So you're going to have f of 2 times, so it's going to be that this height times this base times 1 half, which is just the area of a triangle. But let's look at the second trapezoid, trapezoid 2. Right over here, what is its area going to be? Well, it's going to be f of 2 plus f of 3. f of 2 is this height, f of 3 is this height."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So you're going to have f of 2 times, so it's going to be that this height times this base times 1 half, which is just the area of a triangle. But let's look at the second trapezoid, trapezoid 2. Right over here, what is its area going to be? Well, it's going to be f of 2 plus f of 3. f of 2 is this height, f of 3 is this height. So we're taking the average of those two heights, divided by 2, that's the average of those two heights, times the base, times delta x. And then let's do trapezoid 3. I think you're getting the idea here."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Well, it's going to be f of 2 plus f of 3. f of 2 is this height, f of 3 is this height. So we're taking the average of those two heights, divided by 2, that's the average of those two heights, times the base, times delta x. And then let's do trapezoid 3. I think you're getting the idea here. Trapezoid 3 is going to be f of 3 plus f of 4, divided by 2 times delta x. And then, let's see, I'm running out of colors. This is trapezoid 4 right over here."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "I think you're getting the idea here. Trapezoid 3 is going to be f of 3 plus f of 4, divided by 2 times delta x. And then, let's see, I'm running out of colors. This is trapezoid 4 right over here. So plus f of 4 plus f of 5, all of that over 2 times delta x. And then we have our last trapezoid, which I will do in yellow. So this is trapezoid number 5."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "This is trapezoid 4 right over here. So plus f of 4 plus f of 5, all of that over 2 times delta x. And then we have our last trapezoid, which I will do in yellow. So this is trapezoid number 5. Scroll down a little bit, get some more real estate. So it's going to be plus, I'll just write the plus over here, plus f of 5 plus f of 6 over 2 times our delta x. So let's see how we can simplify this a little bit."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So this is trapezoid number 5. Scroll down a little bit, get some more real estate. So it's going to be plus, I'll just write the plus over here, plus f of 5 plus f of 6 over 2 times our delta x. So let's see how we can simplify this a little bit. All of these terms, we have a 1 half delta x. So let's actually factor that out. So remember, this is our approximation of our area."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So let's see how we can simplify this a little bit. All of these terms, we have a 1 half delta x. So let's actually factor that out. So remember, this is our approximation of our area. So area is approximately, remember, this is just a rough approximation. It's very clear, actually. You might say, this is pretty good using the trapezoids."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So remember, this is our approximation of our area. So area is approximately, remember, this is just a rough approximation. It's very clear, actually. You might say, this is pretty good using the trapezoids. But it is pretty clear that we are letting go of some of the area. We're letting go of that area. We're letting go of some of this right over here."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "You might say, this is pretty good using the trapezoids. But it is pretty clear that we are letting go of some of the area. We're letting go of that area. We're letting go of some of this right over here. You can barely see it. Some of this right over here, you can barely see it. So this is going to be, it looks like an underestimate."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "We're letting go of some of this right over here. You can barely see it. Some of this right over here, you can barely see it. So this is going to be, it looks like an underestimate. But it is a decent approximation. Let's see if we can simplify this expression. So it's approximately going to be equal to, I'm going to factor out a delta x over 2."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be, it looks like an underestimate. But it is a decent approximation. Let's see if we can simplify this expression. So it's approximately going to be equal to, I'm going to factor out a delta x over 2. I'm going to factor out a delta x over 2. And then what I'm left with, and I will switch to a neutral color, if I factor out a delta x over 2, then I have just an f of 1. And then I have 2 f of 2's, so plus 2 times f of 2."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So it's approximately going to be equal to, I'm going to factor out a delta x over 2. I'm going to factor out a delta x over 2. And then what I'm left with, and I will switch to a neutral color, if I factor out a delta x over 2, then I have just an f of 1. And then I have 2 f of 2's, so plus 2 times f of 2. And I'm doing this because you might see a formula that looks something like this in your calculus book. And it's not some mysterious thing. They just really summed up the areas of the trapezoids."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And then I have 2 f of 2's, so plus 2 times f of 2. And I'm doing this because you might see a formula that looks something like this in your calculus book. And it's not some mysterious thing. They just really summed up the areas of the trapezoids. And then we're going to have 2 f of 3's plus 2 times f of 3's plus we're going to have 2 f of 4's plus 2 times f of 4. And then we're going to have 2 f of 5's plus 2 times f of 5. And then finally, we're going to have 1 f of 6 plus f of 6."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "They just really summed up the areas of the trapezoids. And then we're going to have 2 f of 3's plus 2 times f of 3's plus we're going to have 2 f of 4's plus 2 times f of 4. And then we're going to have 2 f of 5's plus 2 times f of 5. And then finally, we're going to have 1 f of 6 plus f of 6. If you were to generalize it, you have 1 of the first endpoint, the function evaluated at the first endpoint, 1 of it at the very last endpoint, and then 2 of all of the rest of them. But this is just the area of trapezoids. I'm not actually a big fan when textbooks write this."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And then finally, we're going to have 1 f of 6 plus f of 6. If you were to generalize it, you have 1 of the first endpoint, the function evaluated at the first endpoint, 1 of it at the very last endpoint, and then 2 of all of the rest of them. But this is just the area of trapezoids. I'm not actually a big fan when textbooks write this. Because when you see this, it's hard to visualize the trapezoids. When you see this, it's much clearer how you might visualize that. But with that out of the way, let's actually evaluate this."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "I'm not actually a big fan when textbooks write this. Because when you see this, it's hard to visualize the trapezoids. When you see this, it's much clearer how you might visualize that. But with that out of the way, let's actually evaluate this. Lucky for us, the math is simple. Our delta x is just 1. Our delta x is just 1."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "But with that out of the way, let's actually evaluate this. Lucky for us, the math is simple. Our delta x is just 1. Our delta x is just 1. And then we just have to evaluate all of this business. f of 1, let's just remind ourselves what our original function was. Our original function was the square root of x minus 1."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Our delta x is just 1. And then we just have to evaluate all of this business. f of 1, let's just remind ourselves what our original function was. Our original function was the square root of x minus 1. So f of 1 is the square root of 1 minus 1. So that is just going to be 0. This expression right over here is going to be 2 times the square root of 2 minus 1."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Our original function was the square root of x minus 1. So f of 1 is the square root of 1 minus 1. So that is just going to be 0. This expression right over here is going to be 2 times the square root of 2 minus 1. The square root of 2 minus 1 is just 1. So this is just going to be 2. Actually, let me do it in that same."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "This expression right over here is going to be 2 times the square root of 2 minus 1. The square root of 2 minus 1 is just 1. So this is just going to be 2. Actually, let me do it in that same. Well, I'm now using the purple for a different purpose than just the first trapezoid. Hopefully you realize that. I was just sticking with that pen color."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Actually, let me do it in that same. Well, I'm now using the purple for a different purpose than just the first trapezoid. Hopefully you realize that. I was just sticking with that pen color. Then f of 3, 3 minus 1 is 2, square root of 2. So the function evaluated at 3 is the square root of 2. So this is going to be 2 times the square root of 2."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "I was just sticking with that pen color. Then f of 3, 3 minus 1 is 2, square root of 2. So the function evaluated at 3 is the square root of 2. So this is going to be 2 times the square root of 2. Then the function evaluated at 4. When you evaluate at 4, you get the square root of 3. So this is going to be 2 times the square root of 3."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be 2 times the square root of 2. Then the function evaluated at 4. When you evaluate at 4, you get the square root of 3. So this is going to be 2 times the square root of 3. And then you get 2 times the square root of 4. 5 minus 1 is 4. 2 times the square root of 4 is just 4."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be 2 times the square root of 3. And then you get 2 times the square root of 4. 5 minus 1 is 4. 2 times the square root of 4 is just 4. And then finally, you get f of 6 is square root of 6 minus 1 is the square root of 5. And I think we are now ready to evaluate. Let me get my handy TI-85 out and calculate this."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "2 times the square root of 4 is just 4. And then finally, you get f of 6 is square root of 6 minus 1 is the square root of 5. And I think we are now ready to evaluate. Let me get my handy TI-85 out and calculate this. So it's going to be, well, I'm just going to calculate. Well, I'll just multiply. 0.5 times, open parentheses, well, it's a 0."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Let me get my handy TI-85 out and calculate this. So it's going to be, well, I'm just going to calculate. Well, I'll just multiply. 0.5 times, open parentheses, well, it's a 0. I'll just write it just so you know what I'm doing. 0 plus 2, whoops, lost my calculator, plus 2 times the square root of 2, plus 2 times the square root of 3, plus 4. I'm almost done."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "0.5 times, open parentheses, well, it's a 0. I'll just write it just so you know what I'm doing. 0 plus 2, whoops, lost my calculator, plus 2 times the square root of 2, plus 2 times the square root of 3, plus 4. I'm almost done. Plus the square root of 5. So let me write that. Plus the square root of 5 gives me, now we are ready for our drum roll, it gives me, and I'll just round it, 7.26."}, {"video_title": "Trapezoidal sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "I'm almost done. Plus the square root of 5. So let me write that. Plus the square root of 5 gives me, now we are ready for our drum roll, it gives me, and I'll just round it, 7.26. So the area is approximately equal to 7.26 under the curve. y is equal to the square root of x minus 1 between x equals 1 and x equals 6. And we did this using trapezoids."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "And just from an inspection, you can see that the function is not defined when x is equal to 3. You get 0 over 0. It's not defined. So to answer this question, let's try to rewrite the same exact function definition slightly differently. So let's say f of x is going to be equal to, and I'm going to think of two cases. I'm going to think of the case when x is greater than 3 and when x is less than 3. So when x is, we'll do this in two different colors actually."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "So to answer this question, let's try to rewrite the same exact function definition slightly differently. So let's say f of x is going to be equal to, and I'm going to think of two cases. I'm going to think of the case when x is greater than 3 and when x is less than 3. So when x is, we'll do this in two different colors actually. When x, do it in green, that's not green. When x is greater than 3, what does this function simplify to? Well, whatever I get up here, I'm going to get a positive value up here."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "So when x is, we'll do this in two different colors actually. When x, do it in green, that's not green. When x is greater than 3, what does this function simplify to? Well, whatever I get up here, I'm going to get a positive value up here. And then if I take the absolute value, it's going to be the exact same thing. So for x is greater than 3, this is going to be the exact same thing as x minus 3 over x minus 3. Because if x is greater than 3, the numerator is going to be positive."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "Well, whatever I get up here, I'm going to get a positive value up here. And then if I take the absolute value, it's going to be the exact same thing. So for x is greater than 3, this is going to be the exact same thing as x minus 3 over x minus 3. Because if x is greater than 3, the numerator is going to be positive. You take the absolute value of that, you're not going to change its value. So you get this right over here. Or if we were to rewrite it, this is equal to, for x is greater than 3, you're going to have f of x is equal to 1 for x is greater than 3."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "Because if x is greater than 3, the numerator is going to be positive. You take the absolute value of that, you're not going to change its value. So you get this right over here. Or if we were to rewrite it, this is equal to, for x is greater than 3, you're going to have f of x is equal to 1 for x is greater than 3. Similarly, let's think about what happens when x is less than 3. When x is less than 3, well, x minus 3 is going to be a negative number. When you take the absolute value of that, you're essentially negating it."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "Or if we were to rewrite it, this is equal to, for x is greater than 3, you're going to have f of x is equal to 1 for x is greater than 3. Similarly, let's think about what happens when x is less than 3. When x is less than 3, well, x minus 3 is going to be a negative number. When you take the absolute value of that, you're essentially negating it. So it's going to be the negative of x minus 3 over x minus 3. Or if you were to simplify these two things, for any value as long as x does not equal 3, this part right over here simplifies to 1, so you're left with a negative 1. Negative 1 for x is less than 3."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "When you take the absolute value of that, you're essentially negating it. So it's going to be the negative of x minus 3 over x minus 3. Or if you were to simplify these two things, for any value as long as x does not equal 3, this part right over here simplifies to 1, so you're left with a negative 1. Negative 1 for x is less than 3. I encourage you, if you don't believe what I just said, try it out with some numbers. Try out some numbers, 3.1, 3.001, 3.5, 4, 7, any number greater than 3, you're going to get 1. You're going to get the same thing divided by the same thing."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "Negative 1 for x is less than 3. I encourage you, if you don't believe what I just said, try it out with some numbers. Try out some numbers, 3.1, 3.001, 3.5, 4, 7, any number greater than 3, you're going to get 1. You're going to get the same thing divided by the same thing. And try values for x less than 3. You're going to get negative 1 no matter what you try. So let's visualize this function now."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "You're going to get the same thing divided by the same thing. And try values for x less than 3. You're going to get negative 1 no matter what you try. So let's visualize this function now. So now let me draw some axes. That's my x-axis. And then this is my f of x-axis."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "So let's visualize this function now. So now let me draw some axes. That's my x-axis. And then this is my f of x-axis. So y is equal to f of x. And what we care about is x is equal to 3. So x is equal to 1, 2, 3, 4, 5."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "And then this is my f of x-axis. So y is equal to f of x. And what we care about is x is equal to 3. So x is equal to 1, 2, 3, 4, 5. We could keep going. And let's say this is positive 1, 2. So that's y is equal to 1."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "So x is equal to 1, 2, 3, 4, 5. We could keep going. And let's say this is positive 1, 2. So that's y is equal to 1. This is y is equal to negative 1 and negative 2, and we can keep going. So this way that we've rewritten the function is the exact same function as this. We've just written it a different way."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "So that's y is equal to 1. This is y is equal to negative 1 and negative 2, and we can keep going. So this way that we've rewritten the function is the exact same function as this. We've just written it a different way. And so what we're saying is our function is undefined at 3. But if our x's are greater than 3, our function is equal to 1. So if our x is greater than 3, our function is equal to 1."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "We've just written it a different way. And so what we're saying is our function is undefined at 3. But if our x's are greater than 3, our function is equal to 1. So if our x is greater than 3, our function is equal to 1. So it looks like that. And it's undefined at 3. And if x is less than 3, our function is equal to negative 1."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "So if our x is greater than 3, our function is equal to 1. So it looks like that. And it's undefined at 3. And if x is less than 3, our function is equal to negative 1. So it looks like that. Let me do it in that same color. It looks like this."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "And if x is less than 3, our function is equal to negative 1. So it looks like that. Let me do it in that same color. It looks like this. Once again, it's undefined at 3. So it looks like that. So now let's try to answer our question."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "It looks like this. Once again, it's undefined at 3. So it looks like that. So now let's try to answer our question. What is the limit as x approaches 3? Well, let's think about the limit as x approaches 3 from the negative direction, from values less than 3. So let's think about first the limit as x approaches 3 at the limit of f of x as x approaches 3 from the negative direction."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "So now let's try to answer our question. What is the limit as x approaches 3? Well, let's think about the limit as x approaches 3 from the negative direction, from values less than 3. So let's think about first the limit as x approaches 3 at the limit of f of x as x approaches 3 from the negative direction. And all this notation here, I wrote this negative as a superscript right after the 3 says. So let's think about the limit as we're approaching from the left. So in this case, if we start with values lower than 3, as we get closer and closer and closer, so say we start at 0, the f of x is equal to negative 1."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "So let's think about first the limit as x approaches 3 at the limit of f of x as x approaches 3 from the negative direction. And all this notation here, I wrote this negative as a superscript right after the 3 says. So let's think about the limit as we're approaching from the left. So in this case, if we start with values lower than 3, as we get closer and closer and closer, so say we start at 0, the f of x is equal to negative 1. We go to 1, f of x is equal to negative 1. We go to 2, f of x is equal to negative 1. If you go to 2.999999, f of x is equal to negative 1."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "So in this case, if we start with values lower than 3, as we get closer and closer and closer, so say we start at 0, the f of x is equal to negative 1. We go to 1, f of x is equal to negative 1. We go to 2, f of x is equal to negative 1. If you go to 2.999999, f of x is equal to negative 1. So it looks like it is approaching negative 1 if you approach from the left-hand side. Now let's think about the limit of f of x as x approaches 3 from the positive direction, from values greater than 3. So here we see when x is equal to 5, f of x is equal to 1."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "If you go to 2.999999, f of x is equal to negative 1. So it looks like it is approaching negative 1 if you approach from the left-hand side. Now let's think about the limit of f of x as x approaches 3 from the positive direction, from values greater than 3. So here we see when x is equal to 5, f of x is equal to 1. When x is equal to 4, f of x is equal to 1. When x is equal to 3.0000001, f of x is equal to 1. So it seems to be approaching positive 1."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "So here we see when x is equal to 5, f of x is equal to 1. When x is equal to 4, f of x is equal to 1. When x is equal to 3.0000001, f of x is equal to 1. So it seems to be approaching positive 1. So now we have something strange. We seem to be approaching a different value when we approach from the left than when we approach from the right. And if we're approaching two different values, then the limit does not exist."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "So it seems to be approaching positive 1. So now we have something strange. We seem to be approaching a different value when we approach from the left than when we approach from the right. And if we're approaching two different values, then the limit does not exist. So this limit right over here does not exist. Or another way of saying it, the limit of a function f of x as x approaches some value c is equal to L if and only if the limit of f of x as x approaches c from the negative direction is equal to the limit of f of x as x approaches c from the positive direction, which is equal to L. This did not happen here. The limit when we approached from the left was negative 1."}, {"video_title": "Limit at a point of discontinuity.mp3", "Sentence": "And if we're approaching two different values, then the limit does not exist. So this limit right over here does not exist. Or another way of saying it, the limit of a function f of x as x approaches some value c is equal to L if and only if the limit of f of x as x approaches c from the negative direction is equal to the limit of f of x as x approaches c from the positive direction, which is equal to L. This did not happen here. The limit when we approached from the left was negative 1. The limit when we approached from the right was positive 1. So we did not get the same limits when we approached from either side. So the limit does not exist in this case."}, {"video_title": "Alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "I'll stop there, and of course we keep going on and on and on, and it's an alternating series, plus minus just keeps going on and on and on and on forever. Now we know from previous tests, in fact the alternating series test, this satisfies the constraints of the alternating series test, and we're able to show that it converges. What we're doing now is actually trying to estimate what things converge to. So we want to converge, so not converge, we want to estimate what this value S is, and we're gonna do that by doing a finite number of calculations, by not having to add this entire thing together. And so let's estimate it by taking, let's say the partial sum of the first four terms. So let's take these four terms right over here, and let's call that, so that's gonna be S sub four, and then you're gonna have a remainder, which is gonna be everything else. So all of this other stuff, I don't want even the brackets to end, that's going to be your remainder, the remainder to get to your actual sum, or whatever's left over when you just take the first four terms."}, {"video_title": "Alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "So we want to converge, so not converge, we want to estimate what this value S is, and we're gonna do that by doing a finite number of calculations, by not having to add this entire thing together. And so let's estimate it by taking, let's say the partial sum of the first four terms. So let's take these four terms right over here, and let's call that, so that's gonna be S sub four, and then you're gonna have a remainder, which is gonna be everything else. So all of this other stuff, I don't want even the brackets to end, that's going to be your remainder, the remainder to get to your actual sum, or whatever's left over when you just take the first four terms. So this is from the fifth term all the way to infinity, and we've seen this before, so the actual sum is going to be equal to this partial sum, plus this remainder. Well we can calculate this, this is going to be, let's see, common denominator here, let's see, nine times 16 is 144, so it's gonna be 144, and then that's gonna be 144 minus 36 over 144, plus 16 over 144, minus nine over 144, let's see, that is 144, negative 36 plus 16 is negative minus 20, so it's 124 minus nine is 115. So this is all going to be equal to 115 over 144, I didn't even need a calculator to figure that out, plus some remainder, plus some remainder."}, {"video_title": "Alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "So all of this other stuff, I don't want even the brackets to end, that's going to be your remainder, the remainder to get to your actual sum, or whatever's left over when you just take the first four terms. So this is from the fifth term all the way to infinity, and we've seen this before, so the actual sum is going to be equal to this partial sum, plus this remainder. Well we can calculate this, this is going to be, let's see, common denominator here, let's see, nine times 16 is 144, so it's gonna be 144, and then that's gonna be 144 minus 36 over 144, plus 16 over 144, minus nine over 144, let's see, that is 144, negative 36 plus 16 is negative minus 20, so it's 124 minus nine is 115. So this is all going to be equal to 115 over 144, I didn't even need a calculator to figure that out, plus some remainder, plus some remainder. And so if we could figure out some bounds on this remainder, we will figure out the bounds on our actual sum. We'll be able to figure out, well how far is this away from this right over here? And there's two ways to think about it, so let's look at it."}, {"video_title": "Alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "So this is all going to be equal to 115 over 144, I didn't even need a calculator to figure that out, plus some remainder, plus some remainder. And so if we could figure out some bounds on this remainder, we will figure out the bounds on our actual sum. We'll be able to figure out, well how far is this away from this right over here? And there's two ways to think about it, so let's look at it. So the first thing I want to see is, I want to show you that this remainder right over here is definitely going to be positive. And I actually encourage you to pause the video and see if you can prove to yourself that this remainder over here is definitely going to be positive. So I'm assuming you've had a go at it, let's write the remainder down."}, {"video_title": "Alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "And there's two ways to think about it, so let's look at it. So the first thing I want to see is, I want to show you that this remainder right over here is definitely going to be positive. And I actually encourage you to pause the video and see if you can prove to yourself that this remainder over here is definitely going to be positive. So I'm assuming you've had a go at it, let's write the remainder down. So, actually I'll just write it, actually I'll write it up here. So R sub four is one over 25, one over 20, actually I don't even have to write it separately, I could show you in just right over here that this is going to be positive. How do I show that?"}, {"video_title": "Alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "So I'm assuming you've had a go at it, let's write the remainder down. So, actually I'll just write it, actually I'll write it up here. So R sub four is one over 25, one over 20, actually I don't even have to write it separately, I could show you in just right over here that this is going to be positive. How do I show that? Well we just pair, let's just put some parentheses in here and just pair these terms like this. So one over 25 minus one over 36. One 36 is less than one over 25."}, {"video_title": "Alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "How do I show that? Well we just pair, let's just put some parentheses in here and just pair these terms like this. So one over 25 minus one over 36. One 36 is less than one over 25. This one's positive, this one's negative. So this is positive. Then you have a positive term, subtracting from that a smaller negative term, so this is going to be positive."}, {"video_title": "Alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "One 36 is less than one over 25. This one's positive, this one's negative. So this is positive. Then you have a positive term, subtracting from that a smaller negative term, so this is going to be positive. And so if you just pair all of these terms up, you're just going to have a whole series of positive terms. So just like that, we have established that R sub four, or R four we could call it, is going to be greater than zero. So R four is going to be greater than zero."}, {"video_title": "Alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "Then you have a positive term, subtracting from that a smaller negative term, so this is going to be positive. And so if you just pair all of these terms up, you're just going to have a whole series of positive terms. So just like that, we have established that R sub four, or R four we could call it, is going to be greater than zero. So R four is going to be greater than zero. And now the other thing I want to prove is that this remainder is going to be less than the first term that we haven't calculated, that the remainder is going to be less than one over 25. And once again, I encourage you to pause the video and see if you can put some parentheses here in a certain way that will convince you that this entire infinite sum here, this remainder, is going to sum up to something that's less than this first term. Once again, I'm assuming you've had a go at it, so let's just write it down."}, {"video_title": "Alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "So R four is going to be greater than zero. And now the other thing I want to prove is that this remainder is going to be less than the first term that we haven't calculated, that the remainder is going to be less than one over 25. And once again, I encourage you to pause the video and see if you can put some parentheses here in a certain way that will convince you that this entire infinite sum here, this remainder, is going to sum up to something that's less than this first term. Once again, I'm assuming you've had a go at it, so let's just write it down. So I'll do that same pink color. So our remainder, when we take the partial sum of the first four terms, so it's one over 25. And the way I'm going to write it, instead of writing minus one over 36, I'm going to write minus, I'm going to put the parentheses now around the second and third terms."}, {"video_title": "Alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "Once again, I'm assuming you've had a go at it, so let's just write it down. So I'll do that same pink color. So our remainder, when we take the partial sum of the first four terms, so it's one over 25. And the way I'm going to write it, instead of writing minus one over 36, I'm going to write minus, I'm going to put the parentheses now around the second and third terms. So this is going to be one over 36 minus one over 49. And then we're going to have minus one over 64 minus, actually the next term is going to be one over nine squared, one over 81. And then minus, and we keep going like that, on and on and on, on and on and on forever."}, {"video_title": "Alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "And the way I'm going to write it, instead of writing minus one over 36, I'm going to write minus, I'm going to put the parentheses now around the second and third terms. So this is going to be one over 36 minus one over 49. And then we're going to have minus one over 64 minus, actually the next term is going to be one over nine squared, one over 81. And then minus, and we keep going like that, on and on and on, on and on and on forever. Now notice what happens. This term right over here is positive. We have a smaller number being subtracted from a larger number."}, {"video_title": "Alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "And then minus, and we keep going like that, on and on and on, on and on and on forever. Now notice what happens. This term right over here is positive. We have a smaller number being subtracted from a larger number. This term right over here is positive. So we're starting with 125, and then we're subtracting a bunch of positive things from it. So this thing has to be less than one over 25."}, {"video_title": "Alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "We have a smaller number being subtracted from a larger number. This term right over here is positive. So we're starting with 125, and then we're subtracting a bunch of positive things from it. So this thing has to be less than one over 25. So R sub four is going to be less than one over 25. Or we could even write that as R sub four is less than 0.04. Same thing as one over 25."}, {"video_title": "Alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "So this thing has to be less than one over 25. So R sub four is going to be less than one over 25. Or we could even write that as R sub four is less than 0.04. Same thing as one over 25. And actually, this logic right over here is the basis for the proof of the alternating series test. This should make you feel pretty good, that hey look, this thing is going to be greater than zero, and it's increasing the more terms that you add to it. But it's bounded from above."}, {"video_title": "Alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "Same thing as one over 25. And actually, this logic right over here is the basis for the proof of the alternating series test. This should make you feel pretty good, that hey look, this thing is going to be greater than zero, and it's increasing the more terms that you add to it. But it's bounded from above. It's bounded from above at one over 25, which is a pretty good sense that hey, this thing is going to converge. But that's not what we're going to concern ourselves with here. Here we just care about this range."}, {"video_title": "Alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "But it's bounded from above. It's bounded from above at one over 25, which is a pretty good sense that hey, this thing is going to converge. But that's not what we're going to concern ourselves with here. Here we just care about this range. The sum is the sum of these two things. And so the entire sum is going to be less than 115 over 144, plus the upper bound on R four, so plus 0.04. And it's going to be greater than our partial sum plus zero, because this remainder is definitely greater than zero."}, {"video_title": "Alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "Here we just care about this range. The sum is the sum of these two things. And so the entire sum is going to be less than 115 over 144, plus the upper bound on R four, so plus 0.04. And it's going to be greater than our partial sum plus zero, because this remainder is definitely greater than zero. So you could just say it's going to be greater than our partial sum. And just like that, just doing a calculation that I was able to do with hand, we're able to get pretty nice bounds around this infinite series. And let's now get the calculator out just to get a little bit better sense of things."}, {"video_title": "Alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "And it's going to be greater than our partial sum plus zero, because this remainder is definitely greater than zero. So you could just say it's going to be greater than our partial sum. And just like that, just doing a calculation that I was able to do with hand, we're able to get pretty nice bounds around this infinite series. And let's now get the calculator out just to get a little bit better sense of things. So if we say 115 divided by 144, that's 0.79861 repeating. So this is 0.79861 repeating is less than S, which is less than this thing plus 0.04. So let me write that down."}, {"video_title": "Alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "And let's now get the calculator out just to get a little bit better sense of things. So if we say 115 divided by 144, that's 0.79861 repeating. So this is 0.79861 repeating is less than S, which is less than this thing plus 0.04. So let me write that down. So plus 0.04 gets us to 0.83861 repeating. So actually I could have done that in my head. I don't know why I resorted to a calculator."}, {"video_title": "Alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let me write that down. So plus 0.04 gets us to 0.83861 repeating. So actually I could have done that in my head. I don't know why I resorted to a calculator. 0.83861 repeating. And just like that, just a calculation we're able to do by hand, we're able to come up with a pretty good approximation for S. And the big takeaway from here, and we're going to build on this, but this was really to give you the intuition of what a very concrete example is, is when you have an alternating series like this, the type of alternating series that satisfies the alternating series test where you can write it as negative 1 to the n or negative 1 to the n plus 1 times a series of positive terms that are decreasing and whose limits go to zero as n approaches infinity, not only do those things converge, but you can estimate your error based on the first term that you are not including. Now this was one example."}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "What I've got here is the graph of y is equal to cosine of x. What I want to do is figure out the area under the curve y is equal to f of x and above the x-axis. I'm going to do it over various intervals. So first, let's think about the area under the curve between x is equal to 0 and x is equal to pi over 2. So we're talking about this area right over here. Well, the way we denote it is the definite integral from 0 to pi over 2 of cosine of x dx. And all this is is kind of reminiscent of taking a sum of a bunch of super thin rectangles with width dx and height f of x for each of those rectangles."}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "So first, let's think about the area under the curve between x is equal to 0 and x is equal to pi over 2. So we're talking about this area right over here. Well, the way we denote it is the definite integral from 0 to pi over 2 of cosine of x dx. And all this is is kind of reminiscent of taking a sum of a bunch of super thin rectangles with width dx and height f of x for each of those rectangles. And then you take an infinite number of those infinitely thin rectangles. And that's kind of what this notation is trying to depict. But we know how to do this already."}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "And all this is is kind of reminiscent of taking a sum of a bunch of super thin rectangles with width dx and height f of x for each of those rectangles. And then you take an infinite number of those infinitely thin rectangles. And that's kind of what this notation is trying to depict. But we know how to do this already. The second fundamental theorem of calculus helps us. We just have to figure out what the antiderivative of cosine of x is or what an antiderivative of cosine of x is evaluated at pi over 2. And from that, subtract it evaluated at 0."}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "But we know how to do this already. The second fundamental theorem of calculus helps us. We just have to figure out what the antiderivative of cosine of x is or what an antiderivative of cosine of x is evaluated at pi over 2. And from that, subtract it evaluated at 0. So what's the antiderivative of cosine of x? Or what's an antiderivative? Well, we know if we take the derivative, let me write this up here."}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "And from that, subtract it evaluated at 0. So what's the antiderivative of cosine of x? Or what's an antiderivative? Well, we know if we take the derivative, let me write this up here. We know that if we take the derivative of sine of x, we get cosine of x. So the antiderivative of cosine of x is sine of x. Now, why do I keep saying sine of x is an antiderivative?"}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we know if we take the derivative, let me write this up here. We know that if we take the derivative of sine of x, we get cosine of x. So the antiderivative of cosine of x is sine of x. Now, why do I keep saying sine of x is an antiderivative? It's not just the antiderivative. Well, I could also take the derivative of sine of x plus any arbitrary constant and still get cosine of x. Because the derivative of a constant is 0."}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "Now, why do I keep saying sine of x is an antiderivative? It's not just the antiderivative. Well, I could also take the derivative of sine of x plus any arbitrary constant and still get cosine of x. Because the derivative of a constant is 0. This could be pi. This could be 5. This could be a million."}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "Because the derivative of a constant is 0. This could be pi. This could be 5. This could be a million. This could be a googol. This could be any crazy number. But the derivative of this is still going to be cosine x."}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "This could be a million. This could be a googol. This could be any crazy number. But the derivative of this is still going to be cosine x. So when I say that we have to find an antiderivative, I'm just saying, look, we just have to find one of the derivatives. Sine of x is probably the simplest, because in this case, the constant is 0. So let's evaluate."}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "But the derivative of this is still going to be cosine x. So when I say that we have to find an antiderivative, I'm just saying, look, we just have to find one of the derivatives. Sine of x is probably the simplest, because in this case, the constant is 0. So let's evaluate. So one way that we can denote this, the antiderivative of cosine of x, or an antiderivative of cosine of x, is sine of x. And we're going to evaluate it at pi over 2. And from that, subtract it, evaluate it at 0."}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "So let's evaluate. So one way that we can denote this, the antiderivative of cosine of x, or an antiderivative of cosine of x, is sine of x. And we're going to evaluate it at pi over 2. And from that, subtract it, evaluate it at 0. So this is going to be equal to sine of pi over 2 minus sine of 0, which is equal to sine of pi over 2 is 1. Sine of 0 is 0. So it's 1 minus 0 is equal to 1."}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "And from that, subtract it, evaluate it at 0. So this is going to be equal to sine of pi over 2 minus sine of 0, which is equal to sine of pi over 2 is 1. Sine of 0 is 0. So it's 1 minus 0 is equal to 1. So the area of this region right over here, this area is equal to 1. Now let's do something interesting. Let's think about the area under the curve between, let's say, pi over 2 and 3 pi over 2."}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "So it's 1 minus 0 is equal to 1. So the area of this region right over here, this area is equal to 1. Now let's do something interesting. Let's think about the area under the curve between, let's say, pi over 2 and 3 pi over 2. So between here and here. So we're talking about this area. We're talking about that area right over here."}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "Let's think about the area under the curve between, let's say, pi over 2 and 3 pi over 2. So between here and here. So we're talking about this area. We're talking about that area right over here. This is 3 pi over 2. So once again, the way we denote the area is the definite integral from pi over 2 to 3 pi over 2 of cosine of x dx. The antiderivative, or an antiderivative of cosine of x is sine of x evaluated at 3 pi over 2 and pi over 2."}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "We're talking about that area right over here. This is 3 pi over 2. So once again, the way we denote the area is the definite integral from pi over 2 to 3 pi over 2 of cosine of x dx. The antiderivative, or an antiderivative of cosine of x is sine of x evaluated at 3 pi over 2 and pi over 2. So this is going to be equal to sine of 3 pi over 2 minus sine of pi over 2. What's sine of 3 pi over 2? If you visualize the unit circle real fast, 3 pi over 2 is going all the way 3 4ths around the unit circle."}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "The antiderivative, or an antiderivative of cosine of x is sine of x evaluated at 3 pi over 2 and pi over 2. So this is going to be equal to sine of 3 pi over 2 minus sine of pi over 2. What's sine of 3 pi over 2? If you visualize the unit circle real fast, 3 pi over 2 is going all the way 3 4ths around the unit circle. So it's right over there. So the sine is the y-coordinate on that unit circle. So it's negative 1."}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "If you visualize the unit circle real fast, 3 pi over 2 is going all the way 3 4ths around the unit circle. So it's right over there. So the sine is the y-coordinate on that unit circle. So it's negative 1. So this right over here is negative 1. This right over here, sine of pi over 2, pi over 2 is just going straight up like that. So sine of pi over 2 is 1."}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "So it's negative 1. So this right over here is negative 1. This right over here, sine of pi over 2, pi over 2 is just going straight up like that. So sine of pi over 2 is 1. So this is interesting. We get negative 1 minus 1, which is equal to negative 2. We've got a negative area here."}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "So sine of pi over 2 is 1. So this is interesting. We get negative 1 minus 1, which is equal to negative 2. We've got a negative area here. We've got a negative area. Now, how does that make sense? We know in the real world, areas are always positive."}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "We've got a negative area here. We've got a negative area. Now, how does that make sense? We know in the real world, areas are always positive. But what is negative 2 really trying to depict? Well, it's trying to sign it based on the idea that now our function is below the x-axis. So we could kind of think that we have an area of 2, but it's all below the x-axis in this case."}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "We know in the real world, areas are always positive. But what is negative 2 really trying to depict? Well, it's trying to sign it based on the idea that now our function is below the x-axis. So we could kind of think that we have an area of 2, but it's all below the x-axis in this case. And so it is signed as negative 2. The actual area is 2, but since it's below the x-axis, we get a negative right over here. Now let's do one more interesting one."}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "So we could kind of think that we have an area of 2, but it's all below the x-axis in this case. And so it is signed as negative 2. The actual area is 2, but since it's below the x-axis, we get a negative right over here. Now let's do one more interesting one. Let's find the definite integral from 0 to 3 pi over 2 of cosine of x dx. Now, this is just denoting this entire area, going from 0 all the way to 3 pi over 2. And what do you think is going to happen?"}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's do one more interesting one. Let's find the definite integral from 0 to 3 pi over 2 of cosine of x dx. Now, this is just denoting this entire area, going from 0 all the way to 3 pi over 2. And what do you think is going to happen? Well, let's evaluate it. It's going to be sine of 3 pi over 2 minus sine of 0, which is equal to negative 1 minus 0, which is equal to negative 1. So what just happened here?"}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "And what do you think is going to happen? Well, let's evaluate it. It's going to be sine of 3 pi over 2 minus sine of 0, which is equal to negative 1 minus 0, which is equal to negative 1. So what just happened here? The area under all this orange area that I just outlined is clearly not negative, or any area isn't a negative number. And the area isn't even 1. But what just happened here?"}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "So what just happened here? The area under all this orange area that I just outlined is clearly not negative, or any area isn't a negative number. And the area isn't even 1. But what just happened here? Well, we saw in the first case that this first area was 1. The area of this first region right over here is 1. And then the area of the second region, we got negative 2."}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "But what just happened here? Well, we saw in the first case that this first area was 1. The area of this first region right over here is 1. And then the area of the second region, we got negative 2. And so one way to interpret it is that your net area above the x-axis is negative 1. Or another way to say it, your net area is negative 1. So it's taking the one region above the x-axis and then subtracting the two below it."}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "And then the area of the second region, we got negative 2. And so one way to interpret it is that your net area above the x-axis is negative 1. Or another way to say it, your net area is negative 1. So it's taking the one region above the x-axis and then subtracting the two below it. So what the definite integral is doing when we evaluate it using the second fundamental theorem of calculus, it's essentially finding the net area above the x-axis. And if we get a negative number, that means that the net area is that actually most of the area is below the x-axis. If we get 0, then that means it all nets out."}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "So it's taking the one region above the x-axis and then subtracting the two below it. So what the definite integral is doing when we evaluate it using the second fundamental theorem of calculus, it's essentially finding the net area above the x-axis. And if we get a negative number, that means that the net area is that actually most of the area is below the x-axis. If we get 0, then that means it all nets out. And if you want to see a case at 0, take the integral from 0 all the way to 2 pi. And this will evaluate to 0 because you have an area of 1 and another area of 1. But it nets out with this area of negative 2."}, {"video_title": "Area between a curve and the x-axis negative area AP Calculus AB Khan Academy.mp3", "Sentence": "If we get 0, then that means it all nets out. And if you want to see a case at 0, take the integral from 0 all the way to 2 pi. And this will evaluate to 0 because you have an area of 1 and another area of 1. But it nets out with this area of negative 2. Let's try that out. So if we go from 0 to 2 pi of cosine of x dx, this is going to be equal to sine of 2 pi minus sine of 0, which is equal to 0 minus 0, which is indeed equal to 0. Now clearly, there was area here."}, {"video_title": "Position vector valued functions Multivariable Calculus Khan Academy.mp3", "Sentence": "So t is greater than or equal to a, and then less than or equal to b. So if I were to just draw this on a, let me see, I could draw it like this. I'm staying very abstract right now. This is not a very specific example. This is the x-axis, this is the y-axis. My curve, let's say this is when t is equal to a, and then the curve might do something like this. I don't know what it does."}, {"video_title": "Position vector valued functions Multivariable Calculus Khan Academy.mp3", "Sentence": "This is not a very specific example. This is the x-axis, this is the y-axis. My curve, let's say this is when t is equal to a, and then the curve might do something like this. I don't know what it does. Let's say it's over there. This is t is equal to b. This actual point right here will be x of b."}, {"video_title": "Position vector valued functions Multivariable Calculus Khan Academy.mp3", "Sentence": "I don't know what it does. Let's say it's over there. This is t is equal to b. This actual point right here will be x of b. That would be the x-coordinate. You evaluate this function at b, and y of b. And this is, of course, when t is equal to a."}, {"video_title": "Position vector valued functions Multivariable Calculus Khan Academy.mp3", "Sentence": "This actual point right here will be x of b. That would be the x-coordinate. You evaluate this function at b, and y of b. And this is, of course, when t is equal to a. The actual coordinate in R2 on the Cartesian coordinates will be x of a, which is this right here, and then y of a, which is that right there. And we've seen that before. That's just a standard way of describing a parametric equation or curve using two parametric equations."}, {"video_title": "Position vector valued functions Multivariable Calculus Khan Academy.mp3", "Sentence": "And this is, of course, when t is equal to a. The actual coordinate in R2 on the Cartesian coordinates will be x of a, which is this right here, and then y of a, which is that right there. And we've seen that before. That's just a standard way of describing a parametric equation or curve using two parametric equations. What I want to do now is describe this same exact curve using a vector-valued function. So if I define a vector-valued function, and if you don't remember what those are, we'll have a little bit of a review here. Let me say I have a vector-valued function R, and I'll put a little vector arrow on top of it."}, {"video_title": "Position vector valued functions Multivariable Calculus Khan Academy.mp3", "Sentence": "That's just a standard way of describing a parametric equation or curve using two parametric equations. What I want to do now is describe this same exact curve using a vector-valued function. So if I define a vector-valued function, and if you don't remember what those are, we'll have a little bit of a review here. Let me say I have a vector-valued function R, and I'll put a little vector arrow on top of it. In a lot of textbooks, they'll just bold it, and they'll leave scalar-valued functions unbolded, but it's hard to draw bold. So I'll put a little vector on top. And let's say that R is a function of t. And these are going to be position vectors."}, {"video_title": "Position vector valued functions Multivariable Calculus Khan Academy.mp3", "Sentence": "Let me say I have a vector-valued function R, and I'll put a little vector arrow on top of it. In a lot of textbooks, they'll just bold it, and they'll leave scalar-valued functions unbolded, but it's hard to draw bold. So I'll put a little vector on top. And let's say that R is a function of t. And these are going to be position vectors. I'm specifying that because, in general, when someone talks about a vector, this vector and this vector are considered equivalent as long as they have the same magnitude and direction. No one really cares about what their start and end points are as long as their direction is the same and their length is the same. But when you talk about position vectors, you're saying, no, these vectors are all going to start at 0 at the origin."}, {"video_title": "Position vector valued functions Multivariable Calculus Khan Academy.mp3", "Sentence": "And let's say that R is a function of t. And these are going to be position vectors. I'm specifying that because, in general, when someone talks about a vector, this vector and this vector are considered equivalent as long as they have the same magnitude and direction. No one really cares about what their start and end points are as long as their direction is the same and their length is the same. But when you talk about position vectors, you're saying, no, these vectors are all going to start at 0 at the origin. And when you say it's a position vector, you're implicitly saying this is specifying a unique position. In this case, it's going to be in two-dimensional space, but it could be in three-dimensional space or really even four, five, whatever, n-dimensional space. So when you say it's a position vector, you're literally saying, OK, this vector literally specifies that point in space."}, {"video_title": "Position vector valued functions Multivariable Calculus Khan Academy.mp3", "Sentence": "But when you talk about position vectors, you're saying, no, these vectors are all going to start at 0 at the origin. And when you say it's a position vector, you're implicitly saying this is specifying a unique position. In this case, it's going to be in two-dimensional space, but it could be in three-dimensional space or really even four, five, whatever, n-dimensional space. So when you say it's a position vector, you're literally saying, OK, this vector literally specifies that point in space. So let's see if we can describe this curve as a position vector-valued function. So we could say r of t, let me switch back to that pink color. I'll just stay in green, is equal to x of t times the unit vector in the x direction."}, {"video_title": "Position vector valued functions Multivariable Calculus Khan Academy.mp3", "Sentence": "So when you say it's a position vector, you're literally saying, OK, this vector literally specifies that point in space. So let's see if we can describe this curve as a position vector-valued function. So we could say r of t, let me switch back to that pink color. I'll just stay in green, is equal to x of t times the unit vector in the x direction. The unit vector gets a little caret on top, a little hat. That's like the arrow for it. That just says it's a unit vector."}, {"video_title": "Position vector valued functions Multivariable Calculus Khan Academy.mp3", "Sentence": "I'll just stay in green, is equal to x of t times the unit vector in the x direction. The unit vector gets a little caret on top, a little hat. That's like the arrow for it. That just says it's a unit vector. Plus y of t times j. If I was dealing with a curve in three dimensions, I would have plus z of t times k, but we're dealing with two dimensions right here. And so the way this works is you're just taking your, for any t, and still we're going to have t is greater than or equal to a and then less than or equal to b."}, {"video_title": "Position vector valued functions Multivariable Calculus Khan Academy.mp3", "Sentence": "That just says it's a unit vector. Plus y of t times j. If I was dealing with a curve in three dimensions, I would have plus z of t times k, but we're dealing with two dimensions right here. And so the way this works is you're just taking your, for any t, and still we're going to have t is greater than or equal to a and then less than or equal to b. And this is the exact same thing as that. Let me just redraw it. So let me draw our coordinates right here, our axes."}, {"video_title": "Position vector valued functions Multivariable Calculus Khan Academy.mp3", "Sentence": "And so the way this works is you're just taking your, for any t, and still we're going to have t is greater than or equal to a and then less than or equal to b. And this is the exact same thing as that. Let me just redraw it. So let me draw our coordinates right here, our axes. So that's the y-axis and this is the x-axis. So when you evaluate r of a, that's our starting point. So let me do that."}, {"video_title": "Position vector valued functions Multivariable Calculus Khan Academy.mp3", "Sentence": "So let me draw our coordinates right here, our axes. So that's the y-axis and this is the x-axis. So when you evaluate r of a, that's our starting point. So let me do that. So r of a, maybe I'll do it right over here. Our position vector-valued function, evaluated at t is equal to a, is going to be equal to x of a times our unit vector in the x direction plus y of a times our unit vector in the vertical direction, or in the y direction. What's that going to look like?"}, {"video_title": "Position vector valued functions Multivariable Calculus Khan Academy.mp3", "Sentence": "So let me do that. So r of a, maybe I'll do it right over here. Our position vector-valued function, evaluated at t is equal to a, is going to be equal to x of a times our unit vector in the x direction plus y of a times our unit vector in the vertical direction, or in the y direction. What's that going to look like? Well, x of a is this thing right here. So it's x of a times the unit vector. So it's really, maybe the unit vector is this long."}, {"video_title": "Position vector valued functions Multivariable Calculus Khan Academy.mp3", "Sentence": "What's that going to look like? Well, x of a is this thing right here. So it's x of a times the unit vector. So it's really, maybe the unit vector is this long. It has length 1. So now we're just going to have a length of x of a in that direction. And then same thing in y of a."}, {"video_title": "Position vector valued functions Multivariable Calculus Khan Academy.mp3", "Sentence": "So it's really, maybe the unit vector is this long. It has length 1. So now we're just going to have a length of x of a in that direction. And then same thing in y of a. It's going to be y of a length in that direction. But the bottom line, this vector right here, if you add these scaled values of these two unit vectors, you're going to get r of a looking something like this. It's going to be a vector that looks something like that."}, {"video_title": "Position vector valued functions Multivariable Calculus Khan Academy.mp3", "Sentence": "And then same thing in y of a. It's going to be y of a length in that direction. But the bottom line, this vector right here, if you add these scaled values of these two unit vectors, you're going to get r of a looking something like this. It's going to be a vector that looks something like that. Just like that. It's a vector. It's a position vector."}, {"video_title": "Position vector valued functions Multivariable Calculus Khan Academy.mp3", "Sentence": "It's going to be a vector that looks something like that. Just like that. It's a vector. It's a position vector. That's why we're kneeling it at the origin, but drawing it in standard position. And that right there is r of a. Now, what happens if a increases a little bit?"}, {"video_title": "Position vector valued functions Multivariable Calculus Khan Academy.mp3", "Sentence": "It's a position vector. That's why we're kneeling it at the origin, but drawing it in standard position. And that right there is r of a. Now, what happens if a increases a little bit? What is r of a plus a little bit? And I don't know, we could call that r of a plus delta, or r of a plus h. We do it in different colors. So let's say we increase a a little bit."}, {"video_title": "Position vector valued functions Multivariable Calculus Khan Academy.mp3", "Sentence": "Now, what happens if a increases a little bit? What is r of a plus a little bit? And I don't know, we could call that r of a plus delta, or r of a plus h. We do it in different colors. So let's say we increase a a little bit. r of a plus some small h. Well, that's just going to be x of a plus h times the unit vector i plus y times a plus h times the unit vector j. And what's that going to look like? Well, we're going to go a little bit further down the curve."}, {"video_title": "Position vector valued functions Multivariable Calculus Khan Academy.mp3", "Sentence": "So let's say we increase a a little bit. r of a plus some small h. Well, that's just going to be x of a plus h times the unit vector i plus y times a plus h times the unit vector j. And what's that going to look like? Well, we're going to go a little bit further down the curve. That's like saying the coordinate x of a plus h and y plus a plus h. It might be that point right there. So it'll be a new unit vector. Sorry, it'll be a new vector, position vector, not a unit vector."}, {"video_title": "Position vector valued functions Multivariable Calculus Khan Academy.mp3", "Sentence": "Well, we're going to go a little bit further down the curve. That's like saying the coordinate x of a plus h and y plus a plus h. It might be that point right there. So it'll be a new unit vector. Sorry, it'll be a new vector, position vector, not a unit vector. These don't necessarily have length 1. That might be right here. Let me do that same color as this."}, {"video_title": "Position vector valued functions Multivariable Calculus Khan Academy.mp3", "Sentence": "Sorry, it'll be a new vector, position vector, not a unit vector. These don't necessarily have length 1. That might be right here. Let me do that same color as this. So it might be just like that. So that right here is r of a plus h. So you see, as you keep increasing your value of t until you get to b, these position vectors are going to keep specifying points along this curve. So the curve, let me draw the curve in a different color."}, {"video_title": "Position vector valued functions Multivariable Calculus Khan Academy.mp3", "Sentence": "Let me do that same color as this. So it might be just like that. So that right here is r of a plus h. So you see, as you keep increasing your value of t until you get to b, these position vectors are going to keep specifying points along this curve. So the curve, let me draw the curve in a different color. The curve looks something like this. It's meant to look exactly like the curve that I have up here. And for example, r of b is going to be a vector that looks like this."}, {"video_title": "Position vector valued functions Multivariable Calculus Khan Academy.mp3", "Sentence": "So the curve, let me draw the curve in a different color. The curve looks something like this. It's meant to look exactly like the curve that I have up here. And for example, r of b is going to be a vector that looks like this. It's going to be a vector that looks like that. Let me draw it relatively straight. That vector right there is r of b."}, {"video_title": "Position vector valued functions Multivariable Calculus Khan Academy.mp3", "Sentence": "And for example, r of b is going to be a vector that looks like this. It's going to be a vector that looks like that. Let me draw it relatively straight. That vector right there is r of b. So hopefully you realize that, look, these position vectors really are specifying the same points on this curve as this original, I guess, straight up parametrization that we did for this curve. And I just want to do that as a little bit of review, because we're now going to break in into the idea of actually taking a derivative of this vector-valued function. And I'll do that in the next video."}, {"video_title": "Analyzing unbounded limits rational function AP Calculus AB Khan Academy.mp3", "Sentence": "And so we can see we have a bunch of choices where we're approaching x from the right-hand side and we're approaching x from the left-hand side. And we're trying to figure out do we get unbounded on either of those in the positive, towards positive infinity or negative infinity. And there's a couple of ways to tackle it. The most straightforward, let's just consider each of these separately. So we could think about the limit of f of x as x approaches one from the positive direction and limit of f of x as x approaches one from the left-hand side. This is from the right-hand side. This is from the left-hand side."}, {"video_title": "Analyzing unbounded limits rational function AP Calculus AB Khan Academy.mp3", "Sentence": "The most straightforward, let's just consider each of these separately. So we could think about the limit of f of x as x approaches one from the positive direction and limit of f of x as x approaches one from the left-hand side. This is from the right-hand side. This is from the left-hand side. So I'm just gonna make a table and try out some values as we approach one as we approach one from the different sides. X, f of x, and I'll do the same thing over here. So we are going to have our x and have our f of x."}, {"video_title": "Analyzing unbounded limits rational function AP Calculus AB Khan Academy.mp3", "Sentence": "This is from the left-hand side. So I'm just gonna make a table and try out some values as we approach one as we approach one from the different sides. X, f of x, and I'll do the same thing over here. So we are going to have our x and have our f of x. And if we approach one from the right-hand side here, that'd be approaching one from above. So we could try 1.1. We could try 1.01."}, {"video_title": "Analyzing unbounded limits rational function AP Calculus AB Khan Academy.mp3", "Sentence": "So we are going to have our x and have our f of x. And if we approach one from the right-hand side here, that'd be approaching one from above. So we could try 1.1. We could try 1.01. Now f of 1.1 is negative one over 1.1 minus one squared. So see, this denominator here is going to be.1 squared. So this is going to be 0.01."}, {"video_title": "Analyzing unbounded limits rational function AP Calculus AB Khan Academy.mp3", "Sentence": "We could try 1.01. Now f of 1.1 is negative one over 1.1 minus one squared. So see, this denominator here is going to be.1 squared. So this is going to be 0.01. And so this is going to be negative 100. So let me just write that down. That's going to be negative 100."}, {"video_title": "Analyzing unbounded limits rational function AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be 0.01. And so this is going to be negative 100. So let me just write that down. That's going to be negative 100. So if x is 1.01, well, this is going to be negative one over 1.01 minus one squared. Well, in this denominator, this is going to be point, this is the same thing as 0.01 squared, which is the same thing as 0.0001, 1 10,000th. And so the negative 1 1 10,000th is going to be negative 10,000."}, {"video_title": "Analyzing unbounded limits rational function AP Calculus AB Khan Academy.mp3", "Sentence": "That's going to be negative 100. So if x is 1.01, well, this is going to be negative one over 1.01 minus one squared. Well, in this denominator, this is going to be point, this is the same thing as 0.01 squared, which is the same thing as 0.0001, 1 10,000th. And so the negative 1 1 10,000th is going to be negative 10,000. So let's just write that down. Negative 10,000. And so this looks like, as we get closer, because notice, as I'm going here, I am approaching one from the positive direction."}, {"video_title": "Analyzing unbounded limits rational function AP Calculus AB Khan Academy.mp3", "Sentence": "And so the negative 1 1 10,000th is going to be negative 10,000. So let's just write that down. Negative 10,000. And so this looks like, as we get closer, because notice, as I'm going here, I am approaching one from the positive direction. I'm getting closer and closer to one from above, and I'm going unbounded towards negative infinity. So this looks like it is negative infinity. Now we could do the same thing from the left-hand side."}, {"video_title": "Analyzing unbounded limits rational function AP Calculus AB Khan Academy.mp3", "Sentence": "And so this looks like, as we get closer, because notice, as I'm going here, I am approaching one from the positive direction. I'm getting closer and closer to one from above, and I'm going unbounded towards negative infinity. So this looks like it is negative infinity. Now we could do the same thing from the left-hand side. I could do 0.9. I could do 0.99. 0.99."}, {"video_title": "Analyzing unbounded limits rational function AP Calculus AB Khan Academy.mp3", "Sentence": "Now we could do the same thing from the left-hand side. I could do 0.9. I could do 0.99. 0.99. Now 0.9 is actually also going to get me negative 100, because 0.9 minus one is going to be negative 0.1. But then when you square it, the negative goes away, so you get 0.01, and then one divided by that is 100, but then you have the negative, so this is also negative 100. And if you don't follow those calculations, I'll do it, let me do it one more time, just so you see it clearly."}, {"video_title": "Analyzing unbounded limits rational function AP Calculus AB Khan Academy.mp3", "Sentence": "0.99. Now 0.9 is actually also going to get me negative 100, because 0.9 minus one is going to be negative 0.1. But then when you square it, the negative goes away, so you get 0.01, and then one divided by that is 100, but then you have the negative, so this is also negative 100. And if you don't follow those calculations, I'll do it, let me do it one more time, just so you see it clearly. This is going to be negative one over, and I'm doing x is equal to 0.99, so I'm getting even closer to one, but I'm approaching from below, from the left-hand side. So this is going to be 0.99 minus one squared. Well, 0.99 minus one is going to be negative 100th, so this is going to be negative 0.01 squared."}, {"video_title": "Analyzing unbounded limits rational function AP Calculus AB Khan Academy.mp3", "Sentence": "And if you don't follow those calculations, I'll do it, let me do it one more time, just so you see it clearly. This is going to be negative one over, and I'm doing x is equal to 0.99, so I'm getting even closer to one, but I'm approaching from below, from the left-hand side. So this is going to be 0.99 minus one squared. Well, 0.99 minus one is going to be negative 100th, so this is going to be negative 0.01 squared. When you square it, the negative goes away, and you're left with one 10,000th, so this is going to be 0.0001, and so when you evaluate this, you get 10,000. So that, or sorry, you get negative 10,000. So in either case, regardless of which direction we approach from, we are approaching negative infinity."}, {"video_title": "Analyzing unbounded limits rational function AP Calculus AB Khan Academy.mp3", "Sentence": "Well, 0.99 minus one is going to be negative 100th, so this is going to be negative 0.01 squared. When you square it, the negative goes away, and you're left with one 10,000th, so this is going to be 0.0001, and so when you evaluate this, you get 10,000. So that, or sorry, you get negative 10,000. So in either case, regardless of which direction we approach from, we are approaching negative infinity. So that is this choice right over here. Now there's other ways you could have tackled this. If you just look at kind of the structure of this expression here, the numerator is a constant, so that's clearly always going to be positive."}, {"video_title": "Analyzing unbounded limits rational function AP Calculus AB Khan Academy.mp3", "Sentence": "So in either case, regardless of which direction we approach from, we are approaching negative infinity. So that is this choice right over here. Now there's other ways you could have tackled this. If you just look at kind of the structure of this expression here, the numerator is a constant, so that's clearly always going to be positive. Let's ignore this negative for the time being. That negative's out front. This numerator, this one, is always going to be positive."}, {"video_title": "Analyzing unbounded limits rational function AP Calculus AB Khan Academy.mp3", "Sentence": "If you just look at kind of the structure of this expression here, the numerator is a constant, so that's clearly always going to be positive. Let's ignore this negative for the time being. That negative's out front. This numerator, this one, is always going to be positive. Down here, we're taking the, at x equals one, well, this becomes zero, and the whole expression becomes undefined, but as we approach one, x minus one could be positive or negative, as we see over here, but then when we square it, this is going to become positive as well. So the denominator's going to be positive for any x other than one, so positive divided by positive is going to be positive, but then you have a negative out front. So this thing is going to be negative for any x other than one, and it's actually not defined at x equals one."}, {"video_title": "Fundamental theorem of calculus (Part 1) AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say I have some function f that is continuous on an interval between a and b. And I have these brackets here. So it also includes a and b in the interval. So let me graph this just so we get a sense of what I'm talking about. So that's my vertical axis. This is my horizontal axis. I'm going to label my horizontal axis t so that we can save x for later."}, {"video_title": "Fundamental theorem of calculus (Part 1) AP Calculus AB Khan Academy.mp3", "Sentence": "So let me graph this just so we get a sense of what I'm talking about. So that's my vertical axis. This is my horizontal axis. I'm going to label my horizontal axis t so that we can save x for later. I can still make this y right over there. And let me graph. This right over here is the graph of y is equal to f of t. Now, our lower endpoint is a."}, {"video_title": "Fundamental theorem of calculus (Part 1) AP Calculus AB Khan Academy.mp3", "Sentence": "I'm going to label my horizontal axis t so that we can save x for later. I can still make this y right over there. And let me graph. This right over here is the graph of y is equal to f of t. Now, our lower endpoint is a. So that's a right over there. Our upper boundary is b. Let me make that clear."}, {"video_title": "Fundamental theorem of calculus (Part 1) AP Calculus AB Khan Academy.mp3", "Sentence": "This right over here is the graph of y is equal to f of t. Now, our lower endpoint is a. So that's a right over there. Our upper boundary is b. Let me make that clear. And actually, just to show that we're including that endpoint, let me make them bold lines, filled in lines. So lower boundary a, upper boundary b. We're just saying, and I've drawn it this way, that f is continuous on that."}, {"video_title": "Fundamental theorem of calculus (Part 1) AP Calculus AB Khan Academy.mp3", "Sentence": "Let me make that clear. And actually, just to show that we're including that endpoint, let me make them bold lines, filled in lines. So lower boundary a, upper boundary b. We're just saying, and I've drawn it this way, that f is continuous on that. Now, let's define some new function. That's the area under the curve between a and some point that's in our interval. Let me pick this right over here, x."}, {"video_title": "Fundamental theorem of calculus (Part 1) AP Calculus AB Khan Academy.mp3", "Sentence": "We're just saying, and I've drawn it this way, that f is continuous on that. Now, let's define some new function. That's the area under the curve between a and some point that's in our interval. Let me pick this right over here, x. So let's define some new function to capture the area under the curve between a and x. Well, how do we denote the area under the curve between two endpoints? Well, we just use our definite integral."}, {"video_title": "Fundamental theorem of calculus (Part 1) AP Calculus AB Khan Academy.mp3", "Sentence": "Let me pick this right over here, x. So let's define some new function to capture the area under the curve between a and x. Well, how do we denote the area under the curve between two endpoints? Well, we just use our definite integral. That's our Riemann integral. It's really that right now, before we come up with the conclusion of this video, it really just represents the area under the curve between two endpoints. So this right over here, we can say, is the definite integral from a to x of f of t dt."}, {"video_title": "Fundamental theorem of calculus (Part 1) AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we just use our definite integral. That's our Riemann integral. It's really that right now, before we come up with the conclusion of this video, it really just represents the area under the curve between two endpoints. So this right over here, we can say, is the definite integral from a to x of f of t dt. Now, this right over here is going to be a function of x. And let me make it clear. Where x is in the interval between a and b."}, {"video_title": "Fundamental theorem of calculus (Part 1) AP Calculus AB Khan Academy.mp3", "Sentence": "So this right over here, we can say, is the definite integral from a to x of f of t dt. Now, this right over here is going to be a function of x. And let me make it clear. Where x is in the interval between a and b. This thing right over here is going to be another function of x. This value is going to depend on what x we actually choose. So let's define this as a function of x."}, {"video_title": "Fundamental theorem of calculus (Part 1) AP Calculus AB Khan Academy.mp3", "Sentence": "Where x is in the interval between a and b. This thing right over here is going to be another function of x. This value is going to depend on what x we actually choose. So let's define this as a function of x. So I'm going to say that this is equal to uppercase f of x. So all fair and good. Uppercase f of x is a function, if you give me an x value that's between a and b, it'll tell you the area under lowercase f of t between a and x."}, {"video_title": "Fundamental theorem of calculus (Part 1) AP Calculus AB Khan Academy.mp3", "Sentence": "So let's define this as a function of x. So I'm going to say that this is equal to uppercase f of x. So all fair and good. Uppercase f of x is a function, if you give me an x value that's between a and b, it'll tell you the area under lowercase f of t between a and x. Now, the cool part. The fundamental theorem of calculus. The fundamental theorem of calculus tells us, let me write this down, because this is a big deal."}, {"video_title": "Fundamental theorem of calculus (Part 1) AP Calculus AB Khan Academy.mp3", "Sentence": "Uppercase f of x is a function, if you give me an x value that's between a and b, it'll tell you the area under lowercase f of t between a and x. Now, the cool part. The fundamental theorem of calculus. The fundamental theorem of calculus tells us, let me write this down, because this is a big deal. Fundamental theorem of calculus tells us that if we were to take the derivative of our capital F, so the derivative, let me make sure I have enough space here. So if I were to take the derivative of capital F with respect to x, which is the same thing as taking the derivative of this with respect to x, which is equal to the derivative of all of this business. Let me copy this."}, {"video_title": "Fundamental theorem of calculus (Part 1) AP Calculus AB Khan Academy.mp3", "Sentence": "The fundamental theorem of calculus tells us, let me write this down, because this is a big deal. Fundamental theorem of calculus tells us that if we were to take the derivative of our capital F, so the derivative, let me make sure I have enough space here. So if I were to take the derivative of capital F with respect to x, which is the same thing as taking the derivative of this with respect to x, which is equal to the derivative of all of this business. Let me copy this. So copy and then paste. Which is the same thing. I've defined capital F as this stuff."}, {"video_title": "Fundamental theorem of calculus (Part 1) AP Calculus AB Khan Academy.mp3", "Sentence": "Let me copy this. So copy and then paste. Which is the same thing. I've defined capital F as this stuff. So if I'm taking the derivative of the left-hand side, the same thing as taking the derivative of the right-hand side, the fundamental theorem of calculus tells us that this is going to be equal to f, lowercase f of x. Now, why is this a big deal? Why does it get such an important title as the fundamental theorem of calculus?"}, {"video_title": "Fundamental theorem of calculus (Part 1) AP Calculus AB Khan Academy.mp3", "Sentence": "I've defined capital F as this stuff. So if I'm taking the derivative of the left-hand side, the same thing as taking the derivative of the right-hand side, the fundamental theorem of calculus tells us that this is going to be equal to f, lowercase f of x. Now, why is this a big deal? Why does it get such an important title as the fundamental theorem of calculus? Well, it tells us that for any continuous function f, if I define a function that is the area under the curve between a and x right over here, that the derivative of that function is going to be f. So let me make it clear. Every continuous function, every continuous f has an antiderivative capital F of x. That by itself is a cool thing."}, {"video_title": "Fundamental theorem of calculus (Part 1) AP Calculus AB Khan Academy.mp3", "Sentence": "Why does it get such an important title as the fundamental theorem of calculus? Well, it tells us that for any continuous function f, if I define a function that is the area under the curve between a and x right over here, that the derivative of that function is going to be f. So let me make it clear. Every continuous function, every continuous f has an antiderivative capital F of x. That by itself is a cool thing. But the other really cool thing, or I guess these are somewhat related, is remember, coming into this, all we did, we just viewed the definite integral as symbolizing the area under the curve between two points. That's where that Riemann definition of integration comes from. But now we see a connection between that and derivatives."}, {"video_title": "Fundamental theorem of calculus (Part 1) AP Calculus AB Khan Academy.mp3", "Sentence": "That by itself is a cool thing. But the other really cool thing, or I guess these are somewhat related, is remember, coming into this, all we did, we just viewed the definite integral as symbolizing the area under the curve between two points. That's where that Riemann definition of integration comes from. But now we see a connection between that and derivatives. When you're taking the definite integral, one way of thinking, especially if you're taking a definite integral between a lower boundary and an x, one way to think about it is you're essentially taking an antiderivative. So we now see a connection. This is why it is the fundamental theorem of calculus."}, {"video_title": "Fundamental theorem of calculus (Part 1) AP Calculus AB Khan Academy.mp3", "Sentence": "But now we see a connection between that and derivatives. When you're taking the definite integral, one way of thinking, especially if you're taking a definite integral between a lower boundary and an x, one way to think about it is you're essentially taking an antiderivative. So we now see a connection. This is why it is the fundamental theorem of calculus. It connects differential calculus and integral calculus. Connection between derivatives, or maybe I should say antiderivatives, and integration, which before this video we just viewed integration as area under curve. Now that we see it has a connection to derivatives."}, {"video_title": "Fundamental theorem of calculus (Part 1) AP Calculus AB Khan Academy.mp3", "Sentence": "This is why it is the fundamental theorem of calculus. It connects differential calculus and integral calculus. Connection between derivatives, or maybe I should say antiderivatives, and integration, which before this video we just viewed integration as area under curve. Now that we see it has a connection to derivatives. Well, how would you actually use the fundamental theorem of calculus? Well, maybe in the context of a calculus class. And we'll do the intuition for why this happens or why this is true and maybe a proof in later videos."}, {"video_title": "Fundamental theorem of calculus (Part 1) AP Calculus AB Khan Academy.mp3", "Sentence": "Now that we see it has a connection to derivatives. Well, how would you actually use the fundamental theorem of calculus? Well, maybe in the context of a calculus class. And we'll do the intuition for why this happens or why this is true and maybe a proof in later videos. But how would you actually apply this right over here? Well, let's say that someone told you that they want to find the derivative. Let me just in a new color just to show this as an example."}, {"video_title": "Fundamental theorem of calculus (Part 1) AP Calculus AB Khan Academy.mp3", "Sentence": "And we'll do the intuition for why this happens or why this is true and maybe a proof in later videos. But how would you actually apply this right over here? Well, let's say that someone told you that they want to find the derivative. Let me just in a new color just to show this as an example. Let's say someone wanted to find the derivative with respect to x of the integral from, I don't know, I'll pick some random number here. So pi to x of, I'll put something crazy here, cosine squared of t over the natural log of t minus the square root of t dt. So they want you to take the derivative with respect to x of this crazy thing."}, {"video_title": "Fundamental theorem of calculus (Part 1) AP Calculus AB Khan Academy.mp3", "Sentence": "Let me just in a new color just to show this as an example. Let's say someone wanted to find the derivative with respect to x of the integral from, I don't know, I'll pick some random number here. So pi to x of, I'll put something crazy here, cosine squared of t over the natural log of t minus the square root of t dt. So they want you to take the derivative with respect to x of this crazy thing. Remember, this thing in the parentheses is a function of x. It's value. It's going to have a value that is dependent on x."}, {"video_title": "Fundamental theorem of calculus (Part 1) AP Calculus AB Khan Academy.mp3", "Sentence": "So they want you to take the derivative with respect to x of this crazy thing. Remember, this thing in the parentheses is a function of x. It's value. It's going to have a value that is dependent on x. You give it a different x, it's going to have a different value. So what's the derivative of this with respect to x? Well, the fundamental theorem of calculus tells us it can be very simple."}, {"video_title": "Fundamental theorem of calculus (Part 1) AP Calculus AB Khan Academy.mp3", "Sentence": "It's going to have a value that is dependent on x. You give it a different x, it's going to have a different value. So what's the derivative of this with respect to x? Well, the fundamental theorem of calculus tells us it can be very simple. We essentially, and you can even pattern match up here, and we'll get more intuition of why this is true in future videos. But essentially, everywhere where you see this right over here is an f of t. Everywhere you see a t, replace it with an x, and it becomes an f of x. So this is going to be equal to cosine squared of x over the natural log of x minus the square root of x."}, {"video_title": "Fundamental theorem of calculus (Part 1) AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the fundamental theorem of calculus tells us it can be very simple. We essentially, and you can even pattern match up here, and we'll get more intuition of why this is true in future videos. But essentially, everywhere where you see this right over here is an f of t. Everywhere you see a t, replace it with an x, and it becomes an f of x. So this is going to be equal to cosine squared of x over the natural log of x minus the square root of x. You take the derivative of the indefinite integral where the upper boundary is x right over here, it just becomes whatever you were taking the integral of, that as a function instead of t, that is now a function of x. So it can really simplify sometimes taking a derivative. And sometimes you'll see on exams these trick problems where you have this really hairy thing that you take a definite integral of and then take the derivative."}, {"video_title": "Fundamental theorem of calculus (Part 1) AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be equal to cosine squared of x over the natural log of x minus the square root of x. You take the derivative of the indefinite integral where the upper boundary is x right over here, it just becomes whatever you were taking the integral of, that as a function instead of t, that is now a function of x. So it can really simplify sometimes taking a derivative. And sometimes you'll see on exams these trick problems where you have this really hairy thing that you take a definite integral of and then take the derivative. And you just have to remember the fundamental theorem of calculus, the thing that ties it all together, connects derivatives and integration, that you can just simplify it by realizing that this is just going to be, instead of a function of lowercase f of t, it's going to be lowercase f of x. Let me make it clear. In this example right over here, this right over here was lowercase f of t, and now it became lowercase f of x."}, {"video_title": "Fundamental theorem of calculus (Part 1) AP Calculus AB Khan Academy.mp3", "Sentence": "And sometimes you'll see on exams these trick problems where you have this really hairy thing that you take a definite integral of and then take the derivative. And you just have to remember the fundamental theorem of calculus, the thing that ties it all together, connects derivatives and integration, that you can just simplify it by realizing that this is just going to be, instead of a function of lowercase f of t, it's going to be lowercase f of x. Let me make it clear. In this example right over here, this right over here was lowercase f of t, and now it became lowercase f of x. This right over here was our a. And notice, it doesn't matter what the lower boundary of a actually is. You don't have anything on the right-hand side that is in some way dependent on a."}, {"video_title": "Identifying composite functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "What we're going to do in this video is review the notion of composite functions and then build some skills recognizing how functions can actually be composed. If you've never heard of the term composite functions or if the first few minutes of this video look unfamiliar to you, I encourage you to watch the algebra videos on composite functions on Khan Academy. The goal of this one is to really be a little bit of a practice before we get into some skills that are necessary in calculus, in particular, the chain rule. So let's just review what a composite function is. So let's say that I have, let's say that I have f of x, f of x being equal to one plus x. And let's say that we have g of x is equal to, let's say g of x is equal to cosine of x. So what would f of g of x be?"}, {"video_title": "Identifying composite functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So let's just review what a composite function is. So let's say that I have, let's say that I have f of x, f of x being equal to one plus x. And let's say that we have g of x is equal to, let's say g of x is equal to cosine of x. So what would f of g of x be? F of g of x. And I encourage you to pause this video and try to work it out on your own. Well, one way to think about it is the input into f of x is no longer x, it is g of x."}, {"video_title": "Identifying composite functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So what would f of g of x be? F of g of x. And I encourage you to pause this video and try to work it out on your own. Well, one way to think about it is the input into f of x is no longer x, it is g of x. So everywhere where we see an x in the definition of f of x, we would replace with a g of x. So this is going to be equal to, this is going to be equal to one plus, instead of the input being x, the input is g of x, so the output is one plus g of x. And g of x, of course, is cosine of x."}, {"video_title": "Identifying composite functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, one way to think about it is the input into f of x is no longer x, it is g of x. So everywhere where we see an x in the definition of f of x, we would replace with a g of x. So this is going to be equal to, this is going to be equal to one plus, instead of the input being x, the input is g of x, so the output is one plus g of x. And g of x, of course, is cosine of x. So instead of writing g of x there, I could write cosine of x. And one way to visualize this is I'm putting my x into g of x first. So x goes into the function g, and it outputs g of x."}, {"video_title": "Identifying composite functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And g of x, of course, is cosine of x. So instead of writing g of x there, I could write cosine of x. And one way to visualize this is I'm putting my x into g of x first. So x goes into the function g, and it outputs g of x. And then we're gonna take that output, g of x, and then input it into f of x, or input it into the function f, I should say. We input into the function f, and that is going to output f of whatever the input was, and the input is g of x, g of x. So now with that review out of the way, let's see if we can go the other way around."}, {"video_title": "Identifying composite functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So x goes into the function g, and it outputs g of x. And then we're gonna take that output, g of x, and then input it into f of x, or input it into the function f, I should say. We input into the function f, and that is going to output f of whatever the input was, and the input is g of x, g of x. So now with that review out of the way, let's see if we can go the other way around. Let's see if we can look at some type of a function definition and say, hey, can we express that as a composition of other functions? So let's start with, let's say that I have a g of x is equal to cosine of sine of x plus one. And I also wanna state, there's oftentimes more than one way to compose a, or to construct a function based on the composition of other ones."}, {"video_title": "Identifying composite functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So now with that review out of the way, let's see if we can go the other way around. Let's see if we can look at some type of a function definition and say, hey, can we express that as a composition of other functions? So let's start with, let's say that I have a g of x is equal to cosine of sine of x plus one. And I also wanna state, there's oftentimes more than one way to compose a, or to construct a function based on the composition of other ones. But with that said, pause this video and say, hey, can I express g of x as a composition of two other functions? Let's say an f and an h of x. So there's a couple of ways that you could think about it."}, {"video_title": "Identifying composite functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And I also wanna state, there's oftentimes more than one way to compose a, or to construct a function based on the composition of other ones. But with that said, pause this video and say, hey, can I express g of x as a composition of two other functions? Let's say an f and an h of x. So there's a couple of ways that you could think about it. You could say, all right, let's see, I have this sine of x right over here. So what if I called that an f of x? So let's say I call that, well, actually, let me use a different variable so we don't get confused here."}, {"video_title": "Identifying composite functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So there's a couple of ways that you could think about it. You could say, all right, let's see, I have this sine of x right over here. So what if I called that an f of x? So let's say I call that, well, actually, let me use a different variable so we don't get confused here. Let me call this u of x, the sine of x right over there. So this would be cosine of u of x plus one. And so if we then divided, if we then defined another function as v of x being equal to cosine of whatever its input is plus one, well, then this looks like the composition of v and u of x."}, {"video_title": "Identifying composite functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say I call that, well, actually, let me use a different variable so we don't get confused here. Let me call this u of x, the sine of x right over there. So this would be cosine of u of x plus one. And so if we then divided, if we then defined another function as v of x being equal to cosine of whatever its input is plus one, well, then this looks like the composition of v and u of x. Instead of v of x, if we did v of u of x, then this would be cosine of u of x plus one. Let me write that down. So if we wrote v of u of x, which is sine of x, if we did v of u of x, that is going to be equal to cosine of, instead of an x plus one, it's going to be a u of x plus one."}, {"video_title": "Identifying composite functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And so if we then divided, if we then defined another function as v of x being equal to cosine of whatever its input is plus one, well, then this looks like the composition of v and u of x. Instead of v of x, if we did v of u of x, then this would be cosine of u of x plus one. Let me write that down. So if we wrote v of u of x, which is sine of x, if we did v of u of x, that is going to be equal to cosine of, instead of an x plus one, it's going to be a u of x plus one. And u of x, let me write this here, u of x is equal to sine of x. That's how we've set this up. So we can either write cosine of u of x plus one or cosine of sine of x plus one, which was exactly what we had before."}, {"video_title": "Identifying composite functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So if we wrote v of u of x, which is sine of x, if we did v of u of x, that is going to be equal to cosine of, instead of an x plus one, it's going to be a u of x plus one. And u of x, let me write this here, u of x is equal to sine of x. That's how we've set this up. So we can either write cosine of u of x plus one or cosine of sine of x plus one, which was exactly what we had before. And so this function g of x, if we say u of x is equal to sine of x, if we say u of x is equal to sine of x and v of x is equal to cosine of x plus one, then we could write g of x as the composition of these two functions. Now, you could even make it a composition of three functions. We could keep u of x to be equal to sine of x."}, {"video_title": "Identifying composite functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So we can either write cosine of u of x plus one or cosine of sine of x plus one, which was exactly what we had before. And so this function g of x, if we say u of x is equal to sine of x, if we say u of x is equal to sine of x and v of x is equal to cosine of x plus one, then we could write g of x as the composition of these two functions. Now, you could even make it a composition of three functions. We could keep u of x to be equal to sine of x. We could define, let's say, a w of x to be equal to x plus one. And so then, let's think about it. W of x, w of u of x, I should say, w of u, let me do the same color, w of u of x is going to be equal to, now my input is no longer x, it's a u of x, so it's going to be a u of x plus one or just sine of x plus one."}, {"video_title": "Identifying composite functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "We could keep u of x to be equal to sine of x. We could define, let's say, a w of x to be equal to x plus one. And so then, let's think about it. W of x, w of u of x, I should say, w of u, let me do the same color, w of u of x is going to be equal to, now my input is no longer x, it's a u of x, so it's going to be a u of x plus one or just sine of x plus one. So that's sine of x plus one. And then if we define a third function, let's say, let's see, I'll call it, let's call it h. I'm running out of variables. Well, there are a lot of letters left."}, {"video_title": "Identifying composite functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "W of x, w of u of x, I should say, w of u, let me do the same color, w of u of x is going to be equal to, now my input is no longer x, it's a u of x, so it's going to be a u of x plus one or just sine of x plus one. So that's sine of x plus one. And then if we define a third function, let's say, let's see, I'll call it, let's call it h. I'm running out of variables. Well, there are a lot of letters left. So if I say h of x is just equal to the cosine of whatever I input, so it's equal to the cosine of x, well then h of w of u of x is gonna be g of x. Let me write that down. H of w of u of x, u of x, is going to be equal to, remember, h of x takes the cosine of whatever its input is."}, {"video_title": "Identifying composite functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, there are a lot of letters left. So if I say h of x is just equal to the cosine of whatever I input, so it's equal to the cosine of x, well then h of w of u of x is gonna be g of x. Let me write that down. H of w of u of x, u of x, is going to be equal to, remember, h of x takes the cosine of whatever its input is. So it's gonna take the cosine. Now, its input is w of u of x. We already figured out w of u of x is going to be this business."}, {"video_title": "Identifying composite functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "H of w of u of x, u of x, is going to be equal to, remember, h of x takes the cosine of whatever its input is. So it's gonna take the cosine. Now, its input is w of u of x. We already figured out w of u of x is going to be this business. So it's going to be sine of x plus one. Where the u of x is sine of x, but then we inputted that into w, so we got sine of x plus one. And then we inputted that into h to get cosine of that, which is our original expression, which is equal to g of x."}, {"video_title": "Identifying composite functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "We already figured out w of u of x is going to be this business. So it's going to be sine of x plus one. Where the u of x is sine of x, but then we inputted that into w, so we got sine of x plus one. And then we inputted that into h to get cosine of that, which is our original expression, which is equal to g of x. So the whole point here is to appreciate how to recognize compositions of functions. Now, I wanna stress, it's not always going to be a composition of a function. For example, if I have some function, let me just clear this out."}, {"video_title": "Identifying composite functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And then we inputted that into h to get cosine of that, which is our original expression, which is equal to g of x. So the whole point here is to appreciate how to recognize compositions of functions. Now, I wanna stress, it's not always going to be a composition of a function. For example, if I have some function, let me just clear this out. If I had some function f of x is equal to cosine of x times sine of x, it would be hard to express this as a composition of functions, but I can represent it as the product of functions. For example, I could say cosine of x, I could say u of x is equal to cosine of x, and I could say v of x, let me use a different color, I could say v of x is equal to sine of x. And so here, f of x wouldn't be the composition of u and v, it would be the product."}, {"video_title": "Identifying composite functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "For example, if I have some function, let me just clear this out. If I had some function f of x is equal to cosine of x times sine of x, it would be hard to express this as a composition of functions, but I can represent it as the product of functions. For example, I could say cosine of x, I could say u of x is equal to cosine of x, and I could say v of x, let me use a different color, I could say v of x is equal to sine of x. And so here, f of x wouldn't be the composition of u and v, it would be the product. F of x is equal to u of x times v of x, v of x. If we were to take the composition, if we were to say u of v of x, pause the video, think about what that is, and that's a little bit of a review. Well, this is going to be, I take u of x, takes the cosine of whatever is input, and now the input is v of x, which would be sine of x."}, {"video_title": "Identifying composite functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And so here, f of x wouldn't be the composition of u and v, it would be the product. F of x is equal to u of x times v of x, v of x. If we were to take the composition, if we were to say u of v of x, pause the video, think about what that is, and that's a little bit of a review. Well, this is going to be, I take u of x, takes the cosine of whatever is input, and now the input is v of x, which would be sine of x. Sine of x. And then if you did v of u of x, well, that'd be the other way around. It would be sine of cosine of x."}, {"video_title": "Identifying composite functions Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, this is going to be, I take u of x, takes the cosine of whatever is input, and now the input is v of x, which would be sine of x. Sine of x. And then if you did v of u of x, well, that'd be the other way around. It would be sine of cosine of x. But anyway, this is, once again, just to help us recognize, hey, do I have, when I look at an expression or a function definition, am I looking at products of functions? Am I looking at compositions of functions? Sometimes you're looking at products of compositions or quotients of compositions, all sorts of different combinations of how you can put functions together to create new functions."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral AP Calculus AB Khan Academy.mp3", "Sentence": "And the goal of this video is to see if we can rewrite this as a definite integral. I encourage you to pause the video and see if you can work through it on your own. So let's remind ourselves how a definite integral can relate to a Riemann sum. So if I have the definite integral from a to b of f of x, f of x dx, we have seen in other videos, this is going to be the limit as n approaches infinity of the sum, capital sigma, going from i equals one to n. And so we're essentially going to sum the areas of a bunch of rectangles, where the width of each of those rectangles, we can write as a delta x. So your width is going to be delta x of each of those rectangles. And then your height is going to be the value of the function evaluated someplace in that delta x. If we're doing a right Riemann sum, we would do the right end of that rectangle or of that subinterval."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral AP Calculus AB Khan Academy.mp3", "Sentence": "So if I have the definite integral from a to b of f of x, f of x dx, we have seen in other videos, this is going to be the limit as n approaches infinity of the sum, capital sigma, going from i equals one to n. And so we're essentially going to sum the areas of a bunch of rectangles, where the width of each of those rectangles, we can write as a delta x. So your width is going to be delta x of each of those rectangles. And then your height is going to be the value of the function evaluated someplace in that delta x. If we're doing a right Riemann sum, we would do the right end of that rectangle or of that subinterval. And so we would start at our lower bound, a, and we would add as many delta x's as our index specifies. So if i is equal to one, we add one delta x. So we'd be at the right of the first rectangle."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral AP Calculus AB Khan Academy.mp3", "Sentence": "If we're doing a right Riemann sum, we would do the right end of that rectangle or of that subinterval. And so we would start at our lower bound, a, and we would add as many delta x's as our index specifies. So if i is equal to one, we add one delta x. So we'd be at the right of the first rectangle. If i is equal to two, we add two delta x's. So this is going to be delta x times our index. So this is the general form that we have seen before."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral AP Calculus AB Khan Academy.mp3", "Sentence": "So we'd be at the right of the first rectangle. If i is equal to two, we add two delta x's. So this is going to be delta x times our index. So this is the general form that we have seen before. And so one possibility, you could even do a little bit of pattern matching right over here. Our function looks like the natural log function. So that looks like our f of x."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral AP Calculus AB Khan Academy.mp3", "Sentence": "So this is the general form that we have seen before. And so one possibility, you could even do a little bit of pattern matching right over here. Our function looks like the natural log function. So that looks like our f of x. It's the natural log function. So I could write that. So f of x looks like the natural log of x."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral AP Calculus AB Khan Academy.mp3", "Sentence": "So that looks like our f of x. It's the natural log function. So I could write that. So f of x looks like the natural log of x. What else do we see? Well, a, that looks like two. A is equal to two."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral AP Calculus AB Khan Academy.mp3", "Sentence": "So f of x looks like the natural log of x. What else do we see? Well, a, that looks like two. A is equal to two. What would our delta x be? Well, you can see this right over here, this thing that we're multiplying that just is divided by n, and it's not multiplying by an i. So this looks like our delta x."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral AP Calculus AB Khan Academy.mp3", "Sentence": "A is equal to two. What would our delta x be? Well, you can see this right over here, this thing that we're multiplying that just is divided by n, and it's not multiplying by an i. So this looks like our delta x. And this right over here looks like delta x times i. So it looks like our delta x is equal to five over n. So what can we tell so far? Well, we could say that, okay, this thing up here, the original thing, is going to be equal to the definite integral."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral AP Calculus AB Khan Academy.mp3", "Sentence": "So this looks like our delta x. And this right over here looks like delta x times i. So it looks like our delta x is equal to five over n. So what can we tell so far? Well, we could say that, okay, this thing up here, the original thing, is going to be equal to the definite integral. We know our lower bound is going from two to, we haven't figured out our upper bound yet. We haven't figured out our b yet. But our function is the natural log of x, and then I will just write a dx here."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we could say that, okay, this thing up here, the original thing, is going to be equal to the definite integral. We know our lower bound is going from two to, we haven't figured out our upper bound yet. We haven't figured out our b yet. But our function is the natural log of x, and then I will just write a dx here. So in order to complete writing this definite integral, I need to be able to write the upper bound. And the way to figure out the upper bound is by looking at our delta x. Because the way that we would figure out a delta x for this Riemann sum here, we would say that delta x is equal to the difference between our bounds divided by how many sections we want to divide it in, divided by n. So it's equal to b minus a, b minus a over n. Over n. And so you can pattern match here."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral AP Calculus AB Khan Academy.mp3", "Sentence": "But our function is the natural log of x, and then I will just write a dx here. So in order to complete writing this definite integral, I need to be able to write the upper bound. And the way to figure out the upper bound is by looking at our delta x. Because the way that we would figure out a delta x for this Riemann sum here, we would say that delta x is equal to the difference between our bounds divided by how many sections we want to divide it in, divided by n. So it's equal to b minus a, b minus a over n. Over n. And so you can pattern match here. If this is delta x is equal to b minus a over n, let me write this down. So this is going to be equal to b, b minus our a, which is two. All of that over n. So b minus two is equal to five, which would make b equal to seven."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral AP Calculus AB Khan Academy.mp3", "Sentence": "Because the way that we would figure out a delta x for this Riemann sum here, we would say that delta x is equal to the difference between our bounds divided by how many sections we want to divide it in, divided by n. So it's equal to b minus a, b minus a over n. Over n. And so you can pattern match here. If this is delta x is equal to b minus a over n, let me write this down. So this is going to be equal to b, b minus our a, which is two. All of that over n. So b minus two is equal to five, which would make b equal to seven. B is equal to seven. So there you have it. We have our original limit, our Riemann limit, or limit of our Riemann sum being rewritten as a definite integral."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral AP Calculus AB Khan Academy.mp3", "Sentence": "All of that over n. So b minus two is equal to five, which would make b equal to seven. B is equal to seven. So there you have it. We have our original limit, our Riemann limit, or limit of our Riemann sum being rewritten as a definite integral. And once again, I want to emphasize why this makes sense. If I wanted to draw this, it would look something like this. I'm gonna try to hand draw the natural log function."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral AP Calculus AB Khan Academy.mp3", "Sentence": "We have our original limit, our Riemann limit, or limit of our Riemann sum being rewritten as a definite integral. And once again, I want to emphasize why this makes sense. If I wanted to draw this, it would look something like this. I'm gonna try to hand draw the natural log function. It looks something like this. This right over here would be one. And so let's say this is two."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral AP Calculus AB Khan Academy.mp3", "Sentence": "I'm gonna try to hand draw the natural log function. It looks something like this. This right over here would be one. And so let's say this is two. And so we're going from two to seven. This isn't exactly right. So our definite integral is concerned with the area under the curve from two until seven."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral AP Calculus AB Khan Academy.mp3", "Sentence": "And so let's say this is two. And so we're going from two to seven. This isn't exactly right. So our definite integral is concerned with the area under the curve from two until seven. And so this Riemann sum you could view as an approximation when n isn't approaching infinity. But what you're saying is, look, when i is equal to one, your first one is going to be of width five over n. So this is essentially saying our difference between two and seven, we're taking that distance five, dividing it into n rectangles. And so this first one is going to have a width of five over n. And then what's the height gonna be?"}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral AP Calculus AB Khan Academy.mp3", "Sentence": "So our definite integral is concerned with the area under the curve from two until seven. And so this Riemann sum you could view as an approximation when n isn't approaching infinity. But what you're saying is, look, when i is equal to one, your first one is going to be of width five over n. So this is essentially saying our difference between two and seven, we're taking that distance five, dividing it into n rectangles. And so this first one is going to have a width of five over n. And then what's the height gonna be? Well, it's a right Riemann sum. So we're using the value of the function right over there, right at two plus five over n. So this value right over here, this is the natural log, the natural log of two plus five over n. And since this is the first rectangle, times one. Times one."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral AP Calculus AB Khan Academy.mp3", "Sentence": "And so this first one is going to have a width of five over n. And then what's the height gonna be? Well, it's a right Riemann sum. So we're using the value of the function right over there, right at two plus five over n. So this value right over here, this is the natural log, the natural log of two plus five over n. And since this is the first rectangle, times one. Times one. And we could keep going. This one right over here, the width is the same, five over n. But what's the height? Well, the height here, this height right over here is gonna be the natural log of two plus five over n times two."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral AP Calculus AB Khan Academy.mp3", "Sentence": "Times one. And we could keep going. This one right over here, the width is the same, five over n. But what's the height? Well, the height here, this height right over here is gonna be the natural log of two plus five over n times two. Times two. This is for i is equal to two. This is i is equal to one."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the height here, this height right over here is gonna be the natural log of two plus five over n times two. Times two. This is for i is equal to two. This is i is equal to one. And so hopefully you are seeing that this makes sense. The area of this first rectangle is going to be natural log of two plus five over n times one. Times five over n. The second one over here, natural log of two plus five over n times two."}, {"video_title": "Worked example Rewriting limit of Riemann sum as definite integral AP Calculus AB Khan Academy.mp3", "Sentence": "This is i is equal to one. And so hopefully you are seeing that this makes sense. The area of this first rectangle is going to be natural log of two plus five over n times one. Times five over n. The second one over here, natural log of two plus five over n times two. Times five over n. And so this is calculating the sum of the areas of these rectangles. But then it's taking the limit as n approaches infinity. So we get better and better approximations going all the way to the exact area."}, {"video_title": "Power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "The limit is delta x approaches 0 of f of x plus delta x minus f of x, all of that over delta x. And it really just comes out of trying to find the slope of a tangent line at any given point. But we're going to see what the power rule is. It simplifies our life. We won't have to take these sometimes complicated limits. And we're not going to prove it in this video. But we'll hopefully get a sense of how to use it."}, {"video_title": "Power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "It simplifies our life. We won't have to take these sometimes complicated limits. And we're not going to prove it in this video. But we'll hopefully get a sense of how to use it. And in future videos, we'll get a sense of why it makes sense and even prove it. So the power rule just tells us that if I have some function, f of x, and it's equal to some power of x, so x to the n power, where n does not equal 0. So n can be anything."}, {"video_title": "Power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "But we'll hopefully get a sense of how to use it. And in future videos, we'll get a sense of why it makes sense and even prove it. So the power rule just tells us that if I have some function, f of x, and it's equal to some power of x, so x to the n power, where n does not equal 0. So n can be anything. It can be positive and negative. It does not have to be an integer. The power rule tells us that the derivative of this, f prime of x, is just going to be equal to n. So you're literally bringing this out front."}, {"video_title": "Power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So n can be anything. It can be positive and negative. It does not have to be an integer. The power rule tells us that the derivative of this, f prime of x, is just going to be equal to n. So you're literally bringing this out front. n times x. And then you just decrement the power. Times x to the n minus 1 power."}, {"video_title": "Power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "The power rule tells us that the derivative of this, f prime of x, is just going to be equal to n. So you're literally bringing this out front. n times x. And then you just decrement the power. Times x to the n minus 1 power. So let's do a couple of examples just to make sure that that actually makes sense. So let's ask ourselves, well, let's say that f of x was equal to x squared. Based on the power rule, what is f prime of x going to be equal to?"}, {"video_title": "Power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Times x to the n minus 1 power. So let's do a couple of examples just to make sure that that actually makes sense. So let's ask ourselves, well, let's say that f of x was equal to x squared. Based on the power rule, what is f prime of x going to be equal to? Well, in this situation, our n is 2. So we bring the 2 out front. 2 times x to the 2 minus 1 power."}, {"video_title": "Power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Based on the power rule, what is f prime of x going to be equal to? Well, in this situation, our n is 2. So we bring the 2 out front. 2 times x to the 2 minus 1 power. So that's going to be 2 times x to the first power, which is just equal to 2x. That was pretty straightforward. Let's think about the situation where, let's say, we have g of x is equal to x to the third power."}, {"video_title": "Power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "2 times x to the 2 minus 1 power. So that's going to be 2 times x to the first power, which is just equal to 2x. That was pretty straightforward. Let's think about the situation where, let's say, we have g of x is equal to x to the third power. What is g prime of x going to be in this scenario? Well, n is 3. So we just literally pattern match here."}, {"video_title": "Power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Let's think about the situation where, let's say, we have g of x is equal to x to the third power. What is g prime of x going to be in this scenario? Well, n is 3. So we just literally pattern match here. You're probably finding this shockingly straightforward. So this is going to be 3 times x to the 3 minus 1 power. Or this is going to be equal to 3x squared."}, {"video_title": "Power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So we just literally pattern match here. You're probably finding this shockingly straightforward. So this is going to be 3 times x to the 3 minus 1 power. Or this is going to be equal to 3x squared. And we're done. In the next video, we'll think about whether this actually makes sense. Let's do one more example just to show it doesn't have to necessarily apply to only these kind of positive integers."}, {"video_title": "Power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Or this is going to be equal to 3x squared. And we're done. In the next video, we'll think about whether this actually makes sense. Let's do one more example just to show it doesn't have to necessarily apply to only these kind of positive integers. We could have a scenario where maybe we have h of x. h of x is equal to x to the negative 100 power. The power rule tells us that h prime of x would be equal to what? Well, n is negative 100."}, {"video_title": "Power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Let's do one more example just to show it doesn't have to necessarily apply to only these kind of positive integers. We could have a scenario where maybe we have h of x. h of x is equal to x to the negative 100 power. The power rule tells us that h prime of x would be equal to what? Well, n is negative 100. So it's negative 100x to the negative 100 minus 1, which is equal to negative 100x to the negative 101. Let's do one more. Let's say we had z of x. z of x is equal to x to the 2.571 power."}, {"video_title": "Power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, n is negative 100. So it's negative 100x to the negative 100 minus 1, which is equal to negative 100x to the negative 101. Let's do one more. Let's say we had z of x. z of x is equal to x to the 2.571 power. And we are concerned with what is z prime of x. Well, once again, power rule simplifies our life. n is 2.571."}, {"video_title": "Power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say we had z of x. z of x is equal to x to the 2.571 power. And we are concerned with what is z prime of x. Well, once again, power rule simplifies our life. n is 2.571. So it's going to be 2.571 times x to the 2.571 minus 1 power. So it's going to be equal to, let me make sure I'm not falling off the bottom of the page, 2.571 times x to the 1.571 power. Hopefully you enjoyed that."}, {"video_title": "Power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "n is 2.571. So it's going to be 2.571 times x to the 2.571 minus 1 power. So it's going to be equal to, let me make sure I'm not falling off the bottom of the page, 2.571 times x to the 1.571 power. Hopefully you enjoyed that. In the next few videos, we will not only expose you to more properties of derivatives. We'll get a sense for why the power rule at least makes intuitive sense. And then also prove the power rule for a few cases."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "And to do that, what we could do is we could look at a coordinate plane. So, let me draw some axes here. So, let me draw a relatively straight line. Alright, so that's my y-axis, and this is my x-axis. Let me do draw, let me mark this as one, that's two, that's negative one, negative two, one, two, negative one, and negative two. And what I could do is, since this differential equation is just in terms of x's and y's and first derivatives of y with respect to x, I could go, I could sample points on the coordinate plane, I could look at the x and y coordinates, substitute them in here, figure out what the slope is going to be, and then I could visualize the slope, if a solution goes to that point, what the slope needs to be there, and I can visualize that with a line segment, a little small line segment that has the same slope as the slope in question. So, let's actually do that."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "Alright, so that's my y-axis, and this is my x-axis. Let me do draw, let me mark this as one, that's two, that's negative one, negative two, one, two, negative one, and negative two. And what I could do is, since this differential equation is just in terms of x's and y's and first derivatives of y with respect to x, I could go, I could sample points on the coordinate plane, I could look at the x and y coordinates, substitute them in here, figure out what the slope is going to be, and then I could visualize the slope, if a solution goes to that point, what the slope needs to be there, and I can visualize that with a line segment, a little small line segment that has the same slope as the slope in question. So, let's actually do that. Let me set up a little table here. Let me do a little table here to do a bunch of x and y values. Once again, I'm just sampling some points on the coordinate plane to be able to visualize."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "So, let's actually do that. Let me set up a little table here. Let me do a little table here to do a bunch of x and y values. Once again, I'm just sampling some points on the coordinate plane to be able to visualize. So, x, y, and this is dy, dx. So, let's say when x is, let's say when x is zero and y is one, what is the derivative of y with respect to x? It's going to be negative zero over one, so it's just going to be zero."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "Once again, I'm just sampling some points on the coordinate plane to be able to visualize. So, x, y, and this is dy, dx. So, let's say when x is, let's say when x is zero and y is one, what is the derivative of y with respect to x? It's going to be negative zero over one, so it's just going to be zero. And so, at the point zero, one, if a solution goes through this point, its slope is going to be zero. And so, we can visualize that by doing a little horizontal line segment right there. So, let's keep going."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "It's going to be negative zero over one, so it's just going to be zero. And so, at the point zero, one, if a solution goes through this point, its slope is going to be zero. And so, we can visualize that by doing a little horizontal line segment right there. So, let's keep going. What about when x is one and y is one? Well then, dy, dx, the derivative of y with respect to x is negative one over one. So, it's going to be negative one."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "So, let's keep going. What about when x is one and y is one? Well then, dy, dx, the derivative of y with respect to x is negative one over one. So, it's going to be negative one. So, at the point one, comma, one, if a solution goes through that point, it would have a slope of negative one. And so, I draw a little line segment that has a slope of negative one. What about when x is, let me do this in a new color, what about when x is one and y is zero?"}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "So, it's going to be negative one. So, at the point one, comma, one, if a solution goes through that point, it would have a slope of negative one. And so, I draw a little line segment that has a slope of negative one. What about when x is, let me do this in a new color, what about when x is one and y is zero? Well then, it's negative one over zero. So, this is actually undefined. But, it's a clue that maybe, maybe the slope there, I guess if you had a tangent line at that point, maybe it's vertical."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "What about when x is, let me do this in a new color, what about when x is one and y is zero? Well then, it's negative one over zero. So, this is actually undefined. But, it's a clue that maybe, maybe the slope there, I guess if you had a tangent line at that point, maybe it's vertical. So, I'll put that as a question mark. Vertical there. And so, maybe it's something like that if you actually did have, I guess it wouldn't be a function if you had some kind of relation that went through it."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "But, it's a clue that maybe, maybe the slope there, I guess if you had a tangent line at that point, maybe it's vertical. So, I'll put that as a question mark. Vertical there. And so, maybe it's something like that if you actually did have, I guess it wouldn't be a function if you had some kind of relation that went through it. But, let's not draw that just yet, but let's try some other points. Let's say that we had, let's try the point negative one, negative one. So, now we have negative negative one, which is one over negative one."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "And so, maybe it's something like that if you actually did have, I guess it wouldn't be a function if you had some kind of relation that went through it. But, let's not draw that just yet, but let's try some other points. Let's say that we had, let's try the point negative one, negative one. So, now we have negative negative one, which is one over negative one. Well, you would have a slope of negative one here. So, negative one, negative one. You would have a slope of, you would have a slope of negative one."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "So, now we have negative negative one, which is one over negative one. Well, you would have a slope of negative one here. So, negative one, negative one. You would have a slope of, you would have a slope of negative one. What about if you had one negative one? Well, now it's negative one over negative one. Your slope is now one."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "You would have a slope of, you would have a slope of negative one. What about if you had one negative one? Well, now it's negative one over negative one. Your slope is now one. So, one negative one. If your solution, if a solution goes through this, its slope would look like that. And we could keep going."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "Your slope is now one. So, one negative one. If your solution, if a solution goes through this, its slope would look like that. And we could keep going. We could even do two negative two. That's going to have a slope of one as well. If you did positive two, positive two, that'd be negative two over two."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "And we could keep going. We could even do two negative two. That's going to have a slope of one as well. If you did positive two, positive two, that'd be negative two over two. You'd have a slope of negative one right over here. And so, we could do a bunch of points. Just keep going."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "If you did positive two, positive two, that'd be negative two over two. You'd have a slope of negative one right over here. And so, we could do a bunch of points. Just keep going. I'm now just doing them in my head. I'm not going on the table. But, you get a sense of what's going on here."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "Just keep going. I'm now just doing them in my head. I'm not going on the table. But, you get a sense of what's going on here. Here, your slope, what if it was negative one, one. It's going to have a slope of one. So, at this point, your slope, negative one, one."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "But, you get a sense of what's going on here. Here, your slope, what if it was negative one, one. It's going to have a slope of one. So, at this point, your slope, negative one, one. So, negative negative one is one over one. So, you're going to have a slope like that. At negative two, two, same exact idea."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "So, at this point, your slope, negative one, one. So, negative negative one is one over one. So, you're going to have a slope like that. At negative two, two, same exact idea. It would look like that. And so, you get a, when you keep drawing these line segments over these kind of, these sampled points in the Cartesian or in the X-Y plane, you start to get a sense of, well, what would a solution have to do? And you can start to visualize that, hey, maybe a solution, a solution would have to do something, something like this."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "At negative two, two, same exact idea. It would look like that. And so, you get a, when you keep drawing these line segments over these kind of, these sampled points in the Cartesian or in the X-Y plane, you start to get a sense of, well, what would a solution have to do? And you can start to visualize that, hey, maybe a solution, a solution would have to do something, something like this. This would be a solution. So, maybe it would have to do something like this. Or, if we're looking, if we're looking only at functions and not relations, I'll only, I'll make it so it's a very clear, so maybe it would have to do something like this."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "And you can start to visualize that, hey, maybe a solution, a solution would have to do something, something like this. This would be a solution. So, maybe it would have to do something like this. Or, if we're looking, if we're looking only at functions and not relations, I'll only, I'll make it so it's a very clear, so maybe it would have to do something like this. Or, if the function started like here, based on what we've seen so far, maybe it would have to do something, maybe it would have to do something like this. Or, if it was, if this were a point on the function over here, it would have to do something like this. And once again, I'm doing this based on what the slope field is telling me."}, {"video_title": "Creating a slope field First order differential equations Khan Academy.mp3", "Sentence": "Or, if we're looking, if we're looking only at functions and not relations, I'll only, I'll make it so it's a very clear, so maybe it would have to do something like this. Or, if the function started like here, based on what we've seen so far, maybe it would have to do something, maybe it would have to do something like this. Or, if it was, if this were a point on the function over here, it would have to do something like this. And once again, I'm doing this based on what the slope field is telling me. So, this field that I'm creating, where I'm taking, I'm sampling a bunch of points and I'm visualizing the slope with the line segment, once again, this is called a slope, slope field. So, hopefully that gives you kind of the basic idea of what a slope field is. In the next few videos, we'll explore this idea even deeper."}, {"video_title": "Functions defined by definite integrals (accumulation functions) AP Calculus AB Khan Academy.mp3", "Sentence": "And we call that corresponding output f of x. So for example, there's many ways of defining functions. You could say something like f of x is equal to x squared. So that means that whatever x, whatever you input into the function, the output is going to be that input squared. You could have something defined like this. F of x is equal to x squared if x odd, and you could say it's equal to x to the third otherwise. So if it's an odd integer, it's an odd integer, you just square it, but otherwise for any other real number, you take it to the third power."}, {"video_title": "Functions defined by definite integrals (accumulation functions) AP Calculus AB Khan Academy.mp3", "Sentence": "So that means that whatever x, whatever you input into the function, the output is going to be that input squared. You could have something defined like this. F of x is equal to x squared if x odd, and you could say it's equal to x to the third otherwise. So if it's an odd integer, it's an odd integer, you just square it, but otherwise for any other real number, you take it to the third power. This is a valid way of defining a function. What we're going to do in this video is explore a new way, or a potentially new way for you, of defining a function, and that's by using a definite integral, but it's the same general idea. So what we have graphed here, this is the t-axis, this is the y-axis, and we have the graph of the function f, or you could view this as the graph of y is equal to f of t. Now what I want, and this is another way of representing what outputs you might get for a given input here if t is one, f of t is five."}, {"video_title": "Functions defined by definite integrals (accumulation functions) AP Calculus AB Khan Academy.mp3", "Sentence": "So if it's an odd integer, it's an odd integer, you just square it, but otherwise for any other real number, you take it to the third power. This is a valid way of defining a function. What we're going to do in this video is explore a new way, or a potentially new way for you, of defining a function, and that's by using a definite integral, but it's the same general idea. So what we have graphed here, this is the t-axis, this is the y-axis, and we have the graph of the function f, or you could view this as the graph of y is equal to f of t. Now what I want, and this is another way of representing what outputs you might get for a given input here if t is one, f of t is five. If t is four, f of t is three. But I'm now going to define a new function based on a definite integral of f of t. Let's define our new function. Let's say g, let's call it g of x."}, {"video_title": "Functions defined by definite integrals (accumulation functions) AP Calculus AB Khan Academy.mp3", "Sentence": "So what we have graphed here, this is the t-axis, this is the y-axis, and we have the graph of the function f, or you could view this as the graph of y is equal to f of t. Now what I want, and this is another way of representing what outputs you might get for a given input here if t is one, f of t is five. If t is four, f of t is three. But I'm now going to define a new function based on a definite integral of f of t. Let's define our new function. Let's say g, let's call it g of x. Let's make it equal to the definite integral from negative two to x of f of t dt. Now pause this video, really take a look at it. This might look really fancy, but what's happening here is given an input x, g of x is going to be based on what the definite integral here would be for that x."}, {"video_title": "Functions defined by definite integrals (accumulation functions) AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say g, let's call it g of x. Let's make it equal to the definite integral from negative two to x of f of t dt. Now pause this video, really take a look at it. This might look really fancy, but what's happening here is given an input x, g of x is going to be based on what the definite integral here would be for that x. And so we can set up a little table here to think about some potential values. So let's say x, and let's say g of x right over here. So if x is one, what is g of x going to be equal to?"}, {"video_title": "Functions defined by definite integrals (accumulation functions) AP Calculus AB Khan Academy.mp3", "Sentence": "This might look really fancy, but what's happening here is given an input x, g of x is going to be based on what the definite integral here would be for that x. And so we can set up a little table here to think about some potential values. So let's say x, and let's say g of x right over here. So if x is one, what is g of x going to be equal to? All right, so g of one is going to be equal to the definite integral going from negative two. Now x is going to be equal to one in this situation. That's what we're inputting into the function."}, {"video_title": "Functions defined by definite integrals (accumulation functions) AP Calculus AB Khan Academy.mp3", "Sentence": "So if x is one, what is g of x going to be equal to? All right, so g of one is going to be equal to the definite integral going from negative two. Now x is going to be equal to one in this situation. That's what we're inputting into the function. So one is our upper bound of f of t dt. And what is that equal to? Well, that's going to be the area under the curve and above the t-axis between t equals negative two and t is equal to one."}, {"video_title": "Functions defined by definite integrals (accumulation functions) AP Calculus AB Khan Academy.mp3", "Sentence": "That's what we're inputting into the function. So one is our upper bound of f of t dt. And what is that equal to? Well, that's going to be the area under the curve and above the t-axis between t equals negative two and t is equal to one. So it's gonna be this area here. And since it's on a grid, we can actually figure this out. We can actually break this up into two sections."}, {"video_title": "Functions defined by definite integrals (accumulation functions) AP Calculus AB Khan Academy.mp3", "Sentence": "Well, that's going to be the area under the curve and above the t-axis between t equals negative two and t is equal to one. So it's gonna be this area here. And since it's on a grid, we can actually figure this out. We can actually break this up into two sections. This rectangular section is three wide and five high, so it has an area of 15 square units. And this little triangular section up here is two wide and one high, two times one times 1 1\u20442, area of a triangle. This is going to be another one."}, {"video_title": "Functions defined by definite integrals (accumulation functions) AP Calculus AB Khan Academy.mp3", "Sentence": "We can actually break this up into two sections. This rectangular section is three wide and five high, so it has an area of 15 square units. And this little triangular section up here is two wide and one high, two times one times 1 1\u20442, area of a triangle. This is going to be another one. So that area is going to be equal to 16. What if x is equal to two? What is g of two going to be equal to?"}, {"video_title": "Functions defined by definite integrals (accumulation functions) AP Calculus AB Khan Academy.mp3", "Sentence": "This is going to be another one. So that area is going to be equal to 16. What if x is equal to two? What is g of two going to be equal to? Pause this video and try to figure that out. Well, g of two is going to be equal to the definite integral from negative two. And now our upper bound's going to be our input into the function to two of f of t dt."}, {"video_title": "Functions defined by definite integrals (accumulation functions) AP Calculus AB Khan Academy.mp3", "Sentence": "What is g of two going to be equal to? Pause this video and try to figure that out. Well, g of two is going to be equal to the definite integral from negative two. And now our upper bound's going to be our input into the function to two of f of t dt. So that's going to be going from here all the way now to here. And so it's the area we just calculated. It's all of this stuff, which we figured out was 16 square units, plus another one, two, three, four, five square units."}, {"video_title": "Functions defined by definite integrals (accumulation functions) AP Calculus AB Khan Academy.mp3", "Sentence": "And now our upper bound's going to be our input into the function to two of f of t dt. So that's going to be going from here all the way now to here. And so it's the area we just calculated. It's all of this stuff, which we figured out was 16 square units, plus another one, two, three, four, five square units. So 16 plus five, this is going to be equal to 21. So hopefully that helps. And the key thing to appreciate here is that we can define valid functions by using definite integrals."}, {"video_title": "Theorem for limits of composite functions Limits and contiuity AP Calculus Khan Academy.mp3", "Sentence": "And we're going to see under certain circumstances, this is going to be equal to f of the limit, the limit as x approaches a of g of x. And what are those circumstances you are asking? Well, this is going to be true if and only if two things are true. First of all, this limit needs to exist. So the limit as x approaches a of g of x needs to exist. So that needs to exist. And then on top of that, the function f needs to be continuous at this point, and f continuous at L. So let's look at some examples and see if we can apply this idea, or see if we can't apply it."}, {"video_title": "Theorem for limits of composite functions Limits and contiuity AP Calculus Khan Academy.mp3", "Sentence": "First of all, this limit needs to exist. So the limit as x approaches a of g of x needs to exist. So that needs to exist. And then on top of that, the function f needs to be continuous at this point, and f continuous at L. So let's look at some examples and see if we can apply this idea, or see if we can't apply it. So here I have two functions that are graphically represented right over here. Let me make sure I have enough space for them. And what we see on the left-hand side is our function f, and what we see on the right-hand side is our function g. So first, let's figure out what is the limit as x approaches negative three of f of g of x."}, {"video_title": "Theorem for limits of composite functions Limits and contiuity AP Calculus Khan Academy.mp3", "Sentence": "And then on top of that, the function f needs to be continuous at this point, and f continuous at L. So let's look at some examples and see if we can apply this idea, or see if we can't apply it. So here I have two functions that are graphically represented right over here. Let me make sure I have enough space for them. And what we see on the left-hand side is our function f, and what we see on the right-hand side is our function g. So first, let's figure out what is the limit as x approaches negative three of f of g of x. Pause this video and see, first of all, does this theorem apply? And if it does apply, what is this limit? So the first thing we need to see is does this theorem apply?"}, {"video_title": "Theorem for limits of composite functions Limits and contiuity AP Calculus Khan Academy.mp3", "Sentence": "And what we see on the left-hand side is our function f, and what we see on the right-hand side is our function g. So first, let's figure out what is the limit as x approaches negative three of f of g of x. Pause this video and see, first of all, does this theorem apply? And if it does apply, what is this limit? So the first thing we need to see is does this theorem apply? So first of all, if we were to find the limit as x approaches negative three of g of x, what is that? Well, when we're approaching negative three from the right, it looks like our function is actually at three. And it looks like when we're approaching negative three from the left, it looks like our function is at three."}, {"video_title": "Theorem for limits of composite functions Limits and contiuity AP Calculus Khan Academy.mp3", "Sentence": "So the first thing we need to see is does this theorem apply? So first of all, if we were to find the limit as x approaches negative three of g of x, what is that? Well, when we're approaching negative three from the right, it looks like our function is actually at three. And it looks like when we're approaching negative three from the left, it looks like our function is at three. So it looks like this limit is three, even though the value g of negative three is negative two, but it's a point discontinuity. As we approach it from either side, the value of the function is at three. So this thing is going to be three."}, {"video_title": "Theorem for limits of composite functions Limits and contiuity AP Calculus Khan Academy.mp3", "Sentence": "And it looks like when we're approaching negative three from the left, it looks like our function is at three. So it looks like this limit is three, even though the value g of negative three is negative two, but it's a point discontinuity. As we approach it from either side, the value of the function is at three. So this thing is going to be three. So it exists, so we meet that first condition. And then the second question is, is our function f continuous at this limit, continuous at three? So when x equals three, yeah, it looks like at that point, our function is definitely continuous."}, {"video_title": "Theorem for limits of composite functions Limits and contiuity AP Calculus Khan Academy.mp3", "Sentence": "So this thing is going to be three. So it exists, so we meet that first condition. And then the second question is, is our function f continuous at this limit, continuous at three? So when x equals three, yeah, it looks like at that point, our function is definitely continuous. And so we could say that this limit is going to be the same thing as this equals f of the limit as x approaches negative three of g of x, close the parentheses. And we know that this is equal to three. And we know that f of three is going to be equal to negative one."}, {"video_title": "Proof of p-series convergence criteria Series AP Calculus BC Khan Academy.mp3", "Sentence": "And what we're going to do in this video is think about under which conditions, under for what p's will this p-series converge. And for it to be a p-series, by definition, p is going to be greater than zero. So I've set up some visualizations to think about how we are going to understand when this p-series converges. So over here, you have the graph, this curve right here, that's y is equal to one over x to the p. And we're saying it in general terms, because p is greater than zero, we know it's going to be a decreasing function like this. Once again, that's y is equal to one over x to the p. And now what we've shaded in ahead of time, underneath that curve, above the positive x-axis, that is the integral from one to infinity, from one to infinity, the improper integral of one over x to the p dx. So that's this area that I have already shaded in. You see it in white in both of these graphs."}, {"video_title": "Proof of p-series convergence criteria Series AP Calculus BC Khan Academy.mp3", "Sentence": "So over here, you have the graph, this curve right here, that's y is equal to one over x to the p. And we're saying it in general terms, because p is greater than zero, we know it's going to be a decreasing function like this. Once again, that's y is equal to one over x to the p. And now what we've shaded in ahead of time, underneath that curve, above the positive x-axis, that is the integral from one to infinity, from one to infinity, the improper integral of one over x to the p dx. So that's this area that I have already shaded in. You see it in white in both of these graphs. And what we're going to hopefully see visually is that there's a very close convergence or divergence relationship between this p-series and this integral right over here. Because when we look at this left-hand graph, we see that this p-series can be viewed as an upper Riemann approximation of that area. What do I mean by that?"}, {"video_title": "Proof of p-series convergence criteria Series AP Calculus BC Khan Academy.mp3", "Sentence": "You see it in white in both of these graphs. And what we're going to hopefully see visually is that there's a very close convergence or divergence relationship between this p-series and this integral right over here. Because when we look at this left-hand graph, we see that this p-series can be viewed as an upper Riemann approximation of that area. What do I mean by that? Well, think about the area of this first, I guess you could say this first rectangle. The width is one, and its height is one over, one over one to the p. So this would be the first term in this p-series. This would just be an area of one."}, {"video_title": "Proof of p-series convergence criteria Series AP Calculus BC Khan Academy.mp3", "Sentence": "What do I mean by that? Well, think about the area of this first, I guess you could say this first rectangle. The width is one, and its height is one over, one over one to the p. So this would be the first term in this p-series. This would just be an area of one. Just the scale, the x and y scales are not the same. This one right over here, its area would be one over two to the p. This area is one over three to the p. So the sum of the areas of these rectangles, that is what this p-series is. And you can see that each of these rectangles, they are covering more than the area under the curve."}, {"video_title": "Proof of p-series convergence criteria Series AP Calculus BC Khan Academy.mp3", "Sentence": "This would just be an area of one. Just the scale, the x and y scales are not the same. This one right over here, its area would be one over two to the p. This area is one over three to the p. So the sum of the areas of these rectangles, that is what this p-series is. And you can see that each of these rectangles, they are covering more than the area under the curve. And so we know the area under the curve, that's gonna be greater than zero. This p-series is going to be greater than this integral, greater than the area under the curve. But if we add one to the area under the curve, so now we're not just talking about the white area, we're also talking about this red area here."}, {"video_title": "Proof of p-series convergence criteria Series AP Calculus BC Khan Academy.mp3", "Sentence": "And you can see that each of these rectangles, they are covering more than the area under the curve. And so we know the area under the curve, that's gonna be greater than zero. This p-series is going to be greater than this integral, greater than the area under the curve. But if we add one to the area under the curve, so now we're not just talking about the white area, we're also talking about this red area here. Well then our p-series is going to be less than that. Because the first term of our p-series is equal to one, and then all of the other terms, you can view it as a lower Riemann approximation of the curve. And you can see, they fit under the curve and they leave some area."}, {"video_title": "Proof of p-series convergence criteria Series AP Calculus BC Khan Academy.mp3", "Sentence": "But if we add one to the area under the curve, so now we're not just talking about the white area, we're also talking about this red area here. Well then our p-series is going to be less than that. Because the first term of our p-series is equal to one, and then all of the other terms, you can view it as a lower Riemann approximation of the curve. And you can see, they fit under the curve and they leave some area. So this is going to be less than that expression there. Now think about what happens. If we know that this right over here diverges, so if this improper integral diverges, it doesn't converge to a finite value, well this, the p-series is greater than that."}, {"video_title": "Proof of p-series convergence criteria Series AP Calculus BC Khan Academy.mp3", "Sentence": "And you can see, they fit under the curve and they leave some area. So this is going to be less than that expression there. Now think about what happens. If we know that this right over here diverges, so if this improper integral diverges, it doesn't converge to a finite value, well this, the p-series is greater than that. So if this diverges, then that's going to diverge. Similarly, if this converges, the same integral right over here, if this converges, it goes to a finite value, well one plus that is still going to converge. And so this, our p-series must also converge, it must go to a finite value."}, {"video_title": "Proof of p-series convergence criteria Series AP Calculus BC Khan Academy.mp3", "Sentence": "If we know that this right over here diverges, so if this improper integral diverges, it doesn't converge to a finite value, well this, the p-series is greater than that. So if this diverges, then that's going to diverge. Similarly, if this converges, the same integral right over here, if this converges, it goes to a finite value, well one plus that is still going to converge. And so this, our p-series must also converge, it must go to a finite value. And what I'm, all I'm talking about right here, this is really just the integral test, when we think about tests of convergence and divergence, but I'm just making sure that we have a nice conceptual understanding, and not just blindly applying the integral test. So, and you could go the other way too, if the p-series converges, then for sure this integral is going to converge. And if the p-series diverges, then for sure this expression right over here is going to diverge, and the integral diverges."}, {"video_title": "Proof of p-series convergence criteria Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so this, our p-series must also converge, it must go to a finite value. And what I'm, all I'm talking about right here, this is really just the integral test, when we think about tests of convergence and divergence, but I'm just making sure that we have a nice conceptual understanding, and not just blindly applying the integral test. So, and you could go the other way too, if the p-series converges, then for sure this integral is going to converge. And if the p-series diverges, then for sure this expression right over here is going to diverge, and the integral diverges. So we can say the p-series converges if, and only if, this integral right over here converges. So, figuring out under what conditions for what p does the p-series converge, well it's boiling down to under what conditions does this integral converge. So let's scroll on down to give us some real estate to think about what has to be true for that integral to converge."}, {"video_title": "Proof of p-series convergence criteria Series AP Calculus BC Khan Academy.mp3", "Sentence": "And if the p-series diverges, then for sure this expression right over here is going to diverge, and the integral diverges. So we can say the p-series converges if, and only if, this integral right over here converges. So, figuring out under what conditions for what p does the p-series converge, well it's boiling down to under what conditions does this integral converge. So let's scroll on down to give us some real estate to think about what has to be true for that integral to converge. So I'm gonna rewrite it. So we've got the integral from one to infinity, improper integral of one over x to the p dx. This is the same thing, this is the limit as, I'll use the variable m, since we're already using n, as m approaches infinity, and the integral from one to m of one over, and actually let me just write that as x to the negative p. X to the negative p dx."}, {"video_title": "Proof of p-series convergence criteria Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's scroll on down to give us some real estate to think about what has to be true for that integral to converge. So I'm gonna rewrite it. So we've got the integral from one to infinity, improper integral of one over x to the p dx. This is the same thing, this is the limit as, I'll use the variable m, since we're already using n, as m approaches infinity, and the integral from one to m of one over, and actually let me just write that as x to the negative p. X to the negative p dx. And let me just focus on this, and we'll just remember that we're gonna have to take the limit as m approaches infinity. I don't wanna have to keep writing that over and over again. So let's think about what this is."}, {"video_title": "Proof of p-series convergence criteria Series AP Calculus BC Khan Academy.mp3", "Sentence": "This is the same thing, this is the limit as, I'll use the variable m, since we're already using n, as m approaches infinity, and the integral from one to m of one over, and actually let me just write that as x to the negative p. X to the negative p dx. And let me just focus on this, and we'll just remember that we're gonna have to take the limit as m approaches infinity. I don't wanna have to keep writing that over and over again. So let's think about what this is. So there's a couple of conditions. We know, we already know that p is greater than zero. We know that p is greater than zero, but there's two situations right over here."}, {"video_title": "Proof of p-series convergence criteria Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's think about what this is. So there's a couple of conditions. We know, we already know that p is greater than zero. We know that p is greater than zero, but there's two situations right over here. There's one situation when p is equal to one. If p is equal to one, then this is just the integral of one over x, and so this thing is going to be the integral of ln of x, and we're gonna go from one to m, and so this would be the natural log of m minus the natural log of one. Well, e to the zero power is one, so the natural log, I'll write it out, the natural log of one, but the natural log of one is just zero."}, {"video_title": "Proof of p-series convergence criteria Series AP Calculus BC Khan Academy.mp3", "Sentence": "We know that p is greater than zero, but there's two situations right over here. There's one situation when p is equal to one. If p is equal to one, then this is just the integral of one over x, and so this thing is going to be the integral of ln of x, and we're gonna go from one to m, and so this would be the natural log of m minus the natural log of one. Well, e to the zero power is one, so the natural log, I'll write it out, the natural log of one, but the natural log of one is just zero. So in the special case, I guess we can say, when p equals one, this integral from one to m comes down to the natural log of m. Now let's think about the situation where p does not equal one. Well, there we're kind of just reversing the power rule that we learned in basic differentiation. So we'd increment that exponent."}, {"video_title": "Proof of p-series convergence criteria Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, e to the zero power is one, so the natural log, I'll write it out, the natural log of one, but the natural log of one is just zero. So in the special case, I guess we can say, when p equals one, this integral from one to m comes down to the natural log of m. Now let's think about the situation where p does not equal one. Well, there we're kind of just reversing the power rule that we learned in basic differentiation. So we'd increment that exponent. So it'd be x to the negative p plus one, and we could even write that as x to the one minus p. That's the same thing as negative p plus one, and then we would divide by that. So one minus p, and we would go, we are gonna go from one to m, and so this is going to be equal to, we could write this as m to the one minus p over one minus p minus one to the one minus p over one minus p. So now let's take the limits. So remember, this integral, we wanna take the antiderivative or the definite integral here, but then we wanna take the limit as m approaches infinity."}, {"video_title": "Proof of p-series convergence criteria Series AP Calculus BC Khan Academy.mp3", "Sentence": "So we'd increment that exponent. So it'd be x to the negative p plus one, and we could even write that as x to the one minus p. That's the same thing as negative p plus one, and then we would divide by that. So one minus p, and we would go, we are gonna go from one to m, and so this is going to be equal to, we could write this as m to the one minus p over one minus p minus one to the one minus p over one minus p. So now let's take the limits. So remember, this integral, we wanna take the antiderivative or the definite integral here, but then we wanna take the limit as m approaches infinity. So what is the limit as m approaches infinity of natural log of natural log of m? Well, if m goes unbounded to infinity, well, the natural log of that is still going to go, two is still going to go to infinity. So when p equals one, this thing doesn't converge."}, {"video_title": "Proof of p-series convergence criteria Series AP Calculus BC Khan Academy.mp3", "Sentence": "So remember, this integral, we wanna take the antiderivative or the definite integral here, but then we wanna take the limit as m approaches infinity. So what is the limit as m approaches infinity of natural log of natural log of m? Well, if m goes unbounded to infinity, well, the natural log of that is still going to go, two is still going to go to infinity. So when p equals one, this thing doesn't converge. This thing is just unbounded. So p equals one, we diverge. So we know that."}, {"video_title": "Proof of p-series convergence criteria Series AP Calculus BC Khan Academy.mp3", "Sentence": "So when p equals one, this thing doesn't converge. This thing is just unbounded. So p equals one, we diverge. So we know that. So now let's look over here. Let's think about the limit as m approaches infinity of this expression right over here. And the only part that's really affected by the limit is the part that has m. So we could even write this as, we could take this one over one minus p out of this."}, {"video_title": "Proof of p-series convergence criteria Series AP Calculus BC Khan Academy.mp3", "Sentence": "So we know that. So now let's look over here. Let's think about the limit as m approaches infinity of this expression right over here. And the only part that's really affected by the limit is the part that has m. So we could even write this as, we could take this one over one minus p out of this. We could say one over one minus p times the limit as m approaches infinity of m to the one minus p. And then separately, we can subtract one to the one minus p. Well, for any exponent, that's just gonna be one over one minus p. Is that right? Yeah, no matter what exponent I put up here, one to any power is going to be one. And so the interesting thing about whether it converges or not is this part of the expression right over here."}, {"video_title": "Proof of p-series convergence criteria Series AP Calculus BC Khan Academy.mp3", "Sentence": "And the only part that's really affected by the limit is the part that has m. So we could even write this as, we could take this one over one minus p out of this. We could say one over one minus p times the limit as m approaches infinity of m to the one minus p. And then separately, we can subtract one to the one minus p. Well, for any exponent, that's just gonna be one over one minus p. Is that right? Yeah, no matter what exponent I put up here, one to any power is going to be one. And so the interesting thing about whether it converges or not is this part of the expression right over here. And it's all going to depend on whether this exponent is positive or negative. If one minus p is greater than zero, well, if I'm going to infinity and I'm taking that thing to a positive exponent, well, then this is going to diverge. And so in this situation, we diverge."}, {"video_title": "Proof of p-series convergence criteria Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so the interesting thing about whether it converges or not is this part of the expression right over here. And it's all going to depend on whether this exponent is positive or negative. If one minus p is greater than zero, well, if I'm going to infinity and I'm taking that thing to a positive exponent, well, then this is going to diverge. And so in this situation, we diverge. And one minus p is greater than zero. We can add p to both sides. That's the situation."}, {"video_title": "Proof of p-series convergence criteria Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so in this situation, we diverge. And one minus p is greater than zero. We can add p to both sides. That's the situation. That's the same thing as one being greater than p or p being less than one. We are going to diverge. So, so far, we know that p is going to be greater than zero."}, {"video_title": "Proof of p-series convergence criteria Series AP Calculus BC Khan Academy.mp3", "Sentence": "That's the situation. That's the same thing as one being greater than p or p being less than one. We are going to diverge. So, so far, we know that p is going to be greater than zero. And so we saw if p is one or if it's less than one, we're gonna diverge. But if this exponent right over here is negative, if one minus p is less than zero, well, think about it. Then it's gonna be one over m to some positive exponent is one way to think about it."}, {"video_title": "Proof of p-series convergence criteria Series AP Calculus BC Khan Academy.mp3", "Sentence": "So, so far, we know that p is going to be greater than zero. And so we saw if p is one or if it's less than one, we're gonna diverge. But if this exponent right over here is negative, if one minus p is less than zero, well, think about it. Then it's gonna be one over m to some positive exponent is one way to think about it. So as m approaches infinity, this whole thing is going to approach zero. So this is actually going to be a situation where we converge, where we get to a finite value. And so we add p to both sides."}, {"video_title": "Proof of p-series convergence criteria Series AP Calculus BC Khan Academy.mp3", "Sentence": "Then it's gonna be one over m to some positive exponent is one way to think about it. So as m approaches infinity, this whole thing is going to approach zero. So this is actually going to be a situation where we converge, where we get to a finite value. And so we add p to both sides. We have one is less than p. We converge. So there you have it. We have established this integral is going to converge only in the situation where p is greater than one."}, {"video_title": "Proof of p-series convergence criteria Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so we add p to both sides. We have one is less than p. We converge. So there you have it. We have established this integral is going to converge only in the situation where p is greater than one. P greater than one, you are going to converge. And if zero is less than p is less than or equal to one, you are going to diverge. And those are then the exact, because this, our p series converges if and only if this integral converges."}, {"video_title": "Proof of p-series convergence criteria Series AP Calculus BC Khan Academy.mp3", "Sentence": "We have established this integral is going to converge only in the situation where p is greater than one. P greater than one, you are going to converge. And if zero is less than p is less than or equal to one, you are going to diverge. And those are then the exact, because this, our p series converges if and only if this integral converges. And so these exact same constraints apply to our original p series. Our original p series converges only in the situation where p is greater than one. Then we converge."}, {"video_title": "Formal and alternate form of the derivative for ln x Differential Calculus Khan Academy.mp3", "Sentence": "Let's say that f of x is equal to the natural log of x. And we want to figure out what the slope of the tangent line to the curve f is when x is equal to the number e. So here x is equal to the number e. The point e,1 is on the curve. f of e is 1. The natural log of e is 1. And I've drawn the slope of the tangent line, or I've drawn the tangent line, and we need to figure out what the slope of it is, or at least come up with an expression for it. And I'm going to come up with an expression using both the formal definition and the alternate definition, and that will allow us to compare them a little bit. So let's think about first the formal definition."}, {"video_title": "Formal and alternate form of the derivative for ln x Differential Calculus Khan Academy.mp3", "Sentence": "The natural log of e is 1. And I've drawn the slope of the tangent line, or I've drawn the tangent line, and we need to figure out what the slope of it is, or at least come up with an expression for it. And I'm going to come up with an expression using both the formal definition and the alternate definition, and that will allow us to compare them a little bit. So let's think about first the formal definition. So the formal definition wants us to find an expression for the derivative of our function at any x. So let's say that this is some arbitrary x right over here. This would be the point x, f of x."}, {"video_title": "Formal and alternate form of the derivative for ln x Differential Calculus Khan Academy.mp3", "Sentence": "So let's think about first the formal definition. So the formal definition wants us to find an expression for the derivative of our function at any x. So let's say that this is some arbitrary x right over here. This would be the point x, f of x. And let's say that this is, let's call this x plus h. So this distance right over here is going to be h. This right over here is going to be the point x plus h, f of x plus h. Now the whole underlying idea of the formal definition of limits is to find the slope of the secant line between these two points, and then take the limit as h approaches 0. As h gets closer and closer, this blue point is going to get closer and closer and closer to x, and this point is going to approach it on the curve, and the secant line is going to become a better and better and better approximation of the tangent line at x. So let's actually do that."}, {"video_title": "Formal and alternate form of the derivative for ln x Differential Calculus Khan Academy.mp3", "Sentence": "This would be the point x, f of x. And let's say that this is, let's call this x plus h. So this distance right over here is going to be h. This right over here is going to be the point x plus h, f of x plus h. Now the whole underlying idea of the formal definition of limits is to find the slope of the secant line between these two points, and then take the limit as h approaches 0. As h gets closer and closer, this blue point is going to get closer and closer and closer to x, and this point is going to approach it on the curve, and the secant line is going to become a better and better and better approximation of the tangent line at x. So let's actually do that. So what's the slope of the secant line? Well, it's the change in your vertical axis, which is going to be f of x plus h minus f of x over the change in your horizontal axis. And that's x plus h minus x, and we see here the difference is just h over h. And we're going to take the limit of that as h approaches 0."}, {"video_title": "Formal and alternate form of the derivative for ln x Differential Calculus Khan Academy.mp3", "Sentence": "So let's actually do that. So what's the slope of the secant line? Well, it's the change in your vertical axis, which is going to be f of x plus h minus f of x over the change in your horizontal axis. And that's x plus h minus x, and we see here the difference is just h over h. And we're going to take the limit of that as h approaches 0. So in the case when f of x is the natural log of x, this will reduce to the limit as h approaches 0, f of x plus h is the natural log of x plus h minus the natural log of x, all of that over h. So this right over here, for our particular f of x, this is equal to f prime of x. So if we wanted to evaluate this, when x is equal to e, then everywhere we see an x, we just have to replace it with an e. This is essentially expressing our derivative as a function of x. It's kind of a crazy-looking function of x."}, {"video_title": "Formal and alternate form of the derivative for ln x Differential Calculus Khan Academy.mp3", "Sentence": "And that's x plus h minus x, and we see here the difference is just h over h. And we're going to take the limit of that as h approaches 0. So in the case when f of x is the natural log of x, this will reduce to the limit as h approaches 0, f of x plus h is the natural log of x plus h minus the natural log of x, all of that over h. So this right over here, for our particular f of x, this is equal to f prime of x. So if we wanted to evaluate this, when x is equal to e, then everywhere we see an x, we just have to replace it with an e. This is essentially expressing our derivative as a function of x. It's kind of a crazy-looking function of x. You have a limit here and all of that, but every place you see an x, like any function definition, you can replace it now with an e. So we can, let me just do that. Whoops, I lost my screen. Here we go."}, {"video_title": "Formal and alternate form of the derivative for ln x Differential Calculus Khan Academy.mp3", "Sentence": "It's kind of a crazy-looking function of x. You have a limit here and all of that, but every place you see an x, like any function definition, you can replace it now with an e. So we can, let me just do that. Whoops, I lost my screen. Here we go. So we could write f prime of e is equal to the limit as h approaches 0 of natural log, I'll do the same color so we can keep track of things, natural log of e plus h, I'll just leave that blank for now, minus the natural log of e, all of that over h. So just like that, this right over here, if we evaluate this limit, if we're able to, and we actually can, if we're able to evaluate this limit, this would give us the slope of the tangent line when x equals e. This is doing the formal definition. Now let's do the alternate definition. The alternate definition, if you don't want to find a general derivative expressed as a function of x like this, and you just want to find the slope at a particular point, the alternate definition kind of just gets straight to the point there."}, {"video_title": "Formal and alternate form of the derivative for ln x Differential Calculus Khan Academy.mp3", "Sentence": "Here we go. So we could write f prime of e is equal to the limit as h approaches 0 of natural log, I'll do the same color so we can keep track of things, natural log of e plus h, I'll just leave that blank for now, minus the natural log of e, all of that over h. So just like that, this right over here, if we evaluate this limit, if we're able to, and we actually can, if we're able to evaluate this limit, this would give us the slope of the tangent line when x equals e. This is doing the formal definition. Now let's do the alternate definition. The alternate definition, if you don't want to find a general derivative expressed as a function of x like this, and you just want to find the slope at a particular point, the alternate definition kind of just gets straight to the point there. So what they say is, hey, look, let's imagine some other x value here. So let's imagine some other x value. This right over here is the point x, well, we could say f of x or we could even say the natural log of x."}, {"video_title": "Formal and alternate form of the derivative for ln x Differential Calculus Khan Academy.mp3", "Sentence": "The alternate definition, if you don't want to find a general derivative expressed as a function of x like this, and you just want to find the slope at a particular point, the alternate definition kind of just gets straight to the point there. So what they say is, hey, look, let's imagine some other x value here. So let's imagine some other x value. This right over here is the point x, well, we could say f of x or we could even say the natural log of x. What is the slope of the secant line between those two points? Well, it's going to be your change in y values, so it's going to be natural log of x minus 1 over your change in x values. That's x minus e. So that's the slope of the secant line between those two points."}, {"video_title": "Formal and alternate form of the derivative for ln x Differential Calculus Khan Academy.mp3", "Sentence": "This right over here is the point x, well, we could say f of x or we could even say the natural log of x. What is the slope of the secant line between those two points? Well, it's going to be your change in y values, so it's going to be natural log of x minus 1 over your change in x values. That's x minus e. So that's the slope of the secant line between those two points. Well, what if you want to get the tangent line? Well, let's just take the limit as x approaches e. As x gets closer and closer and closer, these points are going to get closer and closer and closer, and the secant line is going to better approximate the tangent line. So we're just going to take the limit as x approaches e. So either one of this, this is using the formal definition of a limit."}, {"video_title": "Analyzing concavity (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "So I have the function g here, it's expressed as a fourth degree polynomial, and I want to think about the intervals over which g is either concave upwards or concave downwards. Now let's just remind ourselves what these things look like. So concave, concave upwards, is an interval, an interval where you're concave upwards, is an interval over which the slope is increasing, and it tends to look like an upward opening u like that, and you can see here that the slope over here is negative, and then as x increases it becomes less negative, it actually approaches zero, it becomes zero, then it crosses zero, it becomes slightly positive, more positive, even more positive, so you can see the slope is constantly increasing. And if you think about it in terms of derivatives, it means that your first derivative is increasing over that interval, and in order for your first derivative to be increasing over that interval, your second derivative, f prime prime of x, actually let me write it as g, because we're using g in this example, in order for your first derivative to be increasing, your, let me write this, so g, so concave upward means that your first derivative increasing, increasing, which means, which means that your second derivative is greater than zero. And concave downward is the opposite. Concave downward, downward, is an interval, or you're going to be concave downward over an interval when your slope is decreasing. So g prime of x is decreasing, or we can say that our second derivative, our second derivative is less than zero."}, {"video_title": "Analyzing concavity (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "And if you think about it in terms of derivatives, it means that your first derivative is increasing over that interval, and in order for your first derivative to be increasing over that interval, your second derivative, f prime prime of x, actually let me write it as g, because we're using g in this example, in order for your first derivative to be increasing, your, let me write this, so g, so concave upward means that your first derivative increasing, increasing, which means, which means that your second derivative is greater than zero. And concave downward is the opposite. Concave downward, downward, is an interval, or you're going to be concave downward over an interval when your slope is decreasing. So g prime of x is decreasing, or we can say that our second derivative, our second derivative is less than zero. It's less than zero. And once again, I could draw it on this. So when x is lower, we have a, look, we have a, or it looks like we have a positive slope, then it becomes less positive, and then it becomes less positive, it's approaching zero, it becomes zero, then it becomes negative, and then even more negative, and then even more negative."}, {"video_title": "Analyzing concavity (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "So g prime of x is decreasing, or we can say that our second derivative, our second derivative is less than zero. It's less than zero. And once again, I could draw it on this. So when x is lower, we have a, look, we have a, or it looks like we have a positive slope, then it becomes less positive, and then it becomes less positive, it's approaching zero, it becomes zero, then it becomes negative, and then even more negative, and then even more negative. So as you see, our slope is constantly decreasing as x increases here. So in order to think about the intervals where g is either concave upward or concave downward, what we need to do is let's find the second derivative of g, and then let's think about the points at which the second, over the second, where the second derivative can go from being, from going, from being positive to negative or negative to positive, and those will be places where it's either undefined or where the second derivative is equal to zero. And then let's see what's happening in the interval between, and then we'll know."}, {"video_title": "Analyzing concavity (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "So when x is lower, we have a, look, we have a, or it looks like we have a positive slope, then it becomes less positive, and then it becomes less positive, it's approaching zero, it becomes zero, then it becomes negative, and then even more negative, and then even more negative. So as you see, our slope is constantly decreasing as x increases here. So in order to think about the intervals where g is either concave upward or concave downward, what we need to do is let's find the second derivative of g, and then let's think about the points at which the second, over the second, where the second derivative can go from being, from going, from being positive to negative or negative to positive, and those will be places where it's either undefined or where the second derivative is equal to zero. And then let's see what's happening in the interval between, and then we'll know. What, over what intervals are we concave upward or concave downward? So let's do that. So let's first, let's take the first derivative, g prime of x, it's gonna apply the power rule a lot, four times negative one is negative four, x to the third power, plus, okay, so you're gonna have two times six is plus 12, x to the first power, you just write it as x, and then minus two, I could say minus two, x to the zeroth power, but that's just minus two, and then the derivative of negative three of a constant is just zero."}, {"video_title": "Analyzing concavity (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "And then let's see what's happening in the interval between, and then we'll know. What, over what intervals are we concave upward or concave downward? So let's do that. So let's first, let's take the first derivative, g prime of x, it's gonna apply the power rule a lot, four times negative one is negative four, x to the third power, plus, okay, so you're gonna have two times six is plus 12, x to the first power, you just write it as x, and then minus two, I could say minus two, x to the zeroth power, but that's just minus two, and then the derivative of negative three of a constant is just zero. And now I can take the second derivative, g prime prime of x is going to be equal to three times negative four is negative 12x squared, decrement the exponent, plus 12. And so let's see, where could this be undefined? Well, the second derivative is just a quadratic expression here which would be defined for any x, so it's not gonna be undefined anywhere."}, {"video_title": "Analyzing concavity (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "So let's first, let's take the first derivative, g prime of x, it's gonna apply the power rule a lot, four times negative one is negative four, x to the third power, plus, okay, so you're gonna have two times six is plus 12, x to the first power, you just write it as x, and then minus two, I could say minus two, x to the zeroth power, but that's just minus two, and then the derivative of negative three of a constant is just zero. And now I can take the second derivative, g prime prime of x is going to be equal to three times negative four is negative 12x squared, decrement the exponent, plus 12. And so let's see, where could this be undefined? Well, the second derivative is just a quadratic expression here which would be defined for any x, so it's not gonna be undefined anywhere. So interesting points where we could transition from going from a negative to a positive or a positive to a negative second derivative is where this thing could be equal to zero. So let's figure that out. So let's figure out where negative 12x plus 12 could be equal to zero."}, {"video_title": "Analyzing concavity (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the second derivative is just a quadratic expression here which would be defined for any x, so it's not gonna be undefined anywhere. So interesting points where we could transition from going from a negative to a positive or a positive to a negative second derivative is where this thing could be equal to zero. So let's figure that out. So let's figure out where negative 12x plus 12 could be equal to zero. See, we could subtract 12 from both sides, and we get negative 12x squared is equal to negative 12. Divide both sides by negative 12, you get x squared is equal to one, or x could be equal to the plus or minus, or x could be equal to the plus or minus square root of one, which is, of course, just one. So the second derivative at plus or minus one is equal to zero."}, {"video_title": "Analyzing concavity (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "So let's figure out where negative 12x plus 12 could be equal to zero. See, we could subtract 12 from both sides, and we get negative 12x squared is equal to negative 12. Divide both sides by negative 12, you get x squared is equal to one, or x could be equal to the plus or minus, or x could be equal to the plus or minus square root of one, which is, of course, just one. So the second derivative at plus or minus one is equal to zero. So either between plus or minus one or on either side of them, we are going to be, we could be concave upward or concave downward. So let's think about this. And to think about this, I'm gonna make a number line."}, {"video_title": "Analyzing concavity (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "So the second derivative at plus or minus one is equal to zero. So either between plus or minus one or on either side of them, we are going to be, we could be concave upward or concave downward. So let's think about this. And to think about this, I'm gonna make a number line. Let me find a nice, soothing color here. All right, that's a nice, soothing color. And let's say, I should make the number line a little bit bigger."}, {"video_title": "Analyzing concavity (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "And to think about this, I'm gonna make a number line. Let me find a nice, soothing color here. All right, that's a nice, soothing color. And let's say, I should make the number line a little bit bigger. So there we go. Let's utilize the screen space. And so if this is zero, this is negative one, this is negative two, this is positive one, this is positive two, we know that at x equals negative one and x equals one, our second derivative is equal to zero."}, {"video_title": "Analyzing concavity (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "And let's say, I should make the number line a little bit bigger. So there we go. Let's utilize the screen space. And so if this is zero, this is negative one, this is negative two, this is positive one, this is positive two, we know that at x equals negative one and x equals one, our second derivative is equal to zero. So let's think about what's happening in between those places to see if our second derivative is positive or negative. And from that, we'll be able to say where it's concave upward or concave downward. So on this first interval right over here, so this is the interval from, this is the interval where we're going from negative infinity to negative one."}, {"video_title": "Analyzing concavity (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "And so if this is zero, this is negative one, this is negative two, this is positive one, this is positive two, we know that at x equals negative one and x equals one, our second derivative is equal to zero. So let's think about what's happening in between those places to see if our second derivative is positive or negative. And from that, we'll be able to say where it's concave upward or concave downward. So on this first interval right over here, so this is the interval from, this is the interval where we're going from negative infinity to negative one. Well, let's just try a value in that interval to see whether our second derivative is positive or negative. And let's see, an easy value there could be negative two. It's in that interval."}, {"video_title": "Analyzing concavity (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "So on this first interval right over here, so this is the interval from, this is the interval where we're going from negative infinity to negative one. Well, let's just try a value in that interval to see whether our second derivative is positive or negative. And let's see, an easy value there could be negative two. It's in that interval. So let's take g prime prime of negative two, which is equal to negative 12 times four, because negative two squared is positive four. So it's negative 48 plus 12, so it's equal to negative 36. The important thing to realize then is well, if over here it's negative, then over this whole interval, because it's not crossing through zero or it's not discontinuous at any of these points, that's why we picked this interval, that over this whole interval, g prime prime of x is less than zero, which means that over this interval we are concave downwards."}, {"video_title": "Analyzing concavity (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "It's in that interval. So let's take g prime prime of negative two, which is equal to negative 12 times four, because negative two squared is positive four. So it's negative 48 plus 12, so it's equal to negative 36. The important thing to realize then is well, if over here it's negative, then over this whole interval, because it's not crossing through zero or it's not discontinuous at any of these points, that's why we picked this interval, that over this whole interval, g prime prime of x is less than zero, which means that over this interval we are concave downwards. So concave, concave downward, concave downward. Now let's go to the interval between negative one and one. So this is the open interval between negative one and one."}, {"video_title": "Analyzing concavity (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "The important thing to realize then is well, if over here it's negative, then over this whole interval, because it's not crossing through zero or it's not discontinuous at any of these points, that's why we picked this interval, that over this whole interval, g prime prime of x is less than zero, which means that over this interval we are concave downwards. So concave, concave downward, concave downward. Now let's go to the interval between negative one and one. So this is the open interval between negative one and one. And let's try a value there. Let's just try zero will be easy to compute. G prime prime of zero."}, {"video_title": "Analyzing concavity (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "So this is the open interval between negative one and one. And let's try a value there. Let's just try zero will be easy to compute. G prime prime of zero. Well, when x is zero, this is zero, so it's just going to be equal to 12. The important thing to realize is our second derivative here is greater than zero, so we are concave upward, concave upward on this interval between negative one and one. And then finally, let's look at the interval where x is greater than one."}, {"video_title": "Analyzing concavity (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "G prime prime of zero. Well, when x is zero, this is zero, so it's just going to be equal to 12. The important thing to realize is our second derivative here is greater than zero, so we are concave upward, concave upward on this interval between negative one and one. And then finally, let's look at the interval where x is greater than one. So this is the interval from one to infinity, if we want to view it that way. And let's just try the value. Let's try g prime prime of two, because that's in the interval."}, {"video_title": "Analyzing concavity (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "And then finally, let's look at the interval where x is greater than one. So this is the interval from one to infinity, if we want to view it that way. And let's just try the value. Let's try g prime prime of two, because that's in the interval. And g prime prime of two is going to be the same thing as g prime prime of negative two, because whether you have a negative two or a positive two, you square, it becomes four. So you're going to have four times negative 12, which is negative 48, plus 12, which is negative 36. Which is negative 36."}, {"video_title": "Analyzing concavity (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "Let's try g prime prime of two, because that's in the interval. And g prime prime of two is going to be the same thing as g prime prime of negative two, because whether you have a negative two or a positive two, you square, it becomes four. So you're going to have four times negative 12, which is negative 48, plus 12, which is negative 36. Which is negative 36. And so once again on this interval you are concave, concave downward. Now let's, I graphed this ahead of time. Let's see if what we just established is actually consistent with what the graph actually looks like."}, {"video_title": "Analyzing concavity (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "Which is negative 36. And so once again on this interval you are concave, concave downward. Now let's, I graphed this ahead of time. Let's see if what we just established is actually consistent with what the graph actually looks like. We were able to come up with these insights about the concavity without graphing it. But now it's kind of satisfying to take a look at a graph. And actually let me see if I can match up the intervals."}, {"video_title": "Analyzing concavity (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "Let's see if what we just established is actually consistent with what the graph actually looks like. We were able to come up with these insights about the concavity without graphing it. But now it's kind of satisfying to take a look at a graph. And actually let me see if I can match up the intervals. So actually this is pretty closely matched right over here. And so this is, actually let me make it a little bit smaller, all right. And so let me move my bounding box."}, {"video_title": "Analyzing concavity (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "And actually let me see if I can match up the intervals. So actually this is pretty closely matched right over here. And so this is, actually let me make it a little bit smaller, all right. And so let me move my bounding box. So I'm saying that I'm concave downward between negative infinity, negative infinity, all the way until, all the way until negative one. All the way until this point right over here. So all the way until that point."}, {"video_title": "Analyzing concavity (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "And so let me move my bounding box. So I'm saying that I'm concave downward between negative infinity, negative infinity, all the way until, all the way until negative one. All the way until this point right over here. So all the way until that point. And that looks right. It looks like the slope is constantly decreasing all the way until we get to x equals negative one. And then the slope starts increasing."}, {"video_title": "Analyzing concavity (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "So all the way until that point. And that looks right. It looks like the slope is constantly decreasing all the way until we get to x equals negative one. And then the slope starts increasing. The slope starts increasing from there, from there all the way, and right at x we're transitioning. So I'm gonna leave a little, I won't color in that. And so here our slope is increasing."}, {"video_title": "Analyzing concavity (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "And then the slope starts increasing. The slope starts increasing from there, from there all the way, and right at x we're transitioning. So I'm gonna leave a little, I won't color in that. And so here our slope is increasing. Do that same color. Our slope is increasing, increasing, increasing, increasing, increasing all the way until we get to x equals one. And then our slope starts decreasing again."}, {"video_title": "Analyzing concavity (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "And so here our slope is increasing. Do that same color. Our slope is increasing, increasing, increasing, increasing, increasing all the way until we get to x equals one. And then our slope starts decreasing again. And we get back into concave downwards. Whoops, I wanna do that in that orange color. We get back into concave downwards."}, {"video_title": "Negative powers differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And what I want to do in this video is I want to find, I want to figure out what g prime of x is. And then I also want to evaluate that at x equals two. So I want to figure that out. And I also want to figure out what does that evaluate to when x is equal to two. So what is the slope of the tangent line to the graph of g when x is equal to two? And like always, pause this video and see if you can work this out on your own before I work through it. With you."}, {"video_title": "Negative powers differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And I also want to figure out what does that evaluate to when x is equal to two. So what is the slope of the tangent line to the graph of g when x is equal to two? And like always, pause this video and see if you can work this out on your own before I work through it. With you. And I'll give you some hints. All you really need to do is apply the power rule, a little bit of basic exponent properties, and some basic derivative properties to be able to do this. All right, now let's just do this together."}, {"video_title": "Negative powers differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "With you. And I'll give you some hints. All you really need to do is apply the power rule, a little bit of basic exponent properties, and some basic derivative properties to be able to do this. All right, now let's just do this together. And I'll just rewrite it. G of x is equal to this first term here, two over x to the third. Well, that could be rewritten as two times x to the negative three."}, {"video_title": "Negative powers differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "All right, now let's just do this together. And I'll just rewrite it. G of x is equal to this first term here, two over x to the third. Well, that could be rewritten as two times x to the negative three. We know that one over x to the n is the same thing as x to the negative n. So I just rewrote it, and now this might be ringing a bell of how the power rule might be useful. And then we have minus, well, one over x squared, that is the same thing as x to the negative two. And so this, if we're gonna take the derivative of both sides of this, let's do that."}, {"video_title": "Negative powers differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, that could be rewritten as two times x to the negative three. We know that one over x to the n is the same thing as x to the negative n. So I just rewrote it, and now this might be ringing a bell of how the power rule might be useful. And then we have minus, well, one over x squared, that is the same thing as x to the negative two. And so this, if we're gonna take the derivative of both sides of this, let's do that. Derivative with respect to x, dx. We're gonna do that on the left-hand side. We're also gonna do it on the right-hand side."}, {"video_title": "Negative powers differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And so this, if we're gonna take the derivative of both sides of this, let's do that. Derivative with respect to x, dx. We're gonna do that on the left-hand side. We're also gonna do it on the right-hand side. Well, on the left-hand side, the derivative with respect to x of g of x, we can write that as g prime of x is going to be equal to, well, the derivative of this first, what we have right here written in green, this is going to be, we're just going to apply the power rule. We're going to take our exponent, multiply it by our coefficient out front. Actually, let me write that out."}, {"video_title": "Negative powers differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "We're also gonna do it on the right-hand side. Well, on the left-hand side, the derivative with respect to x of g of x, we can write that as g prime of x is going to be equal to, well, the derivative of this first, what we have right here written in green, this is going to be, we're just going to apply the power rule. We're going to take our exponent, multiply it by our coefficient out front. Actually, let me write that out. So that's going to be, let me finish this equal sign. That is going to be two times negative three, times x, and now we're going to decrement this exponent. You have to be very careful here because sometimes your brain might say, okay, one less than three is two, so maybe this is x to the negative two."}, {"video_title": "Negative powers differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Actually, let me write that out. So that's going to be, let me finish this equal sign. That is going to be two times negative three, times x, and now we're going to decrement this exponent. You have to be very careful here because sometimes your brain might say, okay, one less than three is two, so maybe this is x to the negative two. But remember, you're going down. So if you're at negative three, and you subtract one, we're going to go to the negative three minus one power. Well, that's going to take us to negative four."}, {"video_title": "Negative powers differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "You have to be very careful here because sometimes your brain might say, okay, one less than three is two, so maybe this is x to the negative two. But remember, you're going down. So if you're at negative three, and you subtract one, we're going to go to the negative three minus one power. Well, that's going to take us to negative four. So this is x to the negative four power. So two times negative three x to the negative four, or we could have also written that as negative six x to the negative four power, and then minus, well, we're going to do the same thing again right over here. We take this negative two, multiply it times the coefficient that's implicitly here."}, {"video_title": "Negative powers differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, that's going to take us to negative four. So this is x to the negative four power. So two times negative three x to the negative four, or we could have also written that as negative six x to the negative four power, and then minus, well, we're going to do the same thing again right over here. We take this negative two, multiply it times the coefficient that's implicitly here. You could say there's a one there, so negative two times one. So you have the negative two there, and then you have the x to the, well, what's negative two minus one? Well, that's negative three to the negative three power."}, {"video_title": "Negative powers differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "We take this negative two, multiply it times the coefficient that's implicitly here. You could say there's a one there, so negative two times one. So you have the negative two there, and then you have the x to the, well, what's negative two minus one? Well, that's negative three to the negative three power. And so we can rewrite all of this business as the derivative g prime of x is equal to negative six, negative six x to the negative fourth, and now we're subtracting a negative, so we could just write this as plus two x to the negative three. This negative cancels out with that negative. Subtract a negative, the same thing as adding the positive."}, {"video_title": "Negative powers differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, that's negative three to the negative three power. And so we can rewrite all of this business as the derivative g prime of x is equal to negative six, negative six x to the negative fourth, and now we're subtracting a negative, so we could just write this as plus two x to the negative three. This negative cancels out with that negative. Subtract a negative, the same thing as adding the positive. So we did the first part. We can express g prime of x as a function of x. Now let's just evaluate what g prime of two is."}, {"video_title": "Negative powers differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Subtract a negative, the same thing as adding the positive. So we did the first part. We can express g prime of x as a function of x. Now let's just evaluate what g prime of two is. So g prime of two is going to be equal to negative six times two to the negative fourth power plus two times two to the negative third power. Well, what's this going to be? This is equal to negative six over two to the fourth plus two over two to the third, which is equal to negative six over, two to the fourth is 16, plus two over, two to the third is eight."}, {"video_title": "Negative powers differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's just evaluate what g prime of two is. So g prime of two is going to be equal to negative six times two to the negative fourth power plus two times two to the negative third power. Well, what's this going to be? This is equal to negative six over two to the fourth plus two over two to the third, which is equal to negative six over, two to the fourth is 16, plus two over, two to the third is eight. And so let's see, this is, we could rewrite this. Let's write all of this with a common denominator. I could write this as one fourth, but then this one won't work out as cleanly."}, {"video_title": "Negative powers differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "This is equal to negative six over two to the fourth plus two over two to the third, which is equal to negative six over, two to the fourth is 16, plus two over, two to the third is eight. And so let's see, this is, we could rewrite this. Let's write all of this with a common denominator. I could write this as one fourth, but then this one won't work out as cleanly. I could write them both as eights. This is negative three eighths, negative three eighths. So you have negative three eighths plus two eighths is equal to negative one eighth."}, {"video_title": "Worked example Merging definite integrals over adjacent intervals AP Calculus AB Khan Academy.mp3", "Sentence": "And what we're going to do in this video is do some examples of evaluating definite integrals using this information and some knowledge of definite integral properties. So let's start with an example. Let's say we want to evaluate the definite integral going from negative four to negative two of f of x, d of x, plus the definite integral going from negative two to zero of f of x, dx. Pause this video and see if you can evaluate this entire expression. So this first part of our expression, the definite integral from negative four to negative two of f of x, dx, we're going from x equals negative four to x equals negative two. And so this would evaluate as this area between our curve and our x-axis, but it would be the negative of that area because our curve is below the x-axis. And we could try to estimate it based on the information they've given us, but they haven't given us exactly that value."}, {"video_title": "Worked example Merging definite integrals over adjacent intervals AP Calculus AB Khan Academy.mp3", "Sentence": "Pause this video and see if you can evaluate this entire expression. So this first part of our expression, the definite integral from negative four to negative two of f of x, dx, we're going from x equals negative four to x equals negative two. And so this would evaluate as this area between our curve and our x-axis, but it would be the negative of that area because our curve is below the x-axis. And we could try to estimate it based on the information they've given us, but they haven't given us exactly that value. But we also need to figure out this right over here. And here we're going from x equals negative two to zero of f of x, d of x, so that's gonna be this area. So if you're looking at the sum of these two definite integrals, and notice the upper bound here is the lower bound here, you're really thinking about this is really going to be the same thing as this is equal to the definite integral going from x equals negative four all the way to x equals zero of f of x, dx."}, {"video_title": "Worked example Merging definite integrals over adjacent intervals AP Calculus AB Khan Academy.mp3", "Sentence": "And we could try to estimate it based on the information they've given us, but they haven't given us exactly that value. But we also need to figure out this right over here. And here we're going from x equals negative two to zero of f of x, d of x, so that's gonna be this area. So if you're looking at the sum of these two definite integrals, and notice the upper bound here is the lower bound here, you're really thinking about this is really going to be the same thing as this is equal to the definite integral going from x equals negative four all the way to x equals zero of f of x, dx. And this is indeed one of our integration properties. If our upper bound here is the same as our lower bound here and we are integrating the same thing, well, then you can merge these two definite integrals in this way. And this is just going to be this entire area, but because we are below the x-axis and above our curve here it would be the negative of that area."}, {"video_title": "Worked example Merging definite integrals over adjacent intervals AP Calculus AB Khan Academy.mp3", "Sentence": "So if you're looking at the sum of these two definite integrals, and notice the upper bound here is the lower bound here, you're really thinking about this is really going to be the same thing as this is equal to the definite integral going from x equals negative four all the way to x equals zero of f of x, dx. And this is indeed one of our integration properties. If our upper bound here is the same as our lower bound here and we are integrating the same thing, well, then you can merge these two definite integrals in this way. And this is just going to be this entire area, but because we are below the x-axis and above our curve here it would be the negative of that area. So this is going to be equal to negative seven. Let's do another example. Let's say someone were to ask you, walk up to you on the street and say, quick, here's a graph, what is the value of the expression that I'm about to write down?"}, {"video_title": "Worked example Merging definite integrals over adjacent intervals AP Calculus AB Khan Academy.mp3", "Sentence": "And this is just going to be this entire area, but because we are below the x-axis and above our curve here it would be the negative of that area. So this is going to be equal to negative seven. Let's do another example. Let's say someone were to ask you, walk up to you on the street and say, quick, here's a graph, what is the value of the expression that I'm about to write down? The definite integral going from zero to four of f of x, dx, plus the definite integral going from four to six of f of x, dx. Pause this video and see if you can figure that out. Well, once again, this first part right over here going from zero to four, so what would be is would be this area, it would be this five right over here, but then we would need to subtract this area because this area is below our x-axis and above our curve."}, {"video_title": "Worked example Merging definite integrals over adjacent intervals AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say someone were to ask you, walk up to you on the street and say, quick, here's a graph, what is the value of the expression that I'm about to write down? The definite integral going from zero to four of f of x, dx, plus the definite integral going from four to six of f of x, dx. Pause this video and see if you can figure that out. Well, once again, this first part right over here going from zero to four, so what would be is would be this area, it would be this five right over here, but then we would need to subtract this area because this area is below our x-axis and above our curve. We don't know exactly what this is, but luckily we also need to take the sum of everything I just showed, but plus this right over here. And this we're going from four to six, so it's going to be this area. Well, once again, when you look at it this way, you can see that this expression is going to be equivalent to taking the definite integral all the way from zero to six, zero to six of f of x, dx."}, {"video_title": "Worked example Merging definite integrals over adjacent intervals AP Calculus AB Khan Academy.mp3", "Sentence": "Well, once again, this first part right over here going from zero to four, so what would be is would be this area, it would be this five right over here, but then we would need to subtract this area because this area is below our x-axis and above our curve. We don't know exactly what this is, but luckily we also need to take the sum of everything I just showed, but plus this right over here. And this we're going from four to six, so it's going to be this area. Well, once again, when you look at it this way, you can see that this expression is going to be equivalent to taking the definite integral all the way from zero to six, zero to six of f of x, dx. And once again, even if you didn't see the graph, you would know that because in both cases you're getting the definite integral of f of x, dx, and our upper bound here is our lower bound here. So once again, we're able to merge the integrals. And what is this going to be equal to?"}, {"video_title": "Worked example Merging definite integrals over adjacent intervals AP Calculus AB Khan Academy.mp3", "Sentence": "Well, once again, when you look at it this way, you can see that this expression is going to be equivalent to taking the definite integral all the way from zero to six, zero to six of f of x, dx. And once again, even if you didn't see the graph, you would know that because in both cases you're getting the definite integral of f of x, dx, and our upper bound here is our lower bound here. So once again, we're able to merge the integrals. And what is this going to be equal to? Well, we have this area here, which is five, and then we have this area, which is six. That was given to us. But since it's below the x-axis and above our curve, when we evaluate it as a definite integral, it would evaluate as a negative six."}, {"video_title": "Applied rate of change forgetfulness Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "In 10 days, I will have forgotten every word. The number of words that I remember t days after studying is modeled by, so w of t, so this is the number of words I have in my head as a function of time is going to be equal to 80 times the one minus 0.1t squared, for t is between zero and 10, including the two boundaries. That's why we have brackets right over here. What is the rate of change of the number of words, of the number of known words per day two days after studying for the test? And I encourage you to pause this video and try it on your own. So the key here is we come up with this equation for modeling how many words have retained in my brain every day after I first memorized them, after I got the 80 of them into my head. And that's this expression here."}, {"video_title": "Applied rate of change forgetfulness Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "What is the rate of change of the number of words, of the number of known words per day two days after studying for the test? And I encourage you to pause this video and try it on your own. So the key here is we come up with this equation for modeling how many words have retained in my brain every day after I first memorized them, after I got the 80 of them into my head. And that's this expression here. And they wanna know the rate of change two days after studying. Well, the rate of change, I can take the derivative of this with respect to time. So let's do that."}, {"video_title": "Applied rate of change forgetfulness Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And that's this expression here. And they wanna know the rate of change two days after studying. Well, the rate of change, I can take the derivative of this with respect to time. So let's do that. So let's take the derivative, the derivative of the number of words I know with respect to time is going to be equal to, well, we have this 80 out front. That's just a constant. And now I can apply the chain rule right over here."}, {"video_title": "Applied rate of change forgetfulness Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let's do that. So let's take the derivative, the derivative of the number of words I know with respect to time is going to be equal to, well, we have this 80 out front. That's just a constant. And now I can apply the chain rule right over here. So the derivative of one minus 0.1t whole thing squared with respect to one minus 0.1t is going to be, so I'm essentially taking the derivative of this whole pink thing, with this whole expression squared with respect to the expression. So that's going to be two times one minus 0.1t. And now I can find the derivative of this inner expression with respect to t. So derivative of this inner expression with respect to t is just going to be zero minus 0.1."}, {"video_title": "Applied rate of change forgetfulness Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And now I can apply the chain rule right over here. So the derivative of one minus 0.1t whole thing squared with respect to one minus 0.1t is going to be, so I'm essentially taking the derivative of this whole pink thing, with this whole expression squared with respect to the expression. So that's going to be two times one minus 0.1t. And now I can find the derivative of this inner expression with respect to t. So derivative of this inner expression with respect to t is just going to be zero minus 0.1. So it's just going to be negative 0.1. So it's gonna be negative 0.1. And of course, we can simplify this a little bit."}, {"video_title": "Applied rate of change forgetfulness Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And now I can find the derivative of this inner expression with respect to t. So derivative of this inner expression with respect to t is just going to be zero minus 0.1. So it's just going to be negative 0.1. So it's gonna be negative 0.1. And of course, we can simplify this a little bit. This is going to be equal to, if we take 80 times two is 160, times negative 0.1. That's gonna be negative 16. 160 times 0.1 is 16."}, {"video_title": "Applied rate of change forgetfulness Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And of course, we can simplify this a little bit. This is going to be equal to, if we take 80 times two is 160, times negative 0.1. That's gonna be negative 16. 160 times 0.1 is 16. So negative 16 times one minus 0.1t. And if we want, we could distribute the 16 or we could just leave it like this. But we're ready now to answer our questions."}, {"video_title": "Applied rate of change forgetfulness Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "160 times 0.1 is 16. So negative 16 times one minus 0.1t. And if we want, we could distribute the 16 or we could just leave it like this. But we're ready now to answer our questions. We could write this as the rate of change of the number of words we know with respect to time. Or we could use the alternate notation. We could say this is w prime of t. Either way, it's going to be equal to this thing."}, {"video_title": "Applied rate of change forgetfulness Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "But we're ready now to answer our questions. We could write this as the rate of change of the number of words we know with respect to time. Or we could use the alternate notation. We could say this is w prime of t. Either way, it's going to be equal to this thing. It's negative, let me do that same color. It's equal to negative 16 times one minus 0.1t. So what's this going to be?"}, {"video_title": "Applied rate of change forgetfulness Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We could say this is w prime of t. Either way, it's going to be equal to this thing. It's negative, let me do that same color. It's equal to negative 16 times one minus 0.1t. So what's this going to be? What is the rate of change of the number of words known per day two days after studying for the test? Well, we just have to evaluate this when t is equal to two. So w prime of two, do that in magenta."}, {"video_title": "Applied rate of change forgetfulness Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So what's this going to be? What is the rate of change of the number of words known per day two days after studying for the test? Well, we just have to evaluate this when t is equal to two. So w prime of two, do that in magenta. W prime of two is going to be equal to, whoops, I'll just go with the magenta, is equal to negative 16 times one minus 0.1 times two. Times two, close parentheses. And that's going to be equal to, well let's see, what is this?"}, {"video_title": "Applied rate of change forgetfulness Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So w prime of two, do that in magenta. W prime of two is going to be equal to, whoops, I'll just go with the magenta, is equal to negative 16 times one minus 0.1 times two. Times two, close parentheses. And that's going to be equal to, well let's see, what is this? This is one minus, essentially 0.2, this is going to be 0.8. So this is going to be equal to negative 16 times, is that right, one minus 0.2 is, yep, it's going to be times 0.8. And what is that going to be?"}, {"video_title": "Applied rate of change forgetfulness Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And that's going to be equal to, well let's see, what is this? This is one minus, essentially 0.2, this is going to be 0.8. So this is going to be equal to negative 16 times, is that right, one minus 0.2 is, yep, it's going to be times 0.8. And what is that going to be? If I were to multiply 16 times eight, it would be 128. It's two times eight times eight. So two times 64, 128."}, {"video_title": "Applied rate of change forgetfulness Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And what is that going to be? If I were to multiply 16 times eight, it would be 128. It's two times eight times eight. So two times 64, 128. So this is going to be negative 12.8. And just to really hit the point home of what we're doing, this is, we're going to, it's negative, the number of words, the rate of change is negative 12.8 words per day. So if you believe this model for how many words we know on a given day, this is saying on day two, or right at the day two point, right after two, exactly two days after studying for the test, I'm going to be, right at that moment, I am essentially losing 12.8 words per day."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "And what I have pictures here are some of the most known, or actually this gentleman right over here might be the most known person when people think about population and the limits to growth of population. This is Thomas Malthus. He was a British cleric and writer and scholar at the end of the 1700s, the end of the 18th century, early 19th century. And he really challenged the notion that population could grow indefinitely and that we would always, through technology, be able to feed ourselves. Really, that the environment would eventually put some caps on how much or where the population could grow to. And P.F. Verhulst, and I'm sure I'm mispronouncing his name here, he was a Belgian mathematician who read Malthus' work and tried to model the behavior that Malthus was talking about."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "And he really challenged the notion that population could grow indefinitely and that we would always, through technology, be able to feed ourselves. Really, that the environment would eventually put some caps on how much or where the population could grow to. And P.F. Verhulst, and I'm sure I'm mispronouncing his name here, he was a Belgian mathematician who read Malthus' work and tried to model the behavior that Malthus was talking about. That, okay, when there aren't environmental constraints, maybe population does grow somewhat exponentially, but then as it approaches the limits set by the environment, it's going to essentially asymptote towards some type of population. Malthus, in particular, he actually doesn't think it's just going to be a nice, clean asymptote. He actually thinks that the population would go above the limit and you would have these catastrophes, and then you would go crashing below the limit, and you would kind of oscillate right around the limit through these catastrophes."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "Verhulst, and I'm sure I'm mispronouncing his name here, he was a Belgian mathematician who read Malthus' work and tried to model the behavior that Malthus was talking about. That, okay, when there aren't environmental constraints, maybe population does grow somewhat exponentially, but then as it approaches the limits set by the environment, it's going to essentially asymptote towards some type of population. Malthus, in particular, he actually doesn't think it's just going to be a nice, clean asymptote. He actually thinks that the population would go above the limit and you would have these catastrophes, and then you would go crashing below the limit, and you would kind of oscillate right around the limit through these catastrophes. As you can tell, Malthus was a fairly optimistic guy. Let's go through a little bit of the math and a little bit of the differential equations. Also, these aren't overly hairy differential equations to think about population."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "He actually thinks that the population would go above the limit and you would have these catastrophes, and then you would go crashing below the limit, and you would kind of oscillate right around the limit through these catastrophes. As you can tell, Malthus was a fairly optimistic guy. Let's go through a little bit of the math and a little bit of the differential equations. Also, these aren't overly hairy differential equations to think about population. The first way to think about population, and I'll express it as a differential equation. Actually, let me just set some variables here. Let's say that n is our population."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "Also, these aren't overly hairy differential equations to think about population. The first way to think about population, and I'll express it as a differential equation. Actually, let me just set some variables here. Let's say that n is our population. That's our population. We are going to assume that n is a function of t. n as a function of t is what we're going to be thinking about in this and, frankly, the next series of videos. One way to think about how to model this is just, well, what is the rate of change of population with respect to time?"}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "Let's say that n is our population. That's our population. We are going to assume that n is a function of t. n as a function of t is what we're going to be thinking about in this and, frankly, the next series of videos. One way to think about how to model this is just, well, what is the rate of change of population with respect to time? How does that relate to things? We could say, okay, well, what is the rate of change of population with respect to time? D capital N dt."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "One way to think about how to model this is just, well, what is the rate of change of population with respect to time? How does that relate to things? We could say, okay, well, what is the rate of change of population with respect to time? D capital N dt. One way to think about it is it's going to be proportional to the population. You could say, well, maybe this is going to be some proportionality constant times the population times the population itself. This makes sense."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "D capital N dt. One way to think about it is it's going to be proportional to the population. You could say, well, maybe this is going to be some proportionality constant times the population times the population itself. This makes sense. If the population is smaller, then you're not going to have as much change per unit time as if the population is larger. The larger the population, the more it's going to grow in a particular unit of time. This is actually a fairly simple to solve differential equation."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "This makes sense. If the population is smaller, then you're not going to have as much change per unit time as if the population is larger. The larger the population, the more it's going to grow in a particular unit of time. This is actually a fairly simple to solve differential equation. You might have done it before. I encourage you to pause this video if you feel inspired to do so. I'll solve it right here."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "This is actually a fairly simple to solve differential equation. You might have done it before. I encourage you to pause this video if you feel inspired to do so. I'll solve it right here. You'll see that we get an exponential function here for n. Let's do that. Let's solve this and we'll essentially separate the variables, separate the n from the t's, although we only see a dt here, but I'll do that in a second. If I divide both sides by n, I get 1 over n. If I multiply both sides by dt, if you think about the dt as something that you can multiply, I'm going to divide both sides by n and multiply both sides by dt."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "I'll solve it right here. You'll see that we get an exponential function here for n. Let's do that. Let's solve this and we'll essentially separate the variables, separate the n from the t's, although we only see a dt here, but I'll do that in a second. If I divide both sides by n, I get 1 over n. If I multiply both sides by dt, if you think about the dt as something that you can multiply, I'm going to divide both sides by n and multiply both sides by dt. I'm going to get 1 over n dn on the left-hand side. On the right-hand side, I'm going to get r times dt. Notice I got the dt onto the right-hand side by multiplying both sides of that."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "If I divide both sides by n, I get 1 over n. If I multiply both sides by dt, if you think about the dt as something that you can multiply, I'm going to divide both sides by n and multiply both sides by dt. I'm going to get 1 over n dn on the left-hand side. On the right-hand side, I'm going to get r times dt. Notice I got the dt onto the right-hand side by multiplying both sides of that. Then I divided both sides by n and I got the 1 over n right over here. Now what we can do is we can take the antiderivative of both sides. What do we get on the left-hand side?"}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "Notice I got the dt onto the right-hand side by multiplying both sides of that. Then I divided both sides by n and I got the 1 over n right over here. Now what we can do is we can take the antiderivative of both sides. What do we get on the left-hand side? This is just going to be the natural log of the absolute value of our population. Actually, if we assume that the population is always going to be nonzero, then we can actually take these absolute value off, but I'll do that in a second. That's going to be equal to r times t. We could have added a constant here, but I'm just going to do it on one side."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "What do we get on the left-hand side? This is just going to be the natural log of the absolute value of our population. Actually, if we assume that the population is always going to be nonzero, then we can actually take these absolute value off, but I'll do that in a second. That's going to be equal to r times t. We could have added a constant here, but I'm just going to do it on one side. r times t plus c. Now if we actually want to solve for n, we could take, if this is equal to this, then e to this power should be the same as e to this power. Another way of thinking about it, e, the natural log of the absolute value of n is equal to this, is another way of saying that e to this is going to be equal to that. Actually, let me just do it this way."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "That's going to be equal to r times t. We could have added a constant here, but I'm just going to do it on one side. r times t plus c. Now if we actually want to solve for n, we could take, if this is equal to this, then e to this power should be the same as e to this power. Another way of thinking about it, e, the natural log of the absolute value of n is equal to this, is another way of saying that e to this is going to be equal to that. Actually, let me just do it this way. Let me just take e to this power and to that power. If that's equal to that, then e to that power should be the same as e to that power. We're going to be left with e to the natural log of the absolute value of n. That's going to give you the absolute value of n. Let's just assume n positive."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "Actually, let me just do it this way. Let me just take e to this power and to that power. If that's equal to that, then e to that power should be the same as e to that power. We're going to be left with e to the natural log of the absolute value of n. That's going to give you the absolute value of n. Let's just assume n positive. Let's just assume population is greater than zero. Then we could, this left-hand side right over here, we'll just simplify to n. Then our right-hand side, it's going to be, well, it could be e to the rt plus c, or this is the same thing. This thing right over here is the same thing as e to the rt, e to the r times t times, actually, let me do that e in that same color, e to the rt times e to the c. I'm just taking e to the sum of these two exponents."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "We're going to be left with e to the natural log of the absolute value of n. That's going to give you the absolute value of n. Let's just assume n positive. Let's just assume population is greater than zero. Then we could, this left-hand side right over here, we'll just simplify to n. Then our right-hand side, it's going to be, well, it could be e to the rt plus c, or this is the same thing. This thing right over here is the same thing as e to the rt, e to the r times t times, actually, let me do that e in that same color, e to the rt times e to the c. I'm just taking e to the sum of these two exponents. That's going to be e to the rt times e to the c. If we want to, we could just say, hey, you know what? This is still just going to be some arbitrary constant here. Actually, let's just call this c. It's e to the rt times c, or we could say c times e to the rt."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "This thing right over here is the same thing as e to the rt, e to the r times t times, actually, let me do that e in that same color, e to the rt times e to the c. I'm just taking e to the sum of these two exponents. That's going to be e to the rt times e to the c. If we want to, we could just say, hey, you know what? This is still just going to be some arbitrary constant here. Actually, let's just call this c. It's e to the rt times c, or we could say c times e to the rt. C times e to the rt. Notice, we have solved the differential equation. We haven't gotten to this less than optimistic reality of Malthus where we're limiting it."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "Actually, let's just call this c. It's e to the rt times c, or we could say c times e to the rt. C times e to the rt. Notice, we have solved the differential equation. We haven't gotten to this less than optimistic reality of Malthus where we're limiting it. This is just, hey, if we just assume population is going to, the rate of change of population with respect to time is going to be proportional to population, when we solve that differential equation, we get that population is a function of time. Actually, let me make it explicit that this is a function of time. Let me just move the n over a little bit."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "We haven't gotten to this less than optimistic reality of Malthus where we're limiting it. This is just, hey, if we just assume population is going to, the rate of change of population with respect to time is going to be proportional to population, when we solve that differential equation, we get that population is a function of time. Actually, let me make it explicit that this is a function of time. Let me just move the n over a little bit. Let me write it this way. N of t is going to be equal to this. This was our solution to this differential equation."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "Let me just move the n over a little bit. Let me write it this way. N of t is going to be equal to this. This was our solution to this differential equation. Once again, this is just going to grow forever. If we know the initial conditions, let's say that we knew that n of zero, when time is equal to zero, let's just say that's n sub naught. What would c be?"}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "This was our solution to this differential equation. Once again, this is just going to grow forever. If we know the initial conditions, let's say that we knew that n of zero, when time is equal to zero, let's just say that's n sub naught. What would c be? Well, n of zero is going to be equal to c, c times e to the zero power. E to the zero power is just one, so it's just going to be equal to c. C is equal to n sub naught. Now we can even write it that the solution to this thing right over here is n as a function of t, is going to be equal to c times, be careful, n naught, our initial population, times e to the rt."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "What would c be? Well, n of zero is going to be equal to c, c times e to the zero power. E to the zero power is just one, so it's just going to be equal to c. C is equal to n sub naught. Now we can even write it that the solution to this thing right over here is n as a function of t, is going to be equal to c times, be careful, n naught, our initial population, times e to the rt. Now once again, this is an exponential. Essentially our population is going to look like this. If I were to graph it, it's going to look, if that's my time axis, if that's my n axis right over here, I could say it's y equals n axis, however I want to denote it."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "Now we can even write it that the solution to this thing right over here is n as a function of t, is going to be equal to c times, be careful, n naught, our initial population, times e to the rt. Now once again, this is an exponential. Essentially our population is going to look like this. If I were to graph it, it's going to look, if that's my time axis, if that's my n axis right over here, I could say it's y equals n axis, however I want to denote it. That would be n naught, and it's going to grow exponentially from there. The rate of this exponential function is going to be dictated by this constant right over there, but it's going to look something like this, and it's just going to grow faster and faster and faster forever and ever and ever and ever. Now as I mentioned at the beginning of the video, Malthus does not believe that this is going to be true."}, {"video_title": "Modeling population as an exponential function First order differential equations Khan Academy.mp3", "Sentence": "If I were to graph it, it's going to look, if that's my time axis, if that's my n axis right over here, I could say it's y equals n axis, however I want to denote it. That would be n naught, and it's going to grow exponentially from there. The rate of this exponential function is going to be dictated by this constant right over there, but it's going to look something like this, and it's just going to grow faster and faster and faster forever and ever and ever and ever. Now as I mentioned at the beginning of the video, Malthus does not believe that this is going to be true. He thinks that we're going to hit some natural limits that are going to start to constrain the population. In his mind, a more natural or a more realistic function to model population would look something like this, or even potentially something that you keep crashing around that kind of limit. What we'll see in the next video is that P.F."}, {"video_title": "Another derivation of the sum of an infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "And we'll use a very similar idea to what we used to find the sum of a finite geometric series. So let's say I have a geometric series, an infinite geometric series. So we're going to start at k equals 0. And we're never going to stop. It's going all the way to infinity. So we're never going to stop adding terms here. And it's going to be our first term times our common ratio to the kth power."}, {"video_title": "Another derivation of the sum of an infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "And we're never going to stop. It's going all the way to infinity. So we're never going to stop adding terms here. And it's going to be our first term times our common ratio to the kth power. Actually, let me do k in that color. k equals 0 all the way to infinity. And so let's just call this thing right over here."}, {"video_title": "Another derivation of the sum of an infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "And it's going to be our first term times our common ratio to the kth power. Actually, let me do k in that color. k equals 0 all the way to infinity. And so let's just call this thing right over here. Let's call this s sub infinity. We're going all the way to infinity right over here. And so this, if we were to expand it out, is going to be equal to a times r to the 0."}, {"video_title": "Another derivation of the sum of an infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "And so let's just call this thing right over here. Let's call this s sub infinity. We're going all the way to infinity right over here. And so this, if we were to expand it out, is going to be equal to a times r to the 0. Actually, let me just write it out like that, which is just a. a times r to the 0 power plus a times r to the first power plus a times r to the second power plus. And we could just keep going. Let me do one more."}, {"video_title": "Another derivation of the sum of an infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "And so this, if we were to expand it out, is going to be equal to a times r to the 0. Actually, let me just write it out like that, which is just a. a times r to the 0 power plus a times r to the first power plus a times r to the second power plus. And we could just keep going. Let me do one more. Plus a. Actually, we could just keep going on and on and on. I think you get the general idea."}, {"video_title": "Another derivation of the sum of an infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "Let me do one more. Plus a. Actually, we could just keep going on and on and on. I think you get the general idea. Now, just like when we tried to derive a formula for the sum of a finite geometric series, we just said, well, what happens if you take this sum? And if you were to multiply every term by your common ratio, every term by r. So let's do that. Let's imagine this sum."}, {"video_title": "Another derivation of the sum of an infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "I think you get the general idea. Now, just like when we tried to derive a formula for the sum of a finite geometric series, we just said, well, what happens if you take this sum? And if you were to multiply every term by your common ratio, every term by r. So let's do that. Let's imagine this sum. And we're going to multiply every term by r. And the reason why I said this is proofy is this is not always clear. When you're multiplying something times infinite terms or an infinite sum, at least this will give you the general idea. When you start thinking about infinity, sometimes you have to think about things a little bit deeper."}, {"video_title": "Another derivation of the sum of an infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "Let's imagine this sum. And we're going to multiply every term by r. And the reason why I said this is proofy is this is not always clear. When you're multiplying something times infinite terms or an infinite sum, at least this will give you the general idea. When you start thinking about infinity, sometimes you have to think about things a little bit deeper. So r times this infinite sum, well, that's going to be equal to. We're just going to multiply every term here times r. So ar to the 0th power times r is going to be a times r to the first power. Multiply this one times r, you're going to get a times r to the second power."}, {"video_title": "Another derivation of the sum of an infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "When you start thinking about infinity, sometimes you have to think about things a little bit deeper. So r times this infinite sum, well, that's going to be equal to. We're just going to multiply every term here times r. So ar to the 0th power times r is going to be a times r to the first power. Multiply this one times r, you're going to get a times r to the second power. I think you see where this is going. Multiply this one times r, you're going to get plus a times r to the third power. And we would just keep on going."}, {"video_title": "Another derivation of the sum of an infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "Multiply this one times r, you're going to get a times r to the second power. I think you see where this is going. Multiply this one times r, you're going to get plus a times r to the third power. And we would just keep on going. So let me just show that. So plus dot, dot, dot. Now what happens if we were to subtract this sum from this top sum?"}, {"video_title": "Another derivation of the sum of an infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "And we would just keep on going. So let me just show that. So plus dot, dot, dot. Now what happens if we were to subtract this sum from this top sum? So on the left-hand side, we could express that as our sum. We could express that as our sum, s sub infinity, minus our common ratio times s sub infinity. So when you subtract, you're going to have a times r to the 0th power, which is really just the same thing as a."}, {"video_title": "Another derivation of the sum of an infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "Now what happens if we were to subtract this sum from this top sum? So on the left-hand side, we could express that as our sum. We could express that as our sum, s sub infinity, minus our common ratio times s sub infinity. So when you subtract, you're going to have a times r to the 0th power, which is really just the same thing as a. That's just going to be a. a times r to the 0th is just a times 1, which is a. Let me write that same color. It's equal to a."}, {"video_title": "Another derivation of the sum of an infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "So when you subtract, you're going to have a times r to the 0th power, which is really just the same thing as a. That's just going to be a. a times r to the 0th is just a times 1, which is a. Let me write that same color. It's equal to a. But every other term, you're going to have a times r to the first. So then you're going to subtract a times r to the first. You're going to have a times r to the second, but you're going to subtract a times r to the second."}, {"video_title": "Another derivation of the sum of an infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "It's equal to a. But every other term, you're going to have a times r to the first. So then you're going to subtract a times r to the first. You're going to have a times r to the second, but you're going to subtract a times r to the second. So every other term is going to be subtracted away. And this happens all the way to infinity. It never, never ends."}, {"video_title": "Another derivation of the sum of an infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "You're going to have a times r to the second, but you're going to subtract a times r to the second. So every other term is going to be subtracted away. And this happens all the way to infinity. It never, never ends. So the only term that you're left with is just that first one, is just a. And so now we can actually try to solve for our sum. If you factor out the s sub infinity, you are left with 1 minus r. Our sum times 1 minus r is equal to a. Divide both sides by 1 minus r, and we get that our sum, the thing that we cared about."}, {"video_title": "Another derivation of the sum of an infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "It never, never ends. So the only term that you're left with is just that first one, is just a. And so now we can actually try to solve for our sum. If you factor out the s sub infinity, you are left with 1 minus r. Our sum times 1 minus r is equal to a. Divide both sides by 1 minus r, and we get that our sum, the thing that we cared about. And once again, this is kind of an amazing result, that we're taking the sum of an infinite number of terms. And under the proper constraints, we are going to get a finite value. So this is going to be equal to a over 1 minus r. So once again, it's kind of neat."}, {"video_title": "Another derivation of the sum of an infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "If you factor out the s sub infinity, you are left with 1 minus r. Our sum times 1 minus r is equal to a. Divide both sides by 1 minus r, and we get that our sum, the thing that we cared about. And once again, this is kind of an amazing result, that we're taking the sum of an infinite number of terms. And under the proper constraints, we are going to get a finite value. So this is going to be equal to a over 1 minus r. So once again, it's kind of neat. If I was to say I had the sum, let's say we started with 5. And then each time we were to multiply by 3 fifths. So 5 plus 3 fifths times 5 is 3, times 3 fifths is going to be 9 fifths."}, {"video_title": "Another derivation of the sum of an infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "So this is going to be equal to a over 1 minus r. So once again, it's kind of neat. If I was to say I had the sum, let's say we started with 5. And then each time we were to multiply by 3 fifths. So 5 plus 3 fifths times 5 is 3, times 3 fifths is going to be 9 fifths. Or I multiply by 3 fifths again, then I'm, sorry, not 9 fifths. So 5, my brain isn't working right. 5 times 3 fifths is going to be 3, times 3 fifths is going to be 3 times this."}, {"video_title": "Another derivation of the sum of an infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "So 5 plus 3 fifths times 5 is 3, times 3 fifths is going to be 9 fifths. Or I multiply by 3 fifths again, then I'm, sorry, not 9 fifths. So 5, my brain isn't working right. 5 times 3 fifths is going to be 3, times 3 fifths is going to be 3 times this. It's going to be 9 fifths. Actually, no, that was right. My brain is working right."}, {"video_title": "Another derivation of the sum of an infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "5 times 3 fifths is going to be 3, times 3 fifths is going to be 3 times this. It's going to be 9 fifths. Actually, no, that was right. My brain is working right. Times 3 fifths is going to be 27 over 25, times 3 fifths is going to be 81 over 125. And we keep on going on and on and on forever. And notice, these terms are starting to get smaller and smaller and smaller."}, {"video_title": "Another derivation of the sum of an infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "My brain is working right. Times 3 fifths is going to be 27 over 25, times 3 fifths is going to be 81 over 125. And we keep on going on and on and on forever. And notice, these terms are starting to get smaller and smaller and smaller. Actually, all of them are getting smaller and smaller and smaller. We're multiplying by 3 fifths every time. We now know what the sum is going to be."}, {"video_title": "Another derivation of the sum of an infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "And notice, these terms are starting to get smaller and smaller and smaller. Actually, all of them are getting smaller and smaller and smaller. We're multiplying by 3 fifths every time. We now know what the sum is going to be. It's going to be our first term. It's going to be 5 over 1 minus our common ratio. And our common ratio in this case is 3 fifths."}, {"video_title": "Another derivation of the sum of an infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "We now know what the sum is going to be. It's going to be our first term. It's going to be 5 over 1 minus our common ratio. And our common ratio in this case is 3 fifths. So this is going to be equal to 5 over 2 fifths, which is the same thing as 5, times 5 over 2, which is 25 over 2, which is equal to 12 and 1 half, or 12.5. Once again, amazing result. I'm taking a sum of infinite terms here, and I was able to get a finite result."}, {"video_title": "Another derivation of the sum of an infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "And our common ratio in this case is 3 fifths. So this is going to be equal to 5 over 2 fifths, which is the same thing as 5, times 5 over 2, which is 25 over 2, which is equal to 12 and 1 half, or 12.5. Once again, amazing result. I'm taking a sum of infinite terms here, and I was able to get a finite result. And once again, when does this happen? Well, if the absolute value of our common ratio is less than 1, then these terms are going to get smaller and smaller and smaller. And you'll even see here, it even works out mathematically in this denominator, that you're going to get a reasonable answer."}, {"video_title": "Another derivation of the sum of an infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "I'm taking a sum of infinite terms here, and I was able to get a finite result. And once again, when does this happen? Well, if the absolute value of our common ratio is less than 1, then these terms are going to get smaller and smaller and smaller. And you'll even see here, it even works out mathematically in this denominator, that you're going to get a reasonable answer. And it makes sense, because these terms are getting smaller and smaller and smaller, that this thing will converge. Even if r is 0, we're still not really dealing, we're not anymore dealing strictly with a geometric series anymore. But obviously, if r was 0, then you're really only going to have this, well, even this first term is kind of under debate, depending on how you define what 0 to 0 is."}, {"video_title": "Another derivation of the sum of an infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "And you'll even see here, it even works out mathematically in this denominator, that you're going to get a reasonable answer. And it makes sense, because these terms are getting smaller and smaller and smaller, that this thing will converge. Even if r is 0, we're still not really dealing, we're not anymore dealing strictly with a geometric series anymore. But obviously, if r was 0, then you're really only going to have this, well, even this first term is kind of under debate, depending on how you define what 0 to 0 is. But if your first term, you just said it would be a, then clearly you'd just be left with a as the sum. And a over 1 minus 0 is still a. So this formula that we just derived does hold up for that."}, {"video_title": "Another derivation of the sum of an infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "But obviously, if r was 0, then you're really only going to have this, well, even this first term is kind of under debate, depending on how you define what 0 to 0 is. But if your first term, you just said it would be a, then clearly you'd just be left with a as the sum. And a over 1 minus 0 is still a. So this formula that we just derived does hold up for that. It does start to break down if r is equal to 1 or negative 1. If r is equal to 1, then as you imagine here, you just have a plus a plus a plus a going on and on forever. If r is equal to negative 1, you just keep oscillating."}, {"video_title": "Another derivation of the sum of an infinite geometric series Precalculus Khan Academy.mp3", "Sentence": "So this formula that we just derived does hold up for that. It does start to break down if r is equal to 1 or negative 1. If r is equal to 1, then as you imagine here, you just have a plus a plus a plus a going on and on forever. If r is equal to negative 1, you just keep oscillating. a minus a plus a minus a. And so the sum's value keeps oscillating between two values. So in general, this infinite geometric series is going to converge if the absolute value of your common ratio is less than 1, or another way of saying that, if your common ratio is between 1 and negative 1."}, {"video_title": "Proof of the derivative of sin(x) Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "But what we're gonna do in this video is dig a little bit deeper and actually prove this first derivative. I'm not gonna prove the second one. You can actually use it using the information we're going to do in this one. But it's just to make you feel good that someone's just not making this up, that there is a little bit of mathematical rigor behind it all. So let's try to calculate it. So the derivative with respect to x of sine of x. By definition, this is going to be the limit as delta x approaches zero of sine of x plus delta x minus sine of x, all of that over delta, all of that over delta x."}, {"video_title": "Proof of the derivative of sin(x) Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "But it's just to make you feel good that someone's just not making this up, that there is a little bit of mathematical rigor behind it all. So let's try to calculate it. So the derivative with respect to x of sine of x. By definition, this is going to be the limit as delta x approaches zero of sine of x plus delta x minus sine of x, all of that over delta, all of that over delta x. This is really just the slope of the line between the point x comma sine of x and x plus delta x comma sine of x plus delta x. So how can we evaluate this? Well, we can rewrite sine of x plus delta x using our angle addition formulas that we learned during our trig identities."}, {"video_title": "Proof of the derivative of sin(x) Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "By definition, this is going to be the limit as delta x approaches zero of sine of x plus delta x minus sine of x, all of that over delta, all of that over delta x. This is really just the slope of the line between the point x comma sine of x and x plus delta x comma sine of x plus delta x. So how can we evaluate this? Well, we can rewrite sine of x plus delta x using our angle addition formulas that we learned during our trig identities. So this is going to be the same thing as the limit as delta x approaches zero. I'll rewrite this using our trig identity as cosine of x times sine of delta x plus sine of x times cosine of delta x. And then we're going to subtract this sine of x up here minus sine of x, all of that over, let me see if I can draw a relatively straight line, all of that over delta x."}, {"video_title": "Proof of the derivative of sin(x) Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we can rewrite sine of x plus delta x using our angle addition formulas that we learned during our trig identities. So this is going to be the same thing as the limit as delta x approaches zero. I'll rewrite this using our trig identity as cosine of x times sine of delta x plus sine of x times cosine of delta x. And then we're going to subtract this sine of x up here minus sine of x, all of that over, let me see if I can draw a relatively straight line, all of that over delta x. So this can be rewritten as being equal to the limit as delta x approaches zero of, let me write this part in red, so that would be cosine of x sine of delta x, all of that over delta x, and then that's going to be plus, I'll do all of this in orange, all I'm doing is I have the sum of things up here divided by delta x, I'm just breaking it up a little bit, plus sine of x cosine of delta x minus sine of x, all of that over delta x, and remember I'm taking the limit of this entire expression. Well the limit of a sum is equal to the sum of the limits, so this is going to be equal to, I'll do this first part in red, the limit as delta x approaches zero of, let's see I can rewrite this as cosine of x times sine of delta x over delta x plus the limit as delta x approaches zero of, and let's see I can factor out a sine of x here, so it's times sine of x, I factor that out and I'll be left with a cosine of delta x minus one, all of that over delta x, so that's this limit, and let's see if I can simplify this even more. Let me scroll down a bit."}, {"video_title": "Proof of the derivative of sin(x) Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And then we're going to subtract this sine of x up here minus sine of x, all of that over, let me see if I can draw a relatively straight line, all of that over delta x. So this can be rewritten as being equal to the limit as delta x approaches zero of, let me write this part in red, so that would be cosine of x sine of delta x, all of that over delta x, and then that's going to be plus, I'll do all of this in orange, all I'm doing is I have the sum of things up here divided by delta x, I'm just breaking it up a little bit, plus sine of x cosine of delta x minus sine of x, all of that over delta x, and remember I'm taking the limit of this entire expression. Well the limit of a sum is equal to the sum of the limits, so this is going to be equal to, I'll do this first part in red, the limit as delta x approaches zero of, let's see I can rewrite this as cosine of x times sine of delta x over delta x plus the limit as delta x approaches zero of, and let's see I can factor out a sine of x here, so it's times sine of x, I factor that out and I'll be left with a cosine of delta x minus one, all of that over delta x, so that's this limit, and let's see if I can simplify this even more. Let me scroll down a bit. So this left hand expression I can rewrite, this cosine of x has nothing to do with the limit as delta x approaches zero, so we can actually take that outside of the limit. So we have the cosine of x times the limit as delta x approaches zero of sine of delta x over delta x, and now we need to add this thing, and let's see how I could write this. So I have a sine of x here, actually let me rewrite this a little bit differently, cosine of delta x minus one, that's the same thing as one minus cosine of delta x times negative one, and so you have a sine of x times a negative one, and since the delta x has nothing to do with the sine of x, let me take that out, the negative and the sine of x, so we have minus sine of x times the limit as delta x approaches zero of, what we have left over is one minus cosine of delta x over delta x."}, {"video_title": "Worked example estimating sin(0.4) using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "What is the least degree of the polynomial that assures an error smaller than 0.001? So what are we talking about here? Well, we could take some function and estimate it with an nth degree Maclaurin polynomial. In fact, we could talk more generally about a Taylor polynomial, but let's just say this is an nth degree Maclaurin polynomial, but this isn't going to be a perfect approximation. There's going to be some error, some remainder, and so we could call this the remainder of that nth degree Maclaurin polynomial, and it's going to be dependent for any given x. Now, if we want to use the specifics of this exact problem, we could phrase it this way. We want to say, look, if we're taking the sine of 0.4, this is going to be equal to our Maclaurin, our nth degree Maclaurin polynomial evaluated at 0.4 plus whatever the remainder is for that nth degree Maclaurin polynomial evaluated at 0.4, and what we really want to do is figure out for what n, what is the least degree of the polynomial?"}, {"video_title": "Worked example estimating sin(0.4) using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "In fact, we could talk more generally about a Taylor polynomial, but let's just say this is an nth degree Maclaurin polynomial, but this isn't going to be a perfect approximation. There's going to be some error, some remainder, and so we could call this the remainder of that nth degree Maclaurin polynomial, and it's going to be dependent for any given x. Now, if we want to use the specifics of this exact problem, we could phrase it this way. We want to say, look, if we're taking the sine of 0.4, this is going to be equal to our Maclaurin, our nth degree Maclaurin polynomial evaluated at 0.4 plus whatever the remainder is for that nth degree Maclaurin polynomial evaluated at 0.4, and what we really want to do is figure out for what n, what is the least degree of the polynomial? So what is, let me do this in a different color. So we want to figure out what is the smallest n, what is the smallest n what is the smallest n such that the remainder of our nth degree Maclaurin polynomial evaluated at 0.4 is less than this number, is less than 0.001. So this is just another way of rephrasing the problem, and the way that we can do it is we can use something called the Lagrange error bound, and I have other videos that prove it, this is often also called Taylor's remainder theorem, and I'll first write it out, and I'll try to explain it while I write it out, but it'll actually become a lot more concrete when we work it out."}, {"video_title": "Worked example estimating sin(0.4) using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "We want to say, look, if we're taking the sine of 0.4, this is going to be equal to our Maclaurin, our nth degree Maclaurin polynomial evaluated at 0.4 plus whatever the remainder is for that nth degree Maclaurin polynomial evaluated at 0.4, and what we really want to do is figure out for what n, what is the least degree of the polynomial? So what is, let me do this in a different color. So we want to figure out what is the smallest n, what is the smallest n what is the smallest n such that the remainder of our nth degree Maclaurin polynomial evaluated at 0.4 is less than this number, is less than 0.001. So this is just another way of rephrasing the problem, and the way that we can do it is we can use something called the Lagrange error bound, and I have other videos that prove it, this is often also called Taylor's remainder theorem, and I'll first write it out, and I'll try to explain it while I write it out, but it'll actually become a lot more concrete when we work it out. So Taylor's remainder theorem tells us, or Lagrange error bound, tells us that if the n plus 1th derivative of our function, so f, so this is our n plus 1th derivative of our function, the absolute value of that is less than or equal to some m for an open interval, open interval containing where our polynomial is centered, in this case it's zero, we're gonna use the Maclaurin case, so it's containing zero, and the x that, or zero and x, the x that we care about in this particular video, is 0.4, but I'll say it in general for any x. So if this is true, if our n plus 1th derivative of our function, if the absolute value of it is less than or equal to m, over an open interval containing where we're centered, this would be c if we're talking about the general case, and x, so this x right over here, then, then this is the Lagrange, this is the part that's useful, we can say that the remainder is bounded, the remainder for that nth degree polynomial, so this is the n plus 1th derivative, that's bounded, then we can say the remainder for the nth degree polynomial that approximates our function is going to be less than or equal to that m times x to the n plus 1, or times, yeah, times x to the n plus 1, over n plus 1 factorial. So how do we apply that to this particular problem?"}, {"video_title": "Worked example estimating sin(0.4) using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "So this is just another way of rephrasing the problem, and the way that we can do it is we can use something called the Lagrange error bound, and I have other videos that prove it, this is often also called Taylor's remainder theorem, and I'll first write it out, and I'll try to explain it while I write it out, but it'll actually become a lot more concrete when we work it out. So Taylor's remainder theorem tells us, or Lagrange error bound, tells us that if the n plus 1th derivative of our function, so f, so this is our n plus 1th derivative of our function, the absolute value of that is less than or equal to some m for an open interval, open interval containing where our polynomial is centered, in this case it's zero, we're gonna use the Maclaurin case, so it's containing zero, and the x that, or zero and x, the x that we care about in this particular video, is 0.4, but I'll say it in general for any x. So if this is true, if our n plus 1th derivative of our function, if the absolute value of it is less than or equal to m, over an open interval containing where we're centered, this would be c if we're talking about the general case, and x, so this x right over here, then, then this is the Lagrange, this is the part that's useful, we can say that the remainder is bounded, the remainder for that nth degree polynomial, so this is the n plus 1th derivative, that's bounded, then we can say the remainder for the nth degree polynomial that approximates our function is going to be less than or equal to that m times x to the n plus 1, or times, yeah, times x to the n plus 1, over n plus 1 factorial. So how do we apply that to this particular problem? Well, think about the derivatives of sine. We know that the absolute value of sine is less than or equal to one, its derivative is cosine of x, the absolute value of that is going to be bounded, is going to be less than or equal to one, so no matter how many times we take the derivative of sine of x, the absolute value of that derivative is going to be less than or equal to one. So we could write generally that for this particular f of x, so let's say for this particular f of x right over here, we could say that the absolute value of f, the absolute value of the n plus 1th derivative evaluated at any x is going to be less than or equal to one, and this is the case for where f is sine, where f is sine of x, and this is actually going to be true over any interval."}, {"video_title": "Worked example estimating sin(0.4) using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "So how do we apply that to this particular problem? Well, think about the derivatives of sine. We know that the absolute value of sine is less than or equal to one, its derivative is cosine of x, the absolute value of that is going to be bounded, is going to be less than or equal to one, so no matter how many times we take the derivative of sine of x, the absolute value of that derivative is going to be less than or equal to one. So we could write generally that for this particular f of x, so let's say for this particular f of x right over here, we could say that the absolute value of f, the absolute value of the n plus 1th derivative evaluated at any x is going to be less than or equal to one, and this is the case for where f is sine, where f is sine of x, and this is actually going to be true over any interval. It doesn't even have to be over some type of a restricted interval where we can do this. So we know that this is our m. We know that this is our m, and that sine and its derivatives are all bounded by, or their absolute values are bounded by one. And so then we have our m, and we can apply the Lagrange error bound."}, {"video_title": "Worked example estimating sin(0.4) using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "So we could write generally that for this particular f of x, so let's say for this particular f of x right over here, we could say that the absolute value of f, the absolute value of the n plus 1th derivative evaluated at any x is going to be less than or equal to one, and this is the case for where f is sine, where f is sine of x, and this is actually going to be true over any interval. It doesn't even have to be over some type of a restricted interval where we can do this. So we know that this is our m. We know that this is our m, and that sine and its derivatives are all bounded by, or their absolute values are bounded by one. And so then we have our m, and we can apply the Lagrange error bound. So we can say that the remainder of our nth degree Maclaurin approximation at 0.4, so our x in this particular case is 0.4. We don't have to do it generally for any x here, is going to be less than or equal to our m is one, so I won't even write that down. Our x is 0.4, 0.4 to the n plus one, to the n plus one, over n plus one factorial, and we're taking the absolute value of this whole thing."}, {"video_title": "Worked example estimating sin(0.4) using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "And so then we have our m, and we can apply the Lagrange error bound. So we can say that the remainder of our nth degree Maclaurin approximation at 0.4, so our x in this particular case is 0.4. We don't have to do it generally for any x here, is going to be less than or equal to our m is one, so I won't even write that down. Our x is 0.4, 0.4 to the n plus one, to the n plus one, over n plus one factorial, and we're taking the absolute value of this whole thing. This is Lagrange error bound. And we want to figure out, if we can figure out a situation where this is less than 0.001, then this for sure is going to be less than 0.001 because the remainder is less than this, or less than or equal to this, which is less than that. So how do we do that?"}, {"video_title": "Worked example estimating sin(0.4) using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "Our x is 0.4, 0.4 to the n plus one, to the n plus one, over n plus one factorial, and we're taking the absolute value of this whole thing. This is Lagrange error bound. And we want to figure out, if we can figure out a situation where this is less than 0.001, then this for sure is going to be less than 0.001 because the remainder is less than this, or less than or equal to this, which is less than that. So how do we do that? How do we figure out the smallest n where this is going to be true? Well, we can just try out some n's and keep increasing until this thing actually becomes smaller than that thing. So let's do it."}, {"video_title": "Worked example estimating sin(0.4) using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "So how do we do that? How do we figure out the smallest n where this is going to be true? Well, we can just try out some n's and keep increasing until this thing actually becomes smaller than that thing. So let's do it. Alright, I'm gonna set up a table here. So let me do it relatively cleanly. So let's do, this is going to be our n, and then this is going to be, this is going to be 0.4 to the n plus one over n plus one factorial."}, {"video_title": "Worked example estimating sin(0.4) using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "So let's do it. Alright, I'm gonna set up a table here. So let me do it relatively cleanly. So let's do, this is going to be our n, and then this is going to be, this is going to be 0.4 to the n plus one over n plus one factorial. So let's try it when n is equal to one. Well, then this is going to be 0.4 to the two, so it's 0.4 squared over two factorial. This is 0.16 over two, which is equal to 0.08."}, {"video_title": "Worked example estimating sin(0.4) using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "So let's do, this is going to be our n, and then this is going to be, this is going to be 0.4 to the n plus one over n plus one factorial. So let's try it when n is equal to one. Well, then this is going to be 0.4 to the two, so it's 0.4 squared over two factorial. This is 0.16 over two, which is equal to 0.08. That's definitely not less than 1,000th here. So let's try n equals two. When n equals two, it's gonna be 0.4 to the third power over three factorial, and that's equal to, what is that?"}, {"video_title": "Worked example estimating sin(0.4) using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "This is 0.16 over two, which is equal to 0.08. That's definitely not less than 1,000th here. So let's try n equals two. When n equals two, it's gonna be 0.4 to the third power over three factorial, and that's equal to, what is that? 0. I'm gonna need three digits behind the decimal. 0.064 over six."}, {"video_title": "Worked example estimating sin(0.4) using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "When n equals two, it's gonna be 0.4 to the third power over three factorial, and that's equal to, what is that? 0. I'm gonna need three digits behind the decimal. 0.064 over six. Well, this is a little bit more than 0.01, so this isn't, our n isn't large enough yet. So let's try three. So this is going to be 0.04 to the three plus one, so that's going to be to the fourth power over four factorial, and let's see, that is going to be equal to, this is going to be, let's see, we're gonna have four digits behind the decimal, so 0.0256 over 24."}, {"video_title": "Worked example estimating sin(0.4) using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "0.064 over six. Well, this is a little bit more than 0.01, so this isn't, our n isn't large enough yet. So let's try three. So this is going to be 0.04 to the three plus one, so that's going to be to the fourth power over four factorial, and let's see, that is going to be equal to, this is going to be, let's see, we're gonna have four digits behind the decimal, so 0.0256 over 24. This is, we're almost there. This is a little bit, this is going to be a little bit more than 0.001, so that doesn't do the trick for us. So I'm guessing already that n equals four is gonna do the trick, but let's verify that."}, {"video_title": "Worked example estimating sin(0.4) using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "So this is going to be 0.04 to the three plus one, so that's going to be to the fourth power over four factorial, and let's see, that is going to be equal to, this is going to be, let's see, we're gonna have four digits behind the decimal, so 0.0256 over 24. This is, we're almost there. This is a little bit, this is going to be a little bit more than 0.001, so that doesn't do the trick for us. So I'm guessing already that n equals four is gonna do the trick, but let's verify that. So this is going to be 0.04, or 0.4, I should say, to the fifth power over five factorial, and what is this equal to? Let's see, four to the fifth is 1024. I'm gonna have five numbers behind the decimal, so I'm gonna divide it by five factorial, which is 120, and let's see, this right over here, yes, this for sure is less than 0.001."}, {"video_title": "Worked example estimating sin(0.4) using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "So I'm guessing already that n equals four is gonna do the trick, but let's verify that. So this is going to be 0.04, or 0.4, I should say, to the fifth power over five factorial, and what is this equal to? Let's see, four to the fifth is 1024. I'm gonna have five numbers behind the decimal, so I'm gonna divide it by five factorial, which is 120, and let's see, this right over here, yes, this for sure is less than 0.001. This is definitely less than a thousandth right over here, so we see that when n is equal to four, so we can say that the remainder for our fourth degree polynomial, fourth degree Maclaurin polynomial, evaluated at x equals 0.4, is for sure going to be less than 0.001. So there you go. That is the least degree of the polynomial that is sure is an error smaller than one thousandth."}, {"video_title": "Common chain rule misunderstandings Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say that we are trying to take the derivative of the expression. So let's say we're taking the derivative of the expression, the natural log of sine of x. So the first key misconception or misunderstanding that many people have is when you're dealing with transcendental functions like this. And transcendental functions is just a fancy word for these functions like trigonometric functions, logarithmic functions that don't use standard algebraic operations. But when you see transcendental functions like this or compositions of them, many people confuse this with the product of functions. So at first when they look at this, they might see this as a product of functions. When they look at this, they might see this as being the same as the derivative with respect to x of natural log of x, natural log of x times sine of x."}, {"video_title": "Common chain rule misunderstandings Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And transcendental functions is just a fancy word for these functions like trigonometric functions, logarithmic functions that don't use standard algebraic operations. But when you see transcendental functions like this or compositions of them, many people confuse this with the product of functions. So at first when they look at this, they might see this as a product of functions. When they look at this, they might see this as being the same as the derivative with respect to x of natural log of x, natural log of x times sine of x. And you can see just the way that it's written, they look very similar. But this is the product of two functions. If you said natural log of x is f of x and sine of x is g of x, this is the product of sine and g of x."}, {"video_title": "Common chain rule misunderstandings Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "When they look at this, they might see this as being the same as the derivative with respect to x of natural log of x, natural log of x times sine of x. And you can see just the way that it's written, they look very similar. But this is the product of two functions. If you said natural log of x is f of x and sine of x is g of x, this is the product of sine and g of x. This is the product of f of x and g of x. And here you would use the product rule. So to actually compute this, you would use the product, the product rule."}, {"video_title": "Common chain rule misunderstandings Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "If you said natural log of x is f of x and sine of x is g of x, this is the product of sine and g of x. This is the product of f of x and g of x. And here you would use the product rule. So to actually compute this, you would use the product, the product rule. But this is a composition. Here you have f of g of x, not f of x times g of x. So here you have, that is our g of x, it equals sine of x."}, {"video_title": "Common chain rule misunderstandings Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So to actually compute this, you would use the product, the product rule. But this is a composition. Here you have f of g of x, not f of x times g of x. So here you have, that is our g of x, it equals sine of x. And then our f of g of x is the natural log of sine of x. So this is f of g of x. F of g of x, just like that. If someone asked you just what f of x was, well, that would be natural log of x."}, {"video_title": "Common chain rule misunderstandings Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So here you have, that is our g of x, it equals sine of x. And then our f of g of x is the natural log of sine of x. So this is f of g of x. F of g of x, just like that. If someone asked you just what f of x was, well, that would be natural log of x. But f of g of x is natural log of our g of x, which is natural log of sine of x. So that's the key first thing. Always make sure whether you're gonna use, especially with these transcendental functions, that hey, if this is a composition, you've gotta use the chain rule, not the product rule."}, {"video_title": "Common chain rule misunderstandings Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "If someone asked you just what f of x was, well, that would be natural log of x. But f of g of x is natural log of our g of x, which is natural log of sine of x. So that's the key first thing. Always make sure whether you're gonna use, especially with these transcendental functions, that hey, if this is a composition, you've gotta use the chain rule, not the product rule. It's not the product. Now sometimes you have a combination. You have a product of compositions, and then things get a little bit more involved."}, {"video_title": "Common chain rule misunderstandings Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Always make sure whether you're gonna use, especially with these transcendental functions, that hey, if this is a composition, you've gotta use the chain rule, not the product rule. It's not the product. Now sometimes you have a combination. You have a product of compositions, and then things get a little bit more involved. But pay close attention to make sure that you're not dealing with a composition. Now the next misconception students have is even if they recognize, okay, I've gotta use the chain rule, sometimes it doesn't go fully to completion. So let's continue using this example."}, {"video_title": "Common chain rule misunderstandings Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "You have a product of compositions, and then things get a little bit more involved. But pay close attention to make sure that you're not dealing with a composition. Now the next misconception students have is even if they recognize, okay, I've gotta use the chain rule, sometimes it doesn't go fully to completion. So let's continue using this example. The chain rule here says, look, we have to take the derivative of the outer function with respect to the inner function. So if I were to say, in this case, f of x is natural log of x. f of g of x is this expression here. So if I wanna do this first part, f prime of g of x, f prime of g of x, well, the derivative of the natural log of x is one over x."}, {"video_title": "Common chain rule misunderstandings Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So let's continue using this example. The chain rule here says, look, we have to take the derivative of the outer function with respect to the inner function. So if I were to say, in this case, f of x is natural log of x. f of g of x is this expression here. So if I wanna do this first part, f prime of g of x, f prime of g of x, well, the derivative of the natural log of x is one over x. So the derivative of natural log of x is one over x, but we don't want the derivative where the input is x. We want the derivative when the input is g of x. So instead of it being one over x, it's going to be one over g of x."}, {"video_title": "Common chain rule misunderstandings Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So if I wanna do this first part, f prime of g of x, f prime of g of x, well, the derivative of the natural log of x is one over x. So the derivative of natural log of x is one over x, but we don't want the derivative where the input is x. We want the derivative when the input is g of x. So instead of it being one over x, it's going to be one over g of x. One over g of x. And we know that g of x is equal to sine of x. That's equal to sine of x."}, {"video_title": "Common chain rule misunderstandings Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So instead of it being one over x, it's going to be one over g of x. One over g of x. And we know that g of x is equal to sine of x. That's equal to sine of x. Now one key misunderstanding that the folks at the College Board told us about is many students stop right there. They just do this first part and then they forget to multiply this second part. So here we are not done."}, {"video_title": "Common chain rule misunderstandings Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "That's equal to sine of x. Now one key misunderstanding that the folks at the College Board told us about is many students stop right there. They just do this first part and then they forget to multiply this second part. So here we are not done. We need to take this and multiply it times g prime of x. And let me write this down. G prime of x, what would that be?"}, {"video_title": "Common chain rule misunderstandings Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So here we are not done. We need to take this and multiply it times g prime of x. And let me write this down. G prime of x, what would that be? Well, the derivative of sine of x with respect to x, well, that's just going to be cosine of x. Cosine, cosine of x. Cosine of x. So in this example right over here, the derivative is going to be, let's see if I can squeeze it in over here, it's gonna be one over sine of x, which is this part, times cosine of x. So let me write it down."}, {"video_title": "Common chain rule misunderstandings Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "G prime of x, what would that be? Well, the derivative of sine of x with respect to x, well, that's just going to be cosine of x. Cosine, cosine of x. Cosine of x. So in this example right over here, the derivative is going to be, let's see if I can squeeze it in over here, it's gonna be one over sine of x, which is this part, times cosine of x. So let me write it down. It is going to be one over sine of x. Do that in that other color. One over sine of x and then times cosine of x."}, {"video_title": "Common chain rule misunderstandings Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So let me write it down. It is going to be one over sine of x. Do that in that other color. One over sine of x and then times cosine of x. So once again, just to make sure that you don't fall into one of these misconceptions, let me box this off so it's a little bit cleaner. So to just make sure that you don't fall into one of these misconceptions here, recognize the composition, that this is not the product of natural log of x and sine of x, it's natural log of sine of x. And then when you're actually applying the chain rule, derivative of the outside with respect to the inside, so the derivative of natural log of x is one over x, so that applied when the input is g of x is one over sine of x, and then multiply that times the derivative of the inner function."}, {"video_title": "Common chain rule misunderstandings Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "One over sine of x and then times cosine of x. So once again, just to make sure that you don't fall into one of these misconceptions, let me box this off so it's a little bit cleaner. So to just make sure that you don't fall into one of these misconceptions here, recognize the composition, that this is not the product of natural log of x and sine of x, it's natural log of sine of x. And then when you're actually applying the chain rule, derivative of the outside with respect to the inside, so the derivative of natural log of x is one over x, so that applied when the input is g of x is one over sine of x, and then multiply that times the derivative of the inner function. So don't forget to do this right over here. Now another misconception that students have is instead of doing what we just did, instead of applying the chain rule like this, they take the derivative of the outer function with respect to the derivative of the inner function. So for example, they would compute this, f prime of g prime of x. f prime of g prime of x."}, {"video_title": "Common chain rule misunderstandings Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And then when you're actually applying the chain rule, derivative of the outside with respect to the inside, so the derivative of natural log of x is one over x, so that applied when the input is g of x is one over sine of x, and then multiply that times the derivative of the inner function. So don't forget to do this right over here. Now another misconception that students have is instead of doing what we just did, instead of applying the chain rule like this, they take the derivative of the outer function with respect to the derivative of the inner function. So for example, they would compute this, f prime of g prime of x. f prime of g prime of x. Which in this case, f prime of x is one over x, but if the input is g prime of x, g prime of x is cosine of x. So many students end up doing this, where they take the derivative of the outside and they apply, and the input into that, they use the derivative of the inside function. This, this is not right."}, {"video_title": "Common chain rule misunderstandings Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So for example, they would compute this, f prime of g prime of x. f prime of g prime of x. Which in this case, f prime of x is one over x, but if the input is g prime of x, g prime of x is cosine of x. So many students end up doing this, where they take the derivative of the outside and they apply, and the input into that, they use the derivative of the inside function. This, this is not right. Be very careful that you're not doing that. You do the derivative of the outside function with respect to the inside function, not taking its derivative, and then multiply, don't forget to multiply, times the derivative of the inside function here. So hopefully that helps a little bit."}, {"video_title": "Common chain rule misunderstandings Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "This, this is not right. Be very careful that you're not doing that. You do the derivative of the outside function with respect to the inside function, not taking its derivative, and then multiply, don't forget to multiply, times the derivative of the inside function here. So hopefully that helps a little bit. If all of this looks completely foreign to you, I encourage you to watch the whole series of chain rule introductory videos and worked examples we have. This is just a topping on top of that to make sure that you don't fall into these misconceptions of applying the product rule when you really need to be applying the chain rule, or forgetting to do part of the chain rule, multiplying by g prime of x, or evaluating f prime of g prime of x. So hopefully that helps."}, {"video_title": "Reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "So let's take the derivative here. So this is going to be equal to the derivative of x to the n plus 1 over n plus 1. We can just use the power rule over here. So our exponent is n plus 1. We can bring it out front. So it's going to be n plus 1 times x times x to the, I want to use that same color. Colors are the hard part."}, {"video_title": "Reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "So our exponent is n plus 1. We can bring it out front. So it's going to be n plus 1 times x times x to the, I want to use that same color. Colors are the hard part. Times x to the, instead of n plus 1, we subtract 1 from the exponent. This is just the power rule. So n plus 1 minus 1 is going to be n. And then we can't forget that we have, we were dividing by this n plus 1."}, {"video_title": "Reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "Colors are the hard part. Times x to the, instead of n plus 1, we subtract 1 from the exponent. This is just the power rule. So n plus 1 minus 1 is going to be n. And then we can't forget that we have, we were dividing by this n plus 1. So we have divided by n plus 1. And then we have plus c, the derivative of a constant with respect to x. A constant does not change as x changes."}, {"video_title": "Reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "So n plus 1 minus 1 is going to be n. And then we can't forget that we have, we were dividing by this n plus 1. So we have divided by n plus 1. And then we have plus c, the derivative of a constant with respect to x. A constant does not change as x changes. So it is just going to be 0. So plus 0. And since n is not equal to negative 1, we know that this is going to be defined."}, {"video_title": "Reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "A constant does not change as x changes. So it is just going to be 0. So plus 0. And since n is not equal to negative 1, we know that this is going to be defined. This is just going to be something divided by itself, which is just going to be 1. And this whole thing simplifies to x to the n. So the derivative of this thing, and this is in very general terms, is equal to x to the n. So given that, what is the antiderivative, let me switch colors here, what is the antiderivative of x to the n, and remember this is just the kind of strange looking notation we use. It makes more sense when we start doing definite integrals."}, {"video_title": "Reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "And since n is not equal to negative 1, we know that this is going to be defined. This is just going to be something divided by itself, which is just going to be 1. And this whole thing simplifies to x to the n. So the derivative of this thing, and this is in very general terms, is equal to x to the n. So given that, what is the antiderivative, let me switch colors here, what is the antiderivative of x to the n, and remember this is just the kind of strange looking notation we use. It makes more sense when we start doing definite integrals. But what is the antiderivative of x to the n, and we could say the antiderivative with respect to x if we want to. And another way of calling this is the indefinite integral. Indefinite integral."}, {"video_title": "Reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "It makes more sense when we start doing definite integrals. But what is the antiderivative of x to the n, and we could say the antiderivative with respect to x if we want to. And another way of calling this is the indefinite integral. Indefinite integral. Well we know, this is saying x to the n is the derivative of what? Well we just figured out, it's the derivative of this thing. And we've written it in very general terms."}, {"video_title": "Reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "Indefinite integral. Well we know, this is saying x to the n is the derivative of what? Well we just figured out, it's the derivative of this thing. And we've written it in very general terms. We're actually encapsulating multiple constants here. We could have x to the n plus 1 over n plus 1 plus 0 plus 1 plus 2 plus pi plus a billion. So this is going to be equal to x to the n plus 1 over n plus 1 plus c. So this is pretty powerful."}, {"video_title": "Reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "And we've written it in very general terms. We're actually encapsulating multiple constants here. We could have x to the n plus 1 over n plus 1 plus 0 plus 1 plus 2 plus pi plus a billion. So this is going to be equal to x to the n plus 1 over n plus 1 plus c. So this is pretty powerful. You can kind of view this as the reverse power rule. And it applies for any n as long as n does not equal negative 1. Let me make that very clear."}, {"video_title": "Reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be equal to x to the n plus 1 over n plus 1 plus c. So this is pretty powerful. You can kind of view this as the reverse power rule. And it applies for any n as long as n does not equal negative 1. Let me make that very clear. n does not equal negative 1. Once again, this thing would be undefined if n were equal to negative 1. So let's do a couple of examples just to apply this."}, {"video_title": "Reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "Let me make that very clear. n does not equal negative 1. Once again, this thing would be undefined if n were equal to negative 1. So let's do a couple of examples just to apply this. You could call it the reverse power rule if you want, or the anti-power rule. So let's take the antiderivative of x to the fifth power. What is the antiderivative of x to the fifth?"}, {"video_title": "Reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "So let's do a couple of examples just to apply this. You could call it the reverse power rule if you want, or the anti-power rule. So let's take the antiderivative of x to the fifth power. What is the antiderivative of x to the fifth? Well all we have to say is, well look, the 5 is equal to the n. We just have to increment the exponent by 1. So this is going to be equal to x to the 5 plus 1 power. And then we divide by that same value."}, {"video_title": "Reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "What is the antiderivative of x to the fifth? Well all we have to say is, well look, the 5 is equal to the n. We just have to increment the exponent by 1. So this is going to be equal to x to the 5 plus 1 power. And then we divide by that same value. Whatever the exponent was, when you increment it by 1, we divide by that same value. Divided by 5 plus 1. And of course we want to encapsulate all of the possible antiderivatives."}, {"video_title": "Reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "And then we divide by that same value. Whatever the exponent was, when you increment it by 1, we divide by that same value. Divided by 5 plus 1. And of course we want to encapsulate all of the possible antiderivatives. So you put the c right over there. So this is going to be equal to x to the sixth over 6 plus c. And you can verify. Take the derivative of this using the power rule, you indeed get x to the fifth."}, {"video_title": "Reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "And of course we want to encapsulate all of the possible antiderivatives. So you put the c right over there. So this is going to be equal to x to the sixth over 6 plus c. And you can verify. Take the derivative of this using the power rule, you indeed get x to the fifth. Let's try another one. Now we'll do it in blue. Let's tie the antiderivative of, let's make it interesting, let's make it 5 times x to the negative 2 power dx."}, {"video_title": "Reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "Take the derivative of this using the power rule, you indeed get x to the fifth. Let's try another one. Now we'll do it in blue. Let's tie the antiderivative of, let's make it interesting, let's make it 5 times x to the negative 2 power dx. So how would we evaluate this? Well one simplification you can do, and I haven't rigorously proven it to you just yet, but we know that scalars can go in and out of the derivative operator when you're multiplying by a scalar. So this is indeed equal to 5 times the antiderivative of x to the negative 2 power dx."}, {"video_title": "Reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "Let's tie the antiderivative of, let's make it interesting, let's make it 5 times x to the negative 2 power dx. So how would we evaluate this? Well one simplification you can do, and I haven't rigorously proven it to you just yet, but we know that scalars can go in and out of the derivative operator when you're multiplying by a scalar. So this is indeed equal to 5 times the antiderivative of x to the negative 2 power dx. And now we can just use, I guess we could call it this anti-power rule. So this is going to be equal to 5 times x to the negative 2 power plus 1 over the negative 2 power plus 1 plus some constant right over here. And then we can rewrite this as 5 times negative 2 power plus 1 is x to the negative 1 over negative 2 plus 1 is negative 1 plus some constant."}, {"video_title": "Reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "So this is indeed equal to 5 times the antiderivative of x to the negative 2 power dx. And now we can just use, I guess we could call it this anti-power rule. So this is going to be equal to 5 times x to the negative 2 power plus 1 over the negative 2 power plus 1 plus some constant right over here. And then we can rewrite this as 5 times negative 2 power plus 1 is x to the negative 1 over negative 2 plus 1 is negative 1 plus some constant. And this is equal to 5 times negative x to the negative 1 plus some constant. And then if we want we can distribute the 5. So this is equal to negative 5 x to the negative 1."}, {"video_title": "Reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "And then we can rewrite this as 5 times negative 2 power plus 1 is x to the negative 1 over negative 2 plus 1 is negative 1 plus some constant. And this is equal to 5 times negative x to the negative 1 plus some constant. And then if we want we can distribute the 5. So this is equal to negative 5 x to the negative 1. Now we could write plus 5 times some constant but this is just an arbitrary constant. So this is still just an arbitrary constant. So maybe we could have written this, if you wanted to show this as different constants you could say this is c1 c1 c1."}, {"video_title": "Reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "So this is equal to negative 5 x to the negative 1. Now we could write plus 5 times some constant but this is just an arbitrary constant. So this is still just an arbitrary constant. So maybe we could have written this, if you wanted to show this as different constants you could say this is c1 c1 c1. You multiply 5 times c1 you get another constant. We could just call that c. Which is equal to 5 times c1. But there you have it."}, {"video_title": "Quotient rule from product & chain rules Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So plus f of x times the derivative of the second function. Times the derivative of the second function. So two terms. In each term, we take the derivative of one of the functions and not the other, and then we switch. So over here is the derivative of f, not of g. Here it's the derivative of g, not of f. This is hopefully a little bit of review. This is the product rule. Now what we're essentially going to do is reapply the product rule to do what many of your calculus books might call the quotient rule."}, {"video_title": "Quotient rule from product & chain rules Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "In each term, we take the derivative of one of the functions and not the other, and then we switch. So over here is the derivative of f, not of g. Here it's the derivative of g, not of f. This is hopefully a little bit of review. This is the product rule. Now what we're essentially going to do is reapply the product rule to do what many of your calculus books might call the quotient rule. I have mixed feelings about the quotient rule. If you know it, it might make some operations a little bit faster, but it really comes straight out of the product rule. And I frankly always forget the quotient rule, and I just re-derive it from the product rule."}, {"video_title": "Quotient rule from product & chain rules Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Now what we're essentially going to do is reapply the product rule to do what many of your calculus books might call the quotient rule. I have mixed feelings about the quotient rule. If you know it, it might make some operations a little bit faster, but it really comes straight out of the product rule. And I frankly always forget the quotient rule, and I just re-derive it from the product rule. So let's see what we're talking about. So let's imagine if we had an expression that could be written as f of x divided by g of x. And we want to take the derivative of this business."}, {"video_title": "Quotient rule from product & chain rules Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And I frankly always forget the quotient rule, and I just re-derive it from the product rule. So let's see what we're talking about. So let's imagine if we had an expression that could be written as f of x divided by g of x. And we want to take the derivative of this business. The derivative of f of x over g of x. The key realization is to just recognize that this is the same thing as the derivative of, instead of writing f of x over g of x, we could write this as f of x times g of x to the negative 1 power. g of x to the negative 1 power."}, {"video_title": "Quotient rule from product & chain rules Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And we want to take the derivative of this business. The derivative of f of x over g of x. The key realization is to just recognize that this is the same thing as the derivative of, instead of writing f of x over g of x, we could write this as f of x times g of x to the negative 1 power. g of x to the negative 1 power. And now we can use the product rule with a little bit of the chain rule. What is this going to be equal to? Well, we just use the product rule."}, {"video_title": "Quotient rule from product & chain rules Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "g of x to the negative 1 power. And now we can use the product rule with a little bit of the chain rule. What is this going to be equal to? Well, we just use the product rule. It's the derivative of the first function right over here. So it's going to be f prime of x times just the second function, which is just g of x to the negative 1 power, plus the first function, which is just f of x, times the derivative of the second function. And here we're going to have to use a little bit of the chain rule."}, {"video_title": "Quotient rule from product & chain rules Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we just use the product rule. It's the derivative of the first function right over here. So it's going to be f prime of x times just the second function, which is just g of x to the negative 1 power, plus the first function, which is just f of x, times the derivative of the second function. And here we're going to have to use a little bit of the chain rule. The derivative of the outside, which we could kind of use something to the negative 1 power with respect to that something, is going to be negative 1 times that something, which in this case is g of x, to the negative 2 power. And then we have to take the derivative of the inside function with respect to x, which is just g prime of x. And there you have it."}, {"video_title": "Quotient rule from product & chain rules Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And here we're going to have to use a little bit of the chain rule. The derivative of the outside, which we could kind of use something to the negative 1 power with respect to that something, is going to be negative 1 times that something, which in this case is g of x, to the negative 2 power. And then we have to take the derivative of the inside function with respect to x, which is just g prime of x. And there you have it. We have found the derivative of this using the product rule and the chain rule. Now, this is not the form that you might see when people are talking about the quotient rule in your bath book. So let's see if we can simplify this a little bit."}, {"video_title": "Quotient rule from product & chain rules Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And there you have it. We have found the derivative of this using the product rule and the chain rule. Now, this is not the form that you might see when people are talking about the quotient rule in your bath book. So let's see if we can simplify this a little bit. All of this is going to be equal to, we can write this term right over here, as f prime of x, as f prime of x over g of x. And we can write all of this as, we can put this negative sign out front, we have negative f of x times g prime of x, and then all of that over g of x squared. All of that over g of x squared."}, {"video_title": "Quotient rule from product & chain rules Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So let's see if we can simplify this a little bit. All of this is going to be equal to, we can write this term right over here, as f prime of x, as f prime of x over g of x. And we can write all of this as, we can put this negative sign out front, we have negative f of x times g prime of x, and then all of that over g of x squared. All of that over g of x squared. And it still isn't the form that you typically see in your calculus book. To do that, we just have to add these two fractions. So let's multiply the numerator and the denominator here by g of x so that we have everything in the form of g of x squared in the denominator."}, {"video_title": "Quotient rule from product & chain rules Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "All of that over g of x squared. And it still isn't the form that you typically see in your calculus book. To do that, we just have to add these two fractions. So let's multiply the numerator and the denominator here by g of x so that we have everything in the form of g of x squared in the denominator. So if we multiply the numerator by g of x, we'll get g of x right over here. And then the denominator will be g of x squared. And now we're ready to add."}, {"video_title": "Quotient rule from product & chain rules Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So let's multiply the numerator and the denominator here by g of x so that we have everything in the form of g of x squared in the denominator. So if we multiply the numerator by g of x, we'll get g of x right over here. And then the denominator will be g of x squared. And now we're ready to add. And so we get the derivative of f of x over g of x is equal to the derivative of f of x times g of x minus, not plus anymore, minus, let me write it in white, minus f of x times g prime of x. Times g prime of x, all of that over g of x squared. So once again, you can always derive this from the product rule and the chain rule."}, {"video_title": "Quotient rule from product & chain rules Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And now we're ready to add. And so we get the derivative of f of x over g of x is equal to the derivative of f of x times g of x minus, not plus anymore, minus, let me write it in white, minus f of x times g prime of x. Times g prime of x, all of that over g of x squared. So once again, you can always derive this from the product rule and the chain rule. Sometimes this might be convenient to remember in order to work through some problems of this form a little bit faster. And if you wanted to kind of see the pattern between the product rule and the quotient rule, the derivative of one function just times the other function, and instead of adding the derivative of the second function times the first function, we now subtract it. And all of that is over the second function squared."}, {"video_title": "Quotient rule from product & chain rules Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So once again, you can always derive this from the product rule and the chain rule. Sometimes this might be convenient to remember in order to work through some problems of this form a little bit faster. And if you wanted to kind of see the pattern between the product rule and the quotient rule, the derivative of one function just times the other function, and instead of adding the derivative of the second function times the first function, we now subtract it. And all of that is over the second function squared. Whatever was in the denominator, it's all of that squared. So when we're taking the derivative of the function in the denominator, up here there's a subtraction. And then we are also putting everything over the second function squared."}, {"video_title": "Worked example estimating e_ using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "In general, if you see a situation like this, where we're talking about approximating a function with a Taylor polynomial centered about some value, and we wanna know, well, how many terms do we need? What degree do we need to bound the error? That's a pretty good clue that we're going to be using the Lagrange Error Bound, or Taylor's Remainder Theorem. And just as a reminder of that, this is a review of Taylor's Remainder Theorem, and it tells us that the absolute value of the remainder for the nth degree Taylor polynomial, it's going to be less than this business right over here. Now, n is the degree of our polynomial in question, so that's the n. The x is the x value at which we are calculating that error, in this case, it's going to be this 1.45. And c is where our Taylor polynomial is centered. But what about our m?"}, {"video_title": "Worked example estimating e_ using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "And just as a reminder of that, this is a review of Taylor's Remainder Theorem, and it tells us that the absolute value of the remainder for the nth degree Taylor polynomial, it's going to be less than this business right over here. Now, n is the degree of our polynomial in question, so that's the n. The x is the x value at which we are calculating that error, in this case, it's going to be this 1.45. And c is where our Taylor polynomial is centered. But what about our m? Well, our m is an upper bound on the absolute value of the n plus 1th derivative of our function. And that might seem like a mouthful, but when we actually work through the details of this example, it'll make it a little bit more concrete. So for this particular thing, we're trying to estimate, we're trying to estimate e to the x."}, {"video_title": "Worked example estimating e_ using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "But what about our m? Well, our m is an upper bound on the absolute value of the n plus 1th derivative of our function. And that might seem like a mouthful, but when we actually work through the details of this example, it'll make it a little bit more concrete. So for this particular thing, we're trying to estimate, we're trying to estimate e to the x. So I could write f of x, let me write this this way. So f of x is equal to e to the x, and we're trying to estimate f of 1.45. And let's just to get the bound here, to figure out what m is, let's just remind ourselves that well, the first derivative of this is going to be e to the x, the second derivative is going to be e to the x, the nth derivative is going to be e to the x, the n plus 1th derivative is going to be e to the x."}, {"video_title": "Worked example estimating e_ using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "So for this particular thing, we're trying to estimate, we're trying to estimate e to the x. So I could write f of x, let me write this this way. So f of x is equal to e to the x, and we're trying to estimate f of 1.45. And let's just to get the bound here, to figure out what m is, let's just remind ourselves that well, the first derivative of this is going to be e to the x, the second derivative is going to be e to the x, the nth derivative is going to be e to the x, the n plus 1th derivative is going to be e to the x. So the n plus 1th derivative of f is going to be, is going to be e to the x, which is convenient. These types of problems are very, very hard if it's difficult to bound the n plus 1th derivative. Well this, we know, we know that e to the x, we know that e to the x, and I could even say the absolute value of this, but this is going to be positive, is going to be less than or equal to, let's say this is going to be less than or equal to e squared for zero is less than x, is less than or equal to two."}, {"video_title": "Worked example estimating e_ using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "And let's just to get the bound here, to figure out what m is, let's just remind ourselves that well, the first derivative of this is going to be e to the x, the second derivative is going to be e to the x, the nth derivative is going to be e to the x, the n plus 1th derivative is going to be e to the x. So the n plus 1th derivative of f is going to be, is going to be e to the x, which is convenient. These types of problems are very, very hard if it's difficult to bound the n plus 1th derivative. Well this, we know, we know that e to the x, we know that e to the x, and I could even say the absolute value of this, but this is going to be positive, is going to be less than or equal to, let's say this is going to be less than or equal to e squared for zero is less than x, is less than or equal to two. So e to the x isn't bounded over the entire, over its entire domain, if x goes to infinity, e to the x will also go to infinity. But here I've set up an interval, I've set up an interval that contains the x we care about. Remember, the x we care about is 1.45, and it also contains where our function is centered."}, {"video_title": "Worked example estimating e_ using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "Well this, we know, we know that e to the x, we know that e to the x, and I could even say the absolute value of this, but this is going to be positive, is going to be less than or equal to, let's say this is going to be less than or equal to e squared for zero is less than x, is less than or equal to two. So e to the x isn't bounded over the entire, over its entire domain, if x goes to infinity, e to the x will also go to infinity. But here I've set up an interval, I've set up an interval that contains the x we care about. Remember, the x we care about is 1.45, and it also contains where our function is centered. Our function is centered at two. So we know we're bounded by e squared, so we can say, we can use e squared as our m. We can use e squared as our m. We were able to establish this bound. And so doing that, we can now go straight to Lagrange error bound."}, {"video_title": "Worked example estimating e_ using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "Remember, the x we care about is 1.45, and it also contains where our function is centered. Our function is centered at two. So we know we're bounded by e squared, so we can say, we can use e squared as our m. We can use e squared as our m. We were able to establish this bound. And so doing that, we can now go straight to Lagrange error bound. We can say, we can say that the remainder of our nth degree Taylor polynomial, we want to solve for n, we want to figure out what n gives us the appropriate bound, evaluated at 1.45, when x is 1.45, is going to be less than or equal to the absolute value, where m is e squared, e squared over, over n plus one factorial, times 1.45, that's our x that we care about, that's where we're calculating the error, we're trying to bound the error, minus where we're centered, minus two, to the n plus oneth power. Now 1.45 minus two, that is negative 0.55, so let me just write that. This is, this is negative 0.55 to the n plus oneth power."}, {"video_title": "Worked example estimating e_ using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "And so doing that, we can now go straight to Lagrange error bound. We can say, we can say that the remainder of our nth degree Taylor polynomial, we want to solve for n, we want to figure out what n gives us the appropriate bound, evaluated at 1.45, when x is 1.45, is going to be less than or equal to the absolute value, where m is e squared, e squared over, over n plus one factorial, times 1.45, that's our x that we care about, that's where we're calculating the error, we're trying to bound the error, minus where we're centered, minus two, to the n plus oneth power. Now 1.45 minus two, that is negative 0.55, so let me just write that. This is, this is negative 0.55 to the n plus oneth power. And we want to figure out for what n is all of this business, is all of this business going to be less than 0.001? Well let's do a little bit of algebraic manipulation here. This term is positive, this is gonna be positive, this right over here, or this part of it, it's not an independent term, but this, the e to the square is gonna be positive, so this n plus one factorial is gonna be positive, the negative 0.55 to some power, that's gonna flip between being positive or negative."}, {"video_title": "Worked example estimating e_ using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "This is, this is negative 0.55 to the n plus oneth power. And we want to figure out for what n is all of this business, is all of this business going to be less than 0.001? Well let's do a little bit of algebraic manipulation here. This term is positive, this is gonna be positive, this right over here, or this part of it, it's not an independent term, but this, the e to the square is gonna be positive, so this n plus one factorial is gonna be positive, the negative 0.55 to some power, that's gonna flip between being positive or negative. But since we're taking the absolute value, we could write it this way, we could write e squared, e squared, since we're taking the absolute value, times 0.55 to the n plus one over n plus one factorial has to be less than, has to be less than 0.001, or, since we want to solve for one, since we want to solve for n, let's divide both sides by e squared. So we could write, we could write, let's find the n where 0.55 to the n plus one power over n plus one factorial is less than, is less than 0.001 over, over e squared. Now to play with this, we're gonna have to use a calculator."}, {"video_title": "Worked example estimating e_ using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "This term is positive, this is gonna be positive, this right over here, or this part of it, it's not an independent term, but this, the e to the square is gonna be positive, so this n plus one factorial is gonna be positive, the negative 0.55 to some power, that's gonna flip between being positive or negative. But since we're taking the absolute value, we could write it this way, we could write e squared, e squared, since we're taking the absolute value, times 0.55 to the n plus one over n plus one factorial has to be less than, has to be less than 0.001, or, since we want to solve for one, since we want to solve for n, let's divide both sides by e squared. So we could write, we could write, let's find the n where 0.55 to the n plus one power over n plus one factorial is less than, is less than 0.001 over, over e squared. Now to play with this, we're gonna have to use a calculator. Remember, from this point, we're just gonna try larger and larger n's until we get an n that makes this true, and we find, when we want to find the smallest possible n that makes this true. But let's get out our calculator so that we can actually, so that we can actually do this. So first I'm just gonna figure out what is 1,000th divided by e squared."}, {"video_title": "Worked example estimating e_ using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "Now to play with this, we're gonna have to use a calculator. Remember, from this point, we're just gonna try larger and larger n's until we get an n that makes this true, and we find, when we want to find the smallest possible n that makes this true. But let's get out our calculator so that we can actually, so that we can actually do this. So first I'm just gonna figure out what is 1,000th divided by e squared. So make sure it's cleared out. So let's take e squared, I'm gonna take its reciprocal, and then I'm gonna multiply that times 1,000th. So times 0.001 is equal to, so it's about, so it's about, so I'll say, so it's three zeros, this is a 1,000th, and then three fives."}, {"video_title": "Worked example estimating e_ using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "So first I'm just gonna figure out what is 1,000th divided by e squared. So make sure it's cleared out. So let's take e squared, I'm gonna take its reciprocal, and then I'm gonna multiply that times 1,000th. So times 0.001 is equal to, so it's about, so it's about, so I'll say, so it's three zeros, this is a 1,000th, and then three fives. So it's three zeros, so I'll say one three six. So this needs to be, this needs to be less than 0.123, and I'll say one three six. If I can find an n that is less than this, then I am in, then I am in good shape."}, {"video_title": "Worked example estimating e_ using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "So times 0.001 is equal to, so it's about, so it's about, so I'll say, so it's three zeros, this is a 1,000th, and then three fives. So it's three zeros, so I'll say one three six. So this needs to be, this needs to be less than 0.123, and I'll say one three six. If I can find an n that is less than this, then I am in, then I am in good shape. Actually, let me say that it's less than one three five, I want to be less than that value. Then I can be, then I will be in good shape. This is a little bit more than one three five, but if I can find an n where that is less than this, then I'm in good shape."}, {"video_title": "Worked example estimating e_ using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "If I can find an n that is less than this, then I am in, then I am in good shape. Actually, let me say that it's less than one three five, I want to be less than that value. Then I can be, then I will be in good shape. This is a little bit more than one three five, but if I can find an n where that is less than this, then I'm in good shape. So let me write this, 0.55 to the n plus one over n plus one factorial. So let's try out some n's, and I'm gonna have to get my calculator out. So let's see, did I do that right?"}, {"video_title": "Worked example estimating e_ using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "This is a little bit more than one three five, but if I can find an n where that is less than this, then I'm in good shape. So let me write this, 0.55 to the n plus one over n plus one factorial. So let's try out some n's, and I'm gonna have to get my calculator out. So let's see, did I do that right? Yeah, 0, 0, 0, one three five. If we get something below this, then we're in good shape, because this is even less than that. All right, so let's do it."}, {"video_title": "Worked example estimating e_ using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "So let's see, did I do that right? Yeah, 0, 0, 0, one three five. If we get something below this, then we're in good shape, because this is even less than that. All right, so let's do it. Let's see what this is equal to when, I don't know, when n is equal to two. I could start at n equals one, n equals two, n equals three, but the more n, the further, if n equals two is good enough, then I might try n equals one, but if n equals two isn't good enough, then I'm gonna go to n equals three or n equals four. So let's start with, actually, let's just start with n equals three."}, {"video_title": "Worked example estimating e_ using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "All right, so let's do it. Let's see what this is equal to when, I don't know, when n is equal to two. I could start at n equals one, n equals two, n equals three, but the more n, the further, if n equals two is good enough, then I might try n equals one, but if n equals two isn't good enough, then I'm gonna go to n equals three or n equals four. So let's start with, actually, let's just start with n equals three. So if n equals three, it's gonna be 0.55 to the fourth power divided by four factorial. So let's do that. So 0.55 to the fourth power is equal to that divided by four factorial."}, {"video_title": "Worked example estimating e_ using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "So let's start with, actually, let's just start with n equals three. So if n equals three, it's gonna be 0.55 to the fourth power divided by four factorial. So let's do that. So 0.55 to the fourth power is equal to that divided by four factorial. So divided by four factorial is 24. So that's nowhere near low enough. So let's try n equals four."}, {"video_title": "Worked example estimating e_ using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "So 0.55 to the fourth power is equal to that divided by four factorial. So divided by four factorial is 24. So that's nowhere near low enough. So let's try n equals four. If n equals four, then it's gonna be this to the fifth power divided by five factorial. So 0.55 to the fifth power is equal to, and then divided by five factorial is 120. Divided by 120 is equal to that."}, {"video_title": "Worked example estimating e_ using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "So let's try n equals four. If n equals four, then it's gonna be this to the fifth power divided by five factorial. So 0.55 to the fifth power is equal to, and then divided by five factorial is 120. Divided by 120 is equal to that. We're almost there with n equals four. I'm guessing that n equals five will do the trick. So for n equals five, so let's clear this out."}, {"video_title": "Worked example estimating e_ using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "Divided by 120 is equal to that. We're almost there with n equals four. I'm guessing that n equals five will do the trick. So for n equals five, so let's clear this out. So for n equals five, we're gonna raise to the sixth power and divide by six factorial. And so let's just remind ourselves what six factorial is, 720. I could have actually done that in my head, but anyway."}, {"video_title": "Worked example estimating e_ using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "So for n equals five, so let's clear this out. So for n equals five, we're gonna raise to the sixth power and divide by six factorial. And so let's just remind ourselves what six factorial is, 720. I could have actually done that in my head, but anyway. All right, so let's see. We're gonna go 0.55 to the, remember, our n is five, so we're gonna raise to the sixth power, to the sixth power, and then we're gonna divide by 720. Divided by 720 is equal to, and this number for sure is less than this number right over here."}, {"video_title": "Worked example estimating e_ using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "I could have actually done that in my head, but anyway. All right, so let's see. We're gonna go 0.55 to the, remember, our n is five, so we're gonna raise to the sixth power, to the sixth power, and then we're gonna divide by 720. Divided by 720 is equal to, and this number for sure is less than this number right over here. We got four zeros before this three after the decimal. Here we only have three. So when n equals five, it got us sufficiently low enough."}, {"video_title": "Worked example estimating e_ using Lagrange error bound AP Calculus BC Khan Academy.mp3", "Sentence": "Divided by 720 is equal to, and this number for sure is less than this number right over here. We got four zeros before this three after the decimal. Here we only have three. So when n equals five, it got us sufficiently low enough. This remainder is going to be sufficiently low. It's gonna be less than this value right over here. So what is the least degree of the polynomial that assures an error smaller than 1,000?"}, {"video_title": "Divergent improper integral AP Calculus BC Khan Academy.mp3", "Sentence": "And what I'm curious about is the area under this curve and above the x axis between x equals 1 and infinity. So I want to figure out this area right over here. So let's try to do it. So we could set this up as an improper integral going from 1 to infinity of 1 over x dx. Well, once again we can view this as the limit. Actually, I'm going to do that same yellow color. I like that more."}, {"video_title": "Divergent improper integral AP Calculus BC Khan Academy.mp3", "Sentence": "So we could set this up as an improper integral going from 1 to infinity of 1 over x dx. Well, once again we can view this as the limit. Actually, I'm going to do that same yellow color. I like that more. We can view this as the limit as n approaches infinity of the integral from 1 to n of 1 over x dx, which we can write as the limit as n approaches infinity of the antiderivative of 1 over x, which is the natural log of the absolute value of x. So this is going to be the natural log of the absolute value of x. And the absolute value of x won't really matter so much."}, {"video_title": "Divergent improper integral AP Calculus BC Khan Academy.mp3", "Sentence": "I like that more. We can view this as the limit as n approaches infinity of the integral from 1 to n of 1 over x dx, which we can write as the limit as n approaches infinity of the antiderivative of 1 over x, which is the natural log of the absolute value of x. So this is going to be the natural log of the absolute value of x. And the absolute value of x won't really matter so much. We could just say x because we're dealing with positive values of x. But I'll just write that as the natural log of the absolute value of x between x is 1 and x is n. And so this is going to be equal to the limit as n approaches infinity of you evaluate this at n, so you're going to get the natural log. I could write the absolute value of n, but we know that n is going to be positive."}, {"video_title": "Divergent improper integral AP Calculus BC Khan Academy.mp3", "Sentence": "And the absolute value of x won't really matter so much. We could just say x because we're dealing with positive values of x. But I'll just write that as the natural log of the absolute value of x between x is 1 and x is n. And so this is going to be equal to the limit as n approaches infinity of you evaluate this at n, so you're going to get the natural log. I could write the absolute value of n, but we know that n is going to be positive. So we could just write the natural log of n minus the natural log of the absolute value of 1 or the natural log of 1. Natural log of 1 is just a 0. e to the 0th power is equal to 1. So this boils down to the limit as n approaches infinity of the natural log of n. Now this is interesting."}, {"video_title": "Divergent improper integral AP Calculus BC Khan Academy.mp3", "Sentence": "I could write the absolute value of n, but we know that n is going to be positive. So we could just write the natural log of n minus the natural log of the absolute value of 1 or the natural log of 1. Natural log of 1 is just a 0. e to the 0th power is equal to 1. So this boils down to the limit as n approaches infinity of the natural log of n. Now this is interesting. Natural log function just keeps getting larger and larger and larger. Natural log function looks something like, keeps growing and growing and growing like this, albeit at a slower and slower pace, but it keeps growing. The limit as n approaches infinity of the natural log of n is just equal to infinity."}, {"video_title": "Divergent improper integral AP Calculus BC Khan Academy.mp3", "Sentence": "So this boils down to the limit as n approaches infinity of the natural log of n. Now this is interesting. Natural log function just keeps getting larger and larger and larger. Natural log function looks something like, keeps growing and growing and growing like this, albeit at a slower and slower pace, but it keeps growing. The limit as n approaches infinity of the natural log of n is just equal to infinity. So here we do not have a finite area. This is an infinite area. It's interesting."}, {"video_title": "Divergent improper integral AP Calculus BC Khan Academy.mp3", "Sentence": "The limit as n approaches infinity of the natural log of n is just equal to infinity. So here we do not have a finite area. This is an infinite area. It's interesting. When this function decreased faster, when it was 1 over x squared, we had a finite area. Now we have an infinite area. And so we would say that this integral right over here, this improper integral, is divergent."}, {"video_title": "Worked example problem involving definite integral (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "Once he realized it, he opened the hot tub's drain and the rest of the water rushed out in 40 minutes. The rate at which the water drained from the hot tub in gallons per minute is shown. How many gallons of water were in the hot tub before it started to leak? So let's see what they have over here. So they've plotted time, or they've plotted gallons per minute versus time within minutes. And so we see here, this blue line, this is the rate at which water drained from the tub, rate at which water drained from the hot tub. So at minute zero, the water wasn't really draining from the hot tub, and then more, not just more drained, but the rate at which the water was draining increased."}, {"video_title": "Worked example problem involving definite integral (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "So let's see what they have over here. So they've plotted time, or they've plotted gallons per minute versus time within minutes. And so we see here, this blue line, this is the rate at which water drained from the tub, rate at which water drained from the hot tub. So at minute zero, the water wasn't really draining from the hot tub, and then more, not just more drained, but the rate at which the water was draining increased. So 10 minutes into his bath, the water was draining at a gallon per minute, and then 20 minutes into his bath, water was draining at two gallons per minute. And then he noticed it, and he opens the drain. I guess he wants to accelerate the end of his bath."}, {"video_title": "Worked example problem involving definite integral (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "So at minute zero, the water wasn't really draining from the hot tub, and then more, not just more drained, but the rate at which the water was draining increased. So 10 minutes into his bath, the water was draining at a gallon per minute, and then 20 minutes into his bath, water was draining at two gallons per minute. And then he noticed it, and he opens the drain. I guess he wants to accelerate the end of his bath. So he opens the drain, and then all of a sudden, water starts draining out at a much higher rate, at 20 gallons per minute, but then that decreases. And so we could think about physically why that might decrease. Maybe there was just less pressure or whatever."}, {"video_title": "Worked example problem involving definite integral (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "I guess he wants to accelerate the end of his bath. So he opens the drain, and then all of a sudden, water starts draining out at a much higher rate, at 20 gallons per minute, but then that decreases. And so we could think about physically why that might decrease. Maybe there was just less pressure or whatever. We're not gonna go into the physics of it, but we're just gonna take this chart as fact. But the rate at which the water drains decreases all the way to the 60th minute, which is 40 minutes after he opened the drain. 60th minute, all of the water was drained."}, {"video_title": "Worked example problem involving definite integral (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "Maybe there was just less pressure or whatever. We're not gonna go into the physics of it, but we're just gonna take this chart as fact. But the rate at which the water drains decreases all the way to the 60th minute, which is 40 minutes after he opened the drain. 60th minute, all of the water was drained. All of the water was actually drained out. So given that, how do we think about how many gallons of water were in the hot tub before it started to leak? And I encourage you to pause the video and try to see if you can figure it out yourself before I work through it."}, {"video_title": "Worked example problem involving definite integral (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "60th minute, all of the water was drained. All of the water was actually drained out. So given that, how do we think about how many gallons of water were in the hot tub before it started to leak? And I encourage you to pause the video and try to see if you can figure it out yourself before I work through it. Well, to answer this question, how many gallons of water were in the hot tub before it started to leak, that's answering the same question as how much total hot water drained out. So how much drained out? And to think about that, we can just kind of go back to what we knew before we learned about calculus."}, {"video_title": "Worked example problem involving definite integral (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "And I encourage you to pause the video and try to see if you can figure it out yourself before I work through it. Well, to answer this question, how many gallons of water were in the hot tub before it started to leak, that's answering the same question as how much total hot water drained out. So how much drained out? And to think about that, we can just kind of go back to what we knew before we learned about calculus. If I have something happening at a fixed rate, so let's say that this is gallons per minute. So this is still the same context, I guess you could say. This is time in minutes."}, {"video_title": "Worked example problem involving definite integral (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "And to think about that, we can just kind of go back to what we knew before we learned about calculus. If I have something happening at a fixed rate, so let's say that this is gallons per minute. So this is still the same context, I guess you could say. This is time in minutes. Now let's say things are draining out at a constant rate. If you wanted to figure out how much drains out over a certain interval of time, let's say this interval of time right over here, let's call that delta t over delta t, you would just multiply the rate over that interval of time, the rate over that interval of time, which we could represent by this orange height right over here, times the amount of time that passed by, which essentially would give you the area under the curve over that interval. The area under the curve over that interval."}, {"video_title": "Worked example problem involving definite integral (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "This is time in minutes. Now let's say things are draining out at a constant rate. If you wanted to figure out how much drains out over a certain interval of time, let's say this interval of time right over here, let's call that delta t over delta t, you would just multiply the rate over that interval of time, the rate over that interval of time, which we could represent by this orange height right over here, times the amount of time that passed by, which essentially would give you the area under the curve over that interval. The area under the curve over that interval. This area would tell you the gallons that drained over that delta t. And this doesn't just apply when you have a constant, when you have a constant rate. If your rate looked something like this, as we've seen in other videos, as we've seen in other videos, you could figure out the amount that has drained, the amount that has drained in a specific interval, let's say this interval right over here, by essentially figuring out the area under the curve over that interval. And the units work out."}, {"video_title": "Worked example problem involving definite integral (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "The area under the curve over that interval. This area would tell you the gallons that drained over that delta t. And this doesn't just apply when you have a constant, when you have a constant rate. If your rate looked something like this, as we've seen in other videos, as we've seen in other videos, you could figure out the amount that has drained, the amount that has drained in a specific interval, let's say this interval right over here, by essentially figuring out the area under the curve over that interval. And the units work out. If you multiply gallons per minute times minute, this area is going to be in terms of gallons. This area is going to be in terms of gallons. Another way you could think about it, and this goes back to, well, how do you figure out the area of a trapezoid right over here?"}, {"video_title": "Worked example problem involving definite integral (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "And the units work out. If you multiply gallons per minute times minute, this area is going to be in terms of gallons. This area is going to be in terms of gallons. Another way you could think about it, and this goes back to, well, how do you figure out the area of a trapezoid right over here? Well, the area of a trapezoid, you could find the average height of the trapezoid, which would be the average of the beginning and the end period. You take that average over there, and this would work for a line like this. If you take that average height and multiply by the change in time, you are going to actually, you are going to figure out, you are going to figure out that area."}, {"video_title": "Worked example problem involving definite integral (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "Another way you could think about it, and this goes back to, well, how do you figure out the area of a trapezoid right over here? Well, the area of a trapezoid, you could find the average height of the trapezoid, which would be the average of the beginning and the end period. You take that average over there, and this would work for a line like this. If you take that average height and multiply by the change in time, you are going to actually, you are going to figure out, you are going to figure out that area. That's another way of thinking about it, is you're taking the average rate over that interval times the interval, and that's going to give you the total number of gallons. And so we just have to apply that idea over here. We just literally figure out the area under the curve over the entire interval when the water was actually leaking or draining."}, {"video_title": "Worked example problem involving definite integral (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "If you take that average height and multiply by the change in time, you are going to actually, you are going to figure out, you are going to figure out that area. That's another way of thinking about it, is you're taking the average rate over that interval times the interval, and that's going to give you the total number of gallons. And so we just have to apply that idea over here. We just literally figure out the area under the curve over the entire interval when the water was actually leaking or draining. So essentially the area under the curve between zero minutes and 60 minutes. And so it's going to be this area, let's put that in magenta, plus all of this area under this part as well. And to help us think about that, I'm just going to split that up into some sections."}, {"video_title": "Worked example problem involving definite integral (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "We just literally figure out the area under the curve over the entire interval when the water was actually leaking or draining. So essentially the area under the curve between zero minutes and 60 minutes. And so it's going to be this area, let's put that in magenta, plus all of this area under this part as well. And to help us think about that, I'm just going to split that up into some sections. So I'll have this triangular section up here. So I have this triangular section. I could just think about this as a trapezoid, but I'm just going to split it up into a triangle and a rectangle."}, {"video_title": "Worked example problem involving definite integral (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "And to help us think about that, I'm just going to split that up into some sections. So I'll have this triangular section up here. So I have this triangular section. I could just think about this as a trapezoid, but I'm just going to split it up into a triangle and a rectangle. And then I have this section right over here in green. So what is the area of this entire thing? Well, the area right over here, we have 20 gallons per minute."}, {"video_title": "Worked example problem involving definite integral (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "I could just think about this as a trapezoid, but I'm just going to split it up into a triangle and a rectangle. And then I have this section right over here in green. So what is the area of this entire thing? Well, the area right over here, we have 20 gallons per minute. So 20, oh sorry, not 20 gallons per minute, 20 minutes times two gallons per minute, two gallons per minute times 1 1\u20442. Times 1 1\u20442 gives us the area under this triangle. So that is going to be 20 gallons."}, {"video_title": "Worked example problem involving definite integral (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the area right over here, we have 20 gallons per minute. So 20, oh sorry, not 20 gallons per minute, 20 minutes times two gallons per minute, two gallons per minute times 1 1\u20442. Times 1 1\u20442 gives us the area under this triangle. So that is going to be 20 gallons. 20 gallons, we see that the units work out nicely. And so that's essentially how much has drained out in the first 20 minutes. And then this green area is going to be 40 minutes, 40 minutes times 10 gallons per minute, times 10, and actually maybe the units, since I'm breaking it up in this strange way, I'll just multiply, I'll just figure out this area in a unitless way."}, {"video_title": "Worked example problem involving definite integral (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "So that is going to be 20 gallons. 20 gallons, we see that the units work out nicely. And so that's essentially how much has drained out in the first 20 minutes. And then this green area is going to be 40 minutes, 40 minutes times 10 gallons per minute, times 10, and actually maybe the units, since I'm breaking it up in this strange way, I'll just multiply, I'll just figure out this area in a unitless way. So 40 times 10, which is equal to 400. And then finally in blue, I have 40 times this height right over here between 10 and 20 is another 10, but then I'm gonna multiply that times 1 1\u20442. So it's gonna be 40 times 10 times 1 1\u20442, which is going to be 200."}, {"video_title": "Worked example problem involving definite integral (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "And then this green area is going to be 40 minutes, 40 minutes times 10 gallons per minute, times 10, and actually maybe the units, since I'm breaking it up in this strange way, I'll just multiply, I'll just figure out this area in a unitless way. So 40 times 10, which is equal to 400. And then finally in blue, I have 40 times this height right over here between 10 and 20 is another 10, but then I'm gonna multiply that times 1 1\u20442. So it's gonna be 40 times 10 times 1 1\u20442, which is going to be 200. And so you add all of these together, these areas together, you're going to get 400 plus 200 is 600 plus 20, you're going to have 620 gallons. 620 gallons is how much water in total drained out or how much water was in the tub before it started to leak, or the hot tub. Yeah, this is quite big for a regular tub, but this could be like a jacuzzi or a hot tub or something."}, {"video_title": "Infinite limits and asymptotes Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So let me get that one graphed. And so you can immediately see that something interesting happens at x is equal to one. If you were to just substitute x equals one into this expression, you're going to get two over zero. And whenever you get a non-zero thing over zero, that's a good sign that you might be dealing with a vertical asymptote. In fact, we can draw that vertical asymptote right over here at x equals one. But let's think about how that relates to limits. What if we were to explore the limit as x approaches one of f of x is equal to two over x minus one?"}, {"video_title": "Infinite limits and asymptotes Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And whenever you get a non-zero thing over zero, that's a good sign that you might be dealing with a vertical asymptote. In fact, we can draw that vertical asymptote right over here at x equals one. But let's think about how that relates to limits. What if we were to explore the limit as x approaches one of f of x is equal to two over x minus one? And we could think about it from the left and from the right. So if we approach one from the left, let me zoom in a little bit over here. So we can see as we approach from the left when x is equal to zero, the f of x would be equal to negative two."}, {"video_title": "Infinite limits and asymptotes Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "What if we were to explore the limit as x approaches one of f of x is equal to two over x minus one? And we could think about it from the left and from the right. So if we approach one from the left, let me zoom in a little bit over here. So we can see as we approach from the left when x is equal to zero, the f of x would be equal to negative two. When x is equal to.5, f of x is equal to negative four. And then it just gets more and more negative the closer we get to one from the left. I could really, so I'm not even that close yet."}, {"video_title": "Infinite limits and asymptotes Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So we can see as we approach from the left when x is equal to zero, the f of x would be equal to negative two. When x is equal to.5, f of x is equal to negative four. And then it just gets more and more negative the closer we get to one from the left. I could really, so I'm not even that close yet. If I get to, let's say.91, I'm still 900, so less than one. I'm at negative 22.222 already. And so the limit as we approach one from the left is unbounded."}, {"video_title": "Infinite limits and asymptotes Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "I could really, so I'm not even that close yet. If I get to, let's say.91, I'm still 900, so less than one. I'm at negative 22.222 already. And so the limit as we approach one from the left is unbounded. Some people would say it goes to negative infinity, but it's really an undefined limit. It is unbounded in the negative direction. And likewise, as we approach from the right, we get unbounded in the positive infinity direction, and technically we would say that that limit does not exist."}, {"video_title": "Infinite limits and asymptotes Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And so the limit as we approach one from the left is unbounded. Some people would say it goes to negative infinity, but it's really an undefined limit. It is unbounded in the negative direction. And likewise, as we approach from the right, we get unbounded in the positive infinity direction, and technically we would say that that limit does not exist. And this would be the case when we're dealing with a vertical asymptote like we see over here. Now let's compare that to a horizontal asymptote where it turns out that the limit actually can exist. So let me delete these or just erase them for now."}, {"video_title": "Infinite limits and asymptotes Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And likewise, as we approach from the right, we get unbounded in the positive infinity direction, and technically we would say that that limit does not exist. And this would be the case when we're dealing with a vertical asymptote like we see over here. Now let's compare that to a horizontal asymptote where it turns out that the limit actually can exist. So let me delete these or just erase them for now. And so let's look at this function, which is a pretty neat function. I made it up right before this video started, but it's kind of cool looking. So let's think about the behavior as x approaches infinity."}, {"video_title": "Infinite limits and asymptotes Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So let me delete these or just erase them for now. And so let's look at this function, which is a pretty neat function. I made it up right before this video started, but it's kind of cool looking. So let's think about the behavior as x approaches infinity. So as x approaches infinity, it looks like our y value or the value of the expression, if we said y is equal to that expression, it looks like it's getting closer and closer and closer to three. And so we could say that we have a horizontal asymptote at y is equal to three. And we could also, and there's a more rigorous way of defining it, say that our limit as x approaches infinity is equal of the expression or of the function is equal to three."}, {"video_title": "Infinite limits and asymptotes Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So let's think about the behavior as x approaches infinity. So as x approaches infinity, it looks like our y value or the value of the expression, if we said y is equal to that expression, it looks like it's getting closer and closer and closer to three. And so we could say that we have a horizontal asymptote at y is equal to three. And we could also, and there's a more rigorous way of defining it, say that our limit as x approaches infinity is equal of the expression or of the function is equal to three. Notice my mouse is covering it a little bit. As we get larger and larger, we're getting closer and closer to three. In fact, we're getting so close now that, well, here, you can see it."}, {"video_title": "Infinite limits and asymptotes Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And we could also, and there's a more rigorous way of defining it, say that our limit as x approaches infinity is equal of the expression or of the function is equal to three. Notice my mouse is covering it a little bit. As we get larger and larger, we're getting closer and closer to three. In fact, we're getting so close now that, well, here, you can see it. We're getting closer and closer and closer to three. And you could also think about what happens as x approaches negative infinity. And here you're getting closer and closer and closer to three from below."}, {"video_title": "Infinite limits and asymptotes Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "In fact, we're getting so close now that, well, here, you can see it. We're getting closer and closer and closer to three. And you could also think about what happens as x approaches negative infinity. And here you're getting closer and closer and closer to three from below. Now one thing that's interesting about horizontal asymptotes is you might see that the function actually can cross a horizontal asymptote. It's crossing this horizontal asymptote in this area in between. And even as we approach infinity or negative infinity, you can oscillate around that horizontal asymptote."}, {"video_title": "Infinite limits and asymptotes Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And here you're getting closer and closer and closer to three from below. Now one thing that's interesting about horizontal asymptotes is you might see that the function actually can cross a horizontal asymptote. It's crossing this horizontal asymptote in this area in between. And even as we approach infinity or negative infinity, you can oscillate around that horizontal asymptote. Let me set this up. Let me multiply this times sine of x. And so there you have it."}, {"video_title": "Infinite limits and asymptotes Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And even as we approach infinity or negative infinity, you can oscillate around that horizontal asymptote. Let me set this up. Let me multiply this times sine of x. And so there you have it. We are now oscillating around the horizontal asymptote. And once again, this limit can exist. Even though we keep crossing the horizontal asymptote, we're getting closer and closer and closer to it the larger x gets."}, {"video_title": "Infinite limits and asymptotes Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And so there you have it. We are now oscillating around the horizontal asymptote. And once again, this limit can exist. Even though we keep crossing the horizontal asymptote, we're getting closer and closer and closer to it the larger x gets. And that's actually a key difference between a horizontal and a vertical asymptote. Vertical asymptotes, if you're dealing with a function, you're not going to cross it. While with a horizontal asymptote, you could."}, {"video_title": "Convergent and divergent sequences Series AP Calculus BC Khan Academy.mp3", "Sentence": "And we could graph it. Let me draw our vertical axis. So I'll draft this as our y axis, and I'm gonna graph y is equal to a sub n. And let's make this our horizontal axis where I'm gonna plot our n's. So this right over here is our n's, and let's say this right over here is positive 1, this right over here is negative 1, this would be negative 1 half, this would be positive 1 half. And I'm not drawing the vertical and horizontal axes at the same scale, just so that we can kind of visualize this properly. So let's say this is 1, 2, 3, 4, 5, and I could keep going on and on and on. So we see here that when n is equal to 1, a sub n is equal to 1."}, {"video_title": "Convergent and divergent sequences Series AP Calculus BC Khan Academy.mp3", "Sentence": "So this right over here is our n's, and let's say this right over here is positive 1, this right over here is negative 1, this would be negative 1 half, this would be positive 1 half. And I'm not drawing the vertical and horizontal axes at the same scale, just so that we can kind of visualize this properly. So let's say this is 1, 2, 3, 4, 5, and I could keep going on and on and on. So we see here that when n is equal to 1, a sub n is equal to 1. So this is right over there. So when n is equal to 1, a sub n is equal to 1. So this is y is equal to a sub n. Now, when n is equal to 2, we have a sub n is equal to negative 1 half."}, {"video_title": "Convergent and divergent sequences Series AP Calculus BC Khan Academy.mp3", "Sentence": "So we see here that when n is equal to 1, a sub n is equal to 1. So this is right over there. So when n is equal to 1, a sub n is equal to 1. So this is y is equal to a sub n. Now, when n is equal to 2, we have a sub n is equal to negative 1 half. So when n is equal to 2, a sub n is equal to negative 1 half. When n is equal to 3, a sub n is equal to 1 third, which is right about there. When n is equal to 4, a sub n is equal to negative 1 fourth, which is right about there."}, {"video_title": "Convergent and divergent sequences Series AP Calculus BC Khan Academy.mp3", "Sentence": "So this is y is equal to a sub n. Now, when n is equal to 2, we have a sub n is equal to negative 1 half. So when n is equal to 2, a sub n is equal to negative 1 half. When n is equal to 3, a sub n is equal to 1 third, which is right about there. When n is equal to 4, a sub n is equal to negative 1 fourth, which is right about there. And then when n is equal to 5, a sub n is equal to positive 1 fifth, which is maybe right over there. And we keep going on and on and on. So you see the points, they kind of jump around."}, {"video_title": "Convergent and divergent sequences Series AP Calculus BC Khan Academy.mp3", "Sentence": "When n is equal to 4, a sub n is equal to negative 1 fourth, which is right about there. And then when n is equal to 5, a sub n is equal to positive 1 fifth, which is maybe right over there. And we keep going on and on and on. So you see the points, they kind of jump around. But they seem to be getting closer and closer and closer to 0, which would make us ask a very natural question. What happens to a sub n as n approaches infinity? Or another way of saying that is, what is the limit of a sub n as n approaches infinity?"}, {"video_title": "Convergent and divergent sequences Series AP Calculus BC Khan Academy.mp3", "Sentence": "So you see the points, they kind of jump around. But they seem to be getting closer and closer and closer to 0, which would make us ask a very natural question. What happens to a sub n as n approaches infinity? Or another way of saying that is, what is the limit of a sub n as n approaches infinity? Well, let's think about if we can define a sub n explicitly, if we can define the sequence explicitly. So we can define the sequence as a sub n, where n starts at 1 and goes to infinity, with a sub n equaling. What does it equal?"}, {"video_title": "Convergent and divergent sequences Series AP Calculus BC Khan Academy.mp3", "Sentence": "Or another way of saying that is, what is the limit of a sub n as n approaches infinity? Well, let's think about if we can define a sub n explicitly, if we can define the sequence explicitly. So we can define the sequence as a sub n, where n starts at 1 and goes to infinity, with a sub n equaling. What does it equal? Well, if we ignore sine for a second, it looks like it's just 1 over n. But then we seem like we oscillate in sines. We start with a positive, then a negative, positive, negative. So we could multiply this times negative 1 to the n. Then this one would be negative and this one would be positive."}, {"video_title": "Convergent and divergent sequences Series AP Calculus BC Khan Academy.mp3", "Sentence": "What does it equal? Well, if we ignore sine for a second, it looks like it's just 1 over n. But then we seem like we oscillate in sines. We start with a positive, then a negative, positive, negative. So we could multiply this times negative 1 to the n. Then this one would be negative and this one would be positive. But we don't want it that way. We want the first term to be positive. So we could say negative 1 to the n plus 1 power."}, {"video_title": "Convergent and divergent sequences Series AP Calculus BC Khan Academy.mp3", "Sentence": "So we could multiply this times negative 1 to the n. Then this one would be negative and this one would be positive. But we don't want it that way. We want the first term to be positive. So we could say negative 1 to the n plus 1 power. And you can verify this works. When n is equal to 1, you have 1 times negative 1 squared, which is just 1. And they'll work for all the rest."}, {"video_title": "Convergent and divergent sequences Series AP Calculus BC Khan Academy.mp3", "Sentence": "So we could say negative 1 to the n plus 1 power. And you can verify this works. When n is equal to 1, you have 1 times negative 1 squared, which is just 1. And they'll work for all the rest. So we could write this as equaling negative 1 to the n plus 1 power over n. And so asking what the limit of a sub n as n approaches infinity is is equivalent to asking what is the limit of negative 1 to the n plus 1 power over n as n approaches infinity is going to be equal to. Remember, a sub n, this is just a function of n. It's a function where we're limited right over here to positive integers as our domain. But this is still just a limit as something approaches infinity."}, {"video_title": "Convergent and divergent sequences Series AP Calculus BC Khan Academy.mp3", "Sentence": "And they'll work for all the rest. So we could write this as equaling negative 1 to the n plus 1 power over n. And so asking what the limit of a sub n as n approaches infinity is is equivalent to asking what is the limit of negative 1 to the n plus 1 power over n as n approaches infinity is going to be equal to. Remember, a sub n, this is just a function of n. It's a function where we're limited right over here to positive integers as our domain. But this is still just a limit as something approaches infinity. And I haven't rigorously defined it yet, but you can conceptualize what's going on here. As n approaches infinity, the numerator is going to oscillate between positive and negative 1. But this denominator is just going to get bigger and bigger and bigger and bigger."}, {"video_title": "Convergent and divergent sequences Series AP Calculus BC Khan Academy.mp3", "Sentence": "But this is still just a limit as something approaches infinity. And I haven't rigorously defined it yet, but you can conceptualize what's going on here. As n approaches infinity, the numerator is going to oscillate between positive and negative 1. But this denominator is just going to get bigger and bigger and bigger and bigger. So we're going to get really, really, really, really small numbers. And so this thing right over here is going to approach 0. Now, I have not proved this to you yet."}, {"video_title": "Convergent and divergent sequences Series AP Calculus BC Khan Academy.mp3", "Sentence": "But this denominator is just going to get bigger and bigger and bigger and bigger. So we're going to get really, really, really, really small numbers. And so this thing right over here is going to approach 0. Now, I have not proved this to you yet. I am just claiming that this is true. But if this is true, if the limit of a sub n as n approaches infinity is 0, then we can say that a sub n converges to 0. That's another way of saying this right over here."}, {"video_title": "Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say I have some function f of x that is defined as being equal to x squared minus 6x plus 8 for all x. And what I want to do is show that for this function we can definitely find a c in an interval where the derivative at the point c is equal to the average rate of change over that interval. So let's give ourselves an interval right over here. Let's say we care about the interval between 2 and 5. And this function is clearly both continuous over this closed interval and it's also differentiable over it. It just has to be differentiable over the open interval, but this is differentiable really for all x. And so let's show that we can find a c that's inside the open interval, that's a member of the open interval, such that the derivative at c is equal to the average rate of change over this interval or is equal to the slope of the secant line between the two endpoints of the interval."}, {"video_title": "Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say we care about the interval between 2 and 5. And this function is clearly both continuous over this closed interval and it's also differentiable over it. It just has to be differentiable over the open interval, but this is differentiable really for all x. And so let's show that we can find a c that's inside the open interval, that's a member of the open interval, such that the derivative at c is equal to the average rate of change over this interval or is equal to the slope of the secant line between the two endpoints of the interval. So it's equal to f of 5 minus f of 2 over 5 minus 2. And so I encourage you to pause the video now and try to find a c where this is actually true. Well, to do that, let's just calculate what this has to be."}, {"video_title": "Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "And so let's show that we can find a c that's inside the open interval, that's a member of the open interval, such that the derivative at c is equal to the average rate of change over this interval or is equal to the slope of the secant line between the two endpoints of the interval. So it's equal to f of 5 minus f of 2 over 5 minus 2. And so I encourage you to pause the video now and try to find a c where this is actually true. Well, to do that, let's just calculate what this has to be. Then let's just take the derivative and set them equal and we should be able to solve for our c. So let's see, f of 5 minus f of 2. f of 5 is equal to 25 minus 30 plus 8. So that's negative 5 plus 8 is equal to 3. f of 2 is equal to 2 squared minus 12. So it's 4 minus 12 plus 8, that's going to be 0."}, {"video_title": "Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "Well, to do that, let's just calculate what this has to be. Then let's just take the derivative and set them equal and we should be able to solve for our c. So let's see, f of 5 minus f of 2. f of 5 is equal to 25 minus 30 plus 8. So that's negative 5 plus 8 is equal to 3. f of 2 is equal to 2 squared minus 12. So it's 4 minus 12 plus 8, that's going to be 0. So this is equal to 3 over 3, which is equal to 1. f prime of c needs to be equal to 1. And so what is the derivative of this? Well, let's see, f prime of x is equal to 2x minus 6."}, {"video_title": "Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "So it's 4 minus 12 plus 8, that's going to be 0. So this is equal to 3 over 3, which is equal to 1. f prime of c needs to be equal to 1. And so what is the derivative of this? Well, let's see, f prime of x is equal to 2x minus 6. And so we need to figure out at what x value, especially it has to be within this open interval, at what x value is it equal to 1? So this needs to be equal to 1. So let's add 6 to both sides."}, {"video_title": "Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "Well, let's see, f prime of x is equal to 2x minus 6. And so we need to figure out at what x value, especially it has to be within this open interval, at what x value is it equal to 1? So this needs to be equal to 1. So let's add 6 to both sides. You get 2x is equal to 7. x is equal to 7 halves, which is the same thing as 3 and a half. So it's definitely in this interval right over here. So we've just found our c. c is equal to 7 halves."}, {"video_title": "Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "So let's add 6 to both sides. You get 2x is equal to 7. x is equal to 7 halves, which is the same thing as 3 and a half. So it's definitely in this interval right over here. So we've just found our c. c is equal to 7 halves. And let's just graph this to really make sure that this makes sense. So this right over here is our y-axis. And then this right over here is our x-axis."}, {"video_title": "Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "So we've just found our c. c is equal to 7 halves. And let's just graph this to really make sure that this makes sense. So this right over here is our y-axis. And then this right over here is our x-axis. Looks like all the action is happening in the first and fourth quadrants. So that is our x-axis. Let's say this is 1, 2, 3, 4, and 5."}, {"video_title": "Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "And then this right over here is our x-axis. Looks like all the action is happening in the first and fourth quadrants. So that is our x-axis. Let's say this is 1, 2, 3, 4, and 5. So we already know that 2 is one of the 0s here. So we know that our function, if we wanted to graph it, it intersects the x-axis right over here. And you can factor this out as x minus 2 times x minus 4."}, {"video_title": "Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say this is 1, 2, 3, 4, and 5. So we already know that 2 is one of the 0s here. So we know that our function, if we wanted to graph it, it intersects the x-axis right over here. And you can factor this out as x minus 2 times x minus 4. So the other place where our function hits 0 is when x is equal to 4 right over here. Our vertex is going to be right in between at x is equal to 3. When x is equal to 3, let's see, 9 minus 18 is negative 9 plus 8, so negative 1."}, {"video_title": "Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "And you can factor this out as x minus 2 times x minus 4. So the other place where our function hits 0 is when x is equal to 4 right over here. Our vertex is going to be right in between at x is equal to 3. When x is equal to 3, let's see, 9 minus 18 is negative 9 plus 8, so negative 1. So you have the point 3, negative 1 on it. So that's enough for us to graph it. And we also know at 5 we're at 3."}, {"video_title": "Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "When x is equal to 3, let's see, 9 minus 18 is negative 9 plus 8, so negative 1. So you have the point 3, negative 1 on it. So that's enough for us to graph it. And we also know at 5 we're at 3. So 1, 2, 3. So at 5 we are right over here. So over the interval that we care about, our graph looks something like this."}, {"video_title": "Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "And we also know at 5 we're at 3. So 1, 2, 3. So at 5 we are right over here. So over the interval that we care about, our graph looks something like this. Our graph looks something like this. So that's the interval that we care about. And we're saying that we were looking for a c whose slope is the same as the slope of the secant line, same as the slope of the line between these two points."}, {"video_title": "Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "So over the interval that we care about, our graph looks something like this. Our graph looks something like this. So that's the interval that we care about. And we're saying that we were looking for a c whose slope is the same as the slope of the secant line, same as the slope of the line between these two points. And if I were to just visually look at it, I'd say, well, yeah, it looks like right around there, just based on my drawing, the slope of the tangent line looks like it's parallel. It looks like it has the same slope. It looks like the tangent line is parallel to the secant line."}, {"video_title": "Mean value theorem example polynomial Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "And we're saying that we were looking for a c whose slope is the same as the slope of the secant line, same as the slope of the line between these two points. And if I were to just visually look at it, I'd say, well, yeah, it looks like right around there, just based on my drawing, the slope of the tangent line looks like it's parallel. It looks like it has the same slope. It looks like the tangent line is parallel to the secant line. And that looks like it's right at 3 1\u20442 or 7 halves. So it makes sense. So this right over here is our c. And it's equal to 7 halves."}, {"video_title": "Analyzing related rates problems equations (Pythagoras) AP Calculus AB Khan Academy.mp3", "Sentence": "At a certain instant, t sub zero, the first car is a distance x sub t sub zero, or x of t sub zero, of half a kilometer from the intersection, and the second car is a distance y of t sub zero of 1.2 kilometers from the intersection. What is the rate of change of the distance, d of t, between the cars at that instant, so at t sub zero? Which equation should be used to solve the problem? And they give us a choice of four equations right over here. So you could pause the video and try to work through it on your own, but I'm about to do it as well. So let's just draw what's going on. That's always a healthy thing to do."}, {"video_title": "Analyzing related rates problems equations (Pythagoras) AP Calculus AB Khan Academy.mp3", "Sentence": "And they give us a choice of four equations right over here. So you could pause the video and try to work through it on your own, but I'm about to do it as well. So let's just draw what's going on. That's always a healthy thing to do. So two cars are driving towards an intersection from perpendicular directions. So let's say that this is one car right over here, and it is moving in the x direction towards that intersection, which is right over there. And then you have another car that is moving in the y direction."}, {"video_title": "Analyzing related rates problems equations (Pythagoras) AP Calculus AB Khan Academy.mp3", "Sentence": "That's always a healthy thing to do. So two cars are driving towards an intersection from perpendicular directions. So let's say that this is one car right over here, and it is moving in the x direction towards that intersection, which is right over there. And then you have another car that is moving in the y direction. So let's say it's moving like this. So this is the other car. I should've maybe done a top view."}, {"video_title": "Analyzing related rates problems equations (Pythagoras) AP Calculus AB Khan Academy.mp3", "Sentence": "And then you have another car that is moving in the y direction. So let's say it's moving like this. So this is the other car. I should've maybe done a top view. Oh, here we go. This square represents the car, and it is moving in that direction. Now, they say at a certain instant, t sub zero."}, {"video_title": "Analyzing related rates problems equations (Pythagoras) AP Calculus AB Khan Academy.mp3", "Sentence": "I should've maybe done a top view. Oh, here we go. This square represents the car, and it is moving in that direction. Now, they say at a certain instant, t sub zero. So let's draw that instant. So the first car is a distance x of t sub zero of 0.5 kilometers. So this distance right over here, let's just call this x of t, and let's call this distance right over here y of t. Now, how does the distance between the cars relate to x of t and y of t?"}, {"video_title": "Analyzing related rates problems equations (Pythagoras) AP Calculus AB Khan Academy.mp3", "Sentence": "Now, they say at a certain instant, t sub zero. So let's draw that instant. So the first car is a distance x of t sub zero of 0.5 kilometers. So this distance right over here, let's just call this x of t, and let's call this distance right over here y of t. Now, how does the distance between the cars relate to x of t and y of t? Well, we could just use the distance formula, which is essentially just the Pythagorean theorem, to say, well, the distance between the cars would be the hypotenuse of this right triangle. Remember, they're traveling from perpendicular directions, so that's a right triangle there. So this distance right over here would be x of t squared plus y of t squared and the square root of that, and that's just the Pythagorean theorem right over here."}, {"video_title": "Analyzing related rates problems equations (Pythagoras) AP Calculus AB Khan Academy.mp3", "Sentence": "So this distance right over here, let's just call this x of t, and let's call this distance right over here y of t. Now, how does the distance between the cars relate to x of t and y of t? Well, we could just use the distance formula, which is essentially just the Pythagorean theorem, to say, well, the distance between the cars would be the hypotenuse of this right triangle. Remember, they're traveling from perpendicular directions, so that's a right triangle there. So this distance right over here would be x of t squared plus y of t squared and the square root of that, and that's just the Pythagorean theorem right over here. This would be d of t, or we could say that d of t squared is equal to x of t squared plus y, let me, too many parentheses, plus y of t squared. So that's the relationship between d of t, x of t, and y of t, and it's useful for solving this problem because now we can take the derivative of both sides of this equation with respect to t, and we'd be using various derivative rules, including the chain rule, in order to do it, and then that would give us a relationship between the rate of change of d of t, which would be d prime of t, and the rate of change of x of t, y of t, and x of t and y of t themselves. And so if we look at these choices right over here, we indeed see that d sets up that exact same relationship that we just did ourselves, that it shows that the distance squared between the cars is equal to that x distance from the intersection squared plus the y distance from the intersection squared, and then we can take the derivative of both sides to actually figure out this related rates question."}, {"video_title": "Differentiability at a point algebraic (function is differentiable) AP Calculus AB Khan Academy.mp3", "Sentence": "Now one of these we can knock out right from the get-go. You cannot, in order to be differentiable, you need to be continuous there, so you cannot have differentiable but not continuous, so let's just rule that one out. And now let's think about continuity. So let's first think about continuity. And frankly, if it isn't continuous, then it's not going to be differentiable. So let's think about it a little bit. So in order to be continuous, f of, let me use a darker color, f of three needs to be equal to the limit of f of x as x approaches three."}, {"video_title": "Differentiability at a point algebraic (function is differentiable) AP Calculus AB Khan Academy.mp3", "Sentence": "So let's first think about continuity. And frankly, if it isn't continuous, then it's not going to be differentiable. So let's think about it a little bit. So in order to be continuous, f of, let me use a darker color, f of three needs to be equal to the limit of f of x as x approaches three. Now what is f of three? Well, let's see, we fall into this case right over here, because x is equal to three. So six times three is 18, minus nine is nine, so this is nine."}, {"video_title": "Differentiability at a point algebraic (function is differentiable) AP Calculus AB Khan Academy.mp3", "Sentence": "So in order to be continuous, f of, let me use a darker color, f of three needs to be equal to the limit of f of x as x approaches three. Now what is f of three? Well, let's see, we fall into this case right over here, because x is equal to three. So six times three is 18, minus nine is nine, so this is nine. So the limit of f of x as x approaches three needs to be equal to nine. So let's first think about the limit as we approach from the left-hand side, the limit as x approaches three, as x approaches three from the left-hand side of f of x. Well, when x is less than three, we fall into this case."}, {"video_title": "Differentiability at a point algebraic (function is differentiable) AP Calculus AB Khan Academy.mp3", "Sentence": "So six times three is 18, minus nine is nine, so this is nine. So the limit of f of x as x approaches three needs to be equal to nine. So let's first think about the limit as we approach from the left-hand side, the limit as x approaches three, as x approaches three from the left-hand side of f of x. Well, when x is less than three, we fall into this case. So f of x is just going to be equal to x squared. And so this is defined and continuous for all real numbers, so we can just substitute the three in there. So this is going to be equal to nine."}, {"video_title": "Differentiability at a point algebraic (function is differentiable) AP Calculus AB Khan Academy.mp3", "Sentence": "Well, when x is less than three, we fall into this case. So f of x is just going to be equal to x squared. And so this is defined and continuous for all real numbers, so we can just substitute the three in there. So this is going to be equal to nine. Now what's the limit of, as we approach three from the right-hand side of f of x? Well, as we approach from the right, this one right over here is, f of x is equal to six x minus nine. So we just write six x minus nine."}, {"video_title": "Differentiability at a point algebraic (function is differentiable) AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be equal to nine. Now what's the limit of, as we approach three from the right-hand side of f of x? Well, as we approach from the right, this one right over here is, f of x is equal to six x minus nine. So we just write six x minus nine. And once again, six x minus nine is defined and continuous for all real numbers, so we can just pop a three in there, and you get 18 minus nine. Well, this is also equal to nine. So the right and left-hand, the left and right-hand limits both equal nine, which is equal to the value of the function there, so it is definitely continuous."}, {"video_title": "Differentiability at a point algebraic (function is differentiable) AP Calculus AB Khan Academy.mp3", "Sentence": "So we just write six x minus nine. And once again, six x minus nine is defined and continuous for all real numbers, so we can just pop a three in there, and you get 18 minus nine. Well, this is also equal to nine. So the right and left-hand, the left and right-hand limits both equal nine, which is equal to the value of the function there, so it is definitely continuous. So we can rule out, we can rule out this choice right over there. And now let's think about differentiability. So in order to be differentiable, I'll just, differentiable, in order to be differentiable, the limit as x approaches three of f of x minus f of three over x minus three needs to exist."}, {"video_title": "Differentiability at a point algebraic (function is differentiable) AP Calculus AB Khan Academy.mp3", "Sentence": "So the right and left-hand, the left and right-hand limits both equal nine, which is equal to the value of the function there, so it is definitely continuous. So we can rule out, we can rule out this choice right over there. And now let's think about differentiability. So in order to be differentiable, I'll just, differentiable, in order to be differentiable, the limit as x approaches three of f of x minus f of three over x minus three needs to exist. So let's see if we can evaluate this. So first of all, we know what f of three is. F of three, we've already evaluated this."}, {"video_title": "Differentiability at a point algebraic (function is differentiable) AP Calculus AB Khan Academy.mp3", "Sentence": "So in order to be differentiable, I'll just, differentiable, in order to be differentiable, the limit as x approaches three of f of x minus f of three over x minus three needs to exist. So let's see if we can evaluate this. So first of all, we know what f of three is. F of three, we've already evaluated this. This is going to be nine. And let's see if we can evaluate this limit, or let's see what the limit is as we approach from the left-hand side and the right-hand side. And if they're approaching the same thing, then we know that this, and that same thing that they're approaching is the limit."}, {"video_title": "Differentiability at a point algebraic (function is differentiable) AP Calculus AB Khan Academy.mp3", "Sentence": "F of three, we've already evaluated this. This is going to be nine. And let's see if we can evaluate this limit, or let's see what the limit is as we approach from the left-hand side and the right-hand side. And if they're approaching the same thing, then we know that this, and that same thing that they're approaching is the limit. So let's first think about the limit as x approaches three from the left-hand side. So it's over x minus three, and we have f of x minus nine. But as we approach from the left-hand side, this is f of x."}, {"video_title": "Differentiability at a point algebraic (function is differentiable) AP Calculus AB Khan Academy.mp3", "Sentence": "And if they're approaching the same thing, then we know that this, and that same thing that they're approaching is the limit. So let's first think about the limit as x approaches three from the left-hand side. So it's over x minus three, and we have f of x minus nine. But as we approach from the left-hand side, this is f of x. As x is less than three, f of x is equal to x squared. So this would be, instead of f of x minus nine, I'll write x squared minus nine. And x squared minus nine, this is a difference of squares, so this is x plus three times x minus three."}, {"video_title": "Differentiability at a point algebraic (function is differentiable) AP Calculus AB Khan Academy.mp3", "Sentence": "But as we approach from the left-hand side, this is f of x. As x is less than three, f of x is equal to x squared. So this would be, instead of f of x minus nine, I'll write x squared minus nine. And x squared minus nine, this is a difference of squares, so this is x plus three times x minus three. X plus three times x minus three. And so these would cancel out. We can say that this equivalent to x plus three, as long as x does not equal three, and that's okay, because we're approaching from the left."}, {"video_title": "Differentiability at a point algebraic (function is differentiable) AP Calculus AB Khan Academy.mp3", "Sentence": "And x squared minus nine, this is a difference of squares, so this is x plus three times x minus three. X plus three times x minus three. And so these would cancel out. We can say that this equivalent to x plus three, as long as x does not equal three, and that's okay, because we're approaching from the left. And as we approach from the left, well, x plus three is defined for all real numbers, it's continuous for all real numbers, so we can just substitute the three in there. So we would get a six. So now let's try to evaluate the limit as we approach from the right-hand side."}, {"video_title": "Differentiability at a point algebraic (function is differentiable) AP Calculus AB Khan Academy.mp3", "Sentence": "We can say that this equivalent to x plus three, as long as x does not equal three, and that's okay, because we're approaching from the left. And as we approach from the left, well, x plus three is defined for all real numbers, it's continuous for all real numbers, so we can just substitute the three in there. So we would get a six. So now let's try to evaluate the limit as we approach from the right-hand side. So once again, it's f of x, but as we approach from the right-hand side, f of x is six x minus nine. That's our f of x. And then we have minus f of three, which is nine."}, {"video_title": "Differentiability at a point algebraic (function is differentiable) AP Calculus AB Khan Academy.mp3", "Sentence": "So now let's try to evaluate the limit as we approach from the right-hand side. So once again, it's f of x, but as we approach from the right-hand side, f of x is six x minus nine. That's our f of x. And then we have minus f of three, which is nine. So it's six x minus 18. Six x minus 18. Well, that's the same thing as six times x minus three."}, {"video_title": "Differentiability at a point algebraic (function is differentiable) AP Calculus AB Khan Academy.mp3", "Sentence": "And then we have minus f of three, which is nine. So it's six x minus 18. Six x minus 18. Well, that's the same thing as six times x minus three. And as we approach from the right, well, that's just going to be equal to six. So it looks like our derivative exists there, and it is equal to the limit as x approaches three of all of this business is equal to six, because the limit as we approach from the left and the right is also equal to six. So this looks like we are both continuous and both continuous and differentiable."}, {"video_title": "Volume with cross sections triangle AP Calculus AB Khan Academy.mp3", "Sentence": "But then it's kind of popping out of our screen. What I've drawn here in blue, you could view this as kind of the top ridge of the figure. And if you were to take cross-sections of the figure, and that's what this yellow line is, if you were to take cross-sections of this figure that are vertical, that are, I should say, perpendicular to the x-axis, those cross-sections are going to be isosceles-right triangles. So this cross-section is going to look like this if you were to look at, if you were to flatten it out. So over here it's sitting, it's popping out of your page or out of your screen, but if you were to actually flatten it out, the cross-section would look like this. It's going to be an isosceles-right triangle where the hypotenuse of the isosceles-right triangle sits along the base. So it's isosceles, so that's equal to that."}, {"video_title": "Volume with cross sections triangle AP Calculus AB Khan Academy.mp3", "Sentence": "So this cross-section is going to look like this if you were to look at, if you were to flatten it out. So over here it's sitting, it's popping out of your page or out of your screen, but if you were to actually flatten it out, the cross-section would look like this. It's going to be an isosceles-right triangle where the hypotenuse of the isosceles-right triangle sits along the base. So it's isosceles, so that's equal to that. It's a right triangle. And then this distance between that point and this point is the same as the distance between f of x and g of x for this x value right over there. And obviously that changes as we change our x value."}, {"video_title": "Volume with cross sections triangle AP Calculus AB Khan Academy.mp3", "Sentence": "So it's isosceles, so that's equal to that. It's a right triangle. And then this distance between that point and this point is the same as the distance between f of x and g of x for this x value right over there. And obviously that changes as we change our x value. And to help us visualize the shape here, I've kind of drawn a picture of our coordinate plane. If we view it as an angle, if we're kind of above it, you can kind of start to see how this figure would look. Once again, I've drawn the base of it."}, {"video_title": "Volume with cross sections triangle AP Calculus AB Khan Academy.mp3", "Sentence": "And obviously that changes as we change our x value. And to help us visualize the shape here, I've kind of drawn a picture of our coordinate plane. If we view it as an angle, if we're kind of above it, you can kind of start to see how this figure would look. Once again, I've drawn the base of it. I've drawn the base of it right over there. Maybe I should, to make it clear, let me make it like this. Let me shade it in kind of parallel to these cross-sections."}, {"video_title": "Volume with cross sections triangle AP Calculus AB Khan Academy.mp3", "Sentence": "Once again, I've drawn the base of it. I've drawn the base of it right over there. Maybe I should, to make it clear, let me make it like this. Let me shade it in kind of parallel to these cross-sections. So I've drawn the base right over there. There's two other sides. There's the side that's on, at least the way I've drawn it here, I guess you could view it as its top side or the left side right over there."}, {"video_title": "Volume with cross sections triangle AP Calculus AB Khan Academy.mp3", "Sentence": "Let me shade it in kind of parallel to these cross-sections. So I've drawn the base right over there. There's two other sides. There's the side that's on, at least the way I've drawn it here, I guess you could view it as its top side or the left side right over there. And over on this picture, that would be this when we're looking at it from above. And then you have this other side. I guess on this view, this would be called kind of the right side."}, {"video_title": "Volume with cross sections triangle AP Calculus AB Khan Academy.mp3", "Sentence": "There's the side that's on, at least the way I've drawn it here, I guess you could view it as its top side or the left side right over there. And over on this picture, that would be this when we're looking at it from above. And then you have this other side. I guess on this view, this would be called kind of the right side. And over here, this is kind of the, when you view it over here, this is the bottom side. So the whole reason why I've set this up and we're attempting to visualize this figure is I want to see if you can come up with a definite integral that describes the volume of this figure that kind of almost looks like a football if you cut it in half or a rugby ball, but it's skewed a little bit as well. But what's an expression, a definite integral that expresses the volume of this?"}, {"video_title": "Volume with cross sections triangle AP Calculus AB Khan Academy.mp3", "Sentence": "I guess on this view, this would be called kind of the right side. And over here, this is kind of the, when you view it over here, this is the bottom side. So the whole reason why I've set this up and we're attempting to visualize this figure is I want to see if you can come up with a definite integral that describes the volume of this figure that kind of almost looks like a football if you cut it in half or a rugby ball, but it's skewed a little bit as well. But what's an expression, a definite integral that expresses the volume of this? And I encourage you to use the fact that it intersects at the point, these functions intersect at the point 0, 0 and c, d. So can you come up with some expression, a definite integral in terms of zeros and c's and d's and f's and g's that describe the volume of this figure? So I'm assuming you've paused the video and had a go at it, so let's think about it. So if we want to find the volume, one way to think about it is we could take, we could take the volume of, we could approximate the volume as the volume of these individual triangles."}, {"video_title": "Volume with cross sections triangle AP Calculus AB Khan Academy.mp3", "Sentence": "But what's an expression, a definite integral that expresses the volume of this? And I encourage you to use the fact that it intersects at the point, these functions intersect at the point 0, 0 and c, d. So can you come up with some expression, a definite integral in terms of zeros and c's and d's and f's and g's that describe the volume of this figure? So I'm assuming you've paused the video and had a go at it, so let's think about it. So if we want to find the volume, one way to think about it is we could take, we could take the volume of, we could approximate the volume as the volume of these individual triangles. So that would be the area of each of these triangles times some very small depth, some very small depth, so let me just shade it in to show the depth, some very small depth, which we could call, we could call dx. So once again, we could find the volume of each of these by finding the area, the cross-sectional area there, and then multiplying that times a little bit, a little dx, a little dx, which would give us 3, so this is a little dx, which would give us 3-dimensional, that's our dx, I could write that a little bit neater, dx to give us a little bit of 3-dimensional depth. So how could we, what is the volume of one of these figures going to look like?"}, {"video_title": "Volume with cross sections triangle AP Calculus AB Khan Academy.mp3", "Sentence": "So if we want to find the volume, one way to think about it is we could take, we could take the volume of, we could approximate the volume as the volume of these individual triangles. So that would be the area of each of these triangles times some very small depth, some very small depth, so let me just shade it in to show the depth, some very small depth, which we could call, we could call dx. So once again, we could find the volume of each of these by finding the area, the cross-sectional area there, and then multiplying that times a little bit, a little dx, a little dx, which would give us 3, so this is a little dx, which would give us 3-dimensional, that's our dx, I could write that a little bit neater, dx to give us a little bit of 3-dimensional depth. So how could we, what is the volume of one of these figures going to look like? Well, if we say, let me just call this height h, and we know that h is going to be f of x minus g of x, that's this distance right over here, so let's call that h. We know that h is going to be, and maybe I should say h of x, because it is going to be a function of x, h of x is going to be f of x, f of x minus g of x, minus g of x. But given an h, what is going to be the area of this triangle? Well, this is a 45, 45, 90 triangle."}, {"video_title": "Volume with cross sections triangle AP Calculus AB Khan Academy.mp3", "Sentence": "So how could we, what is the volume of one of these figures going to look like? Well, if we say, let me just call this height h, and we know that h is going to be f of x minus g of x, that's this distance right over here, so let's call that h. We know that h is going to be, and maybe I should say h of x, because it is going to be a function of x, h of x is going to be f of x, f of x minus g of x, minus g of x. But given an h, what is going to be the area of this triangle? Well, this is a 45, 45, 90 triangle. If this is 90, then this is going to have to be 45 degrees, that's going to have to be 45 degrees, and we know that the sides of a 45, 45, 90 triangle are square root of 2 times the hypotenuse. So this is going to be square root of 2, sorry, square root of 2 over 2 times the hypotenuse. Square root of 2 over 2 times the hypotenuse."}, {"video_title": "Volume with cross sections triangle AP Calculus AB Khan Academy.mp3", "Sentence": "Well, this is a 45, 45, 90 triangle. If this is 90, then this is going to have to be 45 degrees, that's going to have to be 45 degrees, and we know that the sides of a 45, 45, 90 triangle are square root of 2 times the hypotenuse. So this is going to be square root of 2, sorry, square root of 2 over 2 times the hypotenuse. Square root of 2 over 2 times the hypotenuse. And you can get that straight from the Pythagorean theorem. If this side, let's say it's length a, then this side has length a, so you're going to have a squared plus a squared is equal to the hypotenuse squared, or 2a squared is equal to the hypotenuse squared, or that a squared is equal to the hypotenuse squared over 2, or that a is equal to h over the square root of 2, which is the same thing as the square root of 2h over 2. I just rationalized the denominator, multiplied both the numerator and the denominator by square root of 2."}, {"video_title": "Volume with cross sections triangle AP Calculus AB Khan Academy.mp3", "Sentence": "Square root of 2 over 2 times the hypotenuse. And you can get that straight from the Pythagorean theorem. If this side, let's say it's length a, then this side has length a, so you're going to have a squared plus a squared is equal to the hypotenuse squared, or 2a squared is equal to the hypotenuse squared, or that a squared is equal to the hypotenuse squared over 2, or that a is equal to h over the square root of 2, which is the same thing as the square root of 2h over 2. I just rationalized the denominator, multiplied both the numerator and the denominator by square root of 2. So that's where I got this from. And so what is the area going to be? Well, the area is just going to be your base times your height times 1 half."}, {"video_title": "Volume with cross sections triangle AP Calculus AB Khan Academy.mp3", "Sentence": "I just rationalized the denominator, multiplied both the numerator and the denominator by square root of 2. So that's where I got this from. And so what is the area going to be? Well, the area is just going to be your base times your height times 1 half. So let me write that down. So the area there, the area is just going to be the base, which is square root of 2 over 2 times our hypotenuse, times the height, which is square root of 2 over 2 times our hypotenuse, times 1 half. If we didn't do that 1 half, we'd be figuring out the area of this entire square."}, {"video_title": "Volume with cross sections triangle AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the area is just going to be your base times your height times 1 half. So let me write that down. So the area there, the area is just going to be the base, which is square root of 2 over 2 times our hypotenuse, times the height, which is square root of 2 over 2 times our hypotenuse, times 1 half. If we didn't do that 1 half, we'd be figuring out the area of this entire square. But this is obviously, we're concerned with the triangle. So what's this going to be? This is going to be square root of 2 over 2 times square root of 2 over 2 is going to be 1 half, and then you're going to have another 1 half."}, {"video_title": "Volume with cross sections triangle AP Calculus AB Khan Academy.mp3", "Sentence": "If we didn't do that 1 half, we'd be figuring out the area of this entire square. But this is obviously, we're concerned with the triangle. So what's this going to be? This is going to be square root of 2 over 2 times square root of 2 over 2 is going to be 1 half, and then you're going to have another 1 half. So it's 1 fourth h squared. Did I do that right? This is, yes, it's going to be 2 over 4, which is 1 half, and then times another 1 half is 1 fourth h squared is the area."}, {"video_title": "Volume with cross sections triangle AP Calculus AB Khan Academy.mp3", "Sentence": "This is going to be square root of 2 over 2 times square root of 2 over 2 is going to be 1 half, and then you're going to have another 1 half. So it's 1 fourth h squared. Did I do that right? This is, yes, it's going to be 2 over 4, which is 1 half, and then times another 1 half is 1 fourth h squared is the area. Now what's going to be the volume of each of these triangles? Well, the volume of each of these triangles right over here, the volume is just going to be our area times dx, or it's just going to be 1 fourth h squared times our depth, dx. And so if we just integrated a bunch of these from our x equals 0 all the way to x equals c, we essentially have our volume of the entire figure."}, {"video_title": "Volume with cross sections triangle AP Calculus AB Khan Academy.mp3", "Sentence": "This is, yes, it's going to be 2 over 4, which is 1 half, and then times another 1 half is 1 fourth h squared is the area. Now what's going to be the volume of each of these triangles? Well, the volume of each of these triangles right over here, the volume is just going to be our area times dx, or it's just going to be 1 fourth h squared times our depth, dx. And so if we just integrated a bunch of these from our x equals 0 all the way to x equals c, we essentially have our volume of the entire figure. So how could we write that? So we want to write, we want to essentially define the volume of the figure, and we're kind of in the home stretch here. Actually, let me write it right."}, {"video_title": "Volume with cross sections triangle AP Calculus AB Khan Academy.mp3", "Sentence": "And so if we just integrated a bunch of these from our x equals 0 all the way to x equals c, we essentially have our volume of the entire figure. So how could we write that? So we want to write, we want to essentially define the volume of the figure, and we're kind of in the home stretch here. Actually, let me write it right. This is volume of a section, volume of a cross section. So what's the volume of the entire thing going to be? Well, the volume of the figure is going to be the definite integral from x equals 0 to x equals c of 1 fourth h squared."}, {"video_title": "Volume with cross sections triangle AP Calculus AB Khan Academy.mp3", "Sentence": "Actually, let me write it right. This is volume of a section, volume of a cross section. So what's the volume of the entire thing going to be? Well, the volume of the figure is going to be the definite integral from x equals 0 to x equals c of 1 fourth h squared. 1 fourth, but we know that h is equal to f of x minus g of x. So instead of h, I'm going to write f of x minus g of x squared dx. And we're done."}, {"video_title": "One-sided limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "This table gives select values of f. We have our table here for these x values. It gives the corresponding f of x. What is a reasonable estimate for the limit of f of x as x approaches one from the left? So pause this video and see if you can figure it out on your own. Alright, now let's work through this together. So the first thing that is really important to realize is when you see this x approaches one and you see this little negative superscript here, this does not mean approaching negative one. So this does not mean negative one."}, {"video_title": "One-sided limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So pause this video and see if you can figure it out on your own. Alright, now let's work through this together. So the first thing that is really important to realize is when you see this x approaches one and you see this little negative superscript here, this does not mean approaching negative one. So this does not mean negative one. Sometimes your brain just sees a one and that little negative sign there, and you're like, oh, this must be a weird way of writing negative one, or you don't even think about it. But it's not saying that. It's saying, this is saying, let me put a little arrow here, this is the limit of f of x as x approaches one from the left."}, {"video_title": "One-sided limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So this does not mean negative one. Sometimes your brain just sees a one and that little negative sign there, and you're like, oh, this must be a weird way of writing negative one, or you don't even think about it. But it's not saying that. It's saying, this is saying, let me put a little arrow here, this is the limit of f of x as x approaches one from the left. From the left. So from the left, how do we know that? Well, that's what that little negative tells us."}, {"video_title": "One-sided limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "It's saying, this is saying, let me put a little arrow here, this is the limit of f of x as x approaches one from the left. From the left. So from the left, how do we know that? Well, that's what that little negative tells us. It tells us we're approaching one from values less than one. If we're approaching one from the right, from values greater than one, that would be a positive sign right over there. So let's think about it."}, {"video_title": "One-sided limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Well, that's what that little negative tells us. It tells us we're approaching one from values less than one. If we're approaching one from the right, from values greater than one, that would be a positive sign right over there. So let's think about it. We want the limit as x approaches one from the left. And lucky for us on this table, we have some values of x approaching one from the left. 0.9, which is already pretty close to one, then we get even closer to one from the left."}, {"video_title": "One-sided limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So let's think about it. We want the limit as x approaches one from the left. And lucky for us on this table, we have some values of x approaching one from the left. 0.9, which is already pretty close to one, then we get even closer to one from the left. Notice, these are all less than one, but they're getting closer and closer to one. And so what we really wanna look at is the value, what does f of x approach as x is getting closer and closer, let me write it, as x is getting closer and closer to one from the left, from the left. And a key realization here is, if we're thinking about general limits, not just from one direction, then we might wanna look at from the left and from the right."}, {"video_title": "One-sided limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "0.9, which is already pretty close to one, then we get even closer to one from the left. Notice, these are all less than one, but they're getting closer and closer to one. And so what we really wanna look at is the value, what does f of x approach as x is getting closer and closer, let me write it, as x is getting closer and closer to one from the left, from the left. And a key realization here is, if we're thinking about general limits, not just from one direction, then we might wanna look at from the left and from the right. But they're asking us only from the left, so we should only be looking at these values right over here. In fact, we shouldn't even let the value of f of x at x equal one confuse us. Sometimes, and oftentimes, the limit is approaching a different value than the value of the function at that point."}, {"video_title": "One-sided limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And a key realization here is, if we're thinking about general limits, not just from one direction, then we might wanna look at from the left and from the right. But they're asking us only from the left, so we should only be looking at these values right over here. In fact, we shouldn't even let the value of f of x at x equal one confuse us. Sometimes, and oftentimes, the limit is approaching a different value than the value of the function at that point. So let's look at this. At 0.9, f of x is 2.5. When we get even closer to one from the left, we go to 2.1."}, {"video_title": "One-sided limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Sometimes, and oftentimes, the limit is approaching a different value than the value of the function at that point. So let's look at this. At 0.9, f of x is 2.5. When we get even closer to one from the left, we go to 2.1. When we get even closer to one from the left, we're getting even closer to two. So a reasonable estimate for the limit as x approaches one from the left of f of x, it looks like x, it looks like f of x, right over here, is approaching two. We don't know for sure."}, {"video_title": "One-sided limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "When we get even closer to one from the left, we go to 2.1. When we get even closer to one from the left, we're getting even closer to two. So a reasonable estimate for the limit as x approaches one from the left of f of x, it looks like x, it looks like f of x, right over here, is approaching two. We don't know for sure. That's why they're saying what is a reasonable estimate. It might be approaching 2.01, or it might be approaching 1.999. On Khan Academy, these will often be multiple choice questions, so you have to pick the most reasonable one."}, {"video_title": "One-sided limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "We don't know for sure. That's why they're saying what is a reasonable estimate. It might be approaching 2.01, or it might be approaching 1.999. On Khan Academy, these will often be multiple choice questions, so you have to pick the most reasonable one. It would not be fair if they gave 1.999 as a choice and 2.01. But if you were saying, hey, maybe this is approaching a whole number, then two could be a reasonable estimate right over here, although it doesn't have to be two. It could be 2.01258."}, {"video_title": "One-sided limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "On Khan Academy, these will often be multiple choice questions, so you have to pick the most reasonable one. It would not be fair if they gave 1.999 as a choice and 2.01. But if you were saying, hey, maybe this is approaching a whole number, then two could be a reasonable estimate right over here, although it doesn't have to be two. It could be 2.01258. It might be what it is actually approaching. So let's try another example here. Here it does look like there's a reasonable estimate for the limit as we approach this value from the left."}, {"video_title": "One-sided limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "It could be 2.01258. It might be what it is actually approaching. So let's try another example here. Here it does look like there's a reasonable estimate for the limit as we approach this value from the left. So now it says the function f is defined over the real numbers. This table gives select values of f, similar to the last question. What is a reasonable estimate for the limit as x approaches negative two from the left?"}, {"video_title": "One-sided limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Here it does look like there's a reasonable estimate for the limit as we approach this value from the left. So now it says the function f is defined over the real numbers. This table gives select values of f, similar to the last question. What is a reasonable estimate for the limit as x approaches negative two from the left? So this is confusing. You see these two negative signs. This first negative sign tells us we're approaching negative two."}, {"video_title": "One-sided limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "What is a reasonable estimate for the limit as x approaches negative two from the left? So this is confusing. You see these two negative signs. This first negative sign tells us we're approaching negative two. We wanna say what happens when we're approaching negative two and we're gonna approach, once again, from the left. So lucky for us, they have values of x that are approaching negative two from the left. So this is x approaches negative two from the left."}, {"video_title": "One-sided limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "This first negative sign tells us we're approaching negative two. We wanna say what happens when we're approaching negative two and we're gonna approach, once again, from the left. So lucky for us, they have values of x that are approaching negative two from the left. So this is x approaches negative two from the left. And so that is happening right over here. So that's these values. So notice, this is negative 2.05."}, {"video_title": "One-sided limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So this is x approaches negative two from the left. And so that is happening right over here. So that's these values. So notice, this is negative 2.05. Then we get even closer, negative 2.01. Then we get even closer, negative 2.002. And these are from the left because these are values less than negative two, but they're getting closer and closer to negative two."}, {"video_title": "One-sided limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So notice, this is negative 2.05. Then we get even closer, negative 2.01. Then we get even closer, negative 2.002. And these are from the left because these are values less than negative two, but they're getting closer and closer to negative two. And so let's see. When we're a little bit further, f of x is negative 20. We get a little bit closer, it's negative 100."}, {"video_title": "One-sided limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And these are from the left because these are values less than negative two, but they're getting closer and closer to negative two. And so let's see. When we're a little bit further, f of x is negative 20. We get a little bit closer, it's negative 100. We get even a little bit closer, it goes to negative 500. So it looks it would be reasonable, and we don't know for sure. This is just giving us a few sample points for this function."}, {"video_title": "One-sided limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "We get a little bit closer, it's negative 100. We get even a little bit closer, it goes to negative 500. So it looks it would be reasonable, and we don't know for sure. This is just giving us a few sample points for this function. But if we follow this trend, as we get closer and closer to two, as we get closer and closer to negative two without getting there, it looks like this is getting unbounded. It looks like it's becoming infinitely negative. And so technically, it looks like this is, I would write this is unbounded."}, {"video_title": "One-sided limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "This is just giving us a few sample points for this function. But if we follow this trend, as we get closer and closer to two, as we get closer and closer to negative two without getting there, it looks like this is getting unbounded. It looks like it's becoming infinitely negative. And so technically, it looks like this is, I would write this is unbounded. Unbounded. And so if this was a multiple choice question, technically you would say this, the limit as x approaches negative two from the left does not exist. Does not exist."}, {"video_title": "One-sided limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And so technically, it looks like this is, I would write this is unbounded. Unbounded. And so if this was a multiple choice question, technically you would say this, the limit as x approaches negative two from the left does not exist. Does not exist. If someone asked the other question, if they said what is the limit as x approaches negative two from the right of f of x, well then you would say, all right, well here are values approaching negative two from the right. So this is x approaching negative two from the right, right over here. And remember, when you're looking at a limit, sometimes it might be distracting to look at the actual value of the function at that point."}, {"video_title": "One-sided limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Does not exist. If someone asked the other question, if they said what is the limit as x approaches negative two from the right of f of x, well then you would say, all right, well here are values approaching negative two from the right. So this is x approaching negative two from the right, right over here. And remember, when you're looking at a limit, sometimes it might be distracting to look at the actual value of the function at that point. So you wanna think about what is the value of the function approaching as your x is approaching that value, as x is approaching, in this case, negative two from the right. So as we're getting closer and closer to negative two from values larger than negative two, it looks like f of x is getting closer and closer to negative four, which is f of negative two. But that actually seems like a reasonable estimate."}, {"video_title": "One-sided limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And remember, when you're looking at a limit, sometimes it might be distracting to look at the actual value of the function at that point. So you wanna think about what is the value of the function approaching as your x is approaching that value, as x is approaching, in this case, negative two from the right. So as we're getting closer and closer to negative two from values larger than negative two, it looks like f of x is getting closer and closer to negative four, which is f of negative two. But that actually seems like a reasonable estimate. Once again, we don't know absolutely for sure just by sampling some points, but this would be a reasonable estimate. And in general, if you are approaching different values from the left than from the right, then you would say at that point, the limit of your function does not exist. And we've seen that in other videos."}, {"video_title": "Derivative of ln(x) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "It is equal to one over x. In a future video, I'm actually going to prove this. It's a little bit involved. But in this one, we're just going to appreciate that this seems like it is actually true. So right here is the graph of y is equal to the natural log of x. And just to feel good about the statement, let's take the slope, let's try to approximate what the slope of the tangent line is at different points. So let's say right over here, when x is equal to one, what does the slope of the tangent line look like?"}, {"video_title": "Derivative of ln(x) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "But in this one, we're just going to appreciate that this seems like it is actually true. So right here is the graph of y is equal to the natural log of x. And just to feel good about the statement, let's take the slope, let's try to approximate what the slope of the tangent line is at different points. So let's say right over here, when x is equal to one, what does the slope of the tangent line look like? Well, it looks like here, the slope looks like it is equal, pretty close to being equal to one, which is consistent with this statement. If x is equal to one, one over one is still one. And that seems like what we see right over there."}, {"video_title": "Derivative of ln(x) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say right over here, when x is equal to one, what does the slope of the tangent line look like? Well, it looks like here, the slope looks like it is equal, pretty close to being equal to one, which is consistent with this statement. If x is equal to one, one over one is still one. And that seems like what we see right over there. What about when x is equal to two? Well, this point right over here is the natural log of two. But more interestingly, what's the slope here?"}, {"video_title": "Derivative of ln(x) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And that seems like what we see right over there. What about when x is equal to two? Well, this point right over here is the natural log of two. But more interestingly, what's the slope here? Well, it looks like, let's see, if I try to draw a tangent line, the slope of the tangent line looks pretty close to 1 1. Well, once again, that is one over x. One over two is 1 1."}, {"video_title": "Derivative of ln(x) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "But more interestingly, what's the slope here? Well, it looks like, let's see, if I try to draw a tangent line, the slope of the tangent line looks pretty close to 1 1. Well, once again, that is one over x. One over two is 1 1. Let's keep doing this. If I go right over here, when x is equal to four, this point is four comma natural log of four. But the slope of the tangent line here looks pretty close to 1 4."}, {"video_title": "Derivative of ln(x) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "One over two is 1 1. Let's keep doing this. If I go right over here, when x is equal to four, this point is four comma natural log of four. But the slope of the tangent line here looks pretty close to 1 4. And if you accept this, it is exactly 1 4. And you could even go to values less than one. Right over here, when x is equal to 1 1, one over 1 1, the slope should be two."}, {"video_title": "Derivative of ln(x) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "But the slope of the tangent line here looks pretty close to 1 4. And if you accept this, it is exactly 1 4. And you could even go to values less than one. Right over here, when x is equal to 1 1, one over 1 1, the slope should be two. And it does indeed, let me do this in a slightly different color, it does indeed look like the slope is two over there. So, once again, you take the derivative with respect to x of the natural log of x, it is one over x. And hopefully you get a sense that that is actually true here."}, {"video_title": "Formal definition of limits Part 4 using the definition AP Calculus AB Khan Academy.mp3", "Sentence": "It's essentially telling us our distance from c such that if x is within delta of c, then f of x is going to be within epsilon of L. If we can find a delta for any epsilon, then we can say that this is indeed the limit of f of x as x approaches c. Now, I know what you're thinking. This seems all very abstract. I want to somehow use this thing. What we will do in this video is use it and to rigorously prove that a limit actually exists. Right over here, I've defined a function f of x. It's equal to 2x everywhere except for x equals 5. It's 2x everywhere for all the other values of x, but when x is equal to 5, it's just equal to x. I could have really just written 5."}, {"video_title": "Formal definition of limits Part 4 using the definition AP Calculus AB Khan Academy.mp3", "Sentence": "What we will do in this video is use it and to rigorously prove that a limit actually exists. Right over here, I've defined a function f of x. It's equal to 2x everywhere except for x equals 5. It's 2x everywhere for all the other values of x, but when x is equal to 5, it's just equal to x. I could have really just written 5. I could have just written 5 there. It's equal to 5 when x is equal to 5. It's equal to itself."}, {"video_title": "Formal definition of limits Part 4 using the definition AP Calculus AB Khan Academy.mp3", "Sentence": "It's 2x everywhere for all the other values of x, but when x is equal to 5, it's just equal to x. I could have really just written 5. I could have just written 5 there. It's equal to 5 when x is equal to 5. It's equal to itself. We've drawn the graph here. Else it looks just like 2x. At x is equal to 5, it's not along the line 2x."}, {"video_title": "Formal definition of limits Part 4 using the definition AP Calculus AB Khan Academy.mp3", "Sentence": "It's equal to itself. We've drawn the graph here. Else it looks just like 2x. At x is equal to 5, it's not along the line 2x. Instead, the function is defined to be that point right over there. If I were to ask you what is the limit of f of x as x approaches 5, you might think of it pretty intuitively. Well, let's see."}, {"video_title": "Formal definition of limits Part 4 using the definition AP Calculus AB Khan Academy.mp3", "Sentence": "At x is equal to 5, it's not along the line 2x. Instead, the function is defined to be that point right over there. If I were to ask you what is the limit of f of x as x approaches 5, you might think of it pretty intuitively. Well, let's see. The closer I get to 5, the closer f of x seems to be getting to 10. The closer I get to 5, the closer f of x seems to be getting to 10. You might fairly intuitively make the claim that the limit of f of x as x approaches 5 really is equal to 10."}, {"video_title": "Formal definition of limits Part 4 using the definition AP Calculus AB Khan Academy.mp3", "Sentence": "Well, let's see. The closer I get to 5, the closer f of x seems to be getting to 10. The closer I get to 5, the closer f of x seems to be getting to 10. You might fairly intuitively make the claim that the limit of f of x as x approaches 5 really is equal to 10. It looks that way. What we're going to do is use the epsilon delta definition to actually prove it. The way that most of these proofs typically go is we define delta in the abstract."}, {"video_title": "Formal definition of limits Part 4 using the definition AP Calculus AB Khan Academy.mp3", "Sentence": "You might fairly intuitively make the claim that the limit of f of x as x approaches 5 really is equal to 10. It looks that way. What we're going to do is use the epsilon delta definition to actually prove it. The way that most of these proofs typically go is we define delta in the abstract. We essentially try to come up with a way that given any epsilon, we can always come up with a delta. Another way is we're going to try to describe our delta as a function of epsilon, not to confuse you too much. Maybe I shouldn't use f again."}, {"video_title": "Formal definition of limits Part 4 using the definition AP Calculus AB Khan Academy.mp3", "Sentence": "The way that most of these proofs typically go is we define delta in the abstract. We essentially try to come up with a way that given any epsilon, we can always come up with a delta. Another way is we're going to try to describe our delta as a function of epsilon, not to confuse you too much. Maybe I shouldn't use f again. Delta equals function of epsilon that is defined for any positive epsilon. You give me an epsilon, I just put it into our little formula, our little function box, and I will always get you a delta. If I can do that for any epsilon, that will always give you a delta where this is true, that if x is within that range of delta of c, then the corresponding f of x is going to be within epsilon of l, then the limit definitely exists."}, {"video_title": "Formal definition of limits Part 4 using the definition AP Calculus AB Khan Academy.mp3", "Sentence": "Maybe I shouldn't use f again. Delta equals function of epsilon that is defined for any positive epsilon. You give me an epsilon, I just put it into our little formula, our little function box, and I will always get you a delta. If I can do that for any epsilon, that will always give you a delta where this is true, that if x is within that range of delta of c, then the corresponding f of x is going to be within epsilon of l, then the limit definitely exists. Let's try to do that. Let's think about being within delta of our c. Let's think about this right over here as 5 plus delta. This is 5 minus delta."}, {"video_title": "Formal definition of limits Part 4 using the definition AP Calculus AB Khan Academy.mp3", "Sentence": "If I can do that for any epsilon, that will always give you a delta where this is true, that if x is within that range of delta of c, then the corresponding f of x is going to be within epsilon of l, then the limit definitely exists. Let's try to do that. Let's think about being within delta of our c. Let's think about this right over here as 5 plus delta. This is 5 minus delta. That's our range we're going to think about. We're going to think about it in the abstract at first, and then we're going to try to come up with a formula for delta in terms of epsilon. How can we describe all of the x's that are in this range but not equal to 5 itself?"}, {"video_title": "Formal definition of limits Part 4 using the definition AP Calculus AB Khan Academy.mp3", "Sentence": "This is 5 minus delta. That's our range we're going to think about. We're going to think about it in the abstract at first, and then we're going to try to come up with a formula for delta in terms of epsilon. How can we describe all of the x's that are in this range but not equal to 5 itself? We really care about the things that are within delta of 5 but not necessarily equal to 5. This is just a strictly less than. They're within a range of c but not equal to c. That's going to be all of the x's that satisfy x minus 5 is less than."}, {"video_title": "Formal definition of limits Part 4 using the definition AP Calculus AB Khan Academy.mp3", "Sentence": "How can we describe all of the x's that are in this range but not equal to 5 itself? We really care about the things that are within delta of 5 but not necessarily equal to 5. This is just a strictly less than. They're within a range of c but not equal to c. That's going to be all of the x's that satisfy x minus 5 is less than. Less than is less than delta. That describes all of these x's right over here. What we're going to do, and the way these proofs typically go, is we're going to try to manipulate the left-hand side of the inequality so it starts to look something like this, or it starts to look exactly like that."}, {"video_title": "Formal definition of limits Part 4 using the definition AP Calculus AB Khan Academy.mp3", "Sentence": "They're within a range of c but not equal to c. That's going to be all of the x's that satisfy x minus 5 is less than. Less than is less than delta. That describes all of these x's right over here. What we're going to do, and the way these proofs typically go, is we're going to try to manipulate the left-hand side of the inequality so it starts to look something like this, or it starts to look exactly like that. As we do that, the right-hand side of the inequality is going to be expressed in terms of delta. Then we can essentially say, well, look, if the right-hand side looks like delta, or is in terms of delta, and the left-hand side looks just like that, that really defines how we can express delta in terms of epsilon. If that doesn't make sense, bear with me."}, {"video_title": "Formal definition of limits Part 4 using the definition AP Calculus AB Khan Academy.mp3", "Sentence": "What we're going to do, and the way these proofs typically go, is we're going to try to manipulate the left-hand side of the inequality so it starts to look something like this, or it starts to look exactly like that. As we do that, the right-hand side of the inequality is going to be expressed in terms of delta. Then we can essentially say, well, look, if the right-hand side looks like delta, or is in terms of delta, and the left-hand side looks just like that, that really defines how we can express delta in terms of epsilon. If that doesn't make sense, bear with me. I'm about to do it. If we want x minus 5 to look a lot more like this, when x is not equal to 5, and all of this whole interval, x is not equal to 5, f of x is equal to 2x. Our limit, our proposed limit, is equal to 10."}, {"video_title": "Formal definition of limits Part 4 using the definition AP Calculus AB Khan Academy.mp3", "Sentence": "If that doesn't make sense, bear with me. I'm about to do it. If we want x minus 5 to look a lot more like this, when x is not equal to 5, and all of this whole interval, x is not equal to 5, f of x is equal to 2x. Our limit, our proposed limit, is equal to 10. If we could somehow get this to be 2x minus 10, then we're in good shape. The easiest way to do that is to multiply both sides of this inequality by 2. You multiply both sides of this inequality by 2."}, {"video_title": "Formal definition of limits Part 4 using the definition AP Calculus AB Khan Academy.mp3", "Sentence": "Our limit, our proposed limit, is equal to 10. If we could somehow get this to be 2x minus 10, then we're in good shape. The easiest way to do that is to multiply both sides of this inequality by 2. You multiply both sides of this inequality by 2. Two times the absolute value of something, that's the same thing as the absolute value of 2 times that thing. If I were to say 2 times the absolute value of a, that's the same thing as the absolute value of 2a. On the left-hand side, right over here, this is just going to be the absolute value of 2x minus 10."}, {"video_title": "Formal definition of limits Part 4 using the definition AP Calculus AB Khan Academy.mp3", "Sentence": "You multiply both sides of this inequality by 2. Two times the absolute value of something, that's the same thing as the absolute value of 2 times that thing. If I were to say 2 times the absolute value of a, that's the same thing as the absolute value of 2a. On the left-hand side, right over here, this is just going to be the absolute value of 2x minus 10. It's going to be less than on the right-hand side. You just end up with a 2 delta. What do we have here on the left-hand side?"}, {"video_title": "Formal definition of limits Part 4 using the definition AP Calculus AB Khan Academy.mp3", "Sentence": "On the left-hand side, right over here, this is just going to be the absolute value of 2x minus 10. It's going to be less than on the right-hand side. You just end up with a 2 delta. What do we have here on the left-hand side? This is f of x as long as x does not equal 5. This is our limit. We can rewrite this as f of x minus L is less than 2 delta."}, {"video_title": "Formal definition of limits Part 4 using the definition AP Calculus AB Khan Academy.mp3", "Sentence": "What do we have here on the left-hand side? This is f of x as long as x does not equal 5. This is our limit. We can rewrite this as f of x minus L is less than 2 delta. This is for x does not equal 5. This is f of x. This literally is our limit."}, {"video_title": "Formal definition of limits Part 4 using the definition AP Calculus AB Khan Academy.mp3", "Sentence": "We can rewrite this as f of x minus L is less than 2 delta. This is for x does not equal 5. This is f of x. This literally is our limit. This is interesting. This statement right over here is almost exactly what we want right over here, except the right sides are just different. This has in terms of epsilon."}, {"video_title": "Formal definition of limits Part 4 using the definition AP Calculus AB Khan Academy.mp3", "Sentence": "This literally is our limit. This is interesting. This statement right over here is almost exactly what we want right over here, except the right sides are just different. This has in terms of epsilon. This has it in terms of delta. How can we define delta so that 2 delta is essentially going to be epsilon? This is our chance."}, {"video_title": "Formal definition of limits Part 4 using the definition AP Calculus AB Khan Academy.mp3", "Sentence": "This has in terms of epsilon. This has it in terms of delta. How can we define delta so that 2 delta is essentially going to be epsilon? This is our chance. We will just define, we will make, and this is where we're defining delta as a function of epsilon. We're going to make 2 delta equal epsilon, or if you divide both sides by 2, we're going to make delta equal to epsilon over 2. If you make delta equal epsilon over 2, if you make, let me switch colors just to ease the monotony, if you make delta equal epsilon over 2, then this statement right over here becomes the absolute value of f of x minus L is less than, instead of 2 delta, it'll be less than 2 times epsilon over 2 is just going to be less than epsilon."}, {"video_title": "Formal definition of limits Part 4 using the definition AP Calculus AB Khan Academy.mp3", "Sentence": "This is our chance. We will just define, we will make, and this is where we're defining delta as a function of epsilon. We're going to make 2 delta equal epsilon, or if you divide both sides by 2, we're going to make delta equal to epsilon over 2. If you make delta equal epsilon over 2, if you make, let me switch colors just to ease the monotony, if you make delta equal epsilon over 2, then this statement right over here becomes the absolute value of f of x minus L is less than, instead of 2 delta, it'll be less than 2 times epsilon over 2 is just going to be less than epsilon. This is the key. If someone gives you any positive number epsilon for this function, as long as you make delta equal epsilon over 2, then any x within that range, that corresponding f of x is going to be within epsilon of our limit. This tells you, say the epsilon, and remember, it has to be true for any positive epsilon, but you could see how the game could go."}, {"video_title": "Formal definition of limits Part 4 using the definition AP Calculus AB Khan Academy.mp3", "Sentence": "If you make delta equal epsilon over 2, if you make, let me switch colors just to ease the monotony, if you make delta equal epsilon over 2, then this statement right over here becomes the absolute value of f of x minus L is less than, instead of 2 delta, it'll be less than 2 times epsilon over 2 is just going to be less than epsilon. This is the key. If someone gives you any positive number epsilon for this function, as long as you make delta equal epsilon over 2, then any x within that range, that corresponding f of x is going to be within epsilon of our limit. This tells you, say the epsilon, and remember, it has to be true for any positive epsilon, but you could see how the game could go. If someone gives you the epsilon, let's say they want to be within 0.5 of our limit. Our limit is up here, so our epsilon is 0.5, so it would literally be the range I want to be between 10 plus epsilon would be 10.5, and then 10 minus epsilon would be 9.5. Well, we just came up with a formula."}, {"video_title": "Formal definition of limits Part 4 using the definition AP Calculus AB Khan Academy.mp3", "Sentence": "This tells you, say the epsilon, and remember, it has to be true for any positive epsilon, but you could see how the game could go. If someone gives you the epsilon, let's say they want to be within 0.5 of our limit. Our limit is up here, so our epsilon is 0.5, so it would literally be the range I want to be between 10 plus epsilon would be 10.5, and then 10 minus epsilon would be 9.5. Well, we just came up with a formula. We just have to make delta equal to epsilon over 2, which is equal to 0.25, so that'll give us a range between 4.75 and 5.25. As long as we pick an x between 4.75 and 5.25, then the corresponding f of x, but not x equals 5, the corresponding f of x will be between 9.5 and 10.5. You give me any epsilon, I can just apply this formula right over here to come up with the delta."}, {"video_title": "Formal definition of limits Part 4 using the definition AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we just came up with a formula. We just have to make delta equal to epsilon over 2, which is equal to 0.25, so that'll give us a range between 4.75 and 5.25. As long as we pick an x between 4.75 and 5.25, then the corresponding f of x, but not x equals 5, the corresponding f of x will be between 9.5 and 10.5. You give me any epsilon, I can just apply this formula right over here to come up with the delta. This would apply for any real number, especially any positive number. For any epsilon you give me, I just get a delta, and then a delta defined this way, and then I can go through this arc. If the absolute value of x minus 5 is less than delta, if delta is defined in this way, which I can define for any epsilon, then it will be the case that f of x will be within epsilon of our limit."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "That is my y-axis. Or maybe I should call it my f of x-axis. y is equal to f of x. And let me draw my x-axis just like that. That is my x-axis. And let me draw a line. So let me draw a line like this."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And let me draw my x-axis just like that. That is my x-axis. And let me draw a line. So let me draw a line like this. And what we want to do is remind ourselves, how do we find the slope of that line? And what we do is we take two points on the line. So let's say we take this point right here."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So let me draw a line like this. And what we want to do is remind ourselves, how do we find the slope of that line? And what we do is we take two points on the line. So let's say we take this point right here. Let's say that that is the point x is equal to a. And then what would this be? This would be the point f of a, where the function is going to be some line."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So let's say we take this point right here. Let's say that that is the point x is equal to a. And then what would this be? This would be the point f of a, where the function is going to be some line. We could write f of x is going to be equal to mx plus b. We don't know what m and b are, but this is all a little bit of a review. So this is a."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "This would be the point f of a, where the function is going to be some line. We could write f of x is going to be equal to mx plus b. We don't know what m and b are, but this is all a little bit of a review. So this is a. And then the y value is what happens to the function when you evaluate it at a. So that's that point right there. And then we could take another point on this line."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So this is a. And then the y value is what happens to the function when you evaluate it at a. So that's that point right there. And then we could take another point on this line. Let's say we take the point b right there. And then this coordinate up here is going to be the point b, bf of b. Because this is just the point when you evaluate the function at b."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And then we could take another point on this line. Let's say we take the point b right there. And then this coordinate up here is going to be the point b, bf of b. Because this is just the point when you evaluate the function at b. You put b in here, you're going to get that point right there. So let me just draw a little line right there. So that is f of b right there."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Because this is just the point when you evaluate the function at b. You put b in here, you're going to get that point right there. So let me just draw a little line right there. So that is f of b right there. Actually, let me make it clear that this coordinate right here is the point a, f of a. So how do we find the slope between these two points? Or more generally, of this entire line?"}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So that is f of b right there. Actually, let me make it clear that this coordinate right here is the point a, f of a. So how do we find the slope between these two points? Or more generally, of this entire line? Because a line has the slope is consistent the whole way through it. And we know that once we find the slope, that's actually going to be the value of this f. That's all a review of your algebra. But how do we do it?"}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Or more generally, of this entire line? Because a line has the slope is consistent the whole way through it. And we know that once we find the slope, that's actually going to be the value of this f. That's all a review of your algebra. But how do we do it? Well, a couple of ways to think about it. Slope is equal to rise over run. You might have seen that when you first learned algebra."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "But how do we do it? Well, a couple of ways to think about it. Slope is equal to rise over run. You might have seen that when you first learned algebra. Or another way of writing it, it's change in y over change in x. So let's figure out what the change in y over the change in x is for this particular case. So the change in y is equal to what?"}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "You might have seen that when you first learned algebra. Or another way of writing it, it's change in y over change in x. So let's figure out what the change in y over the change in x is for this particular case. So the change in y is equal to what? Well, you can take this guy as being the first point or that guy as being the first point. But since this guy has a larger x and a larger y, let's start with him. So the change in y between that guy and that guy is this distance right here."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So the change in y is equal to what? Well, you can take this guy as being the first point or that guy as being the first point. But since this guy has a larger x and a larger y, let's start with him. So the change in y between that guy and that guy is this distance right here. So let me draw a little triangle. That distance right there is the change in y. Or I could just transfer it to the y-axis."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So the change in y between that guy and that guy is this distance right here. So let me draw a little triangle. That distance right there is the change in y. Or I could just transfer it to the y-axis. This is the change in y. That is your change in y, that distance. So what is that distance?"}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Or I could just transfer it to the y-axis. This is the change in y. That is your change in y, that distance. So what is that distance? It's f of b minus f of a. So it equals f of b minus f of a. That is your change in y right here."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So what is that distance? It's f of b minus f of a. So it equals f of b minus f of a. That is your change in y right here. That's your change in y. Now what is your change in x? We have the slope is change in y over change in x."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "That is your change in y right here. That's your change in y. Now what is your change in x? We have the slope is change in y over change in x. Well, what's our change in x? Well, it's this distance. Remember, we're taking this to be the first point."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "We have the slope is change in y over change in x. Well, what's our change in x? Well, it's this distance. Remember, we're taking this to be the first point. So we took its y minus the other point's y. So to be consistent, we're going to have to take this point's x minus this point's x. So this point's x-coordinate is b."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Remember, we're taking this to be the first point. So we took its y minus the other point's y. So to be consistent, we're going to have to take this point's x minus this point's x. So this point's x-coordinate is b. So it's going to be b minus a. And just like that, if you knew the equation of this line or if you had the coordinates of these two points, you would just plug them in right here and you would get your slope. That straightforward."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So this point's x-coordinate is b. So it's going to be b minus a. And just like that, if you knew the equation of this line or if you had the coordinates of these two points, you would just plug them in right here and you would get your slope. That straightforward. And that comes straight out of your Algebra 1 class. And let me just, you know, just to make sure it's concrete for you. If this was the point, let's say this was the point 2, 3."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "That straightforward. And that comes straight out of your Algebra 1 class. And let me just, you know, just to make sure it's concrete for you. If this was the point, let's say this was the point 2, 3. And let's say that this up here was the point 5, 7. Then if we wanted to find the slope of this line, we would do 7 minus 3. So 7 minus 3."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "If this was the point, let's say this was the point 2, 3. And let's say that this up here was the point 5, 7. Then if we wanted to find the slope of this line, we would do 7 minus 3. So 7 minus 3. That would be our change in y. This would be 7. And this would be 3."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So 7 minus 3. That would be our change in y. This would be 7. And this would be 3. And then we would do that over 5 minus 2. Because this would be a 5 and this would be a 2. And so this would be your change in x."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And this would be 3. And then we would do that over 5 minus 2. Because this would be a 5 and this would be a 2. And so this would be your change in x. 5 minus 2. So 7 minus 3 is 4. And 5 minus 2 is 3."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And so this would be your change in x. 5 minus 2. So 7 minus 3 is 4. And 5 minus 2 is 3. So your slope would be 4 over 3. Now let's see if we can generalize this. And this is the new concept that we're going to be learning as we delve into calculus."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And 5 minus 2 is 3. So your slope would be 4 over 3. Now let's see if we can generalize this. And this is the new concept that we're going to be learning as we delve into calculus. Let's see if we can generalize this somehow to a curve. So let's say I have a curve. We have to have a curve before we can generalize it to a curve."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And this is the new concept that we're going to be learning as we delve into calculus. Let's see if we can generalize this somehow to a curve. So let's say I have a curve. We have to have a curve before we can generalize it to a curve. Let me scroll down a little. Actually, I want to leave this up here. Just show you the similarity."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "We have to have a curve before we can generalize it to a curve. Let me scroll down a little. Actually, I want to leave this up here. Just show you the similarity. So let's say I have, I'll keep it pretty general right now. Let's say I have a curve. I'll make it a familiar looking curve."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Just show you the similarity. So let's say I have, I'll keep it pretty general right now. Let's say I have a curve. I'll make it a familiar looking curve. Let's say it's the curve y is equal to x squared, which looks something like that. And I want to find the slope. Let's say I want to find the slope at some point."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "I'll make it a familiar looking curve. Let's say it's the curve y is equal to x squared, which looks something like that. And I want to find the slope. Let's say I want to find the slope at some point. And actually, before even talking about it, let's even think about what it means to find the slope of a curve. Here the slope was the same the whole time. But on a curve, your slope is changing."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Let's say I want to find the slope at some point. And actually, before even talking about it, let's even think about what it means to find the slope of a curve. Here the slope was the same the whole time. But on a curve, your slope is changing. And just to get an intuition for what that means is, what's the slope over here? Your slope over here is the slope of the tangent line. The line just barely touches it."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "But on a curve, your slope is changing. And just to get an intuition for what that means is, what's the slope over here? Your slope over here is the slope of the tangent line. The line just barely touches it. That's the slope over there. It's a negative slope. Then over here, your slope is still negative, but it's a little bit less negative."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "The line just barely touches it. That's the slope over there. It's a negative slope. Then over here, your slope is still negative, but it's a little bit less negative. It goes like that. I don't know if I did that, drew that. Let me do it in a different color."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Then over here, your slope is still negative, but it's a little bit less negative. It goes like that. I don't know if I did that, drew that. Let me do it in a different color. Let me do it in purple. So over here, your slope is slightly less negative. It's a slightly less downward sloping line."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Let me do it in a different color. Let me do it in purple. So over here, your slope is slightly less negative. It's a slightly less downward sloping line. And then when you go over here, at the zero point right here, your slope is pretty much flat, because the horizontal line, y equals 0, is tangent to this curve. And then as you go to more positive x's, then your slope starts increasing. I'm trying to draw a tangent line."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "It's a slightly less downward sloping line. And then when you go over here, at the zero point right here, your slope is pretty much flat, because the horizontal line, y equals 0, is tangent to this curve. And then as you go to more positive x's, then your slope starts increasing. I'm trying to draw a tangent line. And here it's increasing even more. So your slope is changing the entire time. And this is kind of the big change that happens when you go from a line to a curve."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "I'm trying to draw a tangent line. And here it's increasing even more. So your slope is changing the entire time. And this is kind of the big change that happens when you go from a line to a curve. A line, your slope is the same the entire time. You can take any two points in the line, take the change in y over the change in x, and you get the slope for the entire line. But as you can see already, it's going to be a little bit more nuanced when we do it for a curve, because it depends what point we're talking about."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And this is kind of the big change that happens when you go from a line to a curve. A line, your slope is the same the entire time. You can take any two points in the line, take the change in y over the change in x, and you get the slope for the entire line. But as you can see already, it's going to be a little bit more nuanced when we do it for a curve, because it depends what point we're talking about. We can't just say, what is the slope for this curve? The slope is different at every point along the curve. It changes."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "But as you can see already, it's going to be a little bit more nuanced when we do it for a curve, because it depends what point we're talking about. We can't just say, what is the slope for this curve? The slope is different at every point along the curve. It changes. If we go up here, it's going to be even steeper. It's going to look something like that. So let's try a bit of an experiment."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "It changes. If we go up here, it's going to be even steeper. It's going to look something like that. So let's try a bit of an experiment. And I know how this experiment turns out, so it won't be too much of a risk. Let me draw better than that. So that is my y-axis."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So let's try a bit of an experiment. And I know how this experiment turns out, so it won't be too much of a risk. Let me draw better than that. So that is my y-axis. And that's my x-axis. And let's call this, we could call this y, or we could call this the f of x-axis. Either way."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So that is my y-axis. And that's my x-axis. And let's call this, we could call this y, or we could call this the f of x-axis. Either way. And let me draw my curve again. So I'll just draw it in the positive coordinate like that. That's my curve."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Either way. And let me draw my curve again. So I'll just draw it in the positive coordinate like that. That's my curve. And what if I want to find the slope right there? What can I do? Well, based on our definition of a slope, we need two points to find a slope."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "That's my curve. And what if I want to find the slope right there? What can I do? Well, based on our definition of a slope, we need two points to find a slope. Right here, I don't know how to find the slope. Put just one point. So let's just call this one, let's just call this point right here."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Well, based on our definition of a slope, we need two points to find a slope. Right here, I don't know how to find the slope. Put just one point. So let's just call this one, let's just call this point right here. Let's call that, well that's going to be x. We're going to be general. This is going to be our point x."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So let's just call this one, let's just call this point right here. Let's call that, well that's going to be x. We're going to be general. This is going to be our point x. But to find our slope according to our traditional algebra 1 definition of a slope, we need two points. So let's get another point in here. Let's just take a slightly larger version of this x."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "This is going to be our point x. But to find our slope according to our traditional algebra 1 definition of a slope, we need two points. So let's get another point in here. Let's just take a slightly larger version of this x. So let's say we have that point right there. Actually, let me do it even further out, just because it's going to get messy otherwise. So let's say we have this point right here."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Let's just take a slightly larger version of this x. So let's say we have that point right there. Actually, let me do it even further out, just because it's going to get messy otherwise. So let's say we have this point right here. And it's just h bigger than x. Or actually, instead of saying h bigger, let's just, well, let me just say h bigger. So this is x plus h. That's what that point is right there."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So let's say we have this point right here. And it's just h bigger than x. Or actually, instead of saying h bigger, let's just, well, let me just say h bigger. So this is x plus h. That's what that point is right there. So what are going to be their corresponding y-coordinates on the curve? Well, this is the curve of y is equal to f of x. So this point right here is going to be f of our particular x right here."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So this is x plus h. That's what that point is right there. So what are going to be their corresponding y-coordinates on the curve? Well, this is the curve of y is equal to f of x. So this point right here is going to be f of our particular x right here. And maybe to show you that I'm taking a particular x, maybe I'll do a little 0 here. This is x naught. This is x naught plus h. This is f of x naught."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So this point right here is going to be f of our particular x right here. And maybe to show you that I'm taking a particular x, maybe I'll do a little 0 here. This is x naught. This is x naught plus h. This is f of x naught. And then what is this going to be up here? This point up here. That point up here."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "This is x naught plus h. This is f of x naught. And then what is this going to be up here? This point up here. That point up here. Its y-coordinate is going to be f of this x-coordinate, which I shifted over a little bit. It's right there. f of this x-coordinate, which is f of x naught plus h. That's its y-coordinate."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "That point up here. Its y-coordinate is going to be f of this x-coordinate, which I shifted over a little bit. It's right there. f of this x-coordinate, which is f of x naught plus h. That's its y-coordinate. So what is the slope going to be between these two points that are relatively close to each other? Remember, this isn't going to be the slope just at this point. This is the slope of the line between these two points."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "f of this x-coordinate, which is f of x naught plus h. That's its y-coordinate. So what is the slope going to be between these two points that are relatively close to each other? Remember, this isn't going to be the slope just at this point. This is the slope of the line between these two points. And if I were to actually draw it out, it would actually be a secant line to the curve. So it would intersect the curve twice. Once at this point and once at this point."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "This is the slope of the line between these two points. And if I were to actually draw it out, it would actually be a secant line to the curve. So it would intersect the curve twice. Once at this point and once at this point. You can't see it. If I blew it up a little bit, it would look something like this. It would look something like this."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Once at this point and once at this point. You can't see it. If I blew it up a little bit, it would look something like this. It would look something like this. It would look like that, where this is our coordinate x naught. f of x naught. And up here is our coordinate for this point, which would be the x-coordinate would be x naught plus h. And the y-coordinate would be f of x naught plus h. Just whatever this function is, we're evaluating it at this x-coordinate."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "It would look something like this. It would look like that, where this is our coordinate x naught. f of x naught. And up here is our coordinate for this point, which would be the x-coordinate would be x naught plus h. And the y-coordinate would be f of x naught plus h. Just whatever this function is, we're evaluating it at this x-coordinate. That's all it is. So these are the two points. So maybe a good start is just say, hey, what is the slope of this secant line?"}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And up here is our coordinate for this point, which would be the x-coordinate would be x naught plus h. And the y-coordinate would be f of x naught plus h. Just whatever this function is, we're evaluating it at this x-coordinate. That's all it is. So these are the two points. So maybe a good start is just say, hey, what is the slope of this secant line? And just like we did in the previous example, you find the change in x. This is, sorry, the change in y. That'll be your change in y."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So maybe a good start is just say, hey, what is the slope of this secant line? And just like we did in the previous example, you find the change in x. This is, sorry, the change in y. That'll be your change in y. And you divide that by your change in x. Let me draw it here. Your change in y would be that right here."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "That'll be your change in y. And you divide that by your change in x. Let me draw it here. Your change in y would be that right here. Change in y. And then your change in x would be that right there. So what is the slope going to be of the secant line?"}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Your change in y would be that right here. Change in y. And then your change in x would be that right there. So what is the slope going to be of the secant line? The slope is going to be equal to, let's start with this point up here, just because it seems to be larger. So we want a change in y. So this value right here, this y value, is f of x naught plus h. I just evaluated this guy up here."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So what is the slope going to be of the secant line? The slope is going to be equal to, let's start with this point up here, just because it seems to be larger. So we want a change in y. So this value right here, this y value, is f of x naught plus h. I just evaluated this guy up here. It looks like a fancy term. But all it means is, look, the slightly larger x, I'll evaluate its y-coordinate. Where the curve is at that value of x."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So this value right here, this y value, is f of x naught plus h. I just evaluated this guy up here. It looks like a fancy term. But all it means is, look, the slightly larger x, I'll evaluate its y-coordinate. Where the curve is at that value of x. So that is going to be, so the change in y is going to be f of x naught plus h. That's just the y-coordinate up here. Minus this y-coordinate over here. So minus f of x naught."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Where the curve is at that value of x. So that is going to be, so the change in y is going to be f of x naught plus h. That's just the y-coordinate up here. Minus this y-coordinate over here. So minus f of x naught. And that's our change in, so that equals our change in y. And you want to divide that by your change in x. So what is this?"}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So minus f of x naught. And that's our change in, so that equals our change in y. And you want to divide that by your change in x. So what is this? This is the larger x value. We started with this coordinate, so we start with its x-coordinate. So it's x naught plus h. Minus this x-coordinate."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So what is this? This is the larger x value. We started with this coordinate, so we start with its x-coordinate. So it's x naught plus h. Minus this x-coordinate. Well, we just picked a general number. It's x naught. So that is over your change in x."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So it's x naught plus h. Minus this x-coordinate. Well, we just picked a general number. It's x naught. So that is over your change in x. Just like that. So this is a slope of the secant line. We still haven't answered what the slope is right at that point."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So that is over your change in x. Just like that. So this is a slope of the secant line. We still haven't answered what the slope is right at that point. But maybe this will help us get there. So if we simplify this, so let me write it down like this. The slope of the secant, let me write that properly."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "We still haven't answered what the slope is right at that point. But maybe this will help us get there. So if we simplify this, so let me write it down like this. The slope of the secant, let me write that properly. The slope of the secant line is equal to the value of the function at this point, f of x naught plus h, minus the value of the function here, minus f of x naught. So that just tells us the change in y. It's the exact same definition of slope we've always used."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "The slope of the secant, let me write that properly. The slope of the secant line is equal to the value of the function at this point, f of x naught plus h, minus the value of the function here, minus f of x naught. So that just tells us the change in y. It's the exact same definition of slope we've always used. Over the change in x. And we can simplify this. We have x naught plus h, minus x naught."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "It's the exact same definition of slope we've always used. Over the change in x. And we can simplify this. We have x naught plus h, minus x naught. So x naught minus x naught cancel out. So you have that over h. So this is equal to our change in y over change in x. Fair enough."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "We have x naught plus h, minus x naught. So x naught minus x naught cancel out. So you have that over h. So this is equal to our change in y over change in x. Fair enough. But I started off saying, I want to find the slope of the line at that point. At this point right here. This is the zoomed out version of it."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Fair enough. But I started off saying, I want to find the slope of the line at that point. At this point right here. This is the zoomed out version of it. So what can I do? Well, I defined this second point here as just the first point, plus some h. And we have something in our toolkit called a limit. This h is just a general number."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "This is the zoomed out version of it. So what can I do? Well, I defined this second point here as just the first point, plus some h. And we have something in our toolkit called a limit. This h is just a general number. This h is just, it could be 10, it could be 2, it could be 0.02, it could be 1 times 10 to the negative 100. It could be an arbitrarily small number. So what happens, what would happen, at least just theoretically, if I were to take the limit as h approaches 0?"}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "This h is just a general number. This h is just, it could be 10, it could be 2, it could be 0.02, it could be 1 times 10 to the negative 100. It could be an arbitrarily small number. So what happens, what would happen, at least just theoretically, if I were to take the limit as h approaches 0? So if first, maybe h is this fairly large number over here. And then if I take h a little bit smaller, then I'd be finding the slope of this secant line. If I took h to be even a little bit smaller, I'd be finding the slope of that secant line."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So what happens, what would happen, at least just theoretically, if I were to take the limit as h approaches 0? So if first, maybe h is this fairly large number over here. And then if I take h a little bit smaller, then I'd be finding the slope of this secant line. If I took h to be even a little bit smaller, I'd be finding the slope of that secant line. If h is a little bit smaller, I'd be finding the slope of that line. So as h approaches 0, I'll be getting closer and closer to finding the slope of the line right at my point in question. Obviously, if h is a large number, my secant line is going to be way off from the slope at exactly that point right there."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "If I took h to be even a little bit smaller, I'd be finding the slope of that secant line. If h is a little bit smaller, I'd be finding the slope of that line. So as h approaches 0, I'll be getting closer and closer to finding the slope of the line right at my point in question. Obviously, if h is a large number, my secant line is going to be way off from the slope at exactly that point right there. But if h is 0.0000001, if it's an infinitesimally small number, then I'm going to get pretty close. So what happens if I take the limit as h approaches 0 of this? So the limit as h approaches 0 of my secant slope of, let me switch to green, f of x naught plus h minus f of x naught, that was my change in y, over my change in h, which is my change in x."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Obviously, if h is a large number, my secant line is going to be way off from the slope at exactly that point right there. But if h is 0.0000001, if it's an infinitesimally small number, then I'm going to get pretty close. So what happens if I take the limit as h approaches 0 of this? So the limit as h approaches 0 of my secant slope of, let me switch to green, f of x naught plus h minus f of x naught, that was my change in y, over my change in h, which is my change in x. And now just to clarify something, and sometimes you'll see it in different calculus books, sometimes instead of an h, they'll write a delta x here, where this second point would have been defined as x naught plus delta x. And then this would have simplified to just delta x over there, and we'd be taking the limit as delta x approaches 0. The exact same thing."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So the limit as h approaches 0 of my secant slope of, let me switch to green, f of x naught plus h minus f of x naught, that was my change in y, over my change in h, which is my change in x. And now just to clarify something, and sometimes you'll see it in different calculus books, sometimes instead of an h, they'll write a delta x here, where this second point would have been defined as x naught plus delta x. And then this would have simplified to just delta x over there, and we'd be taking the limit as delta x approaches 0. The exact same thing. h delta x doesn't matter. We're taking h as the difference between one x point and then the higher x point, and then we're just going to take the limit as that approaches 0. We could have called that delta x just as easily."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "The exact same thing. h delta x doesn't matter. We're taking h as the difference between one x point and then the higher x point, and then we're just going to take the limit as that approaches 0. We could have called that delta x just as easily. But I'm going to call this thing, which we're saying it equals the slope of the tangent line, and it does equal the slope of the tangent line, I'm going to call this the derivative of f. Let me write that down. The derivative of f. And I'm going to say that this is equal to f prime of x. And this is going to be another function."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "We could have called that delta x just as easily. But I'm going to call this thing, which we're saying it equals the slope of the tangent line, and it does equal the slope of the tangent line, I'm going to call this the derivative of f. Let me write that down. The derivative of f. And I'm going to say that this is equal to f prime of x. And this is going to be another function. Because remember, the slope changes at every x value. No matter what x value you pick, the slope is going to be different. It doesn't have to be, but the way I drew this curve, it is different."}, {"video_title": "Derivative as slope of a tangent line Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And this is going to be another function. Because remember, the slope changes at every x value. No matter what x value you pick, the slope is going to be different. It doesn't have to be, but the way I drew this curve, it is different. It can be different. So now you give me an x value in here. I'll apply this formula over here, and then I can tell you the slope at that point."}, {"video_title": "Definite integral of absolute value function AP Calculus AB Khan Academy.mp3", "Sentence": "So we have f of x being equal to the absolute value of x plus two. And we want to evaluate the definite integral from negative four to zero of f of x dx. And like always, pause this video and see if you could work through this. Now when you first do this, you might stumble around a little bit because how do you take the antiderivative of an absolute value function? And the key here is to, one way to approach it, is to rewrite f of x without the absolute value and we can do that by rewriting it as a piecewise function. And the way I'm gonna do it, I'm gonna think about intervals where whatever we take inside the absolute value is going to be positive, and other intervals where everything that we take inside the absolute value is going to be negative. And the point at which we change is where x plus two is equal to zero or x is equal to negative two."}, {"video_title": "Definite integral of absolute value function AP Calculus AB Khan Academy.mp3", "Sentence": "Now when you first do this, you might stumble around a little bit because how do you take the antiderivative of an absolute value function? And the key here is to, one way to approach it, is to rewrite f of x without the absolute value and we can do that by rewriting it as a piecewise function. And the way I'm gonna do it, I'm gonna think about intervals where whatever we take inside the absolute value is going to be positive, and other intervals where everything that we take inside the absolute value is going to be negative. And the point at which we change is where x plus two is equal to zero or x is equal to negative two. So let's just think about the intervals x is less than negative two and x is greater than or equal to negative two. And this could have been less than or equal, in which case this would have been greater than. Either way, it would have been equal to this absolute value."}, {"video_title": "Definite integral of absolute value function AP Calculus AB Khan Academy.mp3", "Sentence": "And the point at which we change is where x plus two is equal to zero or x is equal to negative two. So let's just think about the intervals x is less than negative two and x is greater than or equal to negative two. And this could have been less than or equal, in which case this would have been greater than. Either way, it would have been equal to this absolute value. This is a continuous function here. And so when, well let's do the easier case. When x is greater than or equal to negative two, then x plus two is going to be positive or it's going to be greater than or equal to zero and so the absolute value of it is just going to be x plus two."}, {"video_title": "Definite integral of absolute value function AP Calculus AB Khan Academy.mp3", "Sentence": "Either way, it would have been equal to this absolute value. This is a continuous function here. And so when, well let's do the easier case. When x is greater than or equal to negative two, then x plus two is going to be positive or it's going to be greater than or equal to zero and so the absolute value of it is just going to be x plus two. So it's going to be x plus two when x is greater than or equal to negative two. And what about when x is less than negative two? Well when x is less than negative two, x plus two is going to be negative and then if you take the absolute value of a negative number, you're going to take the opposite of it."}, {"video_title": "Definite integral of absolute value function AP Calculus AB Khan Academy.mp3", "Sentence": "When x is greater than or equal to negative two, then x plus two is going to be positive or it's going to be greater than or equal to zero and so the absolute value of it is just going to be x plus two. So it's going to be x plus two when x is greater than or equal to negative two. And what about when x is less than negative two? Well when x is less than negative two, x plus two is going to be negative and then if you take the absolute value of a negative number, you're going to take the opposite of it. So this is going to be negative x plus two. And to really help grok this, because frankly this is the hardest part of what we're doing and really this is more algebra than calculus, let me draw the absolute value function to make this clear. So that is my x-axis."}, {"video_title": "Definite integral of absolute value function AP Calculus AB Khan Academy.mp3", "Sentence": "Well when x is less than negative two, x plus two is going to be negative and then if you take the absolute value of a negative number, you're going to take the opposite of it. So this is going to be negative x plus two. And to really help grok this, because frankly this is the hardest part of what we're doing and really this is more algebra than calculus, let me draw the absolute value function to make this clear. So that is my x-axis. That is my y-axis. And let's say we're here at negative two. And so when we are less than negative two, when x is less than negative two, my graph is going to look like this."}, {"video_title": "Definite integral of absolute value function AP Calculus AB Khan Academy.mp3", "Sentence": "So that is my x-axis. That is my y-axis. And let's say we're here at negative two. And so when we are less than negative two, when x is less than negative two, my graph is going to look like this. It is going to, it is going to look, look something, let's see what is, it's going to look like that. And when we are greater than negative two, use that in a different color, when we are greater than negative two, it's going to look like this. It's going to look like that."}, {"video_title": "Definite integral of absolute value function AP Calculus AB Khan Academy.mp3", "Sentence": "And so when we are less than negative two, when x is less than negative two, my graph is going to look like this. It is going to, it is going to look, look something, let's see what is, it's going to look like that. And when we are greater than negative two, use that in a different color, when we are greater than negative two, it's going to look like this. It's going to look like that. And so notice, this is in blue, we have, this is the graph x plus two, we can say this is a graph of y equals x plus two. And what we have in magenta right over here, this is the graph of negative x minus two. It has a negative slope and we intercept the y-axis at negative two."}, {"video_title": "Definite integral of absolute value function AP Calculus AB Khan Academy.mp3", "Sentence": "It's going to look like that. And so notice, this is in blue, we have, this is the graph x plus two, we can say this is a graph of y equals x plus two. And what we have in magenta right over here, this is the graph of negative x minus two. It has a negative slope and we intercept the y-axis at negative two. So it makes sense. There's multiple ways that you could reason through this. Now once we break it up, then we can break up the integral."}, {"video_title": "Definite integral of absolute value function AP Calculus AB Khan Academy.mp3", "Sentence": "It has a negative slope and we intercept the y-axis at negative two. So it makes sense. There's multiple ways that you could reason through this. Now once we break it up, then we can break up the integral. We could say that what we wrote here, this is equal to the integral from negative four to two, sorry, negative four to negative two of f of x, which is in that case, it's going to be negative x minus two. I just distributed the negative sign there. Dx, and then plus, plus the definite integral going from negative two to zero of x plus two dx."}, {"video_title": "Definite integral of absolute value function AP Calculus AB Khan Academy.mp3", "Sentence": "Now once we break it up, then we can break up the integral. We could say that what we wrote here, this is equal to the integral from negative four to two, sorry, negative four to negative two of f of x, which is in that case, it's going to be negative x minus two. I just distributed the negative sign there. Dx, and then plus, plus the definite integral going from negative two to zero of x plus two dx. And just to make sure we know what we're doing here, this, if this is negative four right over here, this is zero, that first integral is going to give us, is going to give us this area right over here. What's the area under the curve negative x minus two, under that curve or under that line and above the x-axis? And the second integral is gonna give us this area right over here between x plus two and the x-axis going from negative two to zero."}, {"video_title": "Definite integral of absolute value function AP Calculus AB Khan Academy.mp3", "Sentence": "Dx, and then plus, plus the definite integral going from negative two to zero of x plus two dx. And just to make sure we know what we're doing here, this, if this is negative four right over here, this is zero, that first integral is going to give us, is going to give us this area right over here. What's the area under the curve negative x minus two, under that curve or under that line and above the x-axis? And the second integral is gonna give us this area right over here between x plus two and the x-axis going from negative two to zero. And so let's evaluate each of these. And you might even be able to just evaluate these with a little bit of triangle areas, but let's just do this analytically or algebraically. And so what's the antiderivative of negative x?"}, {"video_title": "Definite integral of absolute value function AP Calculus AB Khan Academy.mp3", "Sentence": "And the second integral is gonna give us this area right over here between x plus two and the x-axis going from negative two to zero. And so let's evaluate each of these. And you might even be able to just evaluate these with a little bit of triangle areas, but let's just do this analytically or algebraically. And so what's the antiderivative of negative x? Well, that's negative x squared over two, and then we have the negative two, or so the antiderivative is negative two x. We're gonna evaluate that at negative, at negative two and negative four. And so that part is going to be what?"}, {"video_title": "Definite integral of absolute value function AP Calculus AB Khan Academy.mp3", "Sentence": "And so what's the antiderivative of negative x? Well, that's negative x squared over two, and then we have the negative two, or so the antiderivative is negative two x. We're gonna evaluate that at negative, at negative two and negative four. And so that part is going to be what? Negative two squared, so it's the negative of negative two squared. So it's negative four over two minus two times negative two, so plus four. So that's it evaluated at negative two."}, {"video_title": "Definite integral of absolute value function AP Calculus AB Khan Academy.mp3", "Sentence": "And so that part is going to be what? Negative two squared, so it's the negative of negative two squared. So it's negative four over two minus two times negative two, so plus four. So that's it evaluated at negative two. And then minus, if we evaluated it at negative four, so minus, so we're gonna have minus, negative four squared is 16 over two minus two times negative four. So that is plus eight. So what is that?"}, {"video_title": "Definite integral of absolute value function AP Calculus AB Khan Academy.mp3", "Sentence": "So that's it evaluated at negative two. And then minus, if we evaluated it at negative four, so minus, so we're gonna have minus, negative four squared is 16 over two minus two times negative four. So that is plus eight. So what is that? What is that going to give us? So this is negative two. This right over here is negative eight."}, {"video_title": "Definite integral of absolute value function AP Calculus AB Khan Academy.mp3", "Sentence": "So what is that? What is that going to give us? So this is negative two. This right over here is negative eight. So the second term right over here is just going to be equal to zero. Did I do that right? Yeah, 16 over two, it's negative, and it's plus, okay, so this is just going to be zero."}, {"video_title": "Definite integral of absolute value function AP Calculus AB Khan Academy.mp3", "Sentence": "This right over here is negative eight. So the second term right over here is just going to be equal to zero. Did I do that right? Yeah, 16 over two, it's negative, and it's plus, okay, so this is just going to be zero. And this is negative two plus four, which is going to be equal to two. So what we have here in magenta is equal to two. And what we have here in blue, well, let's see, this is the antiderivative of x squared over two plus two x."}, {"video_title": "Definite integral of absolute value function AP Calculus AB Khan Academy.mp3", "Sentence": "Yeah, 16 over two, it's negative, and it's plus, okay, so this is just going to be zero. And this is negative two plus four, which is going to be equal to two. So what we have here in magenta is equal to two. And what we have here in blue, well, let's see, this is the antiderivative of x squared over two plus two x. We're gonna evaluate it at zero and negative two. You evaluate this thing at zero, it's just gonna be zero. And from that, you're going to subtract negative two squared over two."}, {"video_title": "Definite integral of absolute value function AP Calculus AB Khan Academy.mp3", "Sentence": "And what we have here in blue, well, let's see, this is the antiderivative of x squared over two plus two x. We're gonna evaluate it at zero and negative two. You evaluate this thing at zero, it's just gonna be zero. And from that, you're going to subtract negative two squared over two. That is positive four over two, which is positive two. And then plus two times negative two. So minus, minus four."}, {"video_title": "Definite integral of absolute value function AP Calculus AB Khan Academy.mp3", "Sentence": "And from that, you're going to subtract negative two squared over two. That is positive four over two, which is positive two. And then plus two times negative two. So minus, minus four. And so this is going to be, this is going to be the negative of negative two, or positive two. So it's two plus two, and that makes sense. That what we have in magenta here is two, and what we have over here is two."}, {"video_title": "Definite integral of absolute value function AP Calculus AB Khan Academy.mp3", "Sentence": "So minus, minus four. And so this is going to be, this is going to be the negative of negative two, or positive two. So it's two plus two, and that makes sense. That what we have in magenta here is two, and what we have over here is two. There's a symmetry here. There is a symmetry here. And so you add them all together, and you get our integral is going to be equal to four."}, {"video_title": "Definite integral of absolute value function AP Calculus AB Khan Academy.mp3", "Sentence": "That what we have in magenta here is two, and what we have over here is two. There's a symmetry here. There is a symmetry here. And so you add them all together, and you get our integral is going to be equal to four. And once again, just as a reality check, you could say, look, the height here is two, the width, the base here is two. Two times two times 1 1\u20442 is indeed equal to two. Same thing over here."}, {"video_title": "Visualizing Taylor series approximations Series AP Calculus BC Khan Academy.mp3", "Sentence": "I've talked a lot about using polynomials to approximate functions, but what I want to do in this video is actually show you that the approximation is actually happening. So right over here, and I'm using Wolfram Alpha for this, it's a very cool website, you can do all sorts of crazy mathematical things on it, so it's wolframalpha.com, and I got this copied and pasted from them. I met Stephen Wolfram at a conference not too long ago, he said, yes, you should definitely use Wolfram Alpha in your videos, and I said, great, I will. And so that's what I'm doing right here, and it's super useful, because what it does is, and we could have calculated a lot of this on our own, or even done it on a graphing calculator, but you can do it just with one step on Wolfram Alpha, is see how well we can approximate sine of x using a, you could call it a Maclaurin series expansion, or you could call it a Taylor series expansion at x is equal to zero, using more and more terms, and having a good feel for the fact that the more terms we add, the better it hugs the sine curve. So this over here, in orange, is sine of x. That should hopefully look fairly familiar to you. And in previous videos, we figured out what that Maclaurin expansion for sine of x is, and Wolfram Alpha does it for us as well, they actually calculate the factorials, 3 factorial is 6, 5 factorial is 120, so on and so forth."}, {"video_title": "Visualizing Taylor series approximations Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so that's what I'm doing right here, and it's super useful, because what it does is, and we could have calculated a lot of this on our own, or even done it on a graphing calculator, but you can do it just with one step on Wolfram Alpha, is see how well we can approximate sine of x using a, you could call it a Maclaurin series expansion, or you could call it a Taylor series expansion at x is equal to zero, using more and more terms, and having a good feel for the fact that the more terms we add, the better it hugs the sine curve. So this over here, in orange, is sine of x. That should hopefully look fairly familiar to you. And in previous videos, we figured out what that Maclaurin expansion for sine of x is, and Wolfram Alpha does it for us as well, they actually calculate the factorials, 3 factorial is 6, 5 factorial is 120, so on and so forth. But what's interesting here is you can pick how many of the approximations you want to graph. And so what they did is, if you want just one term of the approximation, so if we didn't have this whole thing, if we just said that our polynomial is equal to x, what does that look like? Well, that's going to be this graph right over here."}, {"video_title": "Visualizing Taylor series approximations Series AP Calculus BC Khan Academy.mp3", "Sentence": "And in previous videos, we figured out what that Maclaurin expansion for sine of x is, and Wolfram Alpha does it for us as well, they actually calculate the factorials, 3 factorial is 6, 5 factorial is 120, so on and so forth. But what's interesting here is you can pick how many of the approximations you want to graph. And so what they did is, if you want just one term of the approximation, so if we didn't have this whole thing, if we just said that our polynomial is equal to x, what does that look like? Well, that's going to be this graph right over here. They tell us which term, how many terms we used by how many dots there are right over here, which I think is pretty clever. So this right here, that is the function p of x is equal to x. And so it's a very rough approximation, although for sine of x, it doesn't do a bad job."}, {"video_title": "Visualizing Taylor series approximations Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, that's going to be this graph right over here. They tell us which term, how many terms we used by how many dots there are right over here, which I think is pretty clever. So this right here, that is the function p of x is equal to x. And so it's a very rough approximation, although for sine of x, it doesn't do a bad job. It hugs the sine curve right over there, and then it starts to veer away from the sine curve again. You add another term. So if you have the x minus x to the third over 6, so now you have two terms in the expansion."}, {"video_title": "Visualizing Taylor series approximations Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so it's a very rough approximation, although for sine of x, it doesn't do a bad job. It hugs the sine curve right over there, and then it starts to veer away from the sine curve again. You add another term. So if you have the x minus x to the third over 6, so now you have two terms in the expansion. So, or I guess we should say is that we're up to the third order term, because that's how they're numbering the dots, because they're not talking about the number of terms, they're talking about the order of the terms. So they have one dot here, because we have only one first degree term. When we have two terms here, since we kind of, when you do the expansion for sine of x, it doesn't have a second degree term."}, {"video_title": "Visualizing Taylor series approximations Series AP Calculus BC Khan Academy.mp3", "Sentence": "So if you have the x minus x to the third over 6, so now you have two terms in the expansion. So, or I guess we should say is that we're up to the third order term, because that's how they're numbering the dots, because they're not talking about the number of terms, they're talking about the order of the terms. So they have one dot here, because we have only one first degree term. When we have two terms here, since we kind of, when you do the expansion for sine of x, it doesn't have a second degree term. We now have a third degree polynomial approximation. And so let's look at the third degree. We should look for three dots."}, {"video_title": "Visualizing Taylor series approximations Series AP Calculus BC Khan Academy.mp3", "Sentence": "When we have two terms here, since we kind of, when you do the expansion for sine of x, it doesn't have a second degree term. We now have a third degree polynomial approximation. And so let's look at the third degree. We should look for three dots. That's this curve right over here. So if you just have that first term, you just get that straight line. You add the negative x to the third over 6 to that x."}, {"video_title": "Visualizing Taylor series approximations Series AP Calculus BC Khan Academy.mp3", "Sentence": "We should look for three dots. That's this curve right over here. So if you just have that first term, you just get that straight line. You add the negative x to the third over 6 to that x. You now get a curve that looks like this. Now it gets a curve that looks like this. And notice, it starts hugging sine a little bit earlier, and it keeps hugging it a little bit later."}, {"video_title": "Visualizing Taylor series approximations Series AP Calculus BC Khan Academy.mp3", "Sentence": "You add the negative x to the third over 6 to that x. You now get a curve that looks like this. Now it gets a curve that looks like this. And notice, it starts hugging sine a little bit earlier, and it keeps hugging it a little bit later. So once again, just adding that second term does a pretty good job. It hugs the sine curve pretty well, especially around smaller numbers. You add another term, and now we're up to, now we're at an order 5 polynomial right over here."}, {"video_title": "Visualizing Taylor series approximations Series AP Calculus BC Khan Academy.mp3", "Sentence": "And notice, it starts hugging sine a little bit earlier, and it keeps hugging it a little bit later. So once again, just adding that second term does a pretty good job. It hugs the sine curve pretty well, especially around smaller numbers. You add another term, and now we're up to, now we're at an order 5 polynomial right over here. So x minus x to the third over 6 plus x to the fifth over 120. So let's look for the five dots. So that's this one right over here, 1, 2, 3, 4, 5."}, {"video_title": "Visualizing Taylor series approximations Series AP Calculus BC Khan Academy.mp3", "Sentence": "You add another term, and now we're up to, now we're at an order 5 polynomial right over here. So x minus x to the third over 6 plus x to the fifth over 120. So let's look for the five dots. So that's this one right over here, 1, 2, 3, 4, 5. So that's this curve right over here. And notice, it begins hugging the line a little bit earlier than the magenta version, and it keeps hugging it a little bit longer. So then it keeps hugging it a little bit longer, and then it flips back up like this."}, {"video_title": "Visualizing Taylor series approximations Series AP Calculus BC Khan Academy.mp3", "Sentence": "So that's this one right over here, 1, 2, 3, 4, 5. So that's this curve right over here. And notice, it begins hugging the line a little bit earlier than the magenta version, and it keeps hugging it a little bit longer. So then it keeps hugging it a little bit longer, and then it flips back up like this. So it hugged it a little bit longer. And you can see, I'll keep going. If you have all these first four terms, it gives us a seventh degree polynomial."}, {"video_title": "Visualizing Taylor series approximations Series AP Calculus BC Khan Academy.mp3", "Sentence": "So then it keeps hugging it a little bit longer, and then it flips back up like this. So it hugged it a little bit longer. And you can see, I'll keep going. If you have all these first four terms, it gives us a seventh degree polynomial. So let's look for the seven dots over here. So they come in just like this. And then once again, it hugs the curve sooner than when we just had the first three terms, and it keeps hugging the curve all the way until here."}, {"video_title": "Visualizing Taylor series approximations Series AP Calculus BC Khan Academy.mp3", "Sentence": "If you have all these first four terms, it gives us a seventh degree polynomial. So let's look for the seven dots over here. So they come in just like this. And then once again, it hugs the curve sooner than when we just had the first three terms, and it keeps hugging the curve all the way until here. And then the last one, if you have all of these terms up to x to the ninth, it does it even more. You start here, hugs the curve longer than the others, and goes out. And if you think about it, it makes sense."}, {"video_title": "Visualizing Taylor series approximations Series AP Calculus BC Khan Academy.mp3", "Sentence": "And then once again, it hugs the curve sooner than when we just had the first three terms, and it keeps hugging the curve all the way until here. And then the last one, if you have all of these terms up to x to the ninth, it does it even more. You start here, hugs the curve longer than the others, and goes out. And if you think about it, it makes sense. Because what's happening here is each successive term that we're adding to the expansion, they have a higher degree of x over a much, much, much, much larger number. So for small x values, so when you're close to the origin, for small x values, this denominator is going to overpower the numerator, especially when you're below 1, because when you take something that has absolute value less than 1 to a power, you're actually shrinking it down. So when you're close to the origin, these latter terms don't matter much."}, {"video_title": "Visualizing Taylor series approximations Series AP Calculus BC Khan Academy.mp3", "Sentence": "And if you think about it, it makes sense. Because what's happening here is each successive term that we're adding to the expansion, they have a higher degree of x over a much, much, much, much larger number. So for small x values, so when you're close to the origin, for small x values, this denominator is going to overpower the numerator, especially when you're below 1, because when you take something that has absolute value less than 1 to a power, you're actually shrinking it down. So when you're close to the origin, these latter terms don't matter much. So you're kind of not losing some of the precision of some of the earlier terms. When these tweaking terms come in, these come in when the numerator can start to overpower the denominator. So this last term, it starts to become relevant out here, where all of a sudden x to the ninth can overpower 362,880, and the same thing on the negative side."}, {"video_title": "Visualizing Taylor series approximations Series AP Calculus BC Khan Academy.mp3", "Sentence": "So when you're close to the origin, these latter terms don't matter much. So you're kind of not losing some of the precision of some of the earlier terms. When these tweaking terms come in, these come in when the numerator can start to overpower the denominator. So this last term, it starts to become relevant out here, where all of a sudden x to the ninth can overpower 362,880, and the same thing on the negative side. So hopefully this gives you a sense. We only have 1, 2, 3, 4, 5 terms here. Imagine what would happen if we had an infinite number of terms."}, {"video_title": "Visualizing Taylor series approximations Series AP Calculus BC Khan Academy.mp3", "Sentence": "So this last term, it starts to become relevant out here, where all of a sudden x to the ninth can overpower 362,880, and the same thing on the negative side. So hopefully this gives you a sense. We only have 1, 2, 3, 4, 5 terms here. Imagine what would happen if we had an infinite number of terms. I think you'd get a pretty good sense that it would kind of hug the sine curve out to infinity. So hopefully that makes you feel a little bit better about this. And for fun, you might want to go type in Taylor expansion at 0 and sine of x, or Maclaurin expansion or Maclaurin series for sine of x, cosine of x, e to the x at wolframalpha.com, and try it out for a bunch of different functions."}, {"video_title": "Analyzing mistakes when finding extrema example 2 AP Calculus AB Khan Academy.mp3", "Sentence": "This is her solution. And then they give us her steps, and at the end they say, is Erin's work correct? If not, what's her mistake? So pause this video and see if you can figure it out yourself, is Erin correct? Or did she make a mistake, and where was that mistake? Alright, now let's just do it together. So she says that this is the derivative."}, {"video_title": "Analyzing mistakes when finding extrema example 2 AP Calculus AB Khan Academy.mp3", "Sentence": "So pause this video and see if you can figure it out yourself, is Erin correct? Or did she make a mistake, and where was that mistake? Alright, now let's just do it together. So she says that this is the derivative. I'm just going to reevaluate it here to the right of her work. So let's see, f prime of x is just going to be the chain rule. I'm going to take the derivative of the outside with respect to the inside."}, {"video_title": "Analyzing mistakes when finding extrema example 2 AP Calculus AB Khan Academy.mp3", "Sentence": "So she says that this is the derivative. I'm just going to reevaluate it here to the right of her work. So let's see, f prime of x is just going to be the chain rule. I'm going to take the derivative of the outside with respect to the inside. So this is going to be two-thirds times x squared minus one to the two-thirds minus one, so to the negative one-third power, times the derivative of the inside with respect to x. So the derivative of x squared minus one with respect to x is two x. There's a fire hydrant."}, {"video_title": "Analyzing mistakes when finding extrema example 2 AP Calculus AB Khan Academy.mp3", "Sentence": "I'm going to take the derivative of the outside with respect to the inside. So this is going to be two-thirds times x squared minus one to the two-thirds minus one, so to the negative one-third power, times the derivative of the inside with respect to x. So the derivative of x squared minus one with respect to x is two x. There's a fire hydrant. A fire, fire, fire, not a hydrant. That would be a noisy hydrant. There's a fire truck outside."}, {"video_title": "Analyzing mistakes when finding extrema example 2 AP Calculus AB Khan Academy.mp3", "Sentence": "There's a fire hydrant. A fire, fire, fire, not a hydrant. That would be a noisy hydrant. There's a fire truck outside. But okay, I think it's passed. But this looks like what she got for the derivative, because if you multiply two times two x, you do indeed get four x. You have this three right over here in the denominator, and x squared minus one to the negative one-third, that's the same thing as x squared minus one to the one-third in the denominator, which is the same thing as the cube root of x squared minus one."}, {"video_title": "Analyzing mistakes when finding extrema example 2 AP Calculus AB Khan Academy.mp3", "Sentence": "There's a fire truck outside. But okay, I think it's passed. But this looks like what she got for the derivative, because if you multiply two times two x, you do indeed get four x. You have this three right over here in the denominator, and x squared minus one to the negative one-third, that's the same thing as x squared minus one to the one-third in the denominator, which is the same thing as the cube root of x squared minus one. So all of this is looking good. That is indeed the derivative. Step two, the critical point is x equals zero."}, {"video_title": "Analyzing mistakes when finding extrema example 2 AP Calculus AB Khan Academy.mp3", "Sentence": "You have this three right over here in the denominator, and x squared minus one to the negative one-third, that's the same thing as x squared minus one to the one-third in the denominator, which is the same thing as the cube root of x squared minus one. So all of this is looking good. That is indeed the derivative. Step two, the critical point is x equals zero. So let's see. A critical point is where our first derivative is either equal to zero or it is undefined. And so it does indeed seem that f prime of zero is going to be four times zero."}, {"video_title": "Analyzing mistakes when finding extrema example 2 AP Calculus AB Khan Academy.mp3", "Sentence": "Step two, the critical point is x equals zero. So let's see. A critical point is where our first derivative is either equal to zero or it is undefined. And so it does indeed seem that f prime of zero is going to be four times zero. It's gonna be zero over three times the cube root of zero minus one of negative one. And so this is three times negative one, or zero over negative three. So this is indeed equal to zero."}, {"video_title": "Analyzing mistakes when finding extrema example 2 AP Calculus AB Khan Academy.mp3", "Sentence": "And so it does indeed seem that f prime of zero is going to be four times zero. It's gonna be zero over three times the cube root of zero minus one of negative one. And so this is three times negative one, or zero over negative three. So this is indeed equal to zero. So this is true. A critical point is at x equals zero. But a question is, is this the only critical point?"}, {"video_title": "Analyzing mistakes when finding extrema example 2 AP Calculus AB Khan Academy.mp3", "Sentence": "So this is indeed equal to zero. So this is true. A critical point is at x equals zero. But a question is, is this the only critical point? Well, as we've mentioned, a critical point is where a function's derivative is either equal to zero or it's undefined. This is the only one where the derivative is equal to zero. But can you find some x values where the derivative is undefined?"}, {"video_title": "Analyzing mistakes when finding extrema example 2 AP Calculus AB Khan Academy.mp3", "Sentence": "But a question is, is this the only critical point? Well, as we've mentioned, a critical point is where a function's derivative is either equal to zero or it's undefined. This is the only one where the derivative is equal to zero. But can you find some x values where the derivative is undefined? Well, what if we make the derivative, what would make the denominator of the derivative equal to zero? Well, if x squared minus one is equal to zero, you take the cube root of zero, you're gonna get zero in the denominator. So what would make x squared minus one equal to zero?"}, {"video_title": "Analyzing mistakes when finding extrema example 2 AP Calculus AB Khan Academy.mp3", "Sentence": "But can you find some x values where the derivative is undefined? Well, what if we make the derivative, what would make the denominator of the derivative equal to zero? Well, if x squared minus one is equal to zero, you take the cube root of zero, you're gonna get zero in the denominator. So what would make x squared minus one equal to zero? Well, x is equal to plus or minus one. These are also critical points because they make f prime of x undefined. So I'm not feeling good about step two."}, {"video_title": "Analyzing mistakes when finding extrema example 2 AP Calculus AB Khan Academy.mp3", "Sentence": "So what would make x squared minus one equal to zero? Well, x is equal to plus or minus one. These are also critical points because they make f prime of x undefined. So I'm not feeling good about step two. It is true that a critical point is x equals zero, but it is not the only critical point. So I would put that there. And the reason why it's important, you know, you might say, well, what's the harm in not noticing these other critical points?"}, {"video_title": "Analyzing mistakes when finding extrema example 2 AP Calculus AB Khan Academy.mp3", "Sentence": "So I'm not feeling good about step two. It is true that a critical point is x equals zero, but it is not the only critical point. So I would put that there. And the reason why it's important, you know, you might say, well, what's the harm in not noticing these other critical points? She identified one, maybe this is the relative maximum point. But as we talk about in other videos, in order to use the first derivative test, so to speak, and find this place where the first derivative is zero, in order to test whether it is a maximum or a minimum point is you have to sample values on either side of it to make sure that you have a change in sign of the derivative. But you have to make sure that when you test on either side that you're not going beyond another critical point because critical points are places where you can change direction."}, {"video_title": "Analyzing mistakes when finding extrema example 2 AP Calculus AB Khan Academy.mp3", "Sentence": "And the reason why it's important, you know, you might say, well, what's the harm in not noticing these other critical points? She identified one, maybe this is the relative maximum point. But as we talk about in other videos, in order to use the first derivative test, so to speak, and find this place where the first derivative is zero, in order to test whether it is a maximum or a minimum point is you have to sample values on either side of it to make sure that you have a change in sign of the derivative. But you have to make sure that when you test on either side that you're not going beyond another critical point because critical points are places where you can change direction. And so let's see what she does in step three right over here. Well, it is indeed in step three that she's testing, she's trying to test values on either side of the critical point that she, that the one critical point that she identified. But the problem here, the reason why this is a little shady is this is beyond another critical point that is less than zero, and this is beyond, this is greater than another critical point that is greater than zero."}, {"video_title": "Analyzing mistakes when finding extrema example 2 AP Calculus AB Khan Academy.mp3", "Sentence": "But you have to make sure that when you test on either side that you're not going beyond another critical point because critical points are places where you can change direction. And so let's see what she does in step three right over here. Well, it is indeed in step three that she's testing, she's trying to test values on either side of the critical point that she, that the one critical point that she identified. But the problem here, the reason why this is a little shady is this is beyond another critical point that is less than zero, and this is beyond, this is greater than another critical point that is greater than zero. This is larger than the critical point one, and this is less than the critical point negative one. What she should have tried is x equals 0.5 and x equals negative 0.5. So this is what she should have done, is try maybe negative two, negative one, negative 1.5, zero, 1.5, and then one we know is undefined, and then positive two."}, {"video_title": "Analyzing mistakes when finding extrema example 2 AP Calculus AB Khan Academy.mp3", "Sentence": "But the problem here, the reason why this is a little shady is this is beyond another critical point that is less than zero, and this is beyond, this is greater than another critical point that is greater than zero. This is larger than the critical point one, and this is less than the critical point negative one. What she should have tried is x equals 0.5 and x equals negative 0.5. So this is what she should have done, is try maybe negative two, negative one, negative 1.5, zero, 1.5, and then one we know is undefined, and then positive two. Because this is a candidate extremum, this is a candidate extremum, and this is a candidate extremum right over here. And so you wanna see in which of these situations do you have a sign change of the derivative, and you just wanna test in the intervals between the extremum points. So I would say that really the main mistake she made in step two is not identifying all of the critical points."}, {"video_title": "Introduction to minimum and maximum points Functions Algebra I Khan Academy.mp3", "Sentence": "And I want to think about the maximum and minimum points on this. So we've already talked a little bit about absolute maximum and absolute minimum points on an interval. And those are pretty obvious. We hit a maximum point right over here, right at the beginning of our interval. Looks like when x is equal to 0, this is the absolute maximum point for the interval. And the absolute minimum point for the interval happens at the other end point. So if this is a, this is b, the absolute minimum point is f of b."}, {"video_title": "Introduction to minimum and maximum points Functions Algebra I Khan Academy.mp3", "Sentence": "We hit a maximum point right over here, right at the beginning of our interval. Looks like when x is equal to 0, this is the absolute maximum point for the interval. And the absolute minimum point for the interval happens at the other end point. So if this is a, this is b, the absolute minimum point is f of b. And the absolute maximum point is f of a. And it looks like a is equal to 0. But you're probably thinking, hey, there are other kind of interesting points right over here."}, {"video_title": "Introduction to minimum and maximum points Functions Algebra I Khan Academy.mp3", "Sentence": "So if this is a, this is b, the absolute minimum point is f of b. And the absolute maximum point is f of a. And it looks like a is equal to 0. But you're probably thinking, hey, there are other kind of interesting points right over here. This point right over here, it isn't the largest. We're not taking on this value right over here. It is definitely not the largest value that the function takes on in that interval."}, {"video_title": "Introduction to minimum and maximum points Functions Algebra I Khan Academy.mp3", "Sentence": "But you're probably thinking, hey, there are other kind of interesting points right over here. This point right over here, it isn't the largest. We're not taking on this value right over here. It is definitely not the largest value that the function takes on in that interval. But relative to the other values around it, it seems like a little bit of a hill. It's larger than the other ones. Locally, it looks like a little bit of a maximum."}, {"video_title": "Introduction to minimum and maximum points Functions Algebra I Khan Academy.mp3", "Sentence": "It is definitely not the largest value that the function takes on in that interval. But relative to the other values around it, it seems like a little bit of a hill. It's larger than the other ones. Locally, it looks like a little bit of a maximum. And so that's why this value right over here would be called, this value right over here, let's call this, let's say this right over here is c. This is c, so this is f of c. We would call f of c is a relative maximum value. And we're saying relative because it's obviously the function takes on other values that are larger than it. But for the x values near c, f of c is larger than all of those."}, {"video_title": "Introduction to minimum and maximum points Functions Algebra I Khan Academy.mp3", "Sentence": "Locally, it looks like a little bit of a maximum. And so that's why this value right over here would be called, this value right over here, let's call this, let's say this right over here is c. This is c, so this is f of c. We would call f of c is a relative maximum value. And we're saying relative because it's obviously the function takes on other values that are larger than it. But for the x values near c, f of c is larger than all of those. Similarly, I can never say that word, if this point right over here is d, f of d looks like a relative minimum point or relative minimum value. f of d is a relative minimum or a local minimum value. Once again, over the whole interval, there's definitely points that are lower."}, {"video_title": "Introduction to minimum and maximum points Functions Algebra I Khan Academy.mp3", "Sentence": "But for the x values near c, f of c is larger than all of those. Similarly, I can never say that word, if this point right over here is d, f of d looks like a relative minimum point or relative minimum value. f of d is a relative minimum or a local minimum value. Once again, over the whole interval, there's definitely points that are lower. And we hit an absolute minimum for the interval at x is equal to b. But this is a relative minimum or a local minimum because it's lower than the, if we look at the x values around d, the function at those values is higher than when we get to d. So let's think about, it's fine for me to say, well, you're at a relative maximum if you hit a larger value of your function than any of the surrounding values. And you're at a minimum if you're at a smaller value than any of the surrounding areas."}, {"video_title": "Introduction to minimum and maximum points Functions Algebra I Khan Academy.mp3", "Sentence": "Once again, over the whole interval, there's definitely points that are lower. And we hit an absolute minimum for the interval at x is equal to b. But this is a relative minimum or a local minimum because it's lower than the, if we look at the x values around d, the function at those values is higher than when we get to d. So let's think about, it's fine for me to say, well, you're at a relative maximum if you hit a larger value of your function than any of the surrounding values. And you're at a minimum if you're at a smaller value than any of the surrounding areas. But how could we write that mathematically? So here, I'll just give you the definition that really is just a more formal way of saying what we just said. So we say that f of c is a relative maximum value if f of c is greater than or equal to f of x for all x."}, {"video_title": "Introduction to minimum and maximum points Functions Algebra I Khan Academy.mp3", "Sentence": "And you're at a minimum if you're at a smaller value than any of the surrounding areas. But how could we write that mathematically? So here, I'll just give you the definition that really is just a more formal way of saying what we just said. So we say that f of c is a relative maximum value if f of c is greater than or equal to f of x for all x. We could just say kind of in a casual way, for all x near c. So we could write it like that. But that's not too rigorous because what does it mean to be near c? And so a more rigorous way of saying it, for all x that's within an open interval of c minus h to c plus h, where h is some value greater than 0."}, {"video_title": "Introduction to minimum and maximum points Functions Algebra I Khan Academy.mp3", "Sentence": "So we say that f of c is a relative maximum value if f of c is greater than or equal to f of x for all x. We could just say kind of in a casual way, for all x near c. So we could write it like that. But that's not too rigorous because what does it mean to be near c? And so a more rigorous way of saying it, for all x that's within an open interval of c minus h to c plus h, where h is some value greater than 0. So does that make sense? Well, let's look at it. So let's construct an open interval."}, {"video_title": "Introduction to minimum and maximum points Functions Algebra I Khan Academy.mp3", "Sentence": "And so a more rigorous way of saying it, for all x that's within an open interval of c minus h to c plus h, where h is some value greater than 0. So does that make sense? Well, let's look at it. So let's construct an open interval. So it looks like for all of the x values in, and you just have to find one open interval. There might be many open intervals where this is true. But if we construct an open interval that looks something like that."}, {"video_title": "Introduction to minimum and maximum points Functions Algebra I Khan Academy.mp3", "Sentence": "So let's construct an open interval. So it looks like for all of the x values in, and you just have to find one open interval. There might be many open intervals where this is true. But if we construct an open interval that looks something like that. So this value right over here is c plus h. That value right over here is c minus h. And you see that over that interval, the function at c, f of c, is definitely greater than or equal to the value of the function over any other part of that open interval. And so you could imagine, I encourage you to pause the video and you could write out what the more formal definition of a relative minimum point would be. Well, we would just write, let's take d as our relative minimum."}, {"video_title": "Introduction to minimum and maximum points Functions Algebra I Khan Academy.mp3", "Sentence": "But if we construct an open interval that looks something like that. So this value right over here is c plus h. That value right over here is c minus h. And you see that over that interval, the function at c, f of c, is definitely greater than or equal to the value of the function over any other part of that open interval. And so you could imagine, I encourage you to pause the video and you could write out what the more formal definition of a relative minimum point would be. Well, we would just write, let's take d as our relative minimum. We can say that f of d is a relative minimum point if f of d is less than or equal to f of x for all x in an interval, in an open interval, between d minus h and d plus h for h is greater than 0. So you can find an interval here. So let's say this is d plus h. This is d minus h. The function over that interval, f of d is always less than or equal to any of the other values."}, {"video_title": "Introduction to minimum and maximum points Functions Algebra I Khan Academy.mp3", "Sentence": "Well, we would just write, let's take d as our relative minimum. We can say that f of d is a relative minimum point if f of d is less than or equal to f of x for all x in an interval, in an open interval, between d minus h and d plus h for h is greater than 0. So you can find an interval here. So let's say this is d plus h. This is d minus h. The function over that interval, f of d is always less than or equal to any of the other values. The f is of all of these other x's in that interval. And that's why we say that it's a relative minimum point. So in everyday language, relative max, if the function takes on a larger value at c, then for the x values around c, and you're at a relative minimum value, if the function takes on a lower value at d, then for the x values near d."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "And we're going to rotate the region in between these two functions. So that's this region right over here. And we're not going to rotate it just around the x-axis. We're going to rotate it around the horizontal line y equals 4. So we're going to rotate it around this. And if we do that, we'll get a shape that looks like this. I drew it ahead of time just so I could draw it nicely."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "We're going to rotate it around the horizontal line y equals 4. So we're going to rotate it around this. And if we do that, we'll get a shape that looks like this. I drew it ahead of time just so I could draw it nicely. And as you can see, it looks like some type of a, I don't know, a vase with a hole at the bottom. And so what we're going to do is attempt to do this using, I guess you could call it the washer method, which is a variant of the disk method. So let's construct a washer."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "I drew it ahead of time just so I could draw it nicely. And as you can see, it looks like some type of a, I don't know, a vase with a hole at the bottom. And so what we're going to do is attempt to do this using, I guess you could call it the washer method, which is a variant of the disk method. So let's construct a washer. So let's look at a given x. So let's say an x right over here. So let's say that we're at an x right over there."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "So let's construct a washer. So let's look at a given x. So let's say an x right over here. So let's say that we're at an x right over there. And what we're going to do is we're going to rotate this region. We're going to give it some depth, dx. So that is dx."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say that we're at an x right over there. And what we're going to do is we're going to rotate this region. We're going to give it some depth, dx. So that is dx. And we're going to rotate this around the line y is equal to 4. So if you were to visualize it over here, you have some depth. And when you rotate it around, the inner radius is going to look like the inner radius of our washer is going to look something like that."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "So that is dx. And we're going to rotate this around the line y is equal to 4. So if you were to visualize it over here, you have some depth. And when you rotate it around, the inner radius is going to look like the inner radius of our washer is going to look something like that. And then the outer radius of our washer is going to contour around x squared minus 2x. So it's going to look something like that. And of course, our washer is going to have some depth."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "And when you rotate it around, the inner radius is going to look like the inner radius of our washer is going to look something like that. And then the outer radius of our washer is going to contour around x squared minus 2x. So it's going to look something like that. And of course, our washer is going to have some depth. So let me draw the depth. So it's going to have some depth, dx. So this is my best attempt at drawing some of that to depth."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "And of course, our washer is going to have some depth. So let me draw the depth. So it's going to have some depth, dx. So this is my best attempt at drawing some of that to depth. So this is the depth of our washer. And then just to make the face of the washer a little bit clearer, let me do it in this green color. So the face of the washer is going to be all of this business."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "So this is my best attempt at drawing some of that to depth. So this is the depth of our washer. And then just to make the face of the washer a little bit clearer, let me do it in this green color. So the face of the washer is going to be all of this business. So if we can figure out the volume of one of these washers for a given x, then we just have to sum up all of the washers for all of the x's in our interval. So let's see if we can set up the integral. And maybe then in the next video, we'll just forge ahead and actually evaluate the integral."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "So the face of the washer is going to be all of this business. So if we can figure out the volume of one of these washers for a given x, then we just have to sum up all of the washers for all of the x's in our interval. So let's see if we can set up the integral. And maybe then in the next video, we'll just forge ahead and actually evaluate the integral. So let's think about the volume of the washer. To think about the volume of the washer, we really just have to think about the area of the face of the washer. So area of face is going to be equal to what?"}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "And maybe then in the next video, we'll just forge ahead and actually evaluate the integral. So let's think about the volume of the washer. To think about the volume of the washer, we really just have to think about the area of the face of the washer. So area of face is going to be equal to what? Well, it would be the area of the washer, if it wasn't a washer, if it was just a coin, and then subtract out the area of the part that you're cutting out. So the area of the washer, if it didn't have a hole in the middle, would just be pi times the outer radius squared. It would be pi times this radius squared."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "So area of face is going to be equal to what? Well, it would be the area of the washer, if it wasn't a washer, if it was just a coin, and then subtract out the area of the part that you're cutting out. So the area of the washer, if it didn't have a hole in the middle, would just be pi times the outer radius squared. It would be pi times this radius squared. That we could call the outer radius. And since it's a washer, we need to subtract out the area of this inner circle. So minus pi times inner radius squared."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "It would be pi times this radius squared. That we could call the outer radius. And since it's a washer, we need to subtract out the area of this inner circle. So minus pi times inner radius squared. So we really just have to figure out what the outer and inner radius, or radii, I should say, are. So let's think about it. So our outer radius is going to be equal to what?"}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "So minus pi times inner radius squared. So we really just have to figure out what the outer and inner radius, or radii, I should say, are. So let's think about it. So our outer radius is going to be equal to what? Well, we can visualize it over here. This is our outer radius, which is also going to be equal to that right over there. So that's the distance between y equals 4 and the function that's defining our outside."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "So our outer radius is going to be equal to what? Well, we can visualize it over here. This is our outer radius, which is also going to be equal to that right over there. So that's the distance between y equals 4 and the function that's defining our outside. The distance between y equals 4 and the function that is defining our outside. So this is essentially, this height right over here is going to be equal to 4 minus x squared minus 2x. I'm just finding the distance, or the height, between these two functions."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "So that's the distance between y equals 4 and the function that's defining our outside. The distance between y equals 4 and the function that is defining our outside. So this is essentially, this height right over here is going to be equal to 4 minus x squared minus 2x. I'm just finding the distance, or the height, between these two functions. So the outer radius is going to be 4 minus this. Minus x squared minus 2x, which is just 4 minus x squared plus 2x. Now what is the inner radius?"}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "I'm just finding the distance, or the height, between these two functions. So the outer radius is going to be 4 minus this. Minus x squared minus 2x, which is just 4 minus x squared plus 2x. Now what is the inner radius? Inner radius. What is that going to be? Well, that's just going to be this distance."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "Now what is the inner radius? Inner radius. What is that going to be? Well, that's just going to be this distance. That's just going to be the distance between y equals 4 and y equals x. So this is going to be 4 minus x. It's just going to be 4 minus x."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "Well, that's just going to be this distance. That's just going to be the distance between y equals 4 and y equals x. So this is going to be 4 minus x. It's just going to be 4 minus x. So if we wanted to find the area of the face of one of these washers for a given x, it's going to be, and we can factor out this pi, it's going to be pi times the outer radius squared, which is all of this business squared. So it's going to be 4 minus x squared plus 2x squared. Minus pi times the inner radius, although we factored out the pi."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "It's just going to be 4 minus x. So if we wanted to find the area of the face of one of these washers for a given x, it's going to be, and we can factor out this pi, it's going to be pi times the outer radius squared, which is all of this business squared. So it's going to be 4 minus x squared plus 2x squared. Minus pi times the inner radius, although we factored out the pi. So minus the inner radius squared. So minus 4 minus x squared. So this will give us the surface, the surface there, or the area of the surface or the face of one of these washers."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "Minus pi times the inner radius, although we factored out the pi. So minus the inner radius squared. So minus 4 minus x squared. So this will give us the surface, the surface there, or the area of the surface or the face of one of these washers. If we want the volume of one of those washers, we then just have to multiply times the depth dx. We just multiply times the depth dx. And then if we want to actually find the volume of this entire figure, then we just have to sum up all of these washers for each of our x's."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "So this will give us the surface, the surface there, or the area of the surface or the face of one of these washers. If we want the volume of one of those washers, we then just have to multiply times the depth dx. We just multiply times the depth dx. And then if we want to actually find the volume of this entire figure, then we just have to sum up all of these washers for each of our x's. So let's do that. So we're going to sum up the washers for each of our x's and take the limit as they approach 0. But we have to make sure we got our interval right."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "And then if we want to actually find the volume of this entire figure, then we just have to sum up all of these washers for each of our x's. So let's do that. So we're going to sum up the washers for each of our x's and take the limit as they approach 0. But we have to make sure we got our interval right. So what are these? We care about the entire region between the points where they intersect. So let's make sure we get our interval."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "But we have to make sure we got our interval right. So what are these? We care about the entire region between the points where they intersect. So let's make sure we get our interval. So to figure out our interval, we just say, well, when does y equal x intersect y equal x squared minus 2x? So we just have to say, when does x, let me do this in a different color. We just have to think about when does x equal x squared minus 2x?"}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "So let's make sure we get our interval. So to figure out our interval, we just say, well, when does y equal x intersect y equal x squared minus 2x? So we just have to say, when does x, let me do this in a different color. We just have to think about when does x equal x squared minus 2x? Equal x squared minus 2x. When are our two functions equal to each other? Which is equivalent to, if we just subtract x from both sides, we get, when does x squared minus 3x equal 0?"}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "We just have to think about when does x equal x squared minus 2x? Equal x squared minus 2x. When are our two functions equal to each other? Which is equivalent to, if we just subtract x from both sides, we get, when does x squared minus 3x equal 0? We can factor out an x on the right-hand side. So this is going to be, when does x times x minus 3 equal 0? Well, if the product is equal to 0, at least one of these need to be equal to 0."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "Which is equivalent to, if we just subtract x from both sides, we get, when does x squared minus 3x equal 0? We can factor out an x on the right-hand side. So this is going to be, when does x times x minus 3 equal 0? Well, if the product is equal to 0, at least one of these need to be equal to 0. So x could be equal to 0, or x minus 3 is equal to 0. So x is equal to 0, or x is equal to 3. So this is x is 0, and this right over here is x is equal to 3."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "Well, if the product is equal to 0, at least one of these need to be equal to 0. So x could be equal to 0, or x minus 3 is equal to 0. So x is equal to 0, or x is equal to 3. So this is x is 0, and this right over here is x is equal to 3. So that gives us our interval. We're going to go from x equals 0 to x equals 3 to get our volume. In the next video, we'll actually evaluate this integral."}, {"video_title": "Limits at infinity of quotients with square roots (even power) AP Calculus AB Khan Academy.mp3", "Sentence": "Well, whenever we're trying to find limits at either positive or negative infinity of rational expressions like this, it's useful to look at, well, what is the highest degree term in the numerator or in the denominator, or actually in the numerator and the denominator, and then divide the numerator and the denominator by that highest degree, by x to that degree. Because if we do that, then we're going to end up with some constants and some other things that will go to zero as we approach positive or negative infinity, and we should be able to find this limit. So what I'm talking about, let's divide the numerator by one over x squared, and let's divide the denominator by one over x squared. Now you might be saying, wait, wait, I see an x to the fourth here, that's a higher degree. Remember, it's under the radical here. So if you want to look at it at a very high level, you're saying, okay, well, x to the fourth, but it's under, you're gonna take the square root of this entire expression, so you could really view this as a second degree term. So the highest degree is really second degree, so let's divide the numerator and the denominator by x squared."}, {"video_title": "Limits at infinity of quotients with square roots (even power) AP Calculus AB Khan Academy.mp3", "Sentence": "Now you might be saying, wait, wait, I see an x to the fourth here, that's a higher degree. Remember, it's under the radical here. So if you want to look at it at a very high level, you're saying, okay, well, x to the fourth, but it's under, you're gonna take the square root of this entire expression, so you could really view this as a second degree term. So the highest degree is really second degree, so let's divide the numerator and the denominator by x squared. And if we do that, dividing, so this is going to be the same thing as, so this is going to be the limit, the limit as x approaches negative infinity of, so let me just do a little bit of an aside here. So if I have one over x squared, or let me write it, let me, one over x squared times the square root of four x to the fourth minus x, like we have in the numerator here. This is equal to, this is the same thing as one over the square root of x to the fourth times the square root of four x to the fourth minus x."}, {"video_title": "Limits at infinity of quotients with square roots (even power) AP Calculus AB Khan Academy.mp3", "Sentence": "So the highest degree is really second degree, so let's divide the numerator and the denominator by x squared. And if we do that, dividing, so this is going to be the same thing as, so this is going to be the limit, the limit as x approaches negative infinity of, so let me just do a little bit of an aside here. So if I have one over x squared, or let me write it, let me, one over x squared times the square root of four x to the fourth minus x, like we have in the numerator here. This is equal to, this is the same thing as one over the square root of x to the fourth times the square root of four x to the fourth minus x. And so this is equal to the square root of four x to the fourth minus x over x to the fourth, which is equal to the square root of, and all I did is I brought the radical in here. This is, you could view this as the square root of all this divided by the square root of this, which is equal to, just using our exponent rules, the square root of four x to the fourth minus x over x to the fourth. And then this is the same thing as four minus, x over x to the fourth is one over x to the third."}, {"video_title": "Limits at infinity of quotients with square roots (even power) AP Calculus AB Khan Academy.mp3", "Sentence": "This is equal to, this is the same thing as one over the square root of x to the fourth times the square root of four x to the fourth minus x. And so this is equal to the square root of four x to the fourth minus x over x to the fourth, which is equal to the square root of, and all I did is I brought the radical in here. This is, you could view this as the square root of all this divided by the square root of this, which is equal to, just using our exponent rules, the square root of four x to the fourth minus x over x to the fourth. And then this is the same thing as four minus, x over x to the fourth is one over x to the third. So this numerator is going to be, the numerator is going to be the square root of four minus one x to the third power. And then the denominator is going to be equal to, well you divide two x squared by x squared. You're just going to be left with two."}, {"video_title": "Limits at infinity of quotients with square roots (even power) AP Calculus AB Khan Academy.mp3", "Sentence": "And then this is the same thing as four minus, x over x to the fourth is one over x to the third. So this numerator is going to be, the numerator is going to be the square root of four minus one x to the third power. And then the denominator is going to be equal to, well you divide two x squared by x squared. You're just going to be left with two. And then three divided by x squared is going to be three over x squared. Now let's think about the limit as we approach negative infinity. As we approach negative infinity, this is going to approach zero."}, {"video_title": "Limits at infinity of quotients with square roots (even power) AP Calculus AB Khan Academy.mp3", "Sentence": "You're just going to be left with two. And then three divided by x squared is going to be three over x squared. Now let's think about the limit as we approach negative infinity. As we approach negative infinity, this is going to approach zero. One divided by things that are becoming more and more and more and more and more negative, their magnitude is getting larger, so this is going to approach zero. This over here is also going to be, this thing is also going to be approaching zero. We're dividing by larger and larger and larger values."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Over here I've drawn part of the graph of y is equal to x squared. And what we're going to do is use our powers of definite integrals to find volumes instead of just areas. So let's review what we're doing when we take just a regular definite integral. So if we take the definite integral between say 0 and 2 of x squared dx, what does that represent? Let's look at our endpoints. So this is x is equal to 0. Let's say that this right over here is x is equal to 2."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So if we take the definite integral between say 0 and 2 of x squared dx, what does that represent? Let's look at our endpoints. So this is x is equal to 0. Let's say that this right over here is x is equal to 2. What we're doing is for each x we're finding a little dx around it. So this right over here is a little dx. And we're multiplying that dx times our function, times x squared."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say that this right over here is x is equal to 2. What we're doing is for each x we're finding a little dx around it. So this right over here is a little dx. And we're multiplying that dx times our function, times x squared. So what we're doing is we're multiplying this width times this height. Times this height right over here. The height right over here is x squared."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "And we're multiplying that dx times our function, times x squared. So what we're doing is we're multiplying this width times this height. Times this height right over here. The height right over here is x squared. And we're getting the area of this little rectangle. And the integral sign is literally the sum of all of these rectangles. All of these rectangles for all of the x's between x is equal to 0 and x is equal to 2."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "The height right over here is x squared. And we're getting the area of this little rectangle. And the integral sign is literally the sum of all of these rectangles. All of these rectangles for all of the x's between x is equal to 0 and x is equal to 2. But the limit of that as these dx's get smaller and smaller and smaller, get infinitely small, would not being equal to 0, and we have an infinite number of them. That's the whole power of the definite integral. And so you can imagine as these dx's get smaller and smaller and smaller, these rectangles get narrower and narrower and narrower, and we have more of them, we're getting a better and better approximation of the area under the curve until at the limit we are getting the area under the curve."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "All of these rectangles for all of the x's between x is equal to 0 and x is equal to 2. But the limit of that as these dx's get smaller and smaller and smaller, get infinitely small, would not being equal to 0, and we have an infinite number of them. That's the whole power of the definite integral. And so you can imagine as these dx's get smaller and smaller and smaller, these rectangles get narrower and narrower and narrower, and we have more of them, we're getting a better and better approximation of the area under the curve until at the limit we are getting the area under the curve. Now we're going to apply that same idea, not to find the area under this curve, but to find the volume if we were to rotate this curve around the x axis. So this is going to stretch our powers of visualization here. So let's think about what happens when we rotate this thing around the x axis."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "And so you can imagine as these dx's get smaller and smaller and smaller, these rectangles get narrower and narrower and narrower, and we have more of them, we're getting a better and better approximation of the area under the curve until at the limit we are getting the area under the curve. Now we're going to apply that same idea, not to find the area under this curve, but to find the volume if we were to rotate this curve around the x axis. So this is going to stretch our powers of visualization here. So let's think about what happens when we rotate this thing around the x axis. So let's figure to rotate it, and I'll look at, say we're looking at a little bit from the right. So we get kind of a base that looks something like this. So this is my best attempt to draw it."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So let's think about what happens when we rotate this thing around the x axis. So let's figure to rotate it, and I'll look at, say we're looking at a little bit from the right. So we get kind of a base that looks something like this. So this is my best attempt to draw it. So you have a base that looks something like that, and then the rest of the function. So it looks kind of like, if we just think about it between 0 and 2, it looks like one of those pieces from, I don't know if you ever played the game Sorry, or it looks like a little bit of a kind of a weird hat. So it looks like this, and let me shade it in a little bit."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So this is my best attempt to draw it. So you have a base that looks something like that, and then the rest of the function. So it looks kind of like, if we just think about it between 0 and 2, it looks like one of those pieces from, I don't know if you ever played the game Sorry, or it looks like a little bit of a kind of a weird hat. So it looks like this, and let me shade it in a little bit. So it looks something like that. And just so that we're making sure we can visualize this thing that's being rotated, and we care about the entire volume of the thing, let me draw it from a few different angles. So if I drew it from the top, it would look something like this."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So it looks like this, and let me shade it in a little bit. So it looks something like that. And just so that we're making sure we can visualize this thing that's being rotated, and we care about the entire volume of the thing, let me draw it from a few different angles. So if I drew it from the top, it would look something like this. It would become a little more obvious. It would look something like a hat. Point up like this, and it goes down like that."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So if I drew it from the top, it would look something like this. It would become a little more obvious. It would look something like a hat. Point up like this, and it goes down like that. It would look something like that. So we're not seeing, in this angle, we're not seeing the bottom of it. And if you were to just orient yourself, the axes in this case look like this."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Point up like this, and it goes down like that. It would look something like that. So we're not seeing, in this angle, we're not seeing the bottom of it. And if you were to just orient yourself, the axes in this case look like this. So this is the y-axis, and then the x-axis goes right inside of this thing, and then pops out the other side. And if this thing was transparent, then you could see the back side. It would look something like that."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "And if you were to just orient yourself, the axes in this case look like this. So this is the y-axis, and then the x-axis goes right inside of this thing, and then pops out the other side. And if this thing was transparent, then you could see the back side. It would look something like that. The x-axis, if you could see through it, would pop the base right over there, would go right through the base right over there, and then come out on the other side. So this is one orientation for the same thing. You could visualize it from different angles."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "It would look something like that. The x-axis, if you could see through it, would pop the base right over there, would go right through the base right over there, and then come out on the other side. So this is one orientation for the same thing. You could visualize it from different angles. So let's think about how we can take the volume of it. Well, instead of thinking about the area of each of these rectangles, what happens if we rotate each of these rectangles around the x-axis? So let's do it."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "You could visualize it from different angles. So let's think about how we can take the volume of it. Well, instead of thinking about the area of each of these rectangles, what happens if we rotate each of these rectangles around the x-axis? So let's do it. So let's take each of these. Let's say you have this dx right over here, and you rotate it around the x-axis. So if you were to rotate this thing around the x-axis, so I'm trying my best to, around the x-axis, you rotate it, what do you end up with?"}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So let's do it. So let's take each of these. Let's say you have this dx right over here, and you rotate it around the x-axis. So if you were to rotate this thing around the x-axis, so I'm trying my best to, around the x-axis, you rotate it, what do you end up with? What do you end up with? Well, you get something that looks kind of like a coin, like a disk, like a quarter of some kind. Let me draw it out here."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So if you were to rotate this thing around the x-axis, so I'm trying my best to, around the x-axis, you rotate it, what do you end up with? What do you end up with? Well, you get something that looks kind of like a coin, like a disk, like a quarter of some kind. Let me draw it out here. So if you were to, that same disk out here would look something like this. It would look something like this, and it has a depth of dx. So how can we find the volume of that disk?"}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Let me draw it out here. So if you were to, that same disk out here would look something like this. It would look something like this, and it has a depth of dx. So how can we find the volume of that disk? Let me redraw it out here, too. It's really important to visualize this stuff properly. So this is my x-axis."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So how can we find the volume of that disk? Let me redraw it out here, too. It's really important to visualize this stuff properly. So this is my x-axis. My disk looks something like this. My best attempt. The x-axis hits it right over there, comes out of the center, and then this is the surface of my disk, and then this right over here is my depth, dx."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So this is my x-axis. My disk looks something like this. My best attempt. The x-axis hits it right over there, comes out of the center, and then this is the surface of my disk, and then this right over here is my depth, dx. So that looks pretty good. And then let me just shade it in a little bit to give you a little bit of the depth. So how can we find the volume of this?"}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "The x-axis hits it right over there, comes out of the center, and then this is the surface of my disk, and then this right over here is my depth, dx. So that looks pretty good. And then let me just shade it in a little bit to give you a little bit of the depth. So how can we find the volume of this? Well, like any disk or cylinder, you just have to think about what the area of this face is and then multiply it times the depth. So what's the area of this face? Well, we know that the area of a circle is equal to pi r squared."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So how can we find the volume of this? Well, like any disk or cylinder, you just have to think about what the area of this face is and then multiply it times the depth. So what's the area of this face? Well, we know that the area of a circle is equal to pi r squared. If we know the radius of this face, we can figure out the area of the face. Well, what's the radius? Well, the radius is just the height of that original rectangle, and for any x, the height over here is going to be equal to f of x, and in this case, f of x is x squared."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we know that the area of a circle is equal to pi r squared. If we know the radius of this face, we can figure out the area of the face. Well, what's the radius? Well, the radius is just the height of that original rectangle, and for any x, the height over here is going to be equal to f of x, and in this case, f of x is x squared. So over here, our radius is equal to x squared. So the area of the face for a particular x is going to be equal to pi times f of x squared. In this case, f of x is x squared."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the radius is just the height of that original rectangle, and for any x, the height over here is going to be equal to f of x, and in this case, f of x is x squared. So over here, our radius is equal to x squared. So the area of the face for a particular x is going to be equal to pi times f of x squared. In this case, f of x is x squared. Now what's our volume going to be? Well, our volume is going to be our area times the depth here. It's going to be that times the depth times dx."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "In this case, f of x is x squared. Now what's our volume going to be? Well, our volume is going to be our area times the depth here. It's going to be that times the depth times dx. So the volume of this thing right over here, so the volume just of this coin, I guess you could call it, is going to be equal to my area times dx, which is equal to pi times x squared squared. So it's equal to pi x squared squared is x to the fourth. Pi x to the fourth dx."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "It's going to be that times the depth times dx. So the volume of this thing right over here, so the volume just of this coin, I guess you could call it, is going to be equal to my area times dx, which is equal to pi times x squared squared. So it's equal to pi x squared squared is x to the fourth. Pi x to the fourth dx. Now this expression right over here, this gave us the volume just of one of those disks, but what we want is the volume of this entire hat or this entire kind of bugle or cone-looking, or I guess you could say the front of a trumpet-looking thing. So how could we do that? Well, the exact same technique."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Pi x to the fourth dx. Now this expression right over here, this gave us the volume just of one of those disks, but what we want is the volume of this entire hat or this entire kind of bugle or cone-looking, or I guess you could say the front of a trumpet-looking thing. So how could we do that? Well, the exact same technique. What happens if we were to take the sum of all of these things? So let's do that. Take the sum of all of these things, and I'll switch to one color, pi times x to the fourth dx."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the exact same technique. What happens if we were to take the sum of all of these things? So let's do that. Take the sum of all of these things, and I'll switch to one color, pi times x to the fourth dx. We're going to take the sum of all of these things from x is equal to 0 to 2. Those are the boundaries that we started off with. I just defined them arbitrarily."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Take the sum of all of these things, and I'll switch to one color, pi times x to the fourth dx. We're going to take the sum of all of these things from x is equal to 0 to 2. Those are the boundaries that we started off with. I just defined them arbitrarily. We could do this really for any two x values. Between x is equal to 0 and x equals 2, and we're going to take the sum of all of these, the volumes of all of these coins, but the limit as the depths get smaller and smaller and smaller and we have more and more and more coins, at the limit we're actually going to get the volume of our cone or our bugle or whatever we want to call it. So if we just evaluate this definite integral, we have our volume."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "I just defined them arbitrarily. We could do this really for any two x values. Between x is equal to 0 and x equals 2, and we're going to take the sum of all of these, the volumes of all of these coins, but the limit as the depths get smaller and smaller and smaller and we have more and more and more coins, at the limit we're actually going to get the volume of our cone or our bugle or whatever we want to call it. So if we just evaluate this definite integral, we have our volume. So let's see if we can do that. So now this is just taking a standard definite integral. So this is going to be equal to, and I encourage you to try it out before I do it."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So if we just evaluate this definite integral, we have our volume. So let's see if we can do that. So now this is just taking a standard definite integral. So this is going to be equal to, and I encourage you to try it out before I do it. So we could take the pi out. So it's going to be equal to pi times the integral from 0 to 2 of x to the fourth dx. I don't like that color."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be equal to, and I encourage you to try it out before I do it. So we could take the pi out. So it's going to be equal to pi times the integral from 0 to 2 of x to the fourth dx. I don't like that color. Now the antiderivative of x to the fourth is x to the fifth over 5. So this is going to be equal to pi times x to the fifth over 5. And we're going to go from 0 to 2."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "I don't like that color. Now the antiderivative of x to the fourth is x to the fifth over 5. So this is going to be equal to pi times x to the fifth over 5. And we're going to go from 0 to 2. So this is going to be equal to pi times, this thing evaluated at 2. Let's see, 2 to the third is 8, 2 to the fourth is 16, 2 to the fifth is, let me actually just write it down, 2 to the fifth over 5 minus 0 to the fifth over 5. And this is going to be equal to, 2 to the fifth is 32."}, {"video_title": "Disc method around x-axis Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "And we're going to go from 0 to 2. So this is going to be equal to pi times, this thing evaluated at 2. Let's see, 2 to the third is 8, 2 to the fourth is 16, 2 to the fifth is, let me actually just write it down, 2 to the fifth over 5 minus 0 to the fifth over 5. And this is going to be equal to, 2 to the fifth is 32. So it's going to be equal to pi times 32 over 5 minus, well this is just 0, minus 0, which is equal to 32 pi over 5. And we're done. We're able to figure out the volume of this kind of wacky shape."}, {"video_title": "Planar motion (with integrals) Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "At t equals one, the particle is at the point three comma four so the first part is what is the magnitude of the displacement of the particle between time t equals one and t equals three, and then we need to figure out its position, we need to round to the nearest tenth. So like always, pause this video, and I think you'll have to use a calculator, but pause this video and try to work through it on your own. So we've done questions like this in one dimension, but now we're doing it in two dimensions. But the key is is to just break it up into the component dimensions. So what we really want to do is let's find the displacement in the x direction, so really just the change in x, and then let's just find the displacement in the vertical direction, or our change in y, and then we can use those, essentially using the Pythagorean theorem, to find the magnitude of the total displacement. And also, if we know the change in x and change in y, we just add the change in x to the three, and we add the change in y to the four to find the particle's position at time t equals three. So let's figure it out."}, {"video_title": "Planar motion (with integrals) Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "But the key is is to just break it up into the component dimensions. So what we really want to do is let's find the displacement in the x direction, so really just the change in x, and then let's just find the displacement in the vertical direction, or our change in y, and then we can use those, essentially using the Pythagorean theorem, to find the magnitude of the total displacement. And also, if we know the change in x and change in y, we just add the change in x to the three, and we add the change in y to the four to find the particle's position at time t equals three. So let's figure it out. So change in x from t equals one to t equals three, well that's just going to be the integral of the rate function in the x direction from time equal one to time equals three. So in the x direction, we have one over t plus seven. That's our x velocity as a function of time."}, {"video_title": "Planar motion (with integrals) Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So let's figure it out. So change in x from t equals one to t equals three, well that's just going to be the integral of the rate function in the x direction from time equal one to time equals three. So in the x direction, we have one over t plus seven. That's our x velocity as a function of time. One over t plus seven dt. And what is this going to be equal to? Well, you might want to do u substitution if you're unfamiliar, but you might recognize that the derivative of t plus seven is just one, so you could think of this as one times one over t plus seven."}, {"video_title": "Planar motion (with integrals) Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "That's our x velocity as a function of time. One over t plus seven dt. And what is this going to be equal to? Well, you might want to do u substitution if you're unfamiliar, but you might recognize that the derivative of t plus seven is just one, so you could think of this as one times one over t plus seven. And so we really can just take the antiderivative then with respect to t plus seven. So you get the natural log of the absolute value of t plus seven, and we are going to evaluate that at three and then subtract from that, it evaluated at one. So this is going to be natural log of the absolute value of 10, which is just the natural log of 10, minus the natural log of the absolute value of eight, which is just the natural log of eight, which is equal to the natural log of 10 over eight, just using our logarithm properties, which is equal to the natural log of 1.25."}, {"video_title": "Planar motion (with integrals) Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "Well, you might want to do u substitution if you're unfamiliar, but you might recognize that the derivative of t plus seven is just one, so you could think of this as one times one over t plus seven. And so we really can just take the antiderivative then with respect to t plus seven. So you get the natural log of the absolute value of t plus seven, and we are going to evaluate that at three and then subtract from that, it evaluated at one. So this is going to be natural log of the absolute value of 10, which is just the natural log of 10, minus the natural log of the absolute value of eight, which is just the natural log of eight, which is equal to the natural log of 10 over eight, just using our logarithm properties, which is equal to the natural log of 1.25. So I can get my calculator out in a second to calculate that. Actually, let's just, well, I'll do that in a second. And then let's figure out our change in y."}, {"video_title": "Planar motion (with integrals) Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So this is going to be natural log of the absolute value of 10, which is just the natural log of 10, minus the natural log of the absolute value of eight, which is just the natural log of eight, which is equal to the natural log of 10 over eight, just using our logarithm properties, which is equal to the natural log of 1.25. So I can get my calculator out in a second to calculate that. Actually, let's just, well, I'll do that in a second. And then let's figure out our change in y. Our change in y, once again, we're gonna take the integral from one to three, that's our time, the time over which we're thinking about the change. And then what is the y component of our velocity? Well, it's t to the fourth dt."}, {"video_title": "Planar motion (with integrals) Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "And then let's figure out our change in y. Our change in y, once again, we're gonna take the integral from one to three, that's our time, the time over which we're thinking about the change. And then what is the y component of our velocity? Well, it's t to the fourth dt. Well, this is going to be, take the reverse power rule, t to the fifth over five, at three and one. So this is three to the fifth over five is 243 over five, minus one to the fifth over five, minus 1 5th. So this is equal to 242 over five, which is what, 48.4, 48.4."}, {"video_title": "Planar motion (with integrals) Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "Well, it's t to the fourth dt. Well, this is going to be, take the reverse power rule, t to the fifth over five, at three and one. So this is three to the fifth over five is 243 over five, minus one to the fifth over five, minus 1 5th. So this is equal to 242 over five, which is what, 48.4, 48.4. And let me get my calculator out for this natural log of 1.25. 1.25 natural log, and we'll just round to two decimal places approximately 0.22. So this is approximately 0.22."}, {"video_title": "Planar motion (with integrals) Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So this is equal to 242 over five, which is what, 48.4, 48.4. And let me get my calculator out for this natural log of 1.25. 1.25 natural log, and we'll just round to two decimal places approximately 0.22. So this is approximately 0.22. So I figured out our change in x and our change in y. And actually, just from that, we can answer the second part of the question first. What is the particle's position at t equals three?"}, {"video_title": "Planar motion (with integrals) Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So this is approximately 0.22. So I figured out our change in x and our change in y. And actually, just from that, we can answer the second part of the question first. What is the particle's position at t equals three? Well, it's going to be our position at t equals one, where we, to each of the components, we add the respective change. So we would add, so this would be three plus our change in x from t equals one to t equals three. And it would be four plus our change in y."}, {"video_title": "Planar motion (with integrals) Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "What is the particle's position at t equals three? Well, it's going to be our position at t equals one, where we, to each of the components, we add the respective change. So we would add, so this would be three plus our change in x from t equals one to t equals three. And it would be four plus our change in y. So this is going to be equal to three plus our change in x. Well, that's going to be approximately 3.22. And four plus our change in y, that is what, 52.4."}, {"video_title": "Planar motion (with integrals) Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "And it would be four plus our change in y. So this is going to be equal to three plus our change in x. Well, that's going to be approximately 3.22. And four plus our change in y, that is what, 52.4. This right over here is 52.4. But we still have to answer the first question. What is the magnitude of the displacement?"}, {"video_title": "Planar motion (with integrals) Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "And four plus our change in y, that is what, 52.4. This right over here is 52.4. But we still have to answer the first question. What is the magnitude of the displacement? Well, it's the Pythagorean theorem. I'll draw a very rough sketch of what's going on. Sometimes it's useful to visualize it."}, {"video_title": "Planar motion (with integrals) Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "What is the magnitude of the displacement? Well, it's the Pythagorean theorem. I'll draw a very rough sketch of what's going on. Sometimes it's useful to visualize it. So our initial position is at three comma four. So three comma four. So we're right over there."}, {"video_title": "Planar motion (with integrals) Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "Sometimes it's useful to visualize it. So our initial position is at three comma four. So three comma four. So we're right over there. And we figured out our change in x isn't much. So our change in x is a positive 0.22. So our change in x, we're barely moving in that direction."}, {"video_title": "Planar motion (with integrals) Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So we're right over there. And we figured out our change in x isn't much. So our change in x is a positive 0.22. So our change in x, we're barely moving in that direction. And our change in y is 48.4. So we have a dramatic change. It really goes off the charts over here in that direction."}, {"video_title": "Planar motion (with integrals) Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So our change in x, we're barely moving in that direction. And our change in y is 48.4. So we have a dramatic change. It really goes off the charts over here in that direction. But if we wanted to add them together, if we want to add those vectors together, you could shift over your change in y right over here and then find the hypotenuse. The length of the hypotenuse would be the magnitude of the entire displacement. And so let's do that."}, {"video_title": "Planar motion (with integrals) Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "It really goes off the charts over here in that direction. But if we wanted to add them together, if we want to add those vectors together, you could shift over your change in y right over here and then find the hypotenuse. The length of the hypotenuse would be the magnitude of the entire displacement. And so let's do that. So the magnitude of the displacement is gonna be the square root of our change in x squared plus our change in y squared. Once again, this is just the Pythagorean theorem. And what is this going to be?"}, {"video_title": "Planar motion (with integrals) Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "And so let's do that. So the magnitude of the displacement is gonna be the square root of our change in x squared plus our change in y squared. Once again, this is just the Pythagorean theorem. And what is this going to be? I'll get my calculator out again for this. Okay, that was our change in x. Let me square it."}, {"video_title": "Planar motion (with integrals) Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "And what is this going to be? I'll get my calculator out again for this. Okay, that was our change in x. Let me square it. And then plus, we are going to have 48.4, 8.4 squared is equal to this right over here. And then we take the square root of that. So this is going to be, let's see if we, right over there."}, {"video_title": "Planar motion (with integrals) Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "Let me square it. And then plus, we are going to have 48.4, 8.4 squared is equal to this right over here. And then we take the square root of that. So this is going to be, let's see if we, right over there. And there you go. Our total, the magnitude of our total displacement is 48. If we round to the nearest 10th, 48.4."}, {"video_title": "Planar motion (with integrals) Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So this is going to be, let's see if we, right over there. And there you go. Our total, the magnitude of our total displacement is 48. If we round to the nearest 10th, 48.4. So this is approximately 48.4. And we're done. Now one thing that you might be noting is, hey, it looks like our total displacement, 48.4, is the same as our change in y."}, {"video_title": "Planar motion (with integrals) Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "If we round to the nearest 10th, 48.4. So this is approximately 48.4. And we're done. Now one thing that you might be noting is, hey, it looks like our total displacement, 48.4, is the same as our change in y. Now the reason why it came out this way is because our change in y was exactly 48.4, while the magnitude of our displacement was slightly more than 48.4. But when we round to the nearest 10th, we got to 48.4. The reason why they're so close is because our change in x was so small."}, {"video_title": "Planar motion (with integrals) Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "Now one thing that you might be noting is, hey, it looks like our total displacement, 48.4, is the same as our change in y. Now the reason why it came out this way is because our change in y was exactly 48.4, while the magnitude of our displacement was slightly more than 48.4. But when we round to the nearest 10th, we got to 48.4. The reason why they're so close is because our change in x was so small. We're talking about 0.22 as a change in x. And our change in y was so much that our hypotenuse was only slightly longer than our change in y. So that's why we got this result for this particular instance."}, {"video_title": "Introduction to improper integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So I want to figure out what this entire area is. And one way that we can denote that is with an improper definite integral or an improper integral. And we would denote it as one is our lower boundary, but we're just gonna keep on going forever as our upper boundary. So our upper boundary is infinity and we're taking the integral of one over x squared dx. And so let me be very clear, this right over here is an improper, improper, improper integral. Now, how do we actually deal with this? Well, by definition, this is the same thing as the limit, as the limit as n approaches infinity of the integral from one to n of one over x squared, one over x squared dx."}, {"video_title": "Introduction to improper integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So our upper boundary is infinity and we're taking the integral of one over x squared dx. And so let me be very clear, this right over here is an improper, improper, improper integral. Now, how do we actually deal with this? Well, by definition, this is the same thing as the limit, as the limit as n approaches infinity of the integral from one to n of one over x squared, one over x squared dx. And this is nice because we know how to evaluate this. This is just a definite integral where the upper boundary is n. And then we know how to take limits. We can figure out what the limit is as n approaches infinity."}, {"video_title": "Introduction to improper integrals AP Calculus BC Khan Academy.mp3", "Sentence": "Well, by definition, this is the same thing as the limit, as the limit as n approaches infinity of the integral from one to n of one over x squared, one over x squared dx. And this is nice because we know how to evaluate this. This is just a definite integral where the upper boundary is n. And then we know how to take limits. We can figure out what the limit is as n approaches infinity. So let's figure out if we can actually evaluate this thing. So the second fundamental theorem of calculus or the second part of the fundamental theorem of calculus tells us that this piece right over here, which is, let me write the limit part. So this part, I'll just rewrite the limit as n approaches infinity of, of, and we're gonna use the second fundamental theorem of calculus, we're going to evaluate the antiderivative of x of one over x squared or x to the negative two."}, {"video_title": "Introduction to improper integrals AP Calculus BC Khan Academy.mp3", "Sentence": "We can figure out what the limit is as n approaches infinity. So let's figure out if we can actually evaluate this thing. So the second fundamental theorem of calculus or the second part of the fundamental theorem of calculus tells us that this piece right over here, which is, let me write the limit part. So this part, I'll just rewrite the limit as n approaches infinity of, of, and we're gonna use the second fundamental theorem of calculus, we're going to evaluate the antiderivative of x of one over x squared or x to the negative two. So the antiderivative of x to the negative two is negative x to the negative one. So negative x to the negative one or negative one over x. So negative one over x is the antiderivative and we're gonna evaluate it at n and evaluate it at one."}, {"video_title": "Introduction to improper integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So this part, I'll just rewrite the limit as n approaches infinity of, of, and we're gonna use the second fundamental theorem of calculus, we're going to evaluate the antiderivative of x of one over x squared or x to the negative two. So the antiderivative of x to the negative two is negative x to the negative one. So negative x to the negative one or negative one over x. So negative one over x is the antiderivative and we're gonna evaluate it at n and evaluate it at one. So this is going to be equal to the limit, the limit as n approaches infinity of, let's see, if we evaluate this thing at n, we get negative one over n. So we get negative one over n. And from that, we're going to subtract this thing evaluated at one. So it's negative one over one or it's negative one. So this right over here is negative one."}, {"video_title": "Introduction to improper integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So negative one over x is the antiderivative and we're gonna evaluate it at n and evaluate it at one. So this is going to be equal to the limit, the limit as n approaches infinity of, let's see, if we evaluate this thing at n, we get negative one over n. So we get negative one over n. And from that, we're going to subtract this thing evaluated at one. So it's negative one over one or it's negative one. So this right over here is negative one. And so we're gonna find the limit as n approaches infinity of this business. This stuff right here is just this stuff right here. I haven't found the limit yet."}, {"video_title": "Introduction to improper integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So this right over here is negative one. And so we're gonna find the limit as n approaches infinity of this business. This stuff right here is just this stuff right here. I haven't found the limit yet. So this is going to be equal to the limit as n approaches infinity of, let's see, this is positive one, positive one, and we could even write that minus one over n, of one minus one over n. And lucky for us, this limit actually exists. Limit as n approaches infinity, this term right over here is going to get closer and closer and closer to zero. One over infinity, you can essentially view as zero."}, {"video_title": "Introduction to improper integrals AP Calculus BC Khan Academy.mp3", "Sentence": "I haven't found the limit yet. So this is going to be equal to the limit as n approaches infinity of, let's see, this is positive one, positive one, and we could even write that minus one over n, of one minus one over n. And lucky for us, this limit actually exists. Limit as n approaches infinity, this term right over here is going to get closer and closer and closer to zero. One over infinity, you can essentially view as zero. So this right over here is going to be equal to one, which is pretty neat. We have this area that has no right boundary. It just keeps on going forever, but we still have a finite area and the area is actually exactly equal, exactly equal to one."}, {"video_title": "Introduction to improper integrals AP Calculus BC Khan Academy.mp3", "Sentence": "One over infinity, you can essentially view as zero. So this right over here is going to be equal to one, which is pretty neat. We have this area that has no right boundary. It just keeps on going forever, but we still have a finite area and the area is actually exactly equal, exactly equal to one. So in this case, we had an improper integral and because we were actually able to evaluate it and come up with a number that this limit actually existed, we say that this improper integral right over here is convergent. Convergent. If for whatever reason this was unbounded, we couldn't come up with some type of a finite number here, if the area was infinite, we would say that it is divergent."}, {"video_title": "Maclaurin series of e_ Series AP Calculus BC Khan Academy.mp3", "Sentence": "Now let's do something pretty interesting. This will to some degree be one of the easiest functions to find the Maclaurin series representation of. But let's try to approximate e to the x. f of x is equal to e to the x. And what makes this really simple is when you take the derivative, and this is frankly one of the amazing things about the number e, is that when you take the derivative of e to the x, you get e to the x. So this is equal to f prime of x, this is equal to f, the second derivative of x, this is equal to the third derivative of x, this is equal to the nth derivative of x. It's always equal to e to the x. That's kind of the first mind-blowing thing about the number e. That it's just, you can keep taking its derivative."}, {"video_title": "Maclaurin series of e_ Series AP Calculus BC Khan Academy.mp3", "Sentence": "And what makes this really simple is when you take the derivative, and this is frankly one of the amazing things about the number e, is that when you take the derivative of e to the x, you get e to the x. So this is equal to f prime of x, this is equal to f, the second derivative of x, this is equal to the third derivative of x, this is equal to the nth derivative of x. It's always equal to e to the x. That's kind of the first mind-blowing thing about the number e. That it's just, you can keep taking its derivative. The slope at any point on that curve is the same as the value of that point on that curve. That's kind of crazy. Anyway, with that said, let's take its Maclaurin representation."}, {"video_title": "Maclaurin series of e_ Series AP Calculus BC Khan Academy.mp3", "Sentence": "That's kind of the first mind-blowing thing about the number e. That it's just, you can keep taking its derivative. The slope at any point on that curve is the same as the value of that point on that curve. That's kind of crazy. Anyway, with that said, let's take its Maclaurin representation. So we have to find f of zero, f prime of zero, the second derivative of zero. Well, when you take e to the zero, e to the zero is just equal to one. And so this is going to be equal to f of zero, this is going to be equal to f prime of zero, it's going to be equal to any of the derivatives, any of the derivatives evaluated, any of the derivatives evaluated at zero, the nth derivative evaluated at zero."}, {"video_title": "Maclaurin series of e_ Series AP Calculus BC Khan Academy.mp3", "Sentence": "Anyway, with that said, let's take its Maclaurin representation. So we have to find f of zero, f prime of zero, the second derivative of zero. Well, when you take e to the zero, e to the zero is just equal to one. And so this is going to be equal to f of zero, this is going to be equal to f prime of zero, it's going to be equal to any of the derivatives, any of the derivatives evaluated, any of the derivatives evaluated at zero, the nth derivative evaluated at zero. And that's why it takes, that's why it makes applying the Maclaurin series formula fairly straightforward. If I wanted to approximate e to the x using a Maclaurin series, so e to the x, and I'll put a little approximately over here, and we'll get closer and closer to the real e to the x as we keep adding more and more terms, and especially if we add an infinite number of terms, it would look like this. F of zero, f of zero, let me do it in, what colors did I use for cosine and sine?"}, {"video_title": "Maclaurin series of e_ Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so this is going to be equal to f of zero, this is going to be equal to f prime of zero, it's going to be equal to any of the derivatives, any of the derivatives evaluated, any of the derivatives evaluated at zero, the nth derivative evaluated at zero. And that's why it takes, that's why it makes applying the Maclaurin series formula fairly straightforward. If I wanted to approximate e to the x using a Maclaurin series, so e to the x, and I'll put a little approximately over here, and we'll get closer and closer to the real e to the x as we keep adding more and more terms, and especially if we add an infinite number of terms, it would look like this. F of zero, f of zero, let me do it in, what colors did I use for cosine and sine? So I used pink and I used green. So let me use a non-pink, non-green. I'll use the yellow here."}, {"video_title": "Maclaurin series of e_ Series AP Calculus BC Khan Academy.mp3", "Sentence": "F of zero, f of zero, let me do it in, what colors did I use for cosine and sine? So I used pink and I used green. So let me use a non-pink, non-green. I'll use the yellow here. So f of zero is one, plus f prime of zero times x, f prime of zero is also one. So plus x, plus this is also one, so it's going to be x squared over two factorial, so plus x squared over two factorial. All of these things are going to be one."}, {"video_title": "Maclaurin series of e_ Series AP Calculus BC Khan Academy.mp3", "Sentence": "I'll use the yellow here. So f of zero is one, plus f prime of zero times x, f prime of zero is also one. So plus x, plus this is also one, so it's going to be x squared over two factorial, so plus x squared over two factorial. All of these things are going to be one. This is one, this is one, when we're talking about e to the x. So you go to the third term, this is one, you just have x to the third over three factorial, plus x to the third over three factorial, and I think you see the pattern here. We just keep adding terms, x to the fourth over four factorial, plus x to the fifth over five factorial, plus x to the sixth over six factorial."}, {"video_title": "Maclaurin series of e_ Series AP Calculus BC Khan Academy.mp3", "Sentence": "All of these things are going to be one. This is one, this is one, when we're talking about e to the x. So you go to the third term, this is one, you just have x to the third over three factorial, plus x to the third over three factorial, and I think you see the pattern here. We just keep adding terms, x to the fourth over four factorial, plus x to the fifth over five factorial, plus x to the sixth over six factorial. And something pretty neat is starting to emerge, is that e to the x, one, this is just really cool, that e to the x can be approximated by one, plus x, plus x squared over two factorial, plus x to the third over three factorial. Once again, e to the x is starting to look like a pretty cool thing. This also leads to other interesting results, that if you wanted to approximate e, you just evaluate this at x is equal to one."}, {"video_title": "Maclaurin series of e_ Series AP Calculus BC Khan Academy.mp3", "Sentence": "We just keep adding terms, x to the fourth over four factorial, plus x to the fifth over five factorial, plus x to the sixth over six factorial. And something pretty neat is starting to emerge, is that e to the x, one, this is just really cool, that e to the x can be approximated by one, plus x, plus x squared over two factorial, plus x to the third over three factorial. Once again, e to the x is starting to look like a pretty cool thing. This also leads to other interesting results, that if you wanted to approximate e, you just evaluate this at x is equal to one. This isn't, so this is, so if you wanted to approximate e, you'd say e is approximate to, well, e is e to the first power, and that's going to be approximately equal to this polynomial evaluated at one. If x is one here, we make x one over here. So it'll be one plus one, so it'll be one plus one, plus one over two factorial, plus one over three factorial, plus one over four factorial, so on and so forth, all the way into infinity."}, {"video_title": "Maclaurin series of e_ Series AP Calculus BC Khan Academy.mp3", "Sentence": "This also leads to other interesting results, that if you wanted to approximate e, you just evaluate this at x is equal to one. This isn't, so this is, so if you wanted to approximate e, you'd say e is approximate to, well, e is e to the first power, and that's going to be approximately equal to this polynomial evaluated at one. If x is one here, we make x one over here. So it'll be one plus one, so it'll be one plus one, plus one over two factorial, plus one over three factorial, plus one over four factorial, so on and so forth, all the way into infinity. And you could view this as, you could also view this as one over one factorial as well. One over one factorial. But what's really cool is, it's just another really neat way to represent e. It shows that e, once again, shows up in this kind of neat thing."}, {"video_title": "Maclaurin series of e_ Series AP Calculus BC Khan Academy.mp3", "Sentence": "So it'll be one plus one, so it'll be one plus one, plus one over two factorial, plus one over three factorial, plus one over four factorial, so on and so forth, all the way into infinity. And you could view this as, you could also view this as one over one factorial as well. One over one factorial. But what's really cool is, it's just another really neat way to represent e. It shows that e, once again, shows up in this kind of neat thing. It's kind of two plus one half, plus one sixth, plus, if you just keep doing this, you get close to the number e. But that by itself isn't the only fascinating thing. If we look back at our Maclaurin representations of these other functions, cosine of x, let me copy and paste, cosine of x. So cosine of x right up here."}, {"video_title": "Maclaurin series of e_ Series AP Calculus BC Khan Academy.mp3", "Sentence": "But what's really cool is, it's just another really neat way to represent e. It shows that e, once again, shows up in this kind of neat thing. It's kind of two plus one half, plus one sixth, plus, if you just keep doing this, you get close to the number e. But that by itself isn't the only fascinating thing. If we look back at our Maclaurin representations of these other functions, cosine of x, let me copy and paste, cosine of x. So cosine of x right up here. So let me do my best to, I'll copy and paste the whole thing. So copy and paste. Copy and paste."}, {"video_title": "Maclaurin series of e_ Series AP Calculus BC Khan Academy.mp3", "Sentence": "So cosine of x right up here. So let me do my best to, I'll copy and paste the whole thing. So copy and paste. Copy and paste. So that is cosine of x. And let's do the same thing for the sine of x that we did last video. So the sine of x."}, {"video_title": "Maclaurin series of e_ Series AP Calculus BC Khan Academy.mp3", "Sentence": "Copy and paste. So that is cosine of x. And let's do the same thing for the sine of x that we did last video. So the sine of x. Let me copy and paste that. Copy and then let me paste that. Edit, paste."}, {"video_title": "Maclaurin series of e_ Series AP Calculus BC Khan Academy.mp3", "Sentence": "So the sine of x. Let me copy and paste that. Copy and then let me paste that. Edit, paste. So do we see any relationship between these approximations? So before, you probably would have guessed maybe there's some relationship between cosine and sine, but what about e to the x? And what you see here is that cosine of x looks a lot like this term plus this term, although we would want to put a negative out front here, so it's a negative version of this term right here, plus this term right here, plus a negative version of this term right over here."}, {"video_title": "Maclaurin series of e_ Series AP Calculus BC Khan Academy.mp3", "Sentence": "Edit, paste. So do we see any relationship between these approximations? So before, you probably would have guessed maybe there's some relationship between cosine and sine, but what about e to the x? And what you see here is that cosine of x looks a lot like this term plus this term, although we would want to put a negative out front here, so it's a negative version of this term right here, plus this term right here, plus a negative version of this term right over here. And sine of x, sine of x looks just like this term, plus a negative version of this term, plus this term, plus a negative version of the next term. So if we can somehow reconcile the negatives in some interesting way, it looks like e to the x is somehow, or at least its polynomial representation of e to the x, is somehow related to a combination of the polynomial representations of cosine of x and sine of x. So this is starting to get really, really, really cool."}, {"video_title": "Worked example sequence convergence divergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "So we've explicitly defined four different sequences here. And what I want you to think about is whether these sequences converge or diverge. And remember, converge just means, as n gets larger and larger and larger, that the value of our sequence is approaching some value. And diverge means that it's not approaching some value. So let's look at this, and I encourage you to pause this video and try this on your own before I'm about to explain it. So let's look at this first sequence right over here. So the numerator n plus 8 times n plus 1, denominator n times n minus 10."}, {"video_title": "Worked example sequence convergence divergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "And diverge means that it's not approaching some value. So let's look at this, and I encourage you to pause this video and try this on your own before I'm about to explain it. So let's look at this first sequence right over here. So the numerator n plus 8 times n plus 1, denominator n times n minus 10. So one way to think about what's happening as n gets larger and larger is look at the degree of the numerator and the degree of the denominator. We care about the degree because we want to see, look, is the numerator growing faster than the denominator, in which case this thing is going to go to infinity and this thing's going to diverge? Or is maybe the denominator growing faster, in which case this might approach converge to 0?"}, {"video_title": "Worked example sequence convergence divergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "So the numerator n plus 8 times n plus 1, denominator n times n minus 10. So one way to think about what's happening as n gets larger and larger is look at the degree of the numerator and the degree of the denominator. We care about the degree because we want to see, look, is the numerator growing faster than the denominator, in which case this thing is going to go to infinity and this thing's going to diverge? Or is maybe the denominator growing faster, in which case this might approach converge to 0? Or maybe they're growing at the same level, and maybe it'll converge to a different number. So let's multiply out the numerator and the denominator, figure that out. So n times n is n squared."}, {"video_title": "Worked example sequence convergence divergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "Or is maybe the denominator growing faster, in which case this might approach converge to 0? Or maybe they're growing at the same level, and maybe it'll converge to a different number. So let's multiply out the numerator and the denominator, figure that out. So n times n is n squared. n times 1 is 1n, plus 8n is 9n. So 9n. And then 8 times 1 is 8."}, {"video_title": "Worked example sequence convergence divergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "So n times n is n squared. n times 1 is 1n, plus 8n is 9n. So 9n. And then 8 times 1 is 8. So the numerator is n squared plus 9n plus 8. Denominator is n squared minus 10n. And one way to think about it is n gets really, really, really, really, really large."}, {"video_title": "Worked example sequence convergence divergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "And then 8 times 1 is 8. So the numerator is n squared plus 9n plus 8. Denominator is n squared minus 10n. And one way to think about it is n gets really, really, really, really, really large. What dominates in the numerator, this term is going to represent most of the value, and this term is going to represent most of the value as well. These other terms aren't going to grow. Obviously, this 8 doesn't grow at all."}, {"video_title": "Worked example sequence convergence divergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "And one way to think about it is n gets really, really, really, really, really large. What dominates in the numerator, this term is going to represent most of the value, and this term is going to represent most of the value as well. These other terms aren't going to grow. Obviously, this 8 doesn't grow at all. But the n terms aren't going to grow anywhere near as fast as the n squared terms, especially for large n's. So for very, very large n's, this is really going to be approaching n squared over n squared, or 1. So it's reasonable to say that this converges."}, {"video_title": "Worked example sequence convergence divergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "Obviously, this 8 doesn't grow at all. But the n terms aren't going to grow anywhere near as fast as the n squared terms, especially for large n's. So for very, very large n's, this is really going to be approaching n squared over n squared, or 1. So it's reasonable to say that this converges. So this one converges. And once again, I'm not vigorously proving it here. Or I should say, I'm not rigorously proving it over here."}, {"video_title": "Worked example sequence convergence divergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "So it's reasonable to say that this converges. So this one converges. And once again, I'm not vigorously proving it here. Or I should say, I'm not rigorously proving it over here. But the giveaway is that we have the same degree in the numerator and the denominator. So now let's look at this one right over here. So here in the numerator, I have e to the n power."}, {"video_title": "Worked example sequence convergence divergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "Or I should say, I'm not rigorously proving it over here. But the giveaway is that we have the same degree in the numerator and the denominator. So now let's look at this one right over here. So here in the numerator, I have e to the n power. And here I have e times n. So this grows much faster. I mean, this is e to the n power. Imagine if when you have this as 100, e to the 100th power is a ginormous number."}, {"video_title": "Worked example sequence convergence divergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "So here in the numerator, I have e to the n power. And here I have e times n. So this grows much faster. I mean, this is e to the n power. Imagine if when you have this as 100, e to the 100th power is a ginormous number. e times 100 is, well, that's just 100e. Grows much faster than this right over here. So this thing is just going to balloon."}, {"video_title": "Worked example sequence convergence divergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "Imagine if when you have this as 100, e to the 100th power is a ginormous number. e times 100 is, well, that's just 100e. Grows much faster than this right over here. So this thing is just going to balloon. This is going to go to infinity. So we could say this diverges. Now let's look at this one right over here."}, {"video_title": "Worked example sequence convergence divergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "So this thing is just going to balloon. This is going to go to infinity. So we could say this diverges. Now let's look at this one right over here. Well, we have a higher degree in the numerator than we have in the denominator. n squared, obviously, is going to grow much faster than n. So for the same reason as the b sub n sequence, this thing is going to diverge. The numerator is going to grow much faster than the denominator."}, {"video_title": "Worked example sequence convergence divergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "Now let's look at this one right over here. Well, we have a higher degree in the numerator than we have in the denominator. n squared, obviously, is going to grow much faster than n. So for the same reason as the b sub n sequence, this thing is going to diverge. The numerator is going to grow much faster than the denominator. Or another way to think about it, the limit as n approaches infinity is going to be infinity. This thing's going to go to infinity. Now let's think about this right over here."}, {"video_title": "Worked example sequence convergence divergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "The numerator is going to grow much faster than the denominator. Or another way to think about it, the limit as n approaches infinity is going to be infinity. This thing's going to go to infinity. Now let's think about this right over here. So as we increase n, so we could even think about what the sequence looks like. When n is 0, negative 1 to the 0 is 1. When n is 1, it's going to be negative 1."}, {"video_title": "Worked example sequence convergence divergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "Now let's think about this right over here. So as we increase n, so we could even think about what the sequence looks like. When n is 0, negative 1 to the 0 is 1. When n is 1, it's going to be negative 1. When n is 2, it's going to be 1. And so this thing is just going to keep oscillating between negative 1 and 1. So it's not unbounded."}, {"video_title": "Worked example sequence convergence divergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "When n is 1, it's going to be negative 1. When n is 2, it's going to be 1. And so this thing is just going to keep oscillating between negative 1 and 1. So it's not unbounded. It's not going to go to infinity or negative infinity or something like that. But it just oscillates between these two values. So it doesn't converge to one particular value."}, {"video_title": "Worked example sequence convergence divergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "So it's not unbounded. It's not going to go to infinity or negative infinity or something like that. But it just oscillates between these two values. So it doesn't converge to one particular value. So even though this one isn't unbounded, it doesn't go to infinity, this one still diverges. It doesn't go to one value. So let me write that down."}, {"video_title": "Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "But in all of these theorems, it's always fun to think about the edge cases. Why is it laid out the way it is? And that might give us a little bit of more intuition about it. So the extreme value theorem says if we have some function that is continuous over a closed interval, let's say the closed interval's from a to b. And when we say a closed interval, that means we include the endpoints a and b. That's why we have these brackets here instead of parentheses. Then there will be an absolute maximum value for f and an absolute minimum value for f. So that means there exists, this is a symbol for the logical symbol for there exists, there exists an absolute maximum value of f over interval and absolute minimum value of f over the interval."}, {"video_title": "Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "So the extreme value theorem says if we have some function that is continuous over a closed interval, let's say the closed interval's from a to b. And when we say a closed interval, that means we include the endpoints a and b. That's why we have these brackets here instead of parentheses. Then there will be an absolute maximum value for f and an absolute minimum value for f. So that means there exists, this is a symbol for the logical symbol for there exists, there exists an absolute maximum value of f over interval and absolute minimum value of f over the interval. So let's think about that a little bit and this probably is pretty intuitive for you. You're probably saying, well, why do they even have to write a theorem here? And why do we even have to have this continuity there?"}, {"video_title": "Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "Then there will be an absolute maximum value for f and an absolute minimum value for f. So that means there exists, this is a symbol for the logical symbol for there exists, there exists an absolute maximum value of f over interval and absolute minimum value of f over the interval. So let's think about that a little bit and this probably is pretty intuitive for you. You're probably saying, well, why do they even have to write a theorem here? And why do we even have to have this continuity there? And we'll see in a second why the continuity actually matters. So this is my x-axis. That's my y-axis."}, {"video_title": "Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "And why do we even have to have this continuity there? And we'll see in a second why the continuity actually matters. So this is my x-axis. That's my y-axis. And let's draw the interval. So the interval's from a to b. So let's say that this is a and this is b right over here."}, {"video_title": "Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "That's my y-axis. And let's draw the interval. So the interval's from a to b. So let's say that this is a and this is b right over here. Let's say that this right over here is f of a. So that is f of a. And let's say this right over here is f of b."}, {"video_title": "Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say that this is a and this is b right over here. Let's say that this right over here is f of a. So that is f of a. And let's say this right over here is f of b. So this value right over here is f of b. And let's say the function does something like this. Let's say the function does something like this over the interval."}, {"video_title": "Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "And let's say this right over here is f of b. So this value right over here is f of b. And let's say the function does something like this. Let's say the function does something like this over the interval. And I'm just drawing something somewhat arbitrary right over here. So I've drawn a continuous function. I really didn't have to pick up my pen as I drew this right over here."}, {"video_title": "Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say the function does something like this over the interval. And I'm just drawing something somewhat arbitrary right over here. So I've drawn a continuous function. I really didn't have to pick up my pen as I drew this right over here. And so you can see at least the way this continuous function that I've drawn, it's clear that there's an absolute maximum and absolute minimum point over this interval. The absolute minimum point, well, it seems like we hit it right over here when x is, let's say this is x is c. And this is f of c right over there. And it looks like we hit our absolute maximum point over the interval right over there when x is, let's say that this is x is equal to d. And this right over here is f of d. This right over here is f of d. So another way to say the statement right over here, if f is continuous over the interval, we could say there exists a c and d that are in the interval."}, {"video_title": "Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "I really didn't have to pick up my pen as I drew this right over here. And so you can see at least the way this continuous function that I've drawn, it's clear that there's an absolute maximum and absolute minimum point over this interval. The absolute minimum point, well, it seems like we hit it right over here when x is, let's say this is x is c. And this is f of c right over there. And it looks like we hit our absolute maximum point over the interval right over there when x is, let's say that this is x is equal to d. And this right over here is f of d. This right over here is f of d. So another way to say the statement right over here, if f is continuous over the interval, we could say there exists a c and d that are in the interval. So there are members of this set that are in the interval such that, and I'm just using the logical notation here, such that f of c is less than or equal to f of x, which is less than or equal to f of d for all x in the interval. Just like that. So in this case, they're saying, look, our minimum value when x is equal to c, that's that right over here, our maximum value when f is equal to d. And for all the other x's in the interval, we are between those two values."}, {"video_title": "Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "And it looks like we hit our absolute maximum point over the interval right over there when x is, let's say that this is x is equal to d. And this right over here is f of d. This right over here is f of d. So another way to say the statement right over here, if f is continuous over the interval, we could say there exists a c and d that are in the interval. So there are members of this set that are in the interval such that, and I'm just using the logical notation here, such that f of c is less than or equal to f of x, which is less than or equal to f of d for all x in the interval. Just like that. So in this case, they're saying, look, our minimum value when x is equal to c, that's that right over here, our maximum value when f is equal to d. And for all the other x's in the interval, we are between those two values. Now, one thing, and we could draw other continuous functions. And once again, I'm not doing a proof of the extreme value theorem, but just to make you familiar with it and why it's stated the way it is. And just you could draw a bunch of functions here that are continuous over this closed interval."}, {"video_title": "Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "So in this case, they're saying, look, our minimum value when x is equal to c, that's that right over here, our maximum value when f is equal to d. And for all the other x's in the interval, we are between those two values. Now, one thing, and we could draw other continuous functions. And once again, I'm not doing a proof of the extreme value theorem, but just to make you familiar with it and why it's stated the way it is. And just you could draw a bunch of functions here that are continuous over this closed interval. Here, our maximum point happens right when we hit b, and our minimum point happens at a. For a flat function, we could put any point as a maximum or the minimum point, and we'll see that this would actually be true. But let's dig a little bit deeper as to why f needs to be continuous and why this needs to be a closed interval."}, {"video_title": "Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "And just you could draw a bunch of functions here that are continuous over this closed interval. Here, our maximum point happens right when we hit b, and our minimum point happens at a. For a flat function, we could put any point as a maximum or the minimum point, and we'll see that this would actually be true. But let's dig a little bit deeper as to why f needs to be continuous and why this needs to be a closed interval. So first, let's think about why does f need to be continuous. Well, I could easily construct a function that is not continuous over a closed interval where it is hard to articulate a minimum or a maximum point. And I encourage you, actually, pause this video and try to construct that function on your own."}, {"video_title": "Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "But let's dig a little bit deeper as to why f needs to be continuous and why this needs to be a closed interval. So first, let's think about why does f need to be continuous. Well, I could easily construct a function that is not continuous over a closed interval where it is hard to articulate a minimum or a maximum point. And I encourage you, actually, pause this video and try to construct that function on your own. Try to construct a non-continuous function over a closed interval where it would be very difficult, or you can't really pick out an absolute minimum or an absolute maximum value over that interval. Well, let's see. Let me draw a graph here."}, {"video_title": "Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "And I encourage you, actually, pause this video and try to construct that function on your own. Try to construct a non-continuous function over a closed interval where it would be very difficult, or you can't really pick out an absolute minimum or an absolute maximum value over that interval. Well, let's see. Let me draw a graph here. So let's say that this right over here is my interval. Let's say that's a. That's b."}, {"video_title": "Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "Let me draw a graph here. So let's say that this right over here is my interval. Let's say that's a. That's b. Let's say our function did something like this. Let's say our function did something right where you would have expected to have a maximum value. Let's say the function is not defined."}, {"video_title": "Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "That's b. Let's say our function did something like this. Let's say our function did something right where you would have expected to have a maximum value. Let's say the function is not defined. And right where you would have expected to have a minimum value, the function is not defined. And so right over here, you could say, well, look. The function is clearly approaching."}, {"video_title": "Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say the function is not defined. And right where you would have expected to have a minimum value, the function is not defined. And so right over here, you could say, well, look. The function is clearly approaching. As x approaches this value right over here, the function is clearly approaching this limit. But that limit can't be the maximum value because the function never gets to that. So you could say, well, let's get a little closer here."}, {"video_title": "Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "The function is clearly approaching. As x approaches this value right over here, the function is clearly approaching this limit. But that limit can't be the maximum value because the function never gets to that. So you could say, well, let's get a little closer here. Maybe this number right over here is 5. So you could say maybe the maximum value is 4.9. But then you could get your x even closer to this value and say your y be 4.99 or 4.999."}, {"video_title": "Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "So you could say, well, let's get a little closer here. Maybe this number right over here is 5. So you could say maybe the maximum value is 4.9. But then you could get your x even closer to this value and say your y be 4.99 or 4.999. You could keep adding another 9. So there is no maximum value. Similar here on the minimum."}, {"video_title": "Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "But then you could get your x even closer to this value and say your y be 4.99 or 4.999. You could keep adding another 9. So there is no maximum value. Similar here on the minimum. Let me draw it a little bit so it looks more like a minimum. You can get closer and closer to it, but there's no minimum. Let's say that this value right over here is 1."}, {"video_title": "Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "Similar here on the minimum. Let me draw it a little bit so it looks more like a minimum. You can get closer and closer to it, but there's no minimum. Let's say that this value right over here is 1. So you could get to 1.1 or 1.01 or 1.0001. And so you could keep drawing some 0's between the two 1's, but there is no absolute minimum value there. Now let's think about why it being a closed interval matters, why you have to include your endpoints as kind of candidates for your maximum and minimum values over the interval."}, {"video_title": "Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say that this value right over here is 1. So you could get to 1.1 or 1.01 or 1.0001. And so you could keep drawing some 0's between the two 1's, but there is no absolute minimum value there. Now let's think about why it being a closed interval matters, why you have to include your endpoints as kind of candidates for your maximum and minimum values over the interval. Well, let's imagine that it was an open interval. Let's imagine an open interval. And sometimes if we want to be particular, we could make this is the closed interval right over here, brackets."}, {"video_title": "Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's think about why it being a closed interval matters, why you have to include your endpoints as kind of candidates for your maximum and minimum values over the interval. Well, let's imagine that it was an open interval. Let's imagine an open interval. And sometimes if we want to be particular, we could make this is the closed interval right over here, brackets. And if we want to do an open interval right over here, that's a, that's b. And let's just pick a very simple function. Let's say a function like this."}, {"video_title": "Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "And sometimes if we want to be particular, we could make this is the closed interval right over here, brackets. And if we want to do an open interval right over here, that's a, that's b. And let's just pick a very simple function. Let's say a function like this. So right over here, if a were in our interval, it looks like we hit our minimum value at a. f of a would have been our minimum value. And f of b looks like it would have been our maximum value. So f of a would have been our maximum value."}, {"video_title": "Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say a function like this. So right over here, if a were in our interval, it looks like we hit our minimum value at a. f of a would have been our minimum value. And f of b looks like it would have been our maximum value. So f of a would have been our maximum value. But we're not including a and b in the interval. This is an open interval. So you can keep getting closer and closer and closer to b, and keep getting higher and higher and higher values without ever quite getting to b."}, {"video_title": "Extreme value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "So f of a would have been our maximum value. But we're not including a and b in the interval. This is an open interval. So you can keep getting closer and closer and closer to b, and keep getting higher and higher and higher values without ever quite getting to b. Because once again, we're not including the point b. Similarly, you could get closer and closer and closer to a and get smaller and smaller values, but a is not included in your set, in your consideration. So f of a cannot be your minimum value."}, {"video_title": "Worked example p-series Series AP Calculus BC Khan Academy.mp3", "Sentence": "So we have an infinite series here, one plus one over two to the fifth plus one over three to the fifth, and we just keep on going forever. We could write this as the sum from n equals one to infinity of one over n to the fifth power. One over n to the fifth power. And now you might recognize, notice when n is equal to one, this is one over one to the fifth, that's that over there, and we could keep on going. Now you might immediately recognize this as a p-series. And a p-series has the general form of the sum going from n equals one to infinity of one over n to the p, where p is a positive value. So in this particular case, our p for this p-series is equal to five."}, {"video_title": "Worked example p-series Series AP Calculus BC Khan Academy.mp3", "Sentence": "And now you might recognize, notice when n is equal to one, this is one over one to the fifth, that's that over there, and we could keep on going. Now you might immediately recognize this as a p-series. And a p-series has the general form of the sum going from n equals one to infinity of one over n to the p, where p is a positive value. So in this particular case, our p for this p-series is equal to five. P is equal to five. Now you might already recognize under which conditions for a p-series does it converge or diverge? It's going to converge, it's going to converge when your p is greater than one, which is clearly the case in this scenario right over here."}, {"video_title": "Worked example p-series Series AP Calculus BC Khan Academy.mp3", "Sentence": "So in this particular case, our p for this p-series is equal to five. P is equal to five. Now you might already recognize under which conditions for a p-series does it converge or diverge? It's going to converge, it's going to converge when your p is greater than one, which is clearly the case in this scenario right over here. Our p is clearly greater than one. We would diverge, we would diverge if our p is greater than zero and less than or equal, or less than or equal to one. This would be a divergence."}, {"video_title": "Worked example p-series Series AP Calculus BC Khan Academy.mp3", "Sentence": "It's going to converge, it's going to converge when your p is greater than one, which is clearly the case in this scenario right over here. Our p is clearly greater than one. We would diverge, we would diverge if our p is greater than zero and less than or equal, or less than or equal to one. This would be a divergence. So if this was like.9 here, or if this was a 3 4ths, then we would be diverging. So at least for this one, we are convergent. Let's do another one of these."}, {"video_title": "Worked example p-series Series AP Calculus BC Khan Academy.mp3", "Sentence": "This would be a divergence. So if this was like.9 here, or if this was a 3 4ths, then we would be diverging. So at least for this one, we are convergent. Let's do another one of these. All right. So here, you might again recognize this as a p-series. Let me rewrite this infinite sum."}, {"video_title": "Worked example p-series Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let's do another one of these. All right. So here, you might again recognize this as a p-series. Let me rewrite this infinite sum. So this is the sum from n equals one to infinity of one over, let's see, we have square root of two, square root of three. So you could use this two to the 1 1\u20442, three to the 1 1\u20442, four to the 1 1\u20442. So it's one over n to the 1 1\u20442."}, {"video_title": "Worked example p-series Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let me rewrite this infinite sum. So this is the sum from n equals one to infinity of one over, let's see, we have square root of two, square root of three. So you could use this two to the 1 1\u20442, three to the 1 1\u20442, four to the 1 1\u20442. So it's one over n to the 1 1\u20442. Notice, this is when n is equal to one. One over one to the 1 1\u20442 is one. One over two to the 1 1\u20442, well that's this right over here."}, {"video_title": "Worked example p-series Series AP Calculus BC Khan Academy.mp3", "Sentence": "So it's one over n to the 1 1\u20442. Notice, this is when n is equal to one. One over one to the 1 1\u20442 is one. One over two to the 1 1\u20442, well that's this right over here. And we keep on going on and on and on. Well in this case, we still have a p-series. We have one over n to some power, and that power is positive."}, {"video_title": "Worked example p-series Series AP Calculus BC Khan Academy.mp3", "Sentence": "One over two to the 1 1\u20442, well that's this right over here. And we keep on going on and on and on. Well in this case, we still have a p-series. We have one over n to some power, and that power is positive. But notice, in this case, our p falls between zero and one. So 1 1\u20442 is our p. So p for our p-series is equal to 1 1\u20442, and that's between zero and one. Remember, we're divergent, when our p is greater than zero and less than or equal to one, which was clearly the case right over here."}, {"video_title": "Recognizing concavity exercise Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Highlight an interval where f prime of x, where we could say the first derivative of x, or the first derivative of f with respect to x, is greater than zero, and f double prime of x, or the second derivative of f with respect to x, is less than zero. So let's think about what they're saying. So we're looking for a place where the first derivative is greater than zero. That means that the slope of the tangent line is positive. That means that the function is increasing over that interval. So if we just think about it here, over this whole region right over here, the function is clearly decreasing. Then the slope becomes zero right over here."}, {"video_title": "Recognizing concavity exercise Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "That means that the slope of the tangent line is positive. That means that the function is increasing over that interval. So if we just think about it here, over this whole region right over here, the function is clearly decreasing. Then the slope becomes zero right over here. And then the function starts increasing again, all the way until this point right over here. It hits zero, and then it goes, and the function starts decreasing. So we're going to, just this first constraint right over here, tells us it's going to be something in this interval right over there."}, {"video_title": "Recognizing concavity exercise Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Then the slope becomes zero right over here. And then the function starts increasing again, all the way until this point right over here. It hits zero, and then it goes, and the function starts decreasing. So we're going to, just this first constraint right over here, tells us it's going to be something in this interval right over there. And then they say where the second derivative is less than zero. So this means that the slope itself, whether it's positive or negative, that it's actually decreasing. We are going to be concave downwards right over here."}, {"video_title": "Recognizing concavity exercise Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So we're going to, just this first constraint right over here, tells us it's going to be something in this interval right over there. And then they say where the second derivative is less than zero. So this means that the slope itself, whether it's positive or negative, that it's actually decreasing. We are going to be concave downwards right over here. The slope itself, it could be positive, but it'll be coming less and less and less positive. And so we're looking for a place where the slope is positive, but it's becoming less and less and less positive. If you look right over here, the slope is positive, but the slope is increasing."}, {"video_title": "Recognizing concavity exercise Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "We are going to be concave downwards right over here. The slope itself, it could be positive, but it'll be coming less and less and less positive. And so we're looking for a place where the slope is positive, but it's becoming less and less and less positive. If you look right over here, the slope is positive, but the slope is increasing. It's getting steeper and steeper and steeper as we go. And then all of a sudden, it starts getting less steep, less steep, less steep, less steep, all the way to when the slope gets back to zero. So if we want to select an interval, it would be this interval right over here."}, {"video_title": "Recognizing concavity exercise Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "If you look right over here, the slope is positive, but the slope is increasing. It's getting steeper and steeper and steeper as we go. And then all of a sudden, it starts getting less steep, less steep, less steep, less steep, all the way to when the slope gets back to zero. So if we want to select an interval, it would be this interval right over here. Our slope is positive. Our function is clearly increasing. But it is increasing at a lower and lower rate."}, {"video_title": "Recognizing concavity exercise Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So if we want to select an interval, it would be this interval right over here. Our slope is positive. Our function is clearly increasing. But it is increasing at a lower and lower rate. So I will select that right over there. Let's do one more example. A function f of x is plotted below."}, {"video_title": "Recognizing concavity exercise Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "But it is increasing at a lower and lower rate. So I will select that right over there. Let's do one more example. A function f of x is plotted below. Highlight an interval where f prime of x is greater than zero. So same thing, where our function is increasing, but it's increasing at a slower and slower rate. So our function is increasing."}, {"video_title": "Recognizing concavity exercise Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "A function f of x is plotted below. Highlight an interval where f prime of x is greater than zero. So same thing, where our function is increasing, but it's increasing at a slower and slower rate. So our function is increasing. So our function is increasing in this whole region right over here. And we see it's really steep here. Then it's getting less steep and less steep."}, {"video_title": "Average acceleration over interval AP Calculus BC Khan Academy.mp3", "Sentence": "Let's say that we have a particle that's traveling in one dimension, and its position as a function of time is given as t to the third power plus two over t squared. And what I would like you to do is pause this video and figure out what the average acceleration is of this particle over the interval, the closed interval from t is equal to one to t is equal to two. What is this going to be equal to? So assuming you've given a go at it, and the first thing you might have realized is we're trying to take the average value of a function that we don't know explicitly yet. We know the position function, but not the acceleration function. But luckily, we also know that the acceleration function is the derivative with respect to time of the velocity function, which is the derivative with respect to time of the position function. So the acceleration function is the second derivative of this, and then we have to just find its average value over this interval."}, {"video_title": "Average acceleration over interval AP Calculus BC Khan Academy.mp3", "Sentence": "So assuming you've given a go at it, and the first thing you might have realized is we're trying to take the average value of a function that we don't know explicitly yet. We know the position function, but not the acceleration function. But luckily, we also know that the acceleration function is the derivative with respect to time of the velocity function, which is the derivative with respect to time of the position function. So the acceleration function is the second derivative of this, and then we have to just find its average value over this interval. So let's do that. Let's take the derivative of this twice. And before we do it, let me just even rewrite this so it's going to be a little bit easier to differentiate it."}, {"video_title": "Average acceleration over interval AP Calculus BC Khan Academy.mp3", "Sentence": "So the acceleration function is the second derivative of this, and then we have to just find its average value over this interval. So let's do that. Let's take the derivative of this twice. And before we do it, let me just even rewrite this so it's going to be a little bit easier to differentiate it. So if we just take each of these two terms of the numerator and divide them by t squared, we're going to get t to the third divided by t squared is just t, and then two divided by t squared, we could write that as plus two t to the negative two power. And now let's take the derivative. So the velocity function, the velocity as a function of time, just the derivative of this with respect to time."}, {"video_title": "Average acceleration over interval AP Calculus BC Khan Academy.mp3", "Sentence": "And before we do it, let me just even rewrite this so it's going to be a little bit easier to differentiate it. So if we just take each of these two terms of the numerator and divide them by t squared, we're going to get t to the third divided by t squared is just t, and then two divided by t squared, we could write that as plus two t to the negative two power. And now let's take the derivative. So the velocity function, the velocity as a function of time, just the derivative of this with respect to time. So it's going to be derivative of t with respect to t is one. Derivative of two t to the negative two, let's see, negative two times positive two is negative four t to the, and we just decremented the exponent here, t to the negative negative three power. Now, to find acceleration as a function of time, we just take the derivative of this with respect to time."}, {"video_title": "Average acceleration over interval AP Calculus BC Khan Academy.mp3", "Sentence": "So the velocity function, the velocity as a function of time, just the derivative of this with respect to time. So it's going to be derivative of t with respect to t is one. Derivative of two t to the negative two, let's see, negative two times positive two is negative four t to the, and we just decremented the exponent here, t to the negative negative three power. Now, to find acceleration as a function of time, we just take the derivative of this with respect to time. So acceleration as a function of time is equal to, actually since I've already used that color for the average, let me do a different color now. So acceleration as a function of time is just the derivative of this with respect to t. So derivative of a constant with respect to time, well, it's not changing, so it's a zero, and then over here, negative three times negative four is positive 12 times t to the, let's decrement that exponent, to the negative four power. Now, to find the average value, all we have to do now, average value, is essentially take the definite integral of this over the interval and divide that by the width of the interval."}, {"video_title": "Average acceleration over interval AP Calculus BC Khan Academy.mp3", "Sentence": "Now, to find acceleration as a function of time, we just take the derivative of this with respect to time. So acceleration as a function of time is equal to, actually since I've already used that color for the average, let me do a different color now. So acceleration as a function of time is just the derivative of this with respect to t. So derivative of a constant with respect to time, well, it's not changing, so it's a zero, and then over here, negative three times negative four is positive 12 times t to the, let's decrement that exponent, to the negative four power. Now, to find the average value, all we have to do now, average value, is essentially take the definite integral of this over the interval and divide that by the width of the interval. So, or we could say, we could take, we could divide by the width of the interval, one over two minus one, and this all simplifies to one, times the definite integral over the interval. So one to two of a of t, which is, so this can be 12t to the negative four power dt. So what does this simplify to?"}, {"video_title": "Average acceleration over interval AP Calculus BC Khan Academy.mp3", "Sentence": "Now, to find the average value, all we have to do now, average value, is essentially take the definite integral of this over the interval and divide that by the width of the interval. So, or we could say, we could take, we could divide by the width of the interval, one over two minus one, and this all simplifies to one, times the definite integral over the interval. So one to two of a of t, which is, so this can be 12t to the negative four power dt. So what does this simplify to? Once again, this is one over one. That's just going to be one. If we take the antiderivative of this, and then we actually, well, let me just, so this is going to be equal to, the antiderivative of this is, so we're gonna go t to the negative three power, but then we divide by negative three."}, {"video_title": "Average acceleration over interval AP Calculus BC Khan Academy.mp3", "Sentence": "So what does this simplify to? Once again, this is one over one. That's just going to be one. If we take the antiderivative of this, and then we actually, well, let me just, so this is going to be equal to, the antiderivative of this is, so we're gonna go t to the negative three power, but then we divide by negative three. So an antiderivative of this is going to be, if we don't take the, well, an antiderivative is going to be negative four t to the negative three power, and we saw that over here. Obviously, if you were really just taking an indefinite integral, we would have to put some constant here, but in the definite integral, even if we put a constant here, it would get, assuming the same constant, it would get canceled out when you actually do the calculation. But the antiderivative of this, we increment the exponent, and then we divide by that new exponent."}, {"video_title": "Average acceleration over interval AP Calculus BC Khan Academy.mp3", "Sentence": "If we take the antiderivative of this, and then we actually, well, let me just, so this is going to be equal to, the antiderivative of this is, so we're gonna go t to the negative three power, but then we divide by negative three. So an antiderivative of this is going to be, if we don't take the, well, an antiderivative is going to be negative four t to the negative three power, and we saw that over here. Obviously, if you were really just taking an indefinite integral, we would have to put some constant here, but in the definite integral, even if we put a constant here, it would get, assuming the same constant, it would get canceled out when you actually do the calculation. But the antiderivative of this, we increment the exponent, and then we divide by that new exponent. So 12 divided by negative three is negative four, and we're going to evaluate that from at two and at one. And so this is going to be equal to, when we evaluate it at two, at the upper bound of our interval, it's gonna be negative four times two to the negative three power. So it's negative four times, what is that, two, that's one over two to the third, times 1 1 8th is one way to think about that."}, {"video_title": "Average acceleration over interval AP Calculus BC Khan Academy.mp3", "Sentence": "But the antiderivative of this, we increment the exponent, and then we divide by that new exponent. So 12 divided by negative three is negative four, and we're going to evaluate that from at two and at one. And so this is going to be equal to, when we evaluate it at two, at the upper bound of our interval, it's gonna be negative four times two to the negative three power. So it's negative four times, what is that, two, that's one over two to the third, times 1 1 8th is one way to think about that. And then we're gonna have minus this evaluated at one. So minus negative four times t to the negative three, or one to the negative three is just one. So it's just gonna be negative four times one."}, {"video_title": "Average acceleration over interval AP Calculus BC Khan Academy.mp3", "Sentence": "So it's negative four times, what is that, two, that's one over two to the third, times 1 1 8th is one way to think about that. And then we're gonna have minus this evaluated at one. So minus negative four times t to the negative three, or one to the negative three is just one. So it's just gonna be negative four times one. And this is going to be equal to, we're really in the home stretch now, this is equal to, this part right over here is negative 1 1 2. So this is negative 1 1 2. And this part right over here is positive four."}, {"video_title": "Average acceleration over interval AP Calculus BC Khan Academy.mp3", "Sentence": "So it's just gonna be negative four times one. And this is going to be equal to, we're really in the home stretch now, this is equal to, this part right over here is negative 1 1 2. So this is negative 1 1 2. And this part right over here is positive four. So positive four minus 1 1 2, we could either write that as three and a half, or if we wanted to write it as an improper fraction, we could write this as seven halves. So the average value of our acceleration over this interval is seven halves. And if the position was given in meters and time was in seconds, then this would be seven halves meters per second squared is the average acceleration between time at one second and time at two seconds."}, {"video_title": "Worked example Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We're going to assume that y is a function of x. So let's apply our derivative operator to both sides of this equation. So let's apply our derivative operator. And so first, on the left-hand side, we essentially are just going to apply the chain rule. First, we have the derivative with respect to x of x minus y squared. So the chain rule tells us this is going to be the derivative of the something squared with respect to the something, which is just going to be 2 times x minus y to the first power. I won't write the 1 right over there."}, {"video_title": "Worked example Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And so first, on the left-hand side, we essentially are just going to apply the chain rule. First, we have the derivative with respect to x of x minus y squared. So the chain rule tells us this is going to be the derivative of the something squared with respect to the something, which is just going to be 2 times x minus y to the first power. I won't write the 1 right over there. Times the derivative of the something with respect to x. Well, the derivative of x with respect to x is just 1. And the derivative of y with respect to x, that's what we're trying to solve."}, {"video_title": "Worked example Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "I won't write the 1 right over there. Times the derivative of the something with respect to x. Well, the derivative of x with respect to x is just 1. And the derivative of y with respect to x, that's what we're trying to solve. So it's going to be 1 minus dy dx. Let me make it a little bit clearer what I just did right over here. This right over here is the derivative of x minus y squared with respect to x minus y."}, {"video_title": "Worked example Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And the derivative of y with respect to x, that's what we're trying to solve. So it's going to be 1 minus dy dx. Let me make it a little bit clearer what I just did right over here. This right over here is the derivative of x minus y squared with respect to x minus y. And then this right over here is the derivative of x minus y with respect to x. Just the chain rule. Now let's go to the right-hand side of this equation."}, {"video_title": "Worked example Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "This right over here is the derivative of x minus y squared with respect to x minus y. And then this right over here is the derivative of x minus y with respect to x. Just the chain rule. Now let's go to the right-hand side of this equation. This is going to be equal to the derivative of x with respect to x is 1. The derivative of y with respect to x, we're just going to write that as the derivative of y with respect to x. And then finally, the derivative with respect to x of a constant, that's just going to be equal to 0."}, {"video_title": "Worked example Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's go to the right-hand side of this equation. This is going to be equal to the derivative of x with respect to x is 1. The derivative of y with respect to x, we're just going to write that as the derivative of y with respect to x. And then finally, the derivative with respect to x of a constant, that's just going to be equal to 0. Now let's see if we can solve for the derivative of y with respect to x. So the most obvious thing to do, let's make it clear. This right over here I can rewrite as 2x minus 2y."}, {"video_title": "Worked example Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And then finally, the derivative with respect to x of a constant, that's just going to be equal to 0. Now let's see if we can solve for the derivative of y with respect to x. So the most obvious thing to do, let's make it clear. This right over here I can rewrite as 2x minus 2y. So let me do that. So I can save some space. This is 2x minus 2y if I just distribute the 2."}, {"video_title": "Worked example Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "This right over here I can rewrite as 2x minus 2y. So let me do that. So I can save some space. This is 2x minus 2y if I just distribute the 2. And now I can distribute the 2x minus 2y onto each of these terms. So 2x minus 2y times 1 is just going to be 2x minus 2y. And then 2x minus 2y times negative dy dx, that's just going to be negative 2x minus 2y."}, {"video_title": "Worked example Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "This is 2x minus 2y if I just distribute the 2. And now I can distribute the 2x minus 2y onto each of these terms. So 2x minus 2y times 1 is just going to be 2x minus 2y. And then 2x minus 2y times negative dy dx, that's just going to be negative 2x minus 2y. Or we could write that as 2y minus 2x times dy dx. Is equal to 1 plus dy dx. Is equal to 1 plus."}, {"video_title": "Worked example Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And then 2x minus 2y times negative dy dx, that's just going to be negative 2x minus 2y. Or we could write that as 2y minus 2x times dy dx. Is equal to 1 plus dy dx. Is equal to 1 plus. I'll do all my dy dx's in orange now. 1 plus dy dx. So now there's a couple of things that we could attempt to do."}, {"video_title": "Worked example Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Is equal to 1 plus. I'll do all my dy dx's in orange now. 1 plus dy dx. So now there's a couple of things that we could attempt to do. We could subtract 2x minus 2y from both sides. So let's do that. So let's subtract 2x minus 2y from both sides."}, {"video_title": "Worked example Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So now there's a couple of things that we could attempt to do. We could subtract 2x minus 2y from both sides. So let's do that. So let's subtract 2x minus 2y from both sides. So over here we're going to subtract minus. We're going to subtract 2x minus 2y from that side. And then we could also subtract a dy dx from both sides."}, {"video_title": "Worked example Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let's subtract 2x minus 2y from both sides. So over here we're going to subtract minus. We're going to subtract 2x minus 2y from that side. And then we could also subtract a dy dx from both sides. So that all of our dy dx's are on the left hand side and all of our non-dy dx's are on the right hand side. So let's do that. So we're going to subtract a dy dx on the right and a dy dx here on the left."}, {"video_title": "Worked example Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And then we could also subtract a dy dx from both sides. So that all of our dy dx's are on the left hand side and all of our non-dy dx's are on the right hand side. So let's do that. So we're going to subtract a dy dx on the right and a dy dx here on the left. And so what are we left with? What are we left with? Well, on the left hand side, these cancel out."}, {"video_title": "Worked example Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So we're going to subtract a dy dx on the right and a dy dx here on the left. And so what are we left with? What are we left with? Well, on the left hand side, these cancel out. And we're left with 2y minus 2x dy dx minus 1 dy dx. Or just minus a dy dx. Let me make it clear."}, {"video_title": "Worked example Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, on the left hand side, these cancel out. And we're left with 2y minus 2x dy dx minus 1 dy dx. Or just minus a dy dx. Let me make it clear. We could write this as a minus 1 dy dx. So we can essentially just add these two coefficients. So this simplifies to 2y minus 2x minus 1 times the derivative of y with respect to x, which is going to be equal to, on this side, this cancels out."}, {"video_title": "Worked example Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Let me make it clear. We could write this as a minus 1 dy dx. So we can essentially just add these two coefficients. So this simplifies to 2y minus 2x minus 1 times the derivative of y with respect to x, which is going to be equal to, on this side, this cancels out. We are left with 1 minus 2x plus 2y. So let me write it that way. We could write this as negative 2y is just a positive 2y."}, {"video_title": "Worked example Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So this simplifies to 2y minus 2x minus 1 times the derivative of y with respect to x, which is going to be equal to, on this side, this cancels out. We are left with 1 minus 2x plus 2y. So let me write it that way. We could write this as negative 2y is just a positive 2y. And then we have minus 2x. And then we add that 1 plus 1. And now to solve for dy dx, we just have to divide both sides by 2y minus 2x minus 1."}, {"video_title": "Worked example Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We could write this as negative 2y is just a positive 2y. And then we have minus 2x. And then we add that 1 plus 1. And now to solve for dy dx, we just have to divide both sides by 2y minus 2x minus 1. And we are left with, we deserve a little bit of a drum roll at this point. As you can see, the hardest part was really the algebra to solve for dy dx. We get the derivative of y with respect to x is equal to 2y minus 2x plus 1 over 2y minus 2x minus 1."}, {"video_title": "Integration by parts \u00c2\u00ba___cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "And in this case, neither of them become simpler, and neither of them become dramatically more complicated when I take their antiderivative. So here it's kind of a toss-up which one I assign to f of x and which one I assign to g prime of x. And actually, you can solve this problem either way. So let's just assign this one. Let's assign f of x equaling e to the x. And let's assign g prime of x is equaling cosine of x. So let me write it down."}, {"video_title": "Integration by parts \u00c2\u00ba___cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "So let's just assign this one. Let's assign f of x equaling e to the x. And let's assign g prime of x is equaling cosine of x. So let me write it down. We are saying f of x is equal to e to the x, or f prime of x is equal to e to the x. Derivative of e to the x is just e to the x. And we can say that g, we're making the assignment, g prime of x is equal to cosine of x."}, {"video_title": "Integration by parts \u00c2\u00ba___cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "So let me write it down. We are saying f of x is equal to e to the x, or f prime of x is equal to e to the x. Derivative of e to the x is just e to the x. And we can say that g, we're making the assignment, g prime of x is equal to cosine of x. And the antiderivative of that, g of x, is also, or the antiderivative of cosine of x is just going to be equal to sine of x. So now let's apply integration by parts. So this thing is going to be equal to f of x times g of x, which is equal to e to the x times sine of x, times sine of x minus the antiderivative of f prime of x. f prime of x is e to the x. e to the x times g of x, which is once again sine of x."}, {"video_title": "Integration by parts \u00c2\u00ba___cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "And we can say that g, we're making the assignment, g prime of x is equal to cosine of x. And the antiderivative of that, g of x, is also, or the antiderivative of cosine of x is just going to be equal to sine of x. So now let's apply integration by parts. So this thing is going to be equal to f of x times g of x, which is equal to e to the x times sine of x, times sine of x minus the antiderivative of f prime of x. f prime of x is e to the x. e to the x times g of x, which is once again sine of x. Sine of x dx. Now, it doesn't look like we made a lot of progress. Now we have an indefinite integral that involves a sine of x."}, {"video_title": "Integration by parts \u00c2\u00ba___cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "So this thing is going to be equal to f of x times g of x, which is equal to e to the x times sine of x, times sine of x minus the antiderivative of f prime of x. f prime of x is e to the x. e to the x times g of x, which is once again sine of x. Sine of x dx. Now, it doesn't look like we made a lot of progress. Now we have an indefinite integral that involves a sine of x. So let's see if we can solve this one separately. So let's say if we were trying to find the antiderivative of e to the x sine of x dx. How could we do that?"}, {"video_title": "Integration by parts \u00c2\u00ba___cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "Now we have an indefinite integral that involves a sine of x. So let's see if we can solve this one separately. So let's say if we were trying to find the antiderivative of e to the x sine of x dx. How could we do that? Well, similarly, we can set f of x is equal to e to the x. So now this is we're reassigning, although we're happy to make the exact same reassignment. So we're saying f of x is equal to e to the x. f prime of x is equal to just the derivative of that, which is still e to the x."}, {"video_title": "Integration by parts \u00c2\u00ba___cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "How could we do that? Well, similarly, we can set f of x is equal to e to the x. So now this is we're reassigning, although we're happy to make the exact same reassignment. So we're saying f of x is equal to e to the x. f prime of x is equal to just the derivative of that, which is still e to the x. And then we could say g of x, in this case, is equal to sine of x. We'll put these assignments in the back of our brain for now. And then, let me make this clear, g prime of x. Whoops, there you go."}, {"video_title": "Integration by parts \u00c2\u00ba___cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "So we're saying f of x is equal to e to the x. f prime of x is equal to just the derivative of that, which is still e to the x. And then we could say g of x, in this case, is equal to sine of x. We'll put these assignments in the back of our brain for now. And then, let me make this clear, g prime of x. Whoops, there you go. So we have g prime of x is equal to sine of x, which means that its antiderivative is negative cosine of x. Derivative of cosine is negative sine. Derivative of negative cosine is positive sine."}, {"video_title": "Integration by parts \u00c2\u00ba___cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "And then, let me make this clear, g prime of x. Whoops, there you go. So we have g prime of x is equal to sine of x, which means that its antiderivative is negative cosine of x. Derivative of cosine is negative sine. Derivative of negative cosine is positive sine. So once again, let's apply integration by parts. So we have f of x times g of x. f of x times g of x is negative. I'll put the negative out front."}, {"video_title": "Integration by parts \u00c2\u00ba___cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "Derivative of negative cosine is positive sine. So once again, let's apply integration by parts. So we have f of x times g of x. f of x times g of x is negative. I'll put the negative out front. It's negative e to the x times cosine of x minus the antiderivative of f prime of x g of x. f prime of x is e to the x. And then g of x is negative cosine of x. So I'll put the cosine of x right over here."}, {"video_title": "Integration by parts \u00c2\u00ba___cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "I'll put the negative out front. It's negative e to the x times cosine of x minus the antiderivative of f prime of x g of x. f prime of x is e to the x. And then g of x is negative cosine of x. So I'll put the cosine of x right over here. And then the negative, we can take it out of the integral sign. And so we're subtracting a negative that becomes a positive. And of course, we have our dx right over there."}, {"video_title": "Integration by parts \u00c2\u00ba___cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "So I'll put the cosine of x right over here. And then the negative, we can take it out of the integral sign. And so we're subtracting a negative that becomes a positive. And of course, we have our dx right over there. And you might say, hey, Sal, we're not making any progress. This thing right over here, we now expressed in terms of an integral that was our original integral. We've come back full circle."}, {"video_title": "Integration by parts \u00c2\u00ba___cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "And of course, we have our dx right over there. And you might say, hey, Sal, we're not making any progress. This thing right over here, we now expressed in terms of an integral that was our original integral. We've come back full circle. But let's try to do something interesting. Let's substitute back this. Or let me write it this way."}, {"video_title": "Integration by parts \u00c2\u00ba___cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "We've come back full circle. But let's try to do something interesting. Let's substitute back this. Or let me write it this way. Let's substitute back this thing up here. Or actually, let me write it a different way. Let's substitute this for this in our original equation."}, {"video_title": "Integration by parts \u00c2\u00ba___cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "Or let me write it this way. Let's substitute back this thing up here. Or actually, let me write it a different way. Let's substitute this for this in our original equation. And let's see if we got anything interesting. So what we'll get is our original integral on the left-hand side here. The indefinite integral, or the antiderivative of e to the x cosine of x dx is equal to e to the x sine of x minus all of this business."}, {"video_title": "Integration by parts \u00c2\u00ba___cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "Let's substitute this for this in our original equation. And let's see if we got anything interesting. So what we'll get is our original integral on the left-hand side here. The indefinite integral, or the antiderivative of e to the x cosine of x dx is equal to e to the x sine of x minus all of this business. So let's just subtract all of this. We're subtracting all of this. So if you subtract negative e to the x cosine of x, it's going to be positive e to the x cosine of x."}, {"video_title": "Integration by parts \u00c2\u00ba___cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "The indefinite integral, or the antiderivative of e to the x cosine of x dx is equal to e to the x sine of x minus all of this business. So let's just subtract all of this. We're subtracting all of this. So if you subtract negative e to the x cosine of x, it's going to be positive e to the x cosine of x. And then remember, we're subtracting all of this. So then we're going to subtract. So then we have minus the antiderivative of e to the x cosine of x dx."}, {"video_title": "Integration by parts \u00c2\u00ba___cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "So if you subtract negative e to the x cosine of x, it's going to be positive e to the x cosine of x. And then remember, we're subtracting all of this. So then we're going to subtract. So then we have minus the antiderivative of e to the x cosine of x dx. Now this is interesting. Just remember, all we did is we took this part right over here. We said we used integration by parts to figure out that it's the same thing as this."}, {"video_title": "Integration by parts \u00c2\u00ba___cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "So then we have minus the antiderivative of e to the x cosine of x dx. Now this is interesting. Just remember, all we did is we took this part right over here. We said we used integration by parts to figure out that it's the same thing as this. So we substituted this back in. When you subtracted this from this, we got this business right over here. Now what's interesting here is we have essentially an equation where we have our original expression twice."}, {"video_title": "Integration by parts \u00c2\u00ba___cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "We said we used integration by parts to figure out that it's the same thing as this. So we substituted this back in. When you subtracted this from this, we got this business right over here. Now what's interesting here is we have essentially an equation where we have our original expression twice. We can even assign this to a variable and essentially solve for that variable. So why don't we just add this thing to both sides of the equation? Let me make it clear."}, {"video_title": "Integration by parts \u00c2\u00ba___cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "Now what's interesting here is we have essentially an equation where we have our original expression twice. We can even assign this to a variable and essentially solve for that variable. So why don't we just add this thing to both sides of the equation? Let me make it clear. Let's just add the integral of e to the x cosine of x dx to both sides. e to the x cosine of x dx. And what do you get?"}, {"video_title": "Integration by parts \u00c2\u00ba___cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "Let me make it clear. Let's just add the integral of e to the x cosine of x dx to both sides. e to the x cosine of x dx. And what do you get? Well, on the left-hand side, you have 2 times our original integral, e to the x cosine of x dx is equal to all of this business. Is equal to this. I'll copy and paste it."}, {"video_title": "Integration by parts \u00c2\u00ba___cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "And what do you get? Well, on the left-hand side, you have 2 times our original integral, e to the x cosine of x dx is equal to all of this business. Is equal to this. I'll copy and paste it. So copy and paste. It's equal to all of that. And then this part, this part right over here, cancels out."}, {"video_title": "Integration by parts \u00c2\u00ba___cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "I'll copy and paste it. So copy and paste. It's equal to all of that. And then this part, this part right over here, cancels out. And now we can solve for our original expression, the antiderivative of e to the x cosine of x dx. We just have to divide both sides of this, essentially, an equation by 2. So if you divide the left-hand side by 2, you're left with our original expression, the antiderivative of e to the x cosine of x dx."}, {"video_title": "Integration by parts \u00c2\u00ba___cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "And then this part, this part right over here, cancels out. And now we can solve for our original expression, the antiderivative of e to the x cosine of x dx. We just have to divide both sides of this, essentially, an equation by 2. So if you divide the left-hand side by 2, you're left with our original expression, the antiderivative of e to the x cosine of x dx. And on the right-hand side, you have what it must be equal to. e to the x sine of x plus e to the x cosine of x over 2. And now we want to be careful, because this is an antiderivative of our original expression, but it's not the only one."}, {"video_title": "Integration by parts \u00c2\u00ba___cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "So if you divide the left-hand side by 2, you're left with our original expression, the antiderivative of e to the x cosine of x dx. And on the right-hand side, you have what it must be equal to. e to the x sine of x plus e to the x cosine of x over 2. And now we want to be careful, because this is an antiderivative of our original expression, but it's not the only one. We always have to remember, even though we've worked hard and we've used integration by parts twice and we've had to back substitute in, we have to remember we should still have a constant here. So if you take the derivative of this business, regardless of what the constant is, you will get e to the x cosine of x. And it's actually a pretty neat-looking expression."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds Khan Academy.mp3", "Sentence": "And once again, it looks like you might be able to use the fundamental theorem of calculus. The big giveaway is that you're taking the derivative of a definite integral that gives you a function of x. But here I have x on both the upper and the lower boundary. And the fundamental theorem of calculus, at least from what we've seen, is when we have x's only on the upper boundary. And then, of course, it's an x squared, but we've seen examples of that already when we use the chain rule to do it. But how can we break this up and put this in a form that's a little bit closer to what we're familiar with when we apply the fundamental theorem of calculus? And to realize that, we really just have to attempt to graph what this is representing."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds Khan Academy.mp3", "Sentence": "And the fundamental theorem of calculus, at least from what we've seen, is when we have x's only on the upper boundary. And then, of course, it's an x squared, but we've seen examples of that already when we use the chain rule to do it. But how can we break this up and put this in a form that's a little bit closer to what we're familiar with when we apply the fundamental theorem of calculus? And to realize that, we really just have to attempt to graph what this is representing. So let's say that this is our lowercase f of x, or I should say f of t. So let's call this lowercase f of t. And let's graph it over the interval between x and x squared. So let's say this is my y-axis. This is my t-axis."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds Khan Academy.mp3", "Sentence": "And to realize that, we really just have to attempt to graph what this is representing. So let's say that this is our lowercase f of x, or I should say f of t. So let's call this lowercase f of t. And let's graph it over the interval between x and x squared. So let's say this is my y-axis. This is my t-axis. And let's say that this right over here is y is equal to f of t. I'm drawing it generally. I don't know what this exactly looks like. And we're going to talk about the interval between x and x squared."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds Khan Academy.mp3", "Sentence": "This is my t-axis. And let's say that this right over here is y is equal to f of t. I'm drawing it generally. I don't know what this exactly looks like. And we're going to talk about the interval between x and x squared. So if we're going to talk about the interval between x, which is right over here, it's the lower bound. So x and x squared. It's the lower bound, at least for this definite integral."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds Khan Academy.mp3", "Sentence": "And we're going to talk about the interval between x and x squared. So if we're going to talk about the interval between x, which is right over here, it's the lower bound. So x and x squared. It's the lower bound, at least for this definite integral. We don't know for sure. It depends on what x you choose on which one is actually smaller. But let's just say that for the sake of visualizing, we'll draw x right over here."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds Khan Academy.mp3", "Sentence": "It's the lower bound, at least for this definite integral. We don't know for sure. It depends on what x you choose on which one is actually smaller. But let's just say that for the sake of visualizing, we'll draw x right over here. And we will draw x squared right over here. So this whole expression, this entire definite integral, is essentially asking for, is essentially representing this entire area. The entire area under the curve."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds Khan Academy.mp3", "Sentence": "But let's just say that for the sake of visualizing, we'll draw x right over here. And we will draw x squared right over here. So this whole expression, this entire definite integral, is essentially asking for, is essentially representing this entire area. The entire area under the curve. But what we could do is introduce a constant that's someplace in between x and x squared. Let's say that constant is c. And break this area into two different areas with c as the divider. So that same exact whole area, we can now write it as two separate integrals."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds Khan Academy.mp3", "Sentence": "The entire area under the curve. But what we could do is introduce a constant that's someplace in between x and x squared. Let's say that constant is c. And break this area into two different areas with c as the divider. So that same exact whole area, we can now write it as two separate integrals. So one integral that represents this area right over here. And then another integral that represents this area right over there. And we just say c is some constant between x and x squared."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds Khan Academy.mp3", "Sentence": "So that same exact whole area, we can now write it as two separate integrals. So one integral that represents this area right over here. And then another integral that represents this area right over there. And we just say c is some constant between x and x squared. Well, how can we denote this area in purple? Well, that's going to be, so this thing is going to be equal to the sum of these two areas. The purple area we can show is the definite integral from x to c of our function of t. Cosine t over t dt."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds Khan Academy.mp3", "Sentence": "And we just say c is some constant between x and x squared. Well, how can we denote this area in purple? Well, that's going to be, so this thing is going to be equal to the sum of these two areas. The purple area we can show is the definite integral from x to c of our function of t. Cosine t over t dt. And then to that, we're going to add the green area. And then we'll get the original area. So the green area, for the green area, our lower bound of integration is now our constant c. And our upper bound of integration is x squared."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds Khan Academy.mp3", "Sentence": "The purple area we can show is the definite integral from x to c of our function of t. Cosine t over t dt. And then to that, we're going to add the green area. And then we'll get the original area. So the green area, for the green area, our lower bound of integration is now our constant c. And our upper bound of integration is x squared. It's going to be of cosine t over t dt. And this is a form where if we know how to apply the chain rule, we can apply the fundamental theorem of calculus. And this is almost in a form."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds Khan Academy.mp3", "Sentence": "So the green area, for the green area, our lower bound of integration is now our constant c. And our upper bound of integration is x squared. It's going to be of cosine t over t dt. And this is a form where if we know how to apply the chain rule, we can apply the fundamental theorem of calculus. And this is almost in a form. We're used to seeing it where the x is the upper bound. And we already know what happens. We can swap these two bounds, but it will just be the negative of that integral."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds Khan Academy.mp3", "Sentence": "And this is almost in a form. We're used to seeing it where the x is the upper bound. And we already know what happens. We can swap these two bounds, but it will just be the negative of that integral. So this is going to be equal to, let me rewrite it, the negative of the definite integral from c to x of cosine t over t dt. And then we have plus the definite integral that goes from c to x squared of cosine t over t dt. So all we've done is we've rewritten this thing in a way that we're used to applying the fundamental theorem of calculus."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds Khan Academy.mp3", "Sentence": "We can swap these two bounds, but it will just be the negative of that integral. So this is going to be equal to, let me rewrite it, the negative of the definite integral from c to x of cosine t over t dt. And then we have plus the definite integral that goes from c to x squared of cosine t over t dt. So all we've done is we've rewritten this thing in a way that we're used to applying the fundamental theorem of calculus. So if we want to find f prime of x, well, applying the derivative operator over here, we're going to have a negative out front. It's going to be equal to negative cosine x over x. Once again, just the fundamental theorem of calculus."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds Khan Academy.mp3", "Sentence": "So all we've done is we've rewritten this thing in a way that we're used to applying the fundamental theorem of calculus. So if we want to find f prime of x, well, applying the derivative operator over here, we're going to have a negative out front. It's going to be equal to negative cosine x over x. Once again, just the fundamental theorem of calculus. And then plus, we're first going to take the derivative of this thing with respect to x squared. And that's going to give you cosine of x squared over x squared. Wherever you saw a t, you replace it with an x squared."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds Khan Academy.mp3", "Sentence": "Once again, just the fundamental theorem of calculus. And then plus, we're first going to take the derivative of this thing with respect to x squared. And that's going to give you cosine of x squared over x squared. Wherever you saw a t, you replace it with an x squared. And then you're going to multiply that times the derivative of x squared with respect to x. So that's just going to be derivative of x squared with respect to x is just 2x. And we're done."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on both bounds Khan Academy.mp3", "Sentence": "Wherever you saw a t, you replace it with an x squared. And then you're going to multiply that times the derivative of x squared with respect to x. So that's just going to be derivative of x squared with respect to x is just 2x. And we're done. We just need to simplify this thing. All of this is going to be equal to negative cosine x over x plus, well, this is going to cancel out with just one of those, plus 2 cosine x squared over x. And I guess we could simplify it even more as being equal to, and we can swap these."}, {"video_title": "Average value over a closed interval AP Calculus AB Khan Academy.mp3", "Sentence": "What I want to do in this video is think about the idea of an average value of a function over some closed interval. So what do I mean by that? And how can we think about what average value of a function even means? So let's say that's my y-axis. And let's say that this is my, this right over here is my x-axis. And let me draw a function here. So let's say the function looks something like that."}, {"video_title": "Average value over a closed interval AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say that's my y-axis. And let's say that this is my, this right over here is my x-axis. And let me draw a function here. So let's say the function looks something like that. That's the graph of y is equal to, y is equal to f of x. And now let's think about a closed interval. So we're gonna think about the closed interval between a and b, including a and b."}, {"video_title": "Average value over a closed interval AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say the function looks something like that. That's the graph of y is equal to, y is equal to f of x. And now let's think about a closed interval. So we're gonna think about the closed interval between a and b, including a and b. That's what makes it closed. We're including our endpoints. So we're gonna think about this interval right over here."}, {"video_title": "Average value over a closed interval AP Calculus AB Khan Academy.mp3", "Sentence": "So we're gonna think about the closed interval between a and b, including a and b. That's what makes it closed. We're including our endpoints. So we're gonna think about this interval right over here. So between x is equal to a and x is equal to b, what is the average value of this function? One way to think about it is what is the average height of this function? So how could we, what would that mean?"}, {"video_title": "Average value over a closed interval AP Calculus AB Khan Academy.mp3", "Sentence": "So we're gonna think about this interval right over here. So between x is equal to a and x is equal to b, what is the average value of this function? One way to think about it is what is the average height of this function? So how could we, what would that mean? Well, one way to think about it, it would be some height so that if we multiply it times the width of this interval, we'll get the area under the curve. So another way to think about it is the area under the curve right over here, the area, actually let me do this in a different color. So the area under this curve right over here, I'll shade it in yellow."}, {"video_title": "Average value over a closed interval AP Calculus AB Khan Academy.mp3", "Sentence": "So how could we, what would that mean? Well, one way to think about it, it would be some height so that if we multiply it times the width of this interval, we'll get the area under the curve. So another way to think about it is the area under the curve right over here, the area, actually let me do this in a different color. So the area under this curve right over here, I'll shade it in yellow. We already know that we can express this as the definite integral from a to b of f of x dx. The average value of our function over this closed interval a, b, let me write that, over the closed interval between a and b, including a and b, we could think about it as some height, some height, let me do this in a new color, some value of our function, some height, let me think about it, maybe some height right over here, so that if we multiply this height times this width, we're gonna get the area of a rectangle. The rectangle would be the area of this rectangle right over here, and that rectangle is going to have the same area as the area under the curve, which is a reasonable way if you kind of remember how when you even think about finding the area, or one way to think about the area of a trapezoid, you can multiply, if you have a trapezoid, if you have a trapezoid like this, you have a trapezoid like this, this is, you could kind of turn 90 degrees, but you multiply the height times the average width of the trapezoid, and then that will give you its area, so this would be the average width, which in a trapezoid like this would just be halfway between, this function is not linear, so it's not necessarily going to be halfway in between, but it's that same idea."}, {"video_title": "Average value over a closed interval AP Calculus AB Khan Academy.mp3", "Sentence": "So the area under this curve right over here, I'll shade it in yellow. We already know that we can express this as the definite integral from a to b of f of x dx. The average value of our function over this closed interval a, b, let me write that, over the closed interval between a and b, including a and b, we could think about it as some height, some height, let me do this in a new color, some value of our function, some height, let me think about it, maybe some height right over here, so that if we multiply this height times this width, we're gonna get the area of a rectangle. The rectangle would be the area of this rectangle right over here, and that rectangle is going to have the same area as the area under the curve, which is a reasonable way if you kind of remember how when you even think about finding the area, or one way to think about the area of a trapezoid, you can multiply, if you have a trapezoid, if you have a trapezoid like this, you have a trapezoid like this, this is, you could kind of turn 90 degrees, but you multiply the height times the average width of the trapezoid, and then that will give you its area, so this would be the average width, which in a trapezoid like this would just be halfway between, this function is not linear, so it's not necessarily going to be halfway in between, but it's that same idea. So how could we use this idea, where this is right over here, this height, this height right over here, we could call this, we could call this the function's average, the function's average. How could we use all of this to come up with a formula for the average of a function over this closed interval? Well, let's just express in math what we've already said."}, {"video_title": "Average value over a closed interval AP Calculus AB Khan Academy.mp3", "Sentence": "The rectangle would be the area of this rectangle right over here, and that rectangle is going to have the same area as the area under the curve, which is a reasonable way if you kind of remember how when you even think about finding the area, or one way to think about the area of a trapezoid, you can multiply, if you have a trapezoid, if you have a trapezoid like this, you have a trapezoid like this, this is, you could kind of turn 90 degrees, but you multiply the height times the average width of the trapezoid, and then that will give you its area, so this would be the average width, which in a trapezoid like this would just be halfway between, this function is not linear, so it's not necessarily going to be halfway in between, but it's that same idea. So how could we use this idea, where this is right over here, this height, this height right over here, we could call this, we could call this the function's average, the function's average. How could we use all of this to come up with a formula for the average of a function over this closed interval? Well, let's just express in math what we've already said. We already said that this function average should be some height, so let's say the function average, so that's a height, and if I multiply it times the width of this interval, so this width right over here, this width right over here is just going to be the larger value minus the smaller value, so that's going to be b minus a, so the average value of the function times the width of the interval should give us an area that is equivalent to the area under the curve, so it should be equal to the definite integral from a to b of f of x dx, and so if we just, if we knew all of this other stuff, we could solve for the function's average. The function's average, if we divide both sides by b minus a, the function's average is going to be equal to, just dividing both sides by b minus a, you're going to get one over b minus a times the definite integral, the definite integral from a to b of f of x dx, or another way to think about it, you're going to figure out what the area under the curve is over that interval. You're going to divide that by the width, and then you're going to have the function's average."}, {"video_title": "Average value over a closed interval AP Calculus AB Khan Academy.mp3", "Sentence": "Well, let's just express in math what we've already said. We already said that this function average should be some height, so let's say the function average, so that's a height, and if I multiply it times the width of this interval, so this width right over here, this width right over here is just going to be the larger value minus the smaller value, so that's going to be b minus a, so the average value of the function times the width of the interval should give us an area that is equivalent to the area under the curve, so it should be equal to the definite integral from a to b of f of x dx, and so if we just, if we knew all of this other stuff, we could solve for the function's average. The function's average, if we divide both sides by b minus a, the function's average is going to be equal to, just dividing both sides by b minus a, you're going to get one over b minus a times the definite integral, the definite integral from a to b of f of x dx, or another way to think about it, you're going to figure out what the area under the curve is over that interval. You're going to divide that by the width, and then you're going to have the function's average. One way to think about it, you're going to have the average height, and once again, I'd like to remind you that because you shouldn't just sit there and try to memorize this thing. Just get a conceptual understanding of what this is really just trying to say. Area under the curve divided by the width, well, that's just going to give you the average height."}, {"video_title": "Indefinite integral of 1 x AP Calculus AB Khan Academy.mp3", "Sentence": "And we already know if we somehow tried to apply that anti-power rule, that inverse power rule over here, we would get something that's not defined. We would get x to the 0 over 0. It doesn't make any sense. And you might have been saying, OK, well, I know what to do in this case. When we first learned about derivatives, we know that the derivative, let me do this in yellow, the derivative with respect to x of the natural log of x is equal to 1 over x. So why can't we just say that the antiderivative of this right over here is equal to the natural log of x plus c? And this isn't necessarily wrong."}, {"video_title": "Indefinite integral of 1 x AP Calculus AB Khan Academy.mp3", "Sentence": "And you might have been saying, OK, well, I know what to do in this case. When we first learned about derivatives, we know that the derivative, let me do this in yellow, the derivative with respect to x of the natural log of x is equal to 1 over x. So why can't we just say that the antiderivative of this right over here is equal to the natural log of x plus c? And this isn't necessarily wrong. The problem here is that it's not broad enough. And when I say it's not broad enough, is that the domain over here for our original function, that we're taking the antiderivative of, is all real numbers except for x equals 0. So over here, x cannot be equal to 0."}, {"video_title": "Indefinite integral of 1 x AP Calculus AB Khan Academy.mp3", "Sentence": "And this isn't necessarily wrong. The problem here is that it's not broad enough. And when I say it's not broad enough, is that the domain over here for our original function, that we're taking the antiderivative of, is all real numbers except for x equals 0. So over here, x cannot be equal to 0. While the domain over here is only positive numbers. So over here, x, for this expression, x has to be greater than 0. So it would be nice if we could come up with an antiderivative that has the same domain as the function that we're taking the antiderivative of."}, {"video_title": "Indefinite integral of 1 x AP Calculus AB Khan Academy.mp3", "Sentence": "So over here, x cannot be equal to 0. While the domain over here is only positive numbers. So over here, x, for this expression, x has to be greater than 0. So it would be nice if we could come up with an antiderivative that has the same domain as the function that we're taking the antiderivative of. So it would be nice if we could find an antiderivative that is defined everywhere that our original function is. So pretty much everywhere except for x equaling 0. So how can we rearrange this a little bit so that it could be defined for negative values as well?"}, {"video_title": "Indefinite integral of 1 x AP Calculus AB Khan Academy.mp3", "Sentence": "So it would be nice if we could come up with an antiderivative that has the same domain as the function that we're taking the antiderivative of. So it would be nice if we could find an antiderivative that is defined everywhere that our original function is. So pretty much everywhere except for x equaling 0. So how can we rearrange this a little bit so that it could be defined for negative values as well? Well, one possibility is to think about the natural log of the absolute value of x. So I'll put a little question mark here because we don't really know what the derivative of this thing is going to be. And I'm not going to rigorously prove it here, but I will give you kind of the conceptual understanding."}, {"video_title": "Indefinite integral of 1 x AP Calculus AB Khan Academy.mp3", "Sentence": "So how can we rearrange this a little bit so that it could be defined for negative values as well? Well, one possibility is to think about the natural log of the absolute value of x. So I'll put a little question mark here because we don't really know what the derivative of this thing is going to be. And I'm not going to rigorously prove it here, but I will give you kind of the conceptual understanding. So to understand it, let's plot the natural log of x. And I had done this ahead of time. So that right over there is roughly what the graph of the natural log of x looks like."}, {"video_title": "Indefinite integral of 1 x AP Calculus AB Khan Academy.mp3", "Sentence": "And I'm not going to rigorously prove it here, but I will give you kind of the conceptual understanding. So to understand it, let's plot the natural log of x. And I had done this ahead of time. So that right over there is roughly what the graph of the natural log of x looks like. So what would the natural log of the absolute value of x is going to look like? Well, for positive x's, it's going to look just like this. For positive x's, you take the absolute value of it."}, {"video_title": "Indefinite integral of 1 x AP Calculus AB Khan Academy.mp3", "Sentence": "So that right over there is roughly what the graph of the natural log of x looks like. So what would the natural log of the absolute value of x is going to look like? Well, for positive x's, it's going to look just like this. For positive x's, you take the absolute value of it. It's just the same thing as taking that original value. So it's going to look just like that for positive x's. But now this is also going to be defined for negative x's."}, {"video_title": "Indefinite integral of 1 x AP Calculus AB Khan Academy.mp3", "Sentence": "For positive x's, you take the absolute value of it. It's just the same thing as taking that original value. So it's going to look just like that for positive x's. But now this is also going to be defined for negative x's. If you're taking the absolute value of negative 1, that evaluates to just 1. So it's just the natural log of 1. So you're going to be right there."}, {"video_title": "Indefinite integral of 1 x AP Calculus AB Khan Academy.mp3", "Sentence": "But now this is also going to be defined for negative x's. If you're taking the absolute value of negative 1, that evaluates to just 1. So it's just the natural log of 1. So you're going to be right there. As you get closer and closer and closer to 0 from the negative side, you're just going to take the absolute value. So it's essentially going to be exactly this curve for natural log of x. But the left side of the natural log of the absolute value of x is going to be its mirror image if you were to reflect around the y-axis."}, {"video_title": "Indefinite integral of 1 x AP Calculus AB Khan Academy.mp3", "Sentence": "So you're going to be right there. As you get closer and closer and closer to 0 from the negative side, you're just going to take the absolute value. So it's essentially going to be exactly this curve for natural log of x. But the left side of the natural log of the absolute value of x is going to be its mirror image if you were to reflect around the y-axis. It's going to look something like this. So what's nice about this function is you see it's defined everywhere except for, I'm trying to draw it as symmetrically as possible, it's defined everywhere except for x equals 0. So if you combine this pink part and this part on the right, if you combine both of these, you get y is equal to the natural log of the absolute value of x."}, {"video_title": "Indefinite integral of 1 x AP Calculus AB Khan Academy.mp3", "Sentence": "But the left side of the natural log of the absolute value of x is going to be its mirror image if you were to reflect around the y-axis. It's going to look something like this. So what's nice about this function is you see it's defined everywhere except for, I'm trying to draw it as symmetrically as possible, it's defined everywhere except for x equals 0. So if you combine this pink part and this part on the right, if you combine both of these, you get y is equal to the natural log of the absolute value of x. Now let's think about its derivative. We already know what the derivative of the natural log of x is. And for positive values of x, let me write this down."}, {"video_title": "Indefinite integral of 1 x AP Calculus AB Khan Academy.mp3", "Sentence": "So if you combine this pink part and this part on the right, if you combine both of these, you get y is equal to the natural log of the absolute value of x. Now let's think about its derivative. We already know what the derivative of the natural log of x is. And for positive values of x, let me write this down. For x is greater than 0, we get the natural log of the absolute value of x is equal to the natural log of x. And we would also know, since these two are equal for x is greater than 0, for x is greater than 0, the derivative of the natural log of the absolute value of x is going to be equal to the derivative of the natural log of x, which is equal to 1 over x for x greater than 0. So let's plot that."}, {"video_title": "Indefinite integral of 1 x AP Calculus AB Khan Academy.mp3", "Sentence": "And for positive values of x, let me write this down. For x is greater than 0, we get the natural log of the absolute value of x is equal to the natural log of x. And we would also know, since these two are equal for x is greater than 0, for x is greater than 0, the derivative of the natural log of the absolute value of x is going to be equal to the derivative of the natural log of x, which is equal to 1 over x for x greater than 0. So let's plot that. I'll do that in green. It's equal to 1 over x. So 1 over x, we've seen it before."}, {"video_title": "Indefinite integral of 1 x AP Calculus AB Khan Academy.mp3", "Sentence": "So let's plot that. I'll do that in green. It's equal to 1 over x. So 1 over x, we've seen it before. It looks something like this. So let me, my best attempt to draw it. Both vertical and horizontal asymptotes."}, {"video_title": "Indefinite integral of 1 x AP Calculus AB Khan Academy.mp3", "Sentence": "So 1 over x, we've seen it before. It looks something like this. So let me, my best attempt to draw it. Both vertical and horizontal asymptotes. So it looks something like this. So this right over here is 1 over x for x is greater than 0. So this is 1 over x when x is greater than 0."}, {"video_title": "Indefinite integral of 1 x AP Calculus AB Khan Academy.mp3", "Sentence": "Both vertical and horizontal asymptotes. So it looks something like this. So this right over here is 1 over x for x is greater than 0. So this is 1 over x when x is greater than 0. So all it's saying here, and you can see it pretty clearly, is the slope right over here, the slope of the tangent line is 1. And so you see, when you look at the derivative, the slope right over here, the derivative should be equal to 1 here. When you get close to 0, you have a very, very steep positive slope here."}, {"video_title": "Indefinite integral of 1 x AP Calculus AB Khan Academy.mp3", "Sentence": "So this is 1 over x when x is greater than 0. So all it's saying here, and you can see it pretty clearly, is the slope right over here, the slope of the tangent line is 1. And so you see, when you look at the derivative, the slope right over here, the derivative should be equal to 1 here. When you get close to 0, you have a very, very steep positive slope here. And so you see, you have a very high value for its derivative. And then as you move away from 0, it's still steep, but it becomes less and less and less steep all the way until you get to 1. And then it keeps getting less and less and less steep, but it never quite gets to an absolutely flat slope."}, {"video_title": "Indefinite integral of 1 x AP Calculus AB Khan Academy.mp3", "Sentence": "When you get close to 0, you have a very, very steep positive slope here. And so you see, you have a very high value for its derivative. And then as you move away from 0, it's still steep, but it becomes less and less and less steep all the way until you get to 1. And then it keeps getting less and less and less steep, but it never quite gets to an absolutely flat slope. And that's what you see its derivative doing. Now what is the natural log of absolute value of x doing right over here? When we are out here, our slope is very close to 0."}, {"video_title": "Indefinite integral of 1 x AP Calculus AB Khan Academy.mp3", "Sentence": "And then it keeps getting less and less and less steep, but it never quite gets to an absolutely flat slope. And that's what you see its derivative doing. Now what is the natural log of absolute value of x doing right over here? When we are out here, our slope is very close to 0. It's symmetric. The slope here is essentially the negative of the slope here. I could do it maybe clearer showing it right over here."}, {"video_title": "Indefinite integral of 1 x AP Calculus AB Khan Academy.mp3", "Sentence": "When we are out here, our slope is very close to 0. It's symmetric. The slope here is essentially the negative of the slope here. I could do it maybe clearer showing it right over here. Whatever the slope is right over here, it's the exact negative of whatever the slope is at a symmetric point on the other side. So if on the other side, the slope is right over here, over here, it's going to be the negative of that. So it's going to be right over there."}, {"video_title": "Indefinite integral of 1 x AP Calculus AB Khan Academy.mp3", "Sentence": "I could do it maybe clearer showing it right over here. Whatever the slope is right over here, it's the exact negative of whatever the slope is at a symmetric point on the other side. So if on the other side, the slope is right over here, over here, it's going to be the negative of that. So it's going to be right over there. And then the slope just gets more and more and more negative. Right over here, the slope is a positive 1. Over here, it's going to be a negative 1."}, {"video_title": "Indefinite integral of 1 x AP Calculus AB Khan Academy.mp3", "Sentence": "So it's going to be right over there. And then the slope just gets more and more and more negative. Right over here, the slope is a positive 1. Over here, it's going to be a negative 1. So right over here, our slope is a negative 1. And then as we get closer and closer to 0, it's just going to get more and more and more negative. So the derivative of the natural log of the absolute value of x for x is less than 0 looks something like this."}, {"video_title": "Indefinite integral of 1 x AP Calculus AB Khan Academy.mp3", "Sentence": "Over here, it's going to be a negative 1. So right over here, our slope is a negative 1. And then as we get closer and closer to 0, it's just going to get more and more and more negative. So the derivative of the natural log of the absolute value of x for x is less than 0 looks something like this. And you see, and once again, it's not a ultra rigorous proof, but what you see is that the derivative of the natural log of the absolute value of x is equal to 1 over x for all x's not equaling 0. So what you're seeing, or hopefully you can visualize, that the derivative, let me write it this way, the derivative of the natural log of the absolute value of x is indeed equal to 1 over x for all x does not equal 0. So this is a much more satisfying antiderivative for 1 over x."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "So we have the differential equation, the derivative of y with respect to x is equal to y over six times four minus y. And what we have plotted right over here is the slope field, or a slope field, for this differential equation. And we can verify that this indeed is a slope field for this differential equation. Let's draw a little table here. So let's just verify a few points. So let's say x, y, and dy, dx. So let's say we start with, I don't know, let's start with this point right over here, one comma one."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "Let's draw a little table here. So let's just verify a few points. So let's say x, y, and dy, dx. So let's say we start with, I don't know, let's start with this point right over here, one comma one. When x is one and y is one, well, when I look at the differential equation, 1 6th times four minus one, so it's 1 6th times three, which is 3 6ths, which is 1 1.5. And we see indeed on this slope field, they depicted the slope there. So if a solution goes to that point, right at that point, its slope would be 1 1.5."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "So let's say we start with, I don't know, let's start with this point right over here, one comma one. When x is one and y is one, well, when I look at the differential equation, 1 6th times four minus one, so it's 1 6th times three, which is 3 6ths, which is 1 1.5. And we see indeed on this slope field, they depicted the slope there. So if a solution goes to that point, right at that point, its slope would be 1 1.5. And as you see, it's actually only dependent on the y value. It doesn't matter what x is. As long as y is one, dy, dx is going to be 1 1.5."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "So if a solution goes to that point, right at that point, its slope would be 1 1.5. And as you see, it's actually only dependent on the y value. It doesn't matter what x is. As long as y is one, dy, dx is going to be 1 1.5. And you see that's why when x is 1 1.5 and y is one, you still have a slope of 1 1.5. And as long as y is one, all of these sampled points right over here all have a slope of 1 1.5. So just looking at that, that makes us feel that this slope field is consistent with this differential equation."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "As long as y is one, dy, dx is going to be 1 1.5. And you see that's why when x is 1 1.5 and y is one, you still have a slope of 1 1.5. And as long as y is one, all of these sampled points right over here all have a slope of 1 1.5. So just looking at that, that makes us feel that this slope field is consistent with this differential equation. But let's try a few other points just to feel a little bit better about it. And then we will use the slope field to actually visualize some solutions. So let's do an interesting point."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "So just looking at that, that makes us feel that this slope field is consistent with this differential equation. But let's try a few other points just to feel a little bit better about it. And then we will use the slope field to actually visualize some solutions. So let's do an interesting point. Let's say we have this point. Actually, no, that's at a half point. Let's say we have this, let's see, I want to do, let's say we do this point right over here."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "So let's do an interesting point. Let's say we have this point. Actually, no, that's at a half point. Let's say we have this, let's see, I want to do, let's say we do this point right over here. So that's x is equal to one and y is equal to six. And we see the way the differential equation is defined, it doesn't matter what our x is. It's really dependent on the y that's going to drive the slope."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "Let's say we have this, let's see, I want to do, let's say we do this point right over here. So that's x is equal to one and y is equal to six. And we see the way the differential equation is defined, it doesn't matter what our x is. It's really dependent on the y that's going to drive the slope. But we have six over six, which is one, times four minus six, which is negative two. So it's negative two. So we should have a slope of negative two."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "It's really dependent on the y that's going to drive the slope. But we have six over six, which is one, times four minus six, which is negative two. So it's negative two. So we should have a slope of negative two. And it looks like that's what they depicted. So as long as y is six, we should have a slope of negative two. Have a slope of negative two."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "So we should have a slope of negative two. And it looks like that's what they depicted. So as long as y is six, we should have a slope of negative two. Have a slope of negative two. And you see that in the slope field. So hopefully you feel pretty good that this is the slope field for this differential equation. If you don't, I encourage you to keep, keep verifying these points here."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "Have a slope of negative two. And you see that in the slope field. So hopefully you feel pretty good that this is the slope field for this differential equation. If you don't, I encourage you to keep, keep verifying these points here. But now let's actually use this slope field. Let's actually use this to visualize solutions to this differential equation based on points that the solution might go through. So let's say that we have a solution that goes through this point right over here."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "If you don't, I encourage you to keep, keep verifying these points here. But now let's actually use this slope field. Let's actually use this to visualize solutions to this differential equation based on points that the solution might go through. So let's say that we have a solution that goes through this point right over here. So what is that solution likely to look like? And once again, this is going to be a rough approximation. Well, right at that point, it's going to have a slope, just as the slope field shows."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "So let's say that we have a solution that goes through this point right over here. So what is that solution likely to look like? And once again, this is going to be a rough approximation. Well, right at that point, it's going to have a slope, just as the slope field shows. And as our y increases, it looks like our slope, it looks like our slope, so at this point, I should be, actually let me, let me undo that. So this, if I keep going up at this point, when y is equal to two, I should be parallel to all of these, these segments on the slope field at y is equal to two. And then it looks like the slope starts to decrease as we approach y is equal to four."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "Well, right at that point, it's going to have a slope, just as the slope field shows. And as our y increases, it looks like our slope, it looks like our slope, so at this point, I should be, actually let me, let me undo that. So this, if I keep going up at this point, when y is equal to two, I should be parallel to all of these, these segments on the slope field at y is equal to two. And then it looks like the slope starts to decrease as we approach y is equal to four. And so if I had a solution that went through this point, my guess is that it would look something, and then now the slope decreases again as we approach y is equal to zero. And of course, we see that because if when y equals zero, this whole thing is zero, so our derivative's going to be zero. So a reasonable solution might look something like this."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "And then it looks like the slope starts to decrease as we approach y is equal to four. And so if I had a solution that went through this point, my guess is that it would look something, and then now the slope decreases again as we approach y is equal to zero. And of course, we see that because if when y equals zero, this whole thing is zero, so our derivative's going to be zero. So a reasonable solution might look something like this. So this gives us a clue. Well, look, if a solution goes through this point, this right over here might, might be what it looks like. But what if it goes through, I don't know, what if it goes through this point right over here?"}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "So a reasonable solution might look something like this. So this gives us a clue. Well, look, if a solution goes through this point, this right over here might, might be what it looks like. But what if it goes through, I don't know, what if it goes through this point right over here? Well then, it might look like, it might look like this by the same exact logic. So it might look like this. So just like that, we're starting to get a sense."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "But what if it goes through, I don't know, what if it goes through this point right over here? Well then, it might look like, it might look like this by the same exact logic. So it might look like this. So just like that, we're starting to get a sense. We don't know the actual solution for this differential equation, but we're starting to get a sense of what, what type of functions, what type of functions or the class of functions that might satisfy the differential equation. But what's interesting about this slope field is it looks like there's some, you know, there's some interesting stuff, you know, if we have, if our solution includes points between, where the y value's between zero and four, it looks like we're going to have solutions like this, but what if we had y values that were larger than that or that were less than that or exactly, exactly zero or four? So for example, what if we had a solution that went through this point right over here?"}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "So just like that, we're starting to get a sense. We don't know the actual solution for this differential equation, but we're starting to get a sense of what, what type of functions, what type of functions or the class of functions that might satisfy the differential equation. But what's interesting about this slope field is it looks like there's some, you know, there's some interesting stuff, you know, if we have, if our solution includes points between, where the y value's between zero and four, it looks like we're going to have solutions like this, but what if we had y values that were larger than that or that were less than that or exactly, exactly zero or four? So for example, what if we had a solution that went through this point right over here? Well, at that point right over here, the slope field tells us that our slope is zero, so our y value's not going to change. And as long as our y value doesn't change, our y value's going to stay at four, so our slope is going to stay zero. So we actually already found, this is actually a solution to the differential equation."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "So for example, what if we had a solution that went through this point right over here? Well, at that point right over here, the slope field tells us that our slope is zero, so our y value's not going to change. And as long as our y value doesn't change, our y value's going to stay at four, so our slope is going to stay zero. So we actually already found, this is actually a solution to the differential equation. It's y is equal to four is a solution to this differential equation. So y is equal to four. y is equal to four."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "So we actually already found, this is actually a solution to the differential equation. It's y is equal to four is a solution to this differential equation. So y is equal to four. y is equal to four. And you can verify that that is a solution. When y is equal to four, this right-hand side is going to be zero, and the derivative is zero for y is equal to four. So that is a solution to the differential equation."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "y is equal to four. And you can verify that that is a solution. When y is equal to four, this right-hand side is going to be zero, and the derivative is zero for y is equal to four. So that is a solution to the differential equation. And the same thing for y is equal to zero. That is also a solution to the differential equation. Now, what if we included points, what if we included this point up here?"}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "So that is a solution to the differential equation. And the same thing for y is equal to zero. That is also a solution to the differential equation. Now, what if we included points, what if we included this point up here? And actually, let me do it in a different color so that you could see it. Let's say our solution included that point. Well, then it might look something, it might look something like this."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "Now, what if we included points, what if we included this point up here? And actually, let me do it in a different color so that you could see it. Let's say our solution included that point. Well, then it might look something, it might look something like this. And once again, I'm just using the slope field as a guide to give me an idea of what the slope might be as my curve progresses, as my solution progresses. So a solution that includes the point zero five might look something like this. And once again, it's just another clue."}, {"video_title": "Slope field to visualize solutions First order differential equations Khan Academy.mp3", "Sentence": "Well, then it might look something, it might look something like this. And once again, I'm just using the slope field as a guide to give me an idea of what the slope might be as my curve progresses, as my solution progresses. So a solution that includes the point zero five might look something like this. And once again, it's just another clue. A solution that includes the point zero negative one and a half might look something like, might look something like this. So anyway, hopefully this gives you a better appreciation for why slope fields are interesting. If you have a differential equation that just involves the first derivative and some x's and y's, this one only involves the first derivative and y's, we can plot a slope field like this, not too much trouble."}, {"video_title": "Inflection points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "If you were paying close attention in the last video, an interesting question might have popped up in your brain. We have talked about the intervals over which the function is concave downwards. And then we talked about the interval over which the function is concave upwards. But we see here that there's a point at which we transition from being concave downwards to concave upwards. Before that point, the slope was decreasing, and then the slope starts increasing. The slope was decreasing, and then the slope started increasing. So that's one way to look at it."}, {"video_title": "Inflection points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "But we see here that there's a point at which we transition from being concave downwards to concave upwards. Before that point, the slope was decreasing, and then the slope starts increasing. The slope was decreasing, and then the slope started increasing. So that's one way to look at it. Right here in our function, we go from being concave downwards to concave upwards. When you look at our derivative at that point, our derivative went from decreasing to increasing. And when you look at our second derivative at that point, it went from being negative to positive."}, {"video_title": "Inflection points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So that's one way to look at it. Right here in our function, we go from being concave downwards to concave upwards. When you look at our derivative at that point, our derivative went from decreasing to increasing. And when you look at our second derivative at that point, it went from being negative to positive. So this must have some type of a special name, you're probably thinking, and you'd be thinking correctly. This point at which we transition from being concave downwards to concave upwards, or the point at which our derivative has an extremum point, or the point at which our second derivative switches signs like this, we call it an inflection point. And the most typical way that people think about how could you test for an inflection point, it's a point, well, conceptually, it's where you go from being a downward opening u to an upward opening u, or when you go from being concave downwards to concave upwards."}, {"video_title": "Inflection points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And when you look at our second derivative at that point, it went from being negative to positive. So this must have some type of a special name, you're probably thinking, and you'd be thinking correctly. This point at which we transition from being concave downwards to concave upwards, or the point at which our derivative has an extremum point, or the point at which our second derivative switches signs like this, we call it an inflection point. And the most typical way that people think about how could you test for an inflection point, it's a point, well, conceptually, it's where you go from being a downward opening u to an upward opening u, or when you go from being concave downwards to concave upwards. But the easiest test is it's a point at which your second derivative switches signs. So in this case, we went from negative to positive, but we could have also switched from being positive to negative. So inflection point, your second derivative, f prime prime of x, switches signs, goes from being positive to negative or negative to positive, switches signs."}, {"video_title": "Inflection points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And the most typical way that people think about how could you test for an inflection point, it's a point, well, conceptually, it's where you go from being a downward opening u to an upward opening u, or when you go from being concave downwards to concave upwards. But the easiest test is it's a point at which your second derivative switches signs. So in this case, we went from negative to positive, but we could have also switched from being positive to negative. So inflection point, your second derivative, f prime prime of x, switches signs, goes from being positive to negative or negative to positive, switches signs. So this is a case where we went from concave downwards to concave upwards. If we went from concave upwards to concave downwards like that, then at this inflection point, up until that point, the slope was increasing. So the second derivative would be positive."}, {"video_title": "Inflection points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So inflection point, your second derivative, f prime prime of x, switches signs, goes from being positive to negative or negative to positive, switches signs. So this is a case where we went from concave downwards to concave upwards. If we went from concave upwards to concave downwards like that, then at this inflection point, up until that point, the slope was increasing. So the second derivative would be positive. And then the slope is decreasing. The derivative is decreasing. So your second derivative would be negative."}, {"video_title": "Inflection points introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So the second derivative would be positive. And then the slope is decreasing. The derivative is decreasing. So your second derivative would be negative. So here, your second derivative is going from positive to negative. Here, your second derivative is going from negative to positive. In either case, you are talking about an inflection point."}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum AP Calculus AB Khan Academy.mp3", "Sentence": "And what I want to do is I want to write it as the limit as n approaches infinity of a Riemann sum. So it's gonna take the form of the limit as n approaches infinity, and we can have our sigma notation right over here. And I would say from, let's say, i is equal to one all the way to n. Let me scroll down a little bit so it doesn't get all squenched up at the top. Of, and so let me draw what's actually going on so that we can get a better sense of what to write here within the sigma notation. So, let me do it large. So if this is pi right over here, pi right over here, this would be three pi over two, and this would be two pi right over here, two pi. Now what does the graph of cosine of x do?"}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum AP Calculus AB Khan Academy.mp3", "Sentence": "Of, and so let me draw what's actually going on so that we can get a better sense of what to write here within the sigma notation. So, let me do it large. So if this is pi right over here, pi right over here, this would be three pi over two, and this would be two pi right over here, two pi. Now what does the graph of cosine of x do? Well, at pi, cosine of pi is negative one. We'll assume that's negative one there. And cosine of two pi is one."}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum AP Calculus AB Khan Academy.mp3", "Sentence": "Now what does the graph of cosine of x do? Well, at pi, cosine of pi is negative one. We'll assume that's negative one there. And cosine of two pi is one. And so the graph is gonna do something like this. And this is obviously just a hand-drawn version of it. You have seen cosine functions before."}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum AP Calculus AB Khan Academy.mp3", "Sentence": "And cosine of two pi is one. And so the graph is gonna do something like this. And this is obviously just a hand-drawn version of it. You have seen cosine functions before. This is just part of it. And so this definite integral represents the area from pi to two pi between the curve and the x-axis. And you might already know that this area is going to be, or this part of the definite integral would be negative, and this would be positive and they'll cancel out, and this would all actually end up being zero in this case."}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum AP Calculus AB Khan Academy.mp3", "Sentence": "You have seen cosine functions before. This is just part of it. And so this definite integral represents the area from pi to two pi between the curve and the x-axis. And you might already know that this area is going to be, or this part of the definite integral would be negative, and this would be positive and they'll cancel out, and this would all actually end up being zero in this case. But this exercise for this video is to rewrite this as a limit as n approaches infinity of a Riemann sum. So as a Riemann sum, what we wanna do is think about breaking this up into a bunch of rectangles. So let's say, or I should say n rectangles."}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum AP Calculus AB Khan Academy.mp3", "Sentence": "And you might already know that this area is going to be, or this part of the definite integral would be negative, and this would be positive and they'll cancel out, and this would all actually end up being zero in this case. But this exercise for this video is to rewrite this as a limit as n approaches infinity of a Riemann sum. So as a Riemann sum, what we wanna do is think about breaking this up into a bunch of rectangles. So let's say, or I should say n rectangles. So that's our first one right over there. Then this might be our second one. And let's do right Riemann sum, where the right boundary of our rectangle, what the value of the function is at that point, that's what defines the height."}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say, or I should say n rectangles. So that's our first one right over there. Then this might be our second one. And let's do right Riemann sum, where the right boundary of our rectangle, what the value of the function is at that point, that's what defines the height. So that's our second one, all the way until this one right over here is going to be our nth one. So this is one, let me write it this way. This is i is equal to one, this is i is equal to two, all the way until we get to i is equal to n. And then if we take the limit as n approaches infinity, the sum of the areas of these rectangles are going to get better and better and better."}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum AP Calculus AB Khan Academy.mp3", "Sentence": "And let's do right Riemann sum, where the right boundary of our rectangle, what the value of the function is at that point, that's what defines the height. So that's our second one, all the way until this one right over here is going to be our nth one. So this is one, let me write it this way. This is i is equal to one, this is i is equal to two, all the way until we get to i is equal to n. And then if we take the limit as n approaches infinity, the sum of the areas of these rectangles are going to get better and better and better. And so let's first think about it. What is the width of each of these rectangles going to be? Well, I am taking this interval from pi to two pi, and I'm gonna divide it into n equal intervals."}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum AP Calculus AB Khan Academy.mp3", "Sentence": "This is i is equal to one, this is i is equal to two, all the way until we get to i is equal to n. And then if we take the limit as n approaches infinity, the sum of the areas of these rectangles are going to get better and better and better. And so let's first think about it. What is the width of each of these rectangles going to be? Well, I am taking this interval from pi to two pi, and I'm gonna divide it into n equal intervals. So the width of each of these, the width of each of these, is going to be two pi minus pi. So I'm just taking the difference between my bounds of integration, and I am dividing by n, which is equal to pi over n. So that's the width of each of these, that's pi over n. This is pi over n, this is pi over n. And what's the height of each of these rectangles? And remember, this is a right Riemann sum, so it's going to be the right end of our rectangle is going to define the height."}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum AP Calculus AB Khan Academy.mp3", "Sentence": "Well, I am taking this interval from pi to two pi, and I'm gonna divide it into n equal intervals. So the width of each of these, the width of each of these, is going to be two pi minus pi. So I'm just taking the difference between my bounds of integration, and I am dividing by n, which is equal to pi over n. So that's the width of each of these, that's pi over n. This is pi over n, this is pi over n. And what's the height of each of these rectangles? And remember, this is a right Riemann sum, so it's going to be the right end of our rectangle is going to define the height. So this right over here, what would this height be? Well, this height, this value, I should say, this is going to be equal to f of what? Well, this was pi, and this is going to be pi plus the length of our interval right over here, the base of the rectangle."}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum AP Calculus AB Khan Academy.mp3", "Sentence": "And remember, this is a right Riemann sum, so it's going to be the right end of our rectangle is going to define the height. So this right over here, what would this height be? Well, this height, this value, I should say, this is going to be equal to f of what? Well, this was pi, and this is going to be pi plus the length of our interval right over here, the base of the rectangle. So we started at pi, so it's going to be pi plus, this one's going to be pi over n, I could say, times one. That's this height right over here. What's this one going to be right over here?"}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum AP Calculus AB Khan Academy.mp3", "Sentence": "Well, this was pi, and this is going to be pi plus the length of our interval right over here, the base of the rectangle. So we started at pi, so it's going to be pi plus, this one's going to be pi over n, I could say, times one. That's this height right over here. What's this one going to be right over here? Well, this one is going to be f of pi, our first start, plus pi over n times what? We're going to add pi over n two times. Pi over n times two."}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum AP Calculus AB Khan Academy.mp3", "Sentence": "What's this one going to be right over here? Well, this one is going to be f of pi, our first start, plus pi over n times what? We're going to add pi over n two times. Pi over n times two. So the general form of the right boundary is going to be, so for example, this height right over here, this is going to be f of, we started at pi, plus, we're doing the right Riemann sum, so we're gonna add pi over n n times by this point. Pi over n times n. Or, if we wanted to say it generally, if we're talking about the ith rectangle, remember, we're gonna sum them all up, what's their height? Well, the height is going to be, in this case, it's going to be cosine of pi plus, if we're with the ith rectangle, we are going to add pi over n i times."}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum AP Calculus AB Khan Academy.mp3", "Sentence": "Pi over n times two. So the general form of the right boundary is going to be, so for example, this height right over here, this is going to be f of, we started at pi, plus, we're doing the right Riemann sum, so we're gonna add pi over n n times by this point. Pi over n times n. Or, if we wanted to say it generally, if we're talking about the ith rectangle, remember, we're gonna sum them all up, what's their height? Well, the height is going to be, in this case, it's going to be cosine of pi plus, if we're with the ith rectangle, we are going to add pi over n i times. Pi over n times i. And then, that's the height of each of our rectangles. And then what's the width?"}, {"video_title": "Worked example Rewriting definite integral as limit of Riemann sum AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the height is going to be, in this case, it's going to be cosine of pi plus, if we're with the ith rectangle, we are going to add pi over n i times. Pi over n times i. And then, that's the height of each of our rectangles. And then what's the width? Well, we already figured that out. Times pi over n. And if you wanna be careful and make sure that this sigma notation applies to the whole thing, there you have it. We have just re-expressed this definite integral as the limit of a right Riemann sum."}, {"video_title": "Derivatives of inverse functions Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "If the notion of an inverse function is completely unfamiliar to you, I encourage you to review inverse functions on Khan Academy. Now one of the properties of inverse functions are that if I were to take g of f of x, g of f of x, or I could say the f inverse of f of x, that this is just going to be equal to x. And it comes straight out of what an inverse of a function is. If this is x right over here, the function f would map to some value f of x, so that's f of x right over there, and then the function g, or f inverse, if you input f of x into it, it would take you back, it would take you back to x. So that would be f inverse, or we're saying g is the same thing as f inverse. So all of that so far is a review of inverse functions, but now we're going to apply a little bit of calculus to it using the chain rule. And we're gonna get a pretty interesting result."}, {"video_title": "Derivatives of inverse functions Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "If this is x right over here, the function f would map to some value f of x, so that's f of x right over there, and then the function g, or f inverse, if you input f of x into it, it would take you back, it would take you back to x. So that would be f inverse, or we're saying g is the same thing as f inverse. So all of that so far is a review of inverse functions, but now we're going to apply a little bit of calculus to it using the chain rule. And we're gonna get a pretty interesting result. What I wanna do is take the derivative of both sides of this equation right over here. So let's apply the derivative operator, d dx on the left-hand side, d dx on the right-hand side, and what are we going to get? Well, on the left-hand side, we would apply the chain rule."}, {"video_title": "Derivatives of inverse functions Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And we're gonna get a pretty interesting result. What I wanna do is take the derivative of both sides of this equation right over here. So let's apply the derivative operator, d dx on the left-hand side, d dx on the right-hand side, and what are we going to get? Well, on the left-hand side, we would apply the chain rule. So this is going to be the derivative of g with respect to f of x, so that's going to be g prime of f of x, g prime of f of x, times the derivative of f of x with respect to x, so times f prime of x, and then that is going to be equal to what? Well, the derivative with respect to x of x, that's just equal to one. And this is where we get our interesting result."}, {"video_title": "Derivatives of inverse functions Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, on the left-hand side, we would apply the chain rule. So this is going to be the derivative of g with respect to f of x, so that's going to be g prime of f of x, g prime of f of x, times the derivative of f of x with respect to x, so times f prime of x, and then that is going to be equal to what? Well, the derivative with respect to x of x, that's just equal to one. And this is where we get our interesting result. All we did so far is we used something we knew about inverse functions, and we used the chain rule to take the derivative of the left-hand side, but if you divide both sides by g prime of f of x, what are you going to get? You're going to get a relationship between the derivative of a function and the derivative of its inverse. So you get f prime of x is going to be equal to one over all of this business, one over g prime of f of x, g prime of f of x."}, {"video_title": "Derivatives of inverse functions Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And this is where we get our interesting result. All we did so far is we used something we knew about inverse functions, and we used the chain rule to take the derivative of the left-hand side, but if you divide both sides by g prime of f of x, what are you going to get? You're going to get a relationship between the derivative of a function and the derivative of its inverse. So you get f prime of x is going to be equal to one over all of this business, one over g prime of f of x, g prime of f of x. And this is really neat, because if you know something about the derivative of a function, you can then start to figure out things about the derivative of its inverse. And we can actually see this is true with some classic functions. So let's say that f of x is equal to e to the x, and so g of x would be equal to the inverse of f, so f inverse, which is, what's the inverse of e to the x?"}, {"video_title": "Derivatives of inverse functions Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So you get f prime of x is going to be equal to one over all of this business, one over g prime of f of x, g prime of f of x. And this is really neat, because if you know something about the derivative of a function, you can then start to figure out things about the derivative of its inverse. And we can actually see this is true with some classic functions. So let's say that f of x is equal to e to the x, and so g of x would be equal to the inverse of f, so f inverse, which is, what's the inverse of e to the x? Well, one way to think about it is if you have y is equal to e to the x, if you want the inverse, you can swap the variables and then solve for y again. So you'd get x is equal to e to the y. You take the natural log of both sides, you get natural log of x is equal to y."}, {"video_title": "Derivatives of inverse functions Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say that f of x is equal to e to the x, and so g of x would be equal to the inverse of f, so f inverse, which is, what's the inverse of e to the x? Well, one way to think about it is if you have y is equal to e to the x, if you want the inverse, you can swap the variables and then solve for y again. So you'd get x is equal to e to the y. You take the natural log of both sides, you get natural log of x is equal to y. So the inverse of e to the x is natural log of x. And once again, that's all review of inverse functions. If that's unfamiliar, review it on Khan Academy."}, {"video_title": "Derivatives of inverse functions Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "You take the natural log of both sides, you get natural log of x is equal to y. So the inverse of e to the x is natural log of x. And once again, that's all review of inverse functions. If that's unfamiliar, review it on Khan Academy. So g of x is going to be equal to the natural log of x. Now, let's see if this holds true for these two functions. Well, what is f prime of x going to be?"}, {"video_title": "Derivatives of inverse functions Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "If that's unfamiliar, review it on Khan Academy. So g of x is going to be equal to the natural log of x. Now, let's see if this holds true for these two functions. Well, what is f prime of x going to be? Well, this is one of those amazing results in calculus, one of these neat things about the number e is that the derivative of e to the x is e to the x. And in other videos, we also saw that the derivative of the natural log of x is one over x. So let's see if this holds out."}, {"video_title": "Derivatives of inverse functions Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, what is f prime of x going to be? Well, this is one of those amazing results in calculus, one of these neat things about the number e is that the derivative of e to the x is e to the x. And in other videos, we also saw that the derivative of the natural log of x is one over x. So let's see if this holds out. So we should get a result, f prime of x, e to the x, should be equal to one over g prime of f of x. So g prime of f of x, so g prime is one over our f of x, and f of x is e to the x, one over e to the x. Is this indeed true?"}, {"video_title": "Derivatives of inverse functions Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let's see if this holds out. So we should get a result, f prime of x, e to the x, should be equal to one over g prime of f of x. So g prime of f of x, so g prime is one over our f of x, and f of x is e to the x, one over e to the x. Is this indeed true? Yes, it is. One over one over e to the x is just going to be e to the x. So it all checks out."}, {"video_title": "Derivatives of inverse functions Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Is this indeed true? Yes, it is. One over one over e to the x is just going to be e to the x. So it all checks out. And you could do it the other way because these are inverses of each other. You could say g prime of x is going to be equal to one over f prime of g of x because they're inverses of each other. And actually, what's really neat about this is that you could actually use this to get a sense of what the derivative of an inverse function is even going to be."}, {"video_title": "Partial sums intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "Now what I want to introduce to you is the idea of a partial sum. This right over here is an infinite series. But we could define a partial sum, so if we say S sub six, this notation says, okay, if S is an infinite series, S sub six is the partial sum of the first six terms. So in this case, this is going to be, we're not gonna just keep going on forever, this is going to be a sub one plus a sub two plus a sub three plus a sub four plus a sub five plus a sub six. And I could make this a little bit more tangible if you like. So let's say that S, the infinite series S, is equal to the sum from n equals one to infinity of one over n squared. In this case, it would be, let's see, one over one squared plus one over two squared plus one over three squared, and we would just keep going on and on and on forever."}, {"video_title": "Partial sums intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "So in this case, this is going to be, we're not gonna just keep going on forever, this is going to be a sub one plus a sub two plus a sub three plus a sub four plus a sub five plus a sub six. And I could make this a little bit more tangible if you like. So let's say that S, the infinite series S, is equal to the sum from n equals one to infinity of one over n squared. In this case, it would be, let's see, one over one squared plus one over two squared plus one over three squared, and we would just keep going on and on and on forever. But what would S sub, actually let me do that same color. What would S, I said I would change color and I didn't. What would S sub three be equal to?"}, {"video_title": "Partial sums intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "In this case, it would be, let's see, one over one squared plus one over two squared plus one over three squared, and we would just keep going on and on and on forever. But what would S sub, actually let me do that same color. What would S, I said I would change color and I didn't. What would S sub three be equal to? The partial sum of the first three terms. And I encourage you to pause the video and try to work through it on your own. Well, it's just going to be, this is going to be the first term, one, plus the second term, one fourth, plus the third term, one ninth."}, {"video_title": "Partial sums intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "What would S sub three be equal to? The partial sum of the first three terms. And I encourage you to pause the video and try to work through it on your own. Well, it's just going to be, this is going to be the first term, one, plus the second term, one fourth, plus the third term, one ninth. It's going to be the sum of the first three terms. And we could figure that out. Let's just see if we have a common denominator here."}, {"video_title": "Partial sums intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, it's just going to be, this is going to be the first term, one, plus the second term, one fourth, plus the third term, one ninth. It's going to be the sum of the first three terms. And we could figure that out. Let's just see if we have a common denominator here. It's going to be 36, it's going to be 36 36ths plus nine 36ths plus four 36ths. So this is going to be 49 over 36. 49, 49 over 36."}, {"video_title": "Partial sums intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let's just see if we have a common denominator here. It's going to be 36, it's going to be 36 36ths plus nine 36ths plus four 36ths. So this is going to be 49 over 36. 49, 49 over 36. So the whole point of this video is just to appreciate this idea of a partial sum. And what we'll see is that you can actually express what a partial sum might be algebraically. So for example, for example, let's give ourselves a little bit more real estate here."}, {"video_title": "Partial sums intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "49, 49 over 36. So the whole point of this video is just to appreciate this idea of a partial sum. And what we'll see is that you can actually express what a partial sum might be algebraically. So for example, for example, let's give ourselves a little bit more real estate here. Let's say, let's go back to just saying we have an infinite series S that is equal to the sum from n equals one to infinity of a sub n. And let's say we know the partial sum, the partial sum S sub n, so the sum of the first n terms of this is equal to n squared minus three over, over n to the third plus four. So just as a bit of a reminder of what this is saying, S sub n, S sub n is the same thing as a sub one plus a sub two plus you keep going all the way to a sub n and that's going to be equal to this business, n squared minus three over n to the third plus four. Now given that, if someone were to walk up to you on the street and said, okay, now that you know the notation for a partial sum, I have a little question to ask of you."}, {"video_title": "Partial sums intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "So for example, for example, let's give ourselves a little bit more real estate here. Let's say, let's go back to just saying we have an infinite series S that is equal to the sum from n equals one to infinity of a sub n. And let's say we know the partial sum, the partial sum S sub n, so the sum of the first n terms of this is equal to n squared minus three over, over n to the third plus four. So just as a bit of a reminder of what this is saying, S sub n, S sub n is the same thing as a sub one plus a sub two plus you keep going all the way to a sub n and that's going to be equal to this business, n squared minus three over n to the third plus four. Now given that, if someone were to walk up to you on the street and said, okay, now that you know the notation for a partial sum, I have a little question to ask of you. If this is, if S is the infinite series, and I'm writing it in very general terms right over here, so S is the infinite series from n equals one to infinity of a sub n, and the partial sum S sub n is defined this way, so someone, they tell you these two things, and then they say, find, find what the sum from n equals one to six of a sub n is, and I encourage you to pause the video and try to figure it out. Well, this is just going to be, this is going to be a sub one plus a sub two plus a sub three plus a sub four, and when I say sub, that just means subscript, plus a sub five plus a sub six. Well, that's just the same thing as the partial sum."}, {"video_title": "Partial sums intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "Now given that, if someone were to walk up to you on the street and said, okay, now that you know the notation for a partial sum, I have a little question to ask of you. If this is, if S is the infinite series, and I'm writing it in very general terms right over here, so S is the infinite series from n equals one to infinity of a sub n, and the partial sum S sub n is defined this way, so someone, they tell you these two things, and then they say, find, find what the sum from n equals one to six of a sub n is, and I encourage you to pause the video and try to figure it out. Well, this is just going to be, this is going to be a sub one plus a sub two plus a sub three plus a sub four, and when I say sub, that just means subscript, plus a sub five plus a sub six. Well, that's just the same thing as the partial sum. This is just the same thing as the partial sum of the first six terms for our infinite series. This is going to be the partial sum S sub six, and we know how to algebraically evaluate what S sub six is. We can apply this formula that we were given."}, {"video_title": "Partial sums intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, that's just the same thing as the partial sum. This is just the same thing as the partial sum of the first six terms for our infinite series. This is going to be the partial sum S sub six, and we know how to algebraically evaluate what S sub six is. We can apply this formula that we were given. S sub six is equal to, well, everywhere we see an n, we replace with a six. It's going to be six squared minus three over six to the third plus four, and so what is this going to be? Six squared is 36 minus three, so that's 33, and six to the third, let's see, 36 times six, I always forget."}, {"video_title": "Partial sums intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "We can apply this formula that we were given. S sub six is equal to, well, everywhere we see an n, we replace with a six. It's going to be six squared minus three over six to the third plus four, and so what is this going to be? Six squared is 36 minus three, so that's 33, and six to the third, let's see, 36 times six, I always forget. My brain wants to say 216, but let me make sure that that's actually the case. Six times 30 is 180 plus 36. Yes, it is 216, so I guess I have inadvertently, by seeing six to the third so many times in my life, I have inadvertently memorized six to the third power."}, {"video_title": "Partial sums intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "Six squared is 36 minus three, so that's 33, and six to the third, let's see, 36 times six, I always forget. My brain wants to say 216, but let me make sure that that's actually the case. Six times 30 is 180 plus 36. Yes, it is 216, so I guess I have inadvertently, by seeing six to the third so many times in my life, I have inadvertently memorized six to the third power. Never a horrible thing to have that in your brain. This is going to be 216 plus four, so 200, 220. S sub six, or the sum of the first six terms of this series right over here, is 33,200th, and we're done."}, {"video_title": "Volume with cross sections intro Applications of integration AP Calculus AB Khan Academy.mp3", "Sentence": "For example, we could find this yellow area using a definite integral. But what we're going to do in this video is do something even more interesting. We're gonna find the volume of shapes where the base is defined in some way by the area between two curves. And in this video, we're gonna think about a shape, and I'm gonna draw it in three dimensions. So let me draw this over again, but with a little bit of perspective. So let's make this the y-axis. So that's the y-axis."}, {"video_title": "Volume with cross sections intro Applications of integration AP Calculus AB Khan Academy.mp3", "Sentence": "And in this video, we're gonna think about a shape, and I'm gonna draw it in three dimensions. So let me draw this over again, but with a little bit of perspective. So let's make this the y-axis. So that's the y-axis. This is my x-axis. That is my x-axis. This is the line y is equal to six, right over there."}, {"video_title": "Volume with cross sections intro Applications of integration AP Calculus AB Khan Academy.mp3", "Sentence": "So that's the y-axis. This is my x-axis. That is my x-axis. This is the line y is equal to six, right over there. Y is equal to six. This dotted line, we could just draw it like this. And so this would be the point x equals two."}, {"video_title": "Volume with cross sections intro Applications of integration AP Calculus AB Khan Academy.mp3", "Sentence": "This is the line y is equal to six, right over there. Y is equal to six. This dotted line, we could just draw it like this. And so this would be the point x equals two. And then the graph of y is equal to four times the natural log of three minus x would look something like this. Look something like this. And so this region is this region."}, {"video_title": "Volume with cross sections intro Applications of integration AP Calculus AB Khan Academy.mp3", "Sentence": "And so this would be the point x equals two. And then the graph of y is equal to four times the natural log of three minus x would look something like this. Look something like this. And so this region is this region. But it's going to be the base of a three-dimensional shape where any cross-section, if I were to take a cross-section right over here, is going to be a square. So whatever this length is, we also go that much high. And so the cross-section is a square right over there."}, {"video_title": "Volume with cross sections intro Applications of integration AP Calculus AB Khan Academy.mp3", "Sentence": "And so this region is this region. But it's going to be the base of a three-dimensional shape where any cross-section, if I were to take a cross-section right over here, is going to be a square. So whatever this length is, we also go that much high. And so the cross-section is a square right over there. The cross-section right over here is going to be a square, whatever the difference between these two functions is, that's also how high we are going to go. This length, which is six at this point, this is also going to be the height. It is going to be a square."}, {"video_title": "Volume with cross sections intro Applications of integration AP Calculus AB Khan Academy.mp3", "Sentence": "And so the cross-section is a square right over there. The cross-section right over here is going to be a square, whatever the difference between these two functions is, that's also how high we are going to go. This length, which is six at this point, this is also going to be the height. It is going to be a square. It's going to be quite big. I'm gonna have to scroll down so we can draw the whole thing roughly at the right proportion. So it'll look something like this."}, {"video_title": "Volume with cross sections intro Applications of integration AP Calculus AB Khan Academy.mp3", "Sentence": "It is going to be a square. It's going to be quite big. I'm gonna have to scroll down so we can draw the whole thing roughly at the right proportion. So it'll look something like this. This should be a square. It's gonna look something like this. And so the whole shape would look something like this."}, {"video_title": "Volume with cross sections intro Applications of integration AP Calculus AB Khan Academy.mp3", "Sentence": "So it'll look something like this. This should be a square. It's gonna look something like this. And so the whole shape would look something like this. Would look something like that. Try to shade that in a little bit so that you can appreciate it a little bit more. But hopefully you get the idea."}, {"video_title": "Volume with cross sections intro Applications of integration AP Calculus AB Khan Academy.mp3", "Sentence": "And so the whole shape would look something like this. Would look something like that. Try to shade that in a little bit so that you can appreciate it a little bit more. But hopefully you get the idea. And some of you might be excited and some of you might be a little intimidated. Well, hey, I've been dealing with the two dimensions for so long. What's going on with these three dimensions?"}, {"video_title": "Volume with cross sections intro Applications of integration AP Calculus AB Khan Academy.mp3", "Sentence": "But hopefully you get the idea. And some of you might be excited and some of you might be a little intimidated. Well, hey, I've been dealing with the two dimensions for so long. What's going on with these three dimensions? But you'll quickly appreciate that you already have the powers of integration to solve this and to do that, we just have to break up the shape into a bunch of these, you could view them as these little square tiles that have some depth to them. So let's make that into a little tile, this one into it that also has some depth to it. You could even, I could draw it multiple places."}, {"video_title": "Volume with cross sections intro Applications of integration AP Calculus AB Khan Academy.mp3", "Sentence": "What's going on with these three dimensions? But you'll quickly appreciate that you already have the powers of integration to solve this and to do that, we just have to break up the shape into a bunch of these, you could view them as these little square tiles that have some depth to them. So let's make that into a little tile, this one into it that also has some depth to it. You could even, I could draw it multiple places. You could view it as a, break it up into these things that have a very small depth that we could call dx. And we know how to figure out what their volume is. What would be the volume of one of these things?"}, {"video_title": "Volume with cross sections intro Applications of integration AP Calculus AB Khan Academy.mp3", "Sentence": "You could even, I could draw it multiple places. You could view it as a, break it up into these things that have a very small depth that we could call dx. And we know how to figure out what their volume is. What would be the volume of one of these things? Well, it would be the depth times the area, times the surface area of this cross section right over here. Let me do that in a different color. So what would be the area that I am shading in in pink right over here?"}, {"video_title": "Volume with cross sections intro Applications of integration AP Calculus AB Khan Academy.mp3", "Sentence": "What would be the volume of one of these things? Well, it would be the depth times the area, times the surface area of this cross section right over here. Let me do that in a different color. So what would be the area that I am shading in in pink right over here? Well, that area is going to be the base length squared. What's the base length? What's the difference between these two functions?"}, {"video_title": "Volume with cross sections intro Applications of integration AP Calculus AB Khan Academy.mp3", "Sentence": "So what would be the area that I am shading in in pink right over here? Well, that area is going to be the base length squared. What's the base length? What's the difference between these two functions? It is going to be six minus, our bottom function is four times the natural log of three minus x. And so that would just give us that length. But if we square it, we get this entire area."}, {"video_title": "Volume with cross sections intro Applications of integration AP Calculus AB Khan Academy.mp3", "Sentence": "What's the difference between these two functions? It is going to be six minus, our bottom function is four times the natural log of three minus x. And so that would just give us that length. But if we square it, we get this entire area. We get that entire area, you square it. And then you multiply it times the depth. You multiply it times the depth."}, {"video_title": "Volume with cross sections intro Applications of integration AP Calculus AB Khan Academy.mp3", "Sentence": "But if we square it, we get this entire area. We get that entire area, you square it. And then you multiply it times the depth. You multiply it times the depth. Now you have the volume of just this little section right over here. And I think you might see where this is going. Now what if you were to add up all of these from x equals zero to x equals two?"}, {"video_title": "Volume with cross sections intro Applications of integration AP Calculus AB Khan Academy.mp3", "Sentence": "You multiply it times the depth. Now you have the volume of just this little section right over here. And I think you might see where this is going. Now what if you were to add up all of these from x equals zero to x equals two? Well, then you would have the volume of the entire thing. This is the power of the definite integral. So we could just integrate from x equals zero to x equals two from x equals zero to x equals two."}, {"video_title": "Volume with cross sections intro Applications of integration AP Calculus AB Khan Academy.mp3", "Sentence": "Now what if you were to add up all of these from x equals zero to x equals two? Well, then you would have the volume of the entire thing. This is the power of the definite integral. So we could just integrate from x equals zero to x equals two from x equals zero to x equals two. If you drew where these intersect our base, you would say, all right, this thing right over here would be this thing right over here, where it's dx. Instead of just multiplying dx times the difference between these functions, we're going to square the difference of these functions because we're visualizing this three-dimensional shape, the surface area of this three-dimensional shape as opposed to just the height of this little rectangle. And if you were to evaluate this integral, you would indeed get the volume of this kind of pedestal horn-looking thing."}, {"video_title": "Volume with cross sections intro Applications of integration AP Calculus AB Khan Academy.mp3", "Sentence": "So we could just integrate from x equals zero to x equals two from x equals zero to x equals two. If you drew where these intersect our base, you would say, all right, this thing right over here would be this thing right over here, where it's dx. Instead of just multiplying dx times the difference between these functions, we're going to square the difference of these functions because we're visualizing this three-dimensional shape, the surface area of this three-dimensional shape as opposed to just the height of this little rectangle. And if you were to evaluate this integral, you would indeed get the volume of this kind of pedestal horn-looking thing. This is not an easy definite integral to evaluate by hand, but we can actually use a calculator for that. And so we can hit Math and then hit choice number nine for definite integral. And then we just have to input everything."}, {"video_title": "Volume with cross sections intro Applications of integration AP Calculus AB Khan Academy.mp3", "Sentence": "And if you were to evaluate this integral, you would indeed get the volume of this kind of pedestal horn-looking thing. This is not an easy definite integral to evaluate by hand, but we can actually use a calculator for that. And so we can hit Math and then hit choice number nine for definite integral. And then we just have to input everything. We're going from zero till two of, and then we have, let me open parentheses because I'm gonna have to square everything, six minus four times the natural log of x, or actually the natural log of three minus x. And so let me close the parentheses on the natural log part. And then if I close the parentheses on this whole thing, I want to then square it."}, {"video_title": "Volume with cross sections intro Applications of integration AP Calculus AB Khan Academy.mp3", "Sentence": "And then we just have to input everything. We're going from zero till two of, and then we have, let me open parentheses because I'm gonna have to square everything, six minus four times the natural log of x, or actually the natural log of three minus x. And so let me close the parentheses on the natural log part. And then if I close the parentheses on this whole thing, I want to then square it. And then I'm integrating with respect to x, Enter. I got approximately 26.27. So approximately 26.27."}, {"video_title": "Tangents of polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "What I want to do in this video is think about what is the equation of the tangent line when x is equal to one? So we can visualize that. So this is x equaling one right over here. This is the value of the function when x is equal to one, right over there. And then the tangent line looks something like, will look something like, oh no, I can do a better job than that. It's going to look something like that. And what we want to do is find the equation, the equation of that line."}, {"video_title": "Tangents of polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "This is the value of the function when x is equal to one, right over there. And then the tangent line looks something like, will look something like, oh no, I can do a better job than that. It's going to look something like that. And what we want to do is find the equation, the equation of that line. And if you are inspired, I encourage you to be, pause the video and try to work it out. Well, the way that we could do this is if we find the derivative at x equals one, the derivative is the slope of the tangent line. And so we'll know the slope of the tangent line, and we know that it contains that point, and then we can use that to find the equation of the tangent line."}, {"video_title": "Tangents of polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And what we want to do is find the equation, the equation of that line. And if you are inspired, I encourage you to be, pause the video and try to work it out. Well, the way that we could do this is if we find the derivative at x equals one, the derivative is the slope of the tangent line. And so we'll know the slope of the tangent line, and we know that it contains that point, and then we can use that to find the equation of the tangent line. So let's actually just, let's just, so we want the equation of the tangent line when x is equal to one. So let's just first of all evaluate f of one. So f of one is equal to one to the third power, which is one, minus six times one squared, so it's just minus six, and then plus one, plus one, minus five."}, {"video_title": "Tangents of polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And so we'll know the slope of the tangent line, and we know that it contains that point, and then we can use that to find the equation of the tangent line. So let's actually just, let's just, so we want the equation of the tangent line when x is equal to one. So let's just first of all evaluate f of one. So f of one is equal to one to the third power, which is one, minus six times one squared, so it's just minus six, and then plus one, plus one, minus five. So this is equal to what? Two minus 11, which is equal to negative nine. And that looks about right."}, {"video_title": "Tangents of polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So f of one is equal to one to the third power, which is one, minus six times one squared, so it's just minus six, and then plus one, plus one, minus five. So this is equal to what? Two minus 11, which is equal to negative nine. And that looks about right. That looks like about negative nine right over there. The scales are different on the y and the x-axis. And so that is f of one."}, {"video_title": "Tangents of polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And that looks about right. That looks like about negative nine right over there. The scales are different on the y and the x-axis. And so that is f of one. It is negative nine. Did I do that right? This is negative five, negative nine, yep, negative nine."}, {"video_title": "Tangents of polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And so that is f of one. It is negative nine. Did I do that right? This is negative five, negative nine, yep, negative nine. And now let's evaluate what the derivative is at one. So what is f prime of x? f prime of x."}, {"video_title": "Tangents of polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "This is negative five, negative nine, yep, negative nine. And now let's evaluate what the derivative is at one. So what is f prime of x? f prime of x. Well here it's just a polynomial. Take the derivative of x to the third. Well we apply the power rule."}, {"video_title": "Tangents of polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "f prime of x. Well here it's just a polynomial. Take the derivative of x to the third. Well we apply the power rule. We bring the three out front. So you get three x to the, and then we go one less than three to get the second power. And then you have minus six x squared."}, {"video_title": "Tangents of polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well we apply the power rule. We bring the three out front. So you get three x to the, and then we go one less than three to get the second power. And then you have minus six x squared. So you bring the two times the six to get 12. So minus 12 x to the, well two minus one is one power, so that's the same thing as 12x. And then plus the derivative of x is just one."}, {"video_title": "Tangents of polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And then you have minus six x squared. So you bring the two times the six to get 12. So minus 12 x to the, well two minus one is one power, so that's the same thing as 12x. And then plus the derivative of x is just one. That's just going to be one. And if you view this as x to the first power, we're just bringing the one out front and decrementing the one. So we have one times x to the zero power, which is just one."}, {"video_title": "Tangents of polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And then plus the derivative of x is just one. That's just going to be one. And if you view this as x to the first power, we're just bringing the one out front and decrementing the one. So we have one times x to the zero power, which is just one. And then the derivative of a constant here is just going to be zero. So this is our derivative of f, and if we want to evaluate it at one, f prime of one is going to be three times one squared, which is just three, minus 12 times one, so it's just minus 12, and then we have plus one. So this is three minus 12 is negative nine, and negative nine plus one is equal to negative eight."}, {"video_title": "Tangents of polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So we have one times x to the zero power, which is just one. And then the derivative of a constant here is just going to be zero. So this is our derivative of f, and if we want to evaluate it at one, f prime of one is going to be three times one squared, which is just three, minus 12 times one, so it's just minus 12, and then we have plus one. So this is three minus 12 is negative nine, and negative nine plus one is equal to negative eight. So we know the slope right over here is a slope of negative eight. We know a point on that line, it contains the point one comma negative nine, so we could use that information to find the equation of the line. The line, just to remind ourselves, has the form y is equal to mx plus b, where m is the slope, so we know that y is going to be equal to negative 8x plus b and now we can substitute the x and y value that we know sits on that line to solve for b."}, {"video_title": "Tangents of polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So this is three minus 12 is negative nine, and negative nine plus one is equal to negative eight. So we know the slope right over here is a slope of negative eight. We know a point on that line, it contains the point one comma negative nine, so we could use that information to find the equation of the line. The line, just to remind ourselves, has the form y is equal to mx plus b, where m is the slope, so we know that y is going to be equal to negative 8x plus b and now we can substitute the x and y value that we know sits on that line to solve for b. So we know that y is equal to negative nine, let me just write this here, y is equal to negative nine when x is equal to one. And so we get negative nine is equal to negative eight times one, so negative eight plus b. Well, let's see, we could add eight to both sides and we get negative one is equal to b."}, {"video_title": "Derivative of inverse tangent Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "We already know that the derivative with respect to x of tangent of x is equal to the secant of x squared, which is of course the same thing as 1 over cosine of x squared. Now what we want to do in this video, like we've done in the last few videos, is figure out what the derivative of the inverse function of the tangent of x is. In particular, let's see if we can figure out what the derivative with respect to x of the inverse tangent of x is. I encourage you to pause this video and use a technique similar to the one, or very close to the one, that we've used in the last few videos to figure out what this is. Let's set y equal to the inverse tangent of x. y is equal to inverse tangent of x. That is the same thing as saying that the tangent of y is equal to x. All I've done is, you can kind of think of it as, I've just taken the tangent of both sides right over here."}, {"video_title": "Derivative of inverse tangent Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "I encourage you to pause this video and use a technique similar to the one, or very close to the one, that we've used in the last few videos to figure out what this is. Let's set y equal to the inverse tangent of x. y is equal to inverse tangent of x. That is the same thing as saying that the tangent of y is equal to x. All I've done is, you can kind of think of it as, I've just taken the tangent of both sides right over here. Now we can take the derivative of both sides with respect to x. On the left-hand side, we can just apply the chain rule. Derivative of tangent of y with respect to y is going to be secant squared of y, which is the same thing as 1 over cosine of y squared."}, {"video_title": "Derivative of inverse tangent Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "All I've done is, you can kind of think of it as, I've just taken the tangent of both sides right over here. Now we can take the derivative of both sides with respect to x. On the left-hand side, we can just apply the chain rule. Derivative of tangent of y with respect to y is going to be secant squared of y, which is the same thing as 1 over cosine of y squared. I like to write it this way. It keeps it a little bit simpler, at least in my brain. When we're applying the chain rule, it's going to be the derivative of tangent of y with respect to y times the derivative of y with respect to x."}, {"video_title": "Derivative of inverse tangent Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Derivative of tangent of y with respect to y is going to be secant squared of y, which is the same thing as 1 over cosine of y squared. I like to write it this way. It keeps it a little bit simpler, at least in my brain. When we're applying the chain rule, it's going to be the derivative of tangent of y with respect to y times the derivative of y with respect to x. On the right-hand side, the derivative of x with respect to x is going to be equal to 1. If we want to solve for the derivative of y with respect to x, we just multiply both sides times the cosine of y squared. We get the derivative of y with respect to x is equal to cosine of y squared."}, {"video_title": "Derivative of inverse tangent Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "When we're applying the chain rule, it's going to be the derivative of tangent of y with respect to y times the derivative of y with respect to x. On the right-hand side, the derivative of x with respect to x is going to be equal to 1. If we want to solve for the derivative of y with respect to x, we just multiply both sides times the cosine of y squared. We get the derivative of y with respect to x is equal to cosine of y squared. Like we've seen in previous videos, this isn't that satisfying. I've written the derivative of y with respect to x as a function of y. What we're really interested in is writing it as a function of x."}, {"video_title": "Derivative of inverse tangent Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "We get the derivative of y with respect to x is equal to cosine of y squared. Like we've seen in previous videos, this isn't that satisfying. I've written the derivative of y with respect to x as a function of y. What we're really interested in is writing it as a function of x. To do that, we need to express this somehow in terms of the tangent of y. The reason why the tangent of y is interesting is because we already know that tangent of y is equal to x. If we can rewrite this using a little bit of trigonometric identities, then with tangent of y, we can substitute all the tangent of y's with an x."}, {"video_title": "Derivative of inverse tangent Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "What we're really interested in is writing it as a function of x. To do that, we need to express this somehow in terms of the tangent of y. The reason why the tangent of y is interesting is because we already know that tangent of y is equal to x. If we can rewrite this using a little bit of trigonometric identities, then with tangent of y, we can substitute all the tangent of y's with an x. Let's see if we can do that. This seems a little bit tricky. To introduce a tangent of y, we'd want to have a sine divided by a cosine."}, {"video_title": "Derivative of inverse tangent Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "If we can rewrite this using a little bit of trigonometric identities, then with tangent of y, we can substitute all the tangent of y's with an x. Let's see if we can do that. This seems a little bit tricky. To introduce a tangent of y, we'd want to have a sine divided by a cosine. That's what tangent is. This is just a straight-up cosine squared y. This is really going to take a little bit more experimentation than the last few examples we've done."}, {"video_title": "Derivative of inverse tangent Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "To introduce a tangent of y, we'd want to have a sine divided by a cosine. That's what tangent is. This is just a straight-up cosine squared y. This is really going to take a little bit more experimentation than the last few examples we've done. One thing we could do is say, let's just divide by 1. Dividing by 1 never hurt anyone. We could say that this is the same thing as cosine squared y. I'm really doing this to see if I can start to express it as some type of a rational expression, which might involve, at some point, a sine divided by a cosine, and I could have a tangent."}, {"video_title": "Derivative of inverse tangent Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "This is really going to take a little bit more experimentation than the last few examples we've done. One thing we could do is say, let's just divide by 1. Dividing by 1 never hurt anyone. We could say that this is the same thing as cosine squared y. I'm really doing this to see if I can start to express it as some type of a rational expression, which might involve, at some point, a sine divided by a cosine, and I could have a tangent. Let's divide by 1. We know from the Pythagorean identity that 1 is equal to sine squared of y plus cosine squared of y. Let's write that."}, {"video_title": "Derivative of inverse tangent Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "We could say that this is the same thing as cosine squared y. I'm really doing this to see if I can start to express it as some type of a rational expression, which might involve, at some point, a sine divided by a cosine, and I could have a tangent. Let's divide by 1. We know from the Pythagorean identity that 1 is equal to sine squared of y plus cosine squared of y. Let's write that. We could write cosine squared of y plus sine squared of y. Once again, why was I able to divide by this expression? This expression by the Pythagorean identity, which really comes out of the unit circle definition of trig functions, this is equal to 1, so I have not changed the value of this expression."}, {"video_title": "Derivative of inverse tangent Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Let's write that. We could write cosine squared of y plus sine squared of y. Once again, why was I able to divide by this expression? This expression by the Pythagorean identity, which really comes out of the unit circle definition of trig functions, this is equal to 1, so I have not changed the value of this expression. What makes this interesting is if I want to introduce a sine divided by a cosine, I could just divide the numerator and the denominator by cosine squared. Let's do that. Let's divide the numerator by cosine squared of y and divide the denominator by cosine squared of y, or multiply each of them by 1 over cosine squared of y."}, {"video_title": "Derivative of inverse tangent Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "This expression by the Pythagorean identity, which really comes out of the unit circle definition of trig functions, this is equal to 1, so I have not changed the value of this expression. What makes this interesting is if I want to introduce a sine divided by a cosine, I could just divide the numerator and the denominator by cosine squared. Let's do that. Let's divide the numerator by cosine squared of y and divide the denominator by cosine squared of y, or multiply each of them by 1 over cosine squared of y. What's that going to give us? The numerator, these characters are going to cancel. You're just going to have a 1."}, {"video_title": "Derivative of inverse tangent Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Let's divide the numerator by cosine squared of y and divide the denominator by cosine squared of y, or multiply each of them by 1 over cosine squared of y. What's that going to give us? The numerator, these characters are going to cancel. You're just going to have a 1. The denominator, this times this, that's just going to be equal to 1. Then you're going to have sine squared y over cosine squared y. This is the goal that I was trying to achieve."}, {"video_title": "Derivative of inverse tangent Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "You're just going to have a 1. The denominator, this times this, that's just going to be equal to 1. Then you're going to have sine squared y over cosine squared y. This is the goal that I was trying to achieve. I have a sine divided by cosine squared. This right over here, this is the same thing. Let me write it this way."}, {"video_title": "Derivative of inverse tangent Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "This is the goal that I was trying to achieve. I have a sine divided by cosine squared. This right over here, this is the same thing. Let me write it this way. This is the same thing as sine of y over cosine of y whole thing squared, which is, of course, the same thing as 1 over 1 plus tangent of y squared. This is equal to this. Why is that useful?"}, {"video_title": "Derivative of inverse tangent Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Let me write it this way. This is the same thing as sine of y over cosine of y whole thing squared, which is, of course, the same thing as 1 over 1 plus tangent of y squared. This is equal to this. Why is that useful? We know that x is equal to tangent of y. This is going to be equal to 1 over 1 plus tangent of y is equal to x squared, which is pretty exciting. We just figured out the derivative of y with respect to x."}, {"video_title": "Derivative of inverse tangent Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Why is that useful? We know that x is equal to tangent of y. This is going to be equal to 1 over 1 plus tangent of y is equal to x squared, which is pretty exciting. We just figured out the derivative of y with respect to x. The derivative of this thing with respect to x is 1 over 1 plus x squared. We could write that right up here. This is going to be equal to 1 over 1 plus x squared."}, {"video_title": "Area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "We are now going to then extend this to think about the area between curves. So let's say we care about the region from x equals a to x equals b between y equals f of x and y is equal to g of x. So that would be this area right over here. So based on what you already know about definite integrals, how would you actually try to calculate this? Well, one natural thing that you might say is, well, look, if I were to take the integral from a to b of f of x dx, that would give me the entire area below f of x and above the x-axis, and then if I were to subtract from that, if I were to subtract from that this area right over here, which is equal to, that's the definite integral from a to b of g of x dx, well, then I would net out with the original area that I cared about. I would net out with this area right over here, and that indeed would be the case, and we know from our integration properties that we can rewrite this as the integral from a to b of, let me put some parentheses here, of f of x minus g of x, minus g of x dx, and now I will make a claim to you, and we'll build a little bit more intuition for this as we go through this video, but over an interval from a to b where f of x is greater than g of x, like this interval right over here, this is always going to be the case, that the area between the curves is going to be the integral for the x interval that we care about from a to b of f of x minus g of x, so I know what you're thinking. You're like, okay, well, that worked when both of them were above the x-axis, but what about the case when f of x is above the x-axis and g of x is below the x-axis?"}, {"video_title": "Area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So based on what you already know about definite integrals, how would you actually try to calculate this? Well, one natural thing that you might say is, well, look, if I were to take the integral from a to b of f of x dx, that would give me the entire area below f of x and above the x-axis, and then if I were to subtract from that, if I were to subtract from that this area right over here, which is equal to, that's the definite integral from a to b of g of x dx, well, then I would net out with the original area that I cared about. I would net out with this area right over here, and that indeed would be the case, and we know from our integration properties that we can rewrite this as the integral from a to b of, let me put some parentheses here, of f of x minus g of x, minus g of x dx, and now I will make a claim to you, and we'll build a little bit more intuition for this as we go through this video, but over an interval from a to b where f of x is greater than g of x, like this interval right over here, this is always going to be the case, that the area between the curves is going to be the integral for the x interval that we care about from a to b of f of x minus g of x, so I know what you're thinking. You're like, okay, well, that worked when both of them were above the x-axis, but what about the case when f of x is above the x-axis and g of x is below the x-axis? So, for example, let's say that we were to think about this interval right over here. Let's say this is the point c, and that's x equals c. This is x equals d right over here, so what if we wanted to calculate this area that I am shading in right over here? You might say, well, does this actually work?"}, {"video_title": "Area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "You're like, okay, well, that worked when both of them were above the x-axis, but what about the case when f of x is above the x-axis and g of x is below the x-axis? So, for example, let's say that we were to think about this interval right over here. Let's say this is the point c, and that's x equals c. This is x equals d right over here, so what if we wanted to calculate this area that I am shading in right over here? You might say, well, does this actually work? Well, let's think about now what the integral, let's think about what the integral from c to d of f of x dx represents. Well, that would represent this area right over here, and what would the integral from c to d of g of x dx represent? Well, you might say it is this area right over here, but remember, over this interval, g of x is below the x-axis, so this would give you a negative value, but if you wanted this total area, what you could do is take this blue area, which is positive, and then subtract this negative area, and so then you would get the entire positive area."}, {"video_title": "Area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "You might say, well, does this actually work? Well, let's think about now what the integral, let's think about what the integral from c to d of f of x dx represents. Well, that would represent this area right over here, and what would the integral from c to d of g of x dx represent? Well, you might say it is this area right over here, but remember, over this interval, g of x is below the x-axis, so this would give you a negative value, but if you wanted this total area, what you could do is take this blue area, which is positive, and then subtract this negative area, and so then you would get the entire positive area. Well, this just amounted to, this is equivalent to the integral from c to d of f of x, of f of x minus g of x again, minus g of x. Let me make it clear, we've got parentheses there, and then we have our dx. So once again, even over this interval, when f of x was above the x-axis and g of x was below the x-axis, it still boiled down to the same thing."}, {"video_title": "Area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Well, you might say it is this area right over here, but remember, over this interval, g of x is below the x-axis, so this would give you a negative value, but if you wanted this total area, what you could do is take this blue area, which is positive, and then subtract this negative area, and so then you would get the entire positive area. Well, this just amounted to, this is equivalent to the integral from c to d of f of x, of f of x minus g of x again, minus g of x. Let me make it clear, we've got parentheses there, and then we have our dx. So once again, even over this interval, when f of x was above the x-axis and g of x was below the x-axis, it still boiled down to the same thing. Well, let's take another scenario. Let's take the scenario when they are both below the x-axis. Let's say that we wanted to go from x equals, well, I won't use e since that is a loaded letter in mathematics, and so is f and g. Well, let's just say, well, I'm kind of running out of letters now."}, {"video_title": "Area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So once again, even over this interval, when f of x was above the x-axis and g of x was below the x-axis, it still boiled down to the same thing. Well, let's take another scenario. Let's take the scenario when they are both below the x-axis. Let's say that we wanted to go from x equals, well, I won't use e since that is a loaded letter in mathematics, and so is f and g. Well, let's just say, well, I'm kind of running out of letters now. Let's say that I am gonna go from, I don't know, let's just call this m, and let's call this n right over here. Well, n is getting, let's put n right over here. So what I care about is this area, the area, once again, below f. We're assuming that we're looking at intervals where f is greater than g, so below f and greater than g. Will it still amount to this with now the endpoints being m and n?"}, {"video_title": "Area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say that we wanted to go from x equals, well, I won't use e since that is a loaded letter in mathematics, and so is f and g. Well, let's just say, well, I'm kind of running out of letters now. Let's say that I am gonna go from, I don't know, let's just call this m, and let's call this n right over here. Well, n is getting, let's put n right over here. So what I care about is this area, the area, once again, below f. We're assuming that we're looking at intervals where f is greater than g, so below f and greater than g. Will it still amount to this with now the endpoints being m and n? Well, let's think about it a little bit. If we were to evaluate that integral from m to n of, let's put my dx here, of f of x minus, minus g of x, we already know from our integral properties, this is going to be equal to the integral from m to n of f of x dx minus the integral from m to n of g of x dx. Now let's think about what each of these represent."}, {"video_title": "Area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So what I care about is this area, the area, once again, below f. We're assuming that we're looking at intervals where f is greater than g, so below f and greater than g. Will it still amount to this with now the endpoints being m and n? Well, let's think about it a little bit. If we were to evaluate that integral from m to n of, let's put my dx here, of f of x minus, minus g of x, we already know from our integral properties, this is going to be equal to the integral from m to n of f of x dx minus the integral from m to n of g of x dx. Now let's think about what each of these represent. So this yellow integral right over here, that would give this, the negative of this area. So that would give a negative value here, but the magnitude of it, the absolute value of it, would be this area right over there. Now what would just the integral, I'm not even thinking about the negative sign here, what would the integral of this g of x of this blue integral give?"}, {"video_title": "Area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's think about what each of these represent. So this yellow integral right over here, that would give this, the negative of this area. So that would give a negative value here, but the magnitude of it, the absolute value of it, would be this area right over there. Now what would just the integral, I'm not even thinking about the negative sign here, what would the integral of this g of x of this blue integral give? Well, that would give this the negative of this entire area. But now we're gonna take the negative of that, and so this part right over here, this entire part, including this negative sign, would give us, would give us this entire area, the entire area. This would actually give a positive value because we're taking the negative of a negative."}, {"video_title": "Area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Now what would just the integral, I'm not even thinking about the negative sign here, what would the integral of this g of x of this blue integral give? Well, that would give this the negative of this entire area. But now we're gonna take the negative of that, and so this part right over here, this entire part, including this negative sign, would give us, would give us this entire area, the entire area. This would actually give a positive value because we're taking the negative of a negative. But if with the area that we care about, right over here, the area that we cared about originally, we would want to subtract out this yellow area. Well, this right over here, this yellow integral from the definite integral from m to n of f of x dx, that's exactly that. That is the negative of that yellow area."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "What I want to do in this video is to see whether the power rule is giving us results that at least seem reasonable. This is by no means a proof of the power rule, but at least we'll feel a little bit more comfortable using it. So let's say that f of x is equal to x. The power rule tells us that f prime of x is going to be equal to what? Well, x is the same thing as x to the first power, so n is implicitly 1 right over here. So we bring the 1 out front, it'll be 1 times x to the 1 minus 1 power. So it's going to be 1 times x to the 0 power."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "The power rule tells us that f prime of x is going to be equal to what? Well, x is the same thing as x to the first power, so n is implicitly 1 right over here. So we bring the 1 out front, it'll be 1 times x to the 1 minus 1 power. So it's going to be 1 times x to the 0 power. x to the 0 is just 1, so it's just going to be equal to 1. Now does that make conceptual sense if we actually try to visualize these functions? So let me draw, let me actually try to graph these functions."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So it's going to be 1 times x to the 0 power. x to the 0 is just 1, so it's just going to be equal to 1. Now does that make conceptual sense if we actually try to visualize these functions? So let me draw, let me actually try to graph these functions. So that's my y-axis, this is my x-axis. And let me graph y equals x. So y is equal to f of x here, so y is equal to x."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So let me draw, let me actually try to graph these functions. So that's my y-axis, this is my x-axis. And let me graph y equals x. So y is equal to f of x here, so y is equal to x. So it looks something like that. So y is equal to x, or this is f of x equals to x, or y is equal to this f of x right over there. Now, the derivative, actually let me just call that f of x, just to make it, not to confuse you."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So y is equal to f of x here, so y is equal to x. So it looks something like that. So y is equal to x, or this is f of x equals to x, or y is equal to this f of x right over there. Now, the derivative, actually let me just call that f of x, just to make it, not to confuse you. So this right over here is f of x is equal to x that I graphed right over here. y is equal to f of x, which is equal to x. And now let me graph the derivative, let me graph f prime of x."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Now, the derivative, actually let me just call that f of x, just to make it, not to confuse you. So this right over here is f of x is equal to x that I graphed right over here. y is equal to f of x, which is equal to x. And now let me graph the derivative, let me graph f prime of x. That's saying it's 1, that's saying it's 1 for all x, regardless of what x is, it's going to be equal to 1. Is this consistent with what we know about derivatives and slopes and all the rest? Well, let's look at our function."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And now let me graph the derivative, let me graph f prime of x. That's saying it's 1, that's saying it's 1 for all x, regardless of what x is, it's going to be equal to 1. Is this consistent with what we know about derivatives and slopes and all the rest? Well, let's look at our function. What is the slope of the line, or the tangent line, right at this point? Well, the slope of this right over here, this has slope 1 continuously, or it has a constant slope of 1. Slope is equal to 1 no matter what x is, it's a line."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, let's look at our function. What is the slope of the line, or the tangent line, right at this point? Well, the slope of this right over here, this has slope 1 continuously, or it has a constant slope of 1. Slope is equal to 1 no matter what x is, it's a line. And for a line, the slope is constant. So over here, the slope is indeed 1. If you go to this point over here, the slope is indeed 1."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Slope is equal to 1 no matter what x is, it's a line. And for a line, the slope is constant. So over here, the slope is indeed 1. If you go to this point over here, the slope is indeed 1. If you go over here, the slope is indeed 1. So we got a pretty valid response there. Now let's try something where the slope might change."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "If you go to this point over here, the slope is indeed 1. If you go over here, the slope is indeed 1. So we got a pretty valid response there. Now let's try something where the slope might change. So let's say I have g of x is equal to x squared. The power rule tells us that g prime of x would be equal to what? Well, n is equal to 2, so it's going to be 2 times x to the 2 minus 1."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's try something where the slope might change. So let's say I have g of x is equal to x squared. The power rule tells us that g prime of x would be equal to what? Well, n is equal to 2, so it's going to be 2 times x to the 2 minus 1. Or it's going to be equal to 2x to the first power, it's going to be equal to 2x. So let's see if this makes reasonable sense. I'm going to try to graph this one a little bit more precisely."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, n is equal to 2, so it's going to be 2 times x to the 2 minus 1. Or it's going to be equal to 2x to the first power, it's going to be equal to 2x. So let's see if this makes reasonable sense. I'm going to try to graph this one a little bit more precisely. So let's see how well I can graph it, how precisely I can graph it. So this is x-axis, y-axis, let me mark some stuff off here. So this is 1, 2, 3, 4, 5."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "I'm going to try to graph this one a little bit more precisely. So let's see how well I can graph it, how precisely I can graph it. So this is x-axis, y-axis, let me mark some stuff off here. So this is 1, 2, 3, 4, 5. This is 1, 2, 3, 4. 1, 2, 3, 4. So g of x, when x is 0, it's 0."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So this is 1, 2, 3, 4, 5. This is 1, 2, 3, 4. 1, 2, 3, 4. So g of x, when x is 0, it's 0. When x is 1, it is 1. When x is negative 1, it's 1. When x is 2, it is 4."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So g of x, when x is 0, it's 0. When x is 1, it is 1. When x is negative 1, it's 1. When x is 2, it is 4. So that puts us right over there, 1, 2, 3, 4. Puts us right over there. When x is negative 2, you get to 4."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "When x is 2, it is 4. So that puts us right over there, 1, 2, 3, 4. Puts us right over there. When x is negative 2, you get to 4. It's a parabola. You've seen this for many years. So it looks something like this."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "When x is negative 2, you get to 4. It's a parabola. You've seen this for many years. So it looks something like this. Actually, the last two points I graphed were a little bit weird. So this might be right over here. So it looks something like this."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So it looks something like this. Actually, the last two points I graphed were a little bit weird. So this might be right over here. So it looks something like this. It looks something like that. And then when you come over here, it looks something like that. It's symmetric, so I'm trying my best to draw it reasonably."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So it looks something like this. It looks something like that. And then when you come over here, it looks something like that. It's symmetric, so I'm trying my best to draw it reasonably. So there you go. That's the graph of g of x is equal to x squared. Now let's graph g prime of x, or what the power rule is telling us that g prime of x is."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "It's symmetric, so I'm trying my best to draw it reasonably. So there you go. That's the graph of g of x is equal to x squared. Now let's graph g prime of x, or what the power rule is telling us that g prime of x is. So g prime of x is equal to 2x. So that's just a line that goes through the origin of slope 2. So it looks something like that."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's graph g prime of x, or what the power rule is telling us that g prime of x is. So g prime of x is equal to 2x. So that's just a line that goes through the origin of slope 2. So it looks something like that. When x is equal to 1, y is equal to 2. When x is equal to 2, y or g of x is equal to 4. So it looks something like this."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So it looks something like that. When x is equal to 1, y is equal to 2. When x is equal to 2, y or g of x is equal to 4. So it looks something like this. Let me try my best to draw a straight line. It looks something like this. Now, does this make sense?"}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So it looks something like this. Let me try my best to draw a straight line. It looks something like this. Now, does this make sense? Well, if you just eyeball it really fast, if you look at this point right over here, and you want to think about the slope of the tangent line, the slope, I'm going to try my best to draw, let me do this in a color that pops out a little bit more. So the slope of the tangent line would look something like this. So it looks like it has a reasonably high negative slope."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Now, does this make sense? Well, if you just eyeball it really fast, if you look at this point right over here, and you want to think about the slope of the tangent line, the slope, I'm going to try my best to draw, let me do this in a color that pops out a little bit more. So the slope of the tangent line would look something like this. So it looks like it has a reasonably high negative slope. It's definitely a negative slope, and it's a pretty steep negative slope. Well, for x is equal to 2, g prime of 2, or sorry, for x is equal to negative 2, g prime of negative 2 is equal to 2 times negative 2, which is equal to negative 4. So this is claiming that the slope at this point, so this right over here is negative 4, saying that the slope of this point is negative 4. m is equal to negative 4."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So it looks like it has a reasonably high negative slope. It's definitely a negative slope, and it's a pretty steep negative slope. Well, for x is equal to 2, g prime of 2, or sorry, for x is equal to negative 2, g prime of negative 2 is equal to 2 times negative 2, which is equal to negative 4. So this is claiming that the slope at this point, so this right over here is negative 4, saying that the slope of this point is negative 4. m is equal to negative 4. That looks about right. It's a fairly steep negative slope. Now, what happens if you go right over here when x is equal to 0?"}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So this is claiming that the slope at this point, so this right over here is negative 4, saying that the slope of this point is negative 4. m is equal to negative 4. That looks about right. It's a fairly steep negative slope. Now, what happens if you go right over here when x is equal to 0? Well, our derivative, if you say g prime of 0, is telling us that the slope of our original function g at x is equal to 0 is 2 times 0 is 0. Well, does that make sense? Well, if we go to our original parabola, it does indeed make sense."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Now, what happens if you go right over here when x is equal to 0? Well, our derivative, if you say g prime of 0, is telling us that the slope of our original function g at x is equal to 0 is 2 times 0 is 0. Well, does that make sense? Well, if we go to our original parabola, it does indeed make sense. The slope of the tangent line looks something like this. We're at a minimum point. We're at the vertex."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, if we go to our original parabola, it does indeed make sense. The slope of the tangent line looks something like this. We're at a minimum point. We're at the vertex. The slope really does look to be 0. And what if you go right over here to x equals 2? The slope of the tangent line."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "We're at the vertex. The slope really does look to be 0. And what if you go right over here to x equals 2? The slope of the tangent line. Well, over here, the tangent line looks something like this. It looks like a fairly steep positive slope. What is our derivative telling us based on the power rule?"}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "The slope of the tangent line. Well, over here, the tangent line looks something like this. It looks like a fairly steep positive slope. What is our derivative telling us based on the power rule? It's saying g prime. So this is essentially saying, hey, tell me what the slope of the tangent line for g is when x is equal to 2. Well, we figured it out."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "What is our derivative telling us based on the power rule? It's saying g prime. So this is essentially saying, hey, tell me what the slope of the tangent line for g is when x is equal to 2. Well, we figured it out. It's going to be 2 times x. It's going to be 2 times 2, which is equal to 4. It's telling us that the slope over here is 4."}, {"video_title": "Justifying the power rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we figured it out. It's going to be 2 times x. It's going to be 2 times 2, which is equal to 4. It's telling us that the slope over here is 4. That the slope, and I'm just using m. m is often the letter used to denote slope. They're saying that the slope of the tangent line there is 4, which seems completely, completely reasonable. So I encourage you to try this out yourself."}, {"video_title": "Introduction to limits at infinity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "We now have a lot of experience taking limits of functions, if I'm taking the limit of f of x, we're gonna think about what does f of x approach as x approaches some value a, and this would be equal to some limit. Now everything we've done up till now is where a is a finite value, but when you look at the graph of the function f right over here, you see something interesting happens. As x gets larger and larger, it looks like our function f is getting closer and closer to two. It looks like we have a horizontal asymptote at y equals two. Similarly, as x gets more and more negative, it also seems like we have a horizontal asymptote at y equals two. So is there some type of notation we can use to think about what is the graph approaching as x gets much larger or as x gets smaller and smaller? And the answer there is limits at infinity."}, {"video_title": "Introduction to limits at infinity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "It looks like we have a horizontal asymptote at y equals two. Similarly, as x gets more and more negative, it also seems like we have a horizontal asymptote at y equals two. So is there some type of notation we can use to think about what is the graph approaching as x gets much larger or as x gets smaller and smaller? And the answer there is limits at infinity. So if we wanna think about what is this graph, what is this function approaching as x gets larger and larger, we can think about the limit of f of x as x approaches positive infinity. So that's the notation, and I'm not going to give you the formal definition of this right now. There, in future videos, we might do that, but it's this idea as x gets larger and larger and larger, does it look like that our function is approaching some finite value, that we have a horizontal asymptote there?"}, {"video_title": "Introduction to limits at infinity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And the answer there is limits at infinity. So if we wanna think about what is this graph, what is this function approaching as x gets larger and larger, we can think about the limit of f of x as x approaches positive infinity. So that's the notation, and I'm not going to give you the formal definition of this right now. There, in future videos, we might do that, but it's this idea as x gets larger and larger and larger, does it look like that our function is approaching some finite value, that we have a horizontal asymptote there? And in this situation, it looks like it is. It looks like it's approaching the value two. And for this particular function, the limit of f of x as x approaches negative infinity also looks like it is approaching two."}, {"video_title": "Introduction to limits at infinity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "There, in future videos, we might do that, but it's this idea as x gets larger and larger and larger, does it look like that our function is approaching some finite value, that we have a horizontal asymptote there? And in this situation, it looks like it is. It looks like it's approaching the value two. And for this particular function, the limit of f of x as x approaches negative infinity also looks like it is approaching two. This is not always going to be the same. You could have a situation, maybe we had, you could have another function, so let me draw a little horizontal asymptote right over here. You could imagine a function that looks like this."}, {"video_title": "Introduction to limits at infinity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And for this particular function, the limit of f of x as x approaches negative infinity also looks like it is approaching two. This is not always going to be the same. You could have a situation, maybe we had, you could have another function, so let me draw a little horizontal asymptote right over here. You could imagine a function that looks like this. So let me do it like that, and maybe it does something wacky like this, and then it comes down, and it does something like this. Here, our limit as x approaches infinity is still two, but our limit as x approaches negative infinity right over here would be negative two. And of course, there's many situations where as you approach infinity or negative infinity, you aren't actually approaching some finite value."}, {"video_title": "Introduction to limits at infinity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "You could imagine a function that looks like this. So let me do it like that, and maybe it does something wacky like this, and then it comes down, and it does something like this. Here, our limit as x approaches infinity is still two, but our limit as x approaches negative infinity right over here would be negative two. And of course, there's many situations where as you approach infinity or negative infinity, you aren't actually approaching some finite value. You don't have a horizontal asymptote. But the whole point of this video is just to make you familiar with this notation. And limits at infinity, or you could say limits at negative infinity, they have a different formal definition than some of the limits that we've looked at in the past where we are approaching a finite value, but intuitively, they make sense that these are indeed limits."}, {"video_title": "Inflection points (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "So let's just remind ourselves what a point of inflection is. A point of inflection is where we go from being, where we change our concavity, or you could say where our second derivative, g prime of x, switches signs. Switches, switches signs. So let's study our second derivative. In order to study our second derivative, let's find it. So we know that g of x is equal to 1 4th x to the 4th minus four x to the 3rd power plus 24 x squared. So given that, let's now find g prime of x. G prime of x is going to be equal to, I'm just gonna apply the power rule multiple times, four times 1 4th is just one, I'm not even gonna write the one down, it's gonna be one times x to the four minus one power."}, {"video_title": "Inflection points (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "So let's study our second derivative. In order to study our second derivative, let's find it. So we know that g of x is equal to 1 4th x to the 4th minus four x to the 3rd power plus 24 x squared. So given that, let's now find g prime of x. G prime of x is going to be equal to, I'm just gonna apply the power rule multiple times, four times 1 4th is just one, I'm not even gonna write the one down, it's gonna be one times x to the four minus one power. So four to the 3rd power minus three times four is 12, x to the three minus one power, or x to the 2nd power, plus two times 24, or 48, x to the two minus one, or x to the 1st power, I could just write that as x. So there you have it, I have our first derivative, now we wanna find our second derivative. So g prime prime of x is just the derivative of the first derivative with respect to x, and so more of the power rule, three x squared minus 24 x to the first, or just 24 x, plus 48."}, {"video_title": "Inflection points (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "So given that, let's now find g prime of x. G prime of x is going to be equal to, I'm just gonna apply the power rule multiple times, four times 1 4th is just one, I'm not even gonna write the one down, it's gonna be one times x to the four minus one power. So four to the 3rd power minus three times four is 12, x to the three minus one power, or x to the 2nd power, plus two times 24, or 48, x to the two minus one, or x to the 1st power, I could just write that as x. So there you have it, I have our first derivative, now we wanna find our second derivative. So g prime prime of x is just the derivative of the first derivative with respect to x, and so more of the power rule, three x squared minus 24 x to the first, or just 24 x, plus 48. So let's think about where this switches signs. And this is a continuous function, it's going to be defined for all x's, so the only potential candidates of where it could switch signs are when this thing equals zero. So let's see where it equals zero."}, {"video_title": "Inflection points (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "So g prime prime of x is just the derivative of the first derivative with respect to x, and so more of the power rule, three x squared minus 24 x to the first, or just 24 x, plus 48. So let's think about where this switches signs. And this is a continuous function, it's going to be defined for all x's, so the only potential candidates of where it could switch signs are when this thing equals zero. So let's see where it equals zero. So let's set it equal to zero. Three x squared minus 24 x plus 48 is equal to zero. Let's see, everything is divisible by three, so let's divide everything by three."}, {"video_title": "Inflection points (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "So let's see where it equals zero. So let's set it equal to zero. Three x squared minus 24 x plus 48 is equal to zero. Let's see, everything is divisible by three, so let's divide everything by three. So you get x squared minus eight x plus 16, plus 16 is equal to zero, and let's see, can I factor this? Yeah, this would be x minus four times x minus four. Or I could just do this as x minus four squared is equal to zero, or x minus four is equal to zero, so, or where x equals four."}, {"video_title": "Inflection points (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "Let's see, everything is divisible by three, so let's divide everything by three. So you get x squared minus eight x plus 16, plus 16 is equal to zero, and let's see, can I factor this? Yeah, this would be x minus four times x minus four. Or I could just do this as x minus four squared is equal to zero, or x minus four is equal to zero, so, or where x equals four. So g prime prime of four is equal to zero. So let's see what's happening on either side of that. Let's see if g, if we're actually, if we're actually switching signs or not."}, {"video_title": "Inflection points (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "Or I could just do this as x minus four squared is equal to zero, or x minus four is equal to zero, so, or where x equals four. So g prime prime of four is equal to zero. So let's see what's happening on either side of that. Let's see if g, if we're actually, if we're actually switching signs or not. So let me draw a number line here. And so this is, so this is two, three, four, five, and I could keep going. And so we know that something interesting is happening right over here."}, {"video_title": "Inflection points (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "Let's see if g, if we're actually, if we're actually switching signs or not. So let me draw a number line here. And so this is, so this is two, three, four, five, and I could keep going. And so we know that something interesting is happening right over here. G prime prime of four is equal to zero. G prime prime of four is equal to zero. So let's think about what the second derivative is when we are less than four."}, {"video_title": "Inflection points (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "And so we know that something interesting is happening right over here. G prime prime of four is equal to zero. G prime prime of four is equal to zero. So let's think about what the second derivative is when we are less than four. And so, actually, let me just try g prime prime of zero, since that'll be easy to evaluate. G prime prime of zero, well, it's just going to be equal to 48. So when we are less than four, our second derivative, g prime, the second derivative is greater than zero."}, {"video_title": "Inflection points (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "So let's think about what the second derivative is when we are less than four. And so, actually, let me just try g prime prime of zero, since that'll be easy to evaluate. G prime prime of zero, well, it's just going to be equal to 48. So when we are less than four, our second derivative, g prime, the second derivative is greater than zero. So we are actually going to be concave upwards over this interval to the left of four. Now let's think about to the right of four. Two is a different color."}, {"video_title": "Inflection points (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "So when we are less than four, our second derivative, g prime, the second derivative is greater than zero. So we are actually going to be concave upwards over this interval to the left of four. Now let's think about to the right of four. Two is a different color. So what about to the right of four? And so let me just evaluate, what would be an easy thing to evaluate? Well, I could evaluate g prime of, well, why not do g, or the second derivative, g prime prime, I should say, of 10."}, {"video_title": "Inflection points (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "Two is a different color. So what about to the right of four? And so let me just evaluate, what would be an easy thing to evaluate? Well, I could evaluate g prime of, well, why not do g, or the second derivative, g prime prime, I should say, of 10. So g, I'll do it right over here. Let me do, well, I'm running a little bit out of space, so I'll just scroll down. So g prime prime of 10 is going to be equal to three times 10 squared, so it's 300 minus 24 times 10, so minus 240 plus 48."}, {"video_title": "Inflection points (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "Well, I could evaluate g prime of, well, why not do g, or the second derivative, g prime prime, I should say, of 10. So g, I'll do it right over here. Let me do, well, I'm running a little bit out of space, so I'll just scroll down. So g prime prime of 10 is going to be equal to three times 10 squared, so it's 300 minus 24 times 10, so minus 240 plus 48. So let's see, this is 60. This is, so 300 minus 240 is 60 plus 48, so this is equal to 108, so it's still positive. So on either side of four, g prime prime of x is greater than zero."}, {"video_title": "Inflection points (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "So g prime prime of 10 is going to be equal to three times 10 squared, so it's 300 minus 24 times 10, so minus 240 plus 48. So let's see, this is 60. This is, so 300 minus 240 is 60 plus 48, so this is equal to 108, so it's still positive. So on either side of four, g prime prime of x is greater than zero. So even though the second derivative at x equals four is equal to zero, on either side, we are concave upwards. On either side, the second derivative is positive, and so, and that was the only potential candidate, so there are no values of x for which g has a point of inflection. X equals four would have been a value of x at which g had a point of inflection if we switch, if the second derivative switched signs here, if it went from positive to negative or negative to positive, but it's just staying from positive to positive."}, {"video_title": "Inflection points (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "So on either side of four, g prime prime of x is greater than zero. So even though the second derivative at x equals four is equal to zero, on either side, we are concave upwards. On either side, the second derivative is positive, and so, and that was the only potential candidate, so there are no values of x for which g has a point of inflection. X equals four would have been a value of x at which g had a point of inflection if we switch, if the second derivative switched signs here, if it went from positive to negative or negative to positive, but it's just staying from positive to positive. So the second derivative is positive. It just touches zero right here, and then it goes positive again. So going back to the question, for what x values does the graph of g have a point of inflection?"}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Let's say that we're interested in approximating what the square root of 4.36 is equal to. So we want to figure out an approximation of this, and we don't have a calculator at hand. Well, one way to think about it is we know what the square root of four is. We know that this is positive two. The principal root of four is positive two. So you say, okay, this is going to be a little bit more than two. But let's say that we want to get a little bit more accurate."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "We know that this is positive two. The principal root of four is positive two. So you say, okay, this is going to be a little bit more than two. But let's say that we want to get a little bit more accurate. And so what I'm going to show you in this video is a method for doing that, for approximating the value of a function near a value where we already know the value. So what am I talking about? So let's just imagine that we had the function."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "But let's say that we want to get a little bit more accurate. And so what I'm going to show you in this video is a method for doing that, for approximating the value of a function near a value where we already know the value. So what am I talking about? So let's just imagine that we had the function. We have the function f of x is equal to the square root of x, which is, of course, the same thing as x to the 1 1\u20442 power. So we know what f of two is. We know that f of two, oh, sorry, we know that f of four is."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So let's just imagine that we had the function. We have the function f of x is equal to the square root of x, which is, of course, the same thing as x to the 1 1\u20442 power. So we know what f of two is. We know that f of two, oh, sorry, we know that f of four is. We know that f of four is the square root of four, which is going to be equal to two, or the principal root of four, which is equal to positive two. And what we want to approximate, we want to figure out what f of, we want to figure out what f of 4.36 is equal to. This is just another way of framing the exact same question that we started off this video."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "We know that f of two, oh, sorry, we know that f of four is. We know that f of four is the square root of four, which is going to be equal to two, or the principal root of four, which is equal to positive two. And what we want to approximate, we want to figure out what f of, we want to figure out what f of 4.36 is equal to. This is just another way of framing the exact same question that we started off this video. So let's just imagine our function. Let's just imagine it for a second. So let me draw some axes."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "This is just another way of framing the exact same question that we started off this video. So let's just imagine our function. Let's just imagine it for a second. So let me draw some axes. This is my y-axis. This is my x-axis. And let's graph y is equal to f of x."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So let me draw some axes. This is my y-axis. This is my x-axis. And let's graph y is equal to f of x. So let's say it looks something like this. Y equals f of x looks something like that. So that's pretty decent."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And let's graph y is equal to f of x. So let's say it looks something like this. Y equals f of x looks something like that. So that's pretty decent. All right, so that right there is y is equal to f of x. And we know f of four is equal to two. F of four is equal to two, so this is when x is equal to four."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So that's pretty decent. All right, so that right there is y is equal to f of x. And we know f of four is equal to two. F of four is equal to two, so this is when x is equal to four. I haven't drawn it really to scale, but hopefully this is clear enough. So that right over here is going to be two. That's f of four."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "F of four is equal to two, so this is when x is equal to four. I haven't drawn it really to scale, but hopefully this is clear enough. So that right over here is going to be two. That's f of four. And what we want to approximate is f of 4.36. So 4.36 might be right around there. And so we want to approximate this y value right over here."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "That's f of four. And what we want to approximate is f of 4.36. So 4.36 might be right around there. And so we want to approximate this y value right over here. We want to approximate that right over here is f of 4.36. And once again, we're assuming we don't have a calculator at hand. So how can we do that using what we know about derivatives?"}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And so we want to approximate this y value right over here. We want to approximate that right over here is f of 4.36. And once again, we're assuming we don't have a calculator at hand. So how can we do that using what we know about derivatives? Well, what if we were to figure out an equation for the line that is tangent to the point, to tangent to this point right over here. So the equation of the tangent line at x is equal to four. And then we use that linearization, that linearization to find, to approximate values local to it."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So how can we do that using what we know about derivatives? Well, what if we were to figure out an equation for the line that is tangent to the point, to tangent to this point right over here. So the equation of the tangent line at x is equal to four. And then we use that linearization, that linearization to find, to approximate values local to it. And this technique is called local linearization. So what I'm saying is let's figure out what this equation of this line is. Let's call that L of x."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And then we use that linearization, that linearization to find, to approximate values local to it. And this technique is called local linearization. So what I'm saying is let's figure out what this equation of this line is. Let's call that L of x. And then we can use that to, then we can evaluate that at 4.36. And hopefully that'll be a little bit easier to do than to try to figure out this right over here. So how would we do that?"}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Let's call that L of x. And then we can use that to, then we can evaluate that at 4.36. And hopefully that'll be a little bit easier to do than to try to figure out this right over here. So how would we do that? Well, one way to think about it, obviously there's many ways to express a line, but one way to think about it is, okay, it's going to, L of x is going to be, it's going to be, it's going to be f of four. It's going to be f of four, which is two. It's going to be f of four plus the slope, the slope at x equals four, which is of course the derivative, f prime of four."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So how would we do that? Well, one way to think about it, obviously there's many ways to express a line, but one way to think about it is, okay, it's going to, L of x is going to be, it's going to be, it's going to be f of four. It's going to be f of four, which is two. It's going to be f of four plus the slope, the slope at x equals four, which is of course the derivative, f prime of four. So that's going to be the slope of this line, of L of x is f prime of four. Let me make that clear. So this right over here is the slope, the slope when x is, at x equals four."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "It's going to be f of four plus the slope, the slope at x equals four, which is of course the derivative, f prime of four. So that's going to be the slope of this line, of L of x is f prime of four. Let me make that clear. So this right over here is the slope, the slope when x is, at x equals four. So it's the slope of this entire line. And so any other point on this, it's going to be f of four plus the slope times how far you are away from x equals four. So it's going to be times x minus four."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So this right over here is the slope, the slope when x is, at x equals four. So it's the slope of this entire line. And so any other point on this, it's going to be f of four plus the slope times how far you are away from x equals four. So it's going to be times x minus four. Let's just validate that this makes sense. When we put 4.36 here, when we put 4.36 here, actually let me zoom in on this graph just to make things a little bit clearer. So if this is, so I'm going to do a zoom in, I'm going to do a zoom in, I'm going to try to zoom in into this region right over here."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So it's going to be times x minus four. Let's just validate that this makes sense. When we put 4.36 here, when we put 4.36 here, actually let me zoom in on this graph just to make things a little bit clearer. So if this is, so I'm going to do a zoom in, I'm going to do a zoom in, I'm going to try to zoom in into this region right over here. So this is the point, this is the point four comma f of four. And we are going to graph L of x. So let me do that."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So if this is, so I'm going to do a zoom in, I'm going to do a zoom in, I'm going to try to zoom in into this region right over here. So this is the point, this is the point four comma f of four. And we are going to graph L of x. So let me do that. So this right over here is L of x. That's L of x. And let's say this right over here, this right over here is the point 4.36 comma f of 4.36."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So let me do that. So this right over here is L of x. That's L of x. And let's say this right over here, this right over here is the point 4.36 comma f of 4.36. And the way we're going to approximate this value is to figure out what, to figure out what this value is right over here. And what is this one going to be? This right over here is going to be, this is going to be 4.36 comma L of 4.36."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And let's say this right over here, this right over here is the point 4.36 comma f of 4.36. And the way we're going to approximate this value is to figure out what, to figure out what this value is right over here. And what is this one going to be? This right over here is going to be, this is going to be 4.36 comma L of 4.36. This line evaluated when x is equal to 4.36. And what is that going to be? What is that going to be equal to?"}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "This right over here is going to be, this is going to be 4.36 comma L of 4.36. This line evaluated when x is equal to 4.36. And what is that going to be? What is that going to be equal to? Well let's see, let's just evaluate it. L of 4.36 is going to be f of four. So it's going to be two plus the derivative, so the slope of this line, plus f prime of four times x minus four."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "What is that going to be equal to? Well let's see, let's just evaluate it. L of 4.36 is going to be f of four. So it's going to be two plus the derivative, so the slope of this line, plus f prime of four times x minus four. So 4.36 minus four is going to be times 0.36. And that makes sense, you're starting at two, and you're saying okay, my change in x, my change in x is 4.36, so my change in y is going to be my slope times that change in x to get me that value, to get me that value right over there. So let's figure out, let's figure out what this, let's figure out what this thing, what this thing actually is."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So it's going to be two plus the derivative, so the slope of this line, plus f prime of four times x minus four. So 4.36 minus four is going to be times 0.36. And that makes sense, you're starting at two, and you're saying okay, my change in x, my change in x is 4.36, so my change in y is going to be my slope times that change in x to get me that value, to get me that value right over there. So let's figure out, let's figure out what this, let's figure out what this thing, what this thing actually is. So to do that we need to figure out f prime of four. So let's go back up here. I'll try to leave, actually, I'll leave this little visualization here."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So let's figure out, let's figure out what this, let's figure out what this thing, what this thing actually is. So to do that we need to figure out f prime of four. So let's go back up here. I'll try to leave, actually, I'll leave this little visualization here. So let's see, f prime, f prime of x is going to be 1.5 x to the negative 1.5, just using the power rule over here. So f prime of four, f prime of four is equal to 1.5 times four to the negative 1.5, which is of course equal to 1.5 times 1.5. Four to the 1.5 would be two, four to the negative 1.5 is going to be 1.5."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "I'll try to leave, actually, I'll leave this little visualization here. So let's see, f prime, f prime of x is going to be 1.5 x to the negative 1.5, just using the power rule over here. So f prime of four, f prime of four is equal to 1.5 times four to the negative 1.5, which is of course equal to 1.5 times 1.5. Four to the 1.5 would be two, four to the negative 1.5 is going to be 1.5. So this is equal to 1.4. So L of, we deserve a little bit of a drum roll now, L of 4.36 is equal to f of four, it's equal to f of four, which is, let me just rewrite it, it's f of four plus f prime of four plus, gee, why am I switching to that color? Let me do the yellow."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Four to the 1.5 would be two, four to the negative 1.5 is going to be 1.5. So this is equal to 1.4. So L of, we deserve a little bit of a drum roll now, L of 4.36 is equal to f of four, it's equal to f of four, which is, let me just rewrite it, it's f of four plus f prime of four plus, gee, why am I switching to that color? Let me do the yellow. Plus f prime of four times, times, times 4.36, 4.36, let me make this actually in a new color just so we see it. So 4.36, so times 4.36 minus four, minus four. Actually, let me make all the fours one color too so you see it's the same."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Let me do the yellow. Plus f prime of four times, times, times 4.36, 4.36, let me make this actually in a new color just so we see it. So 4.36, so times 4.36 minus four, minus four. Actually, let me make all the fours one color too so you see it's the same. So just like that. So what is this going to be? Well, this we already established is positive two."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Actually, let me make all the fours one color too so you see it's the same. So just like that. So what is this going to be? Well, this we already established is positive two. This we already established, let me do this in the yellow color. This we already established is 1.4, and this part right over here is 0.36. So this is going to be equal to two plus 1.4 times 0.36 is 0.9, or 0.09."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Well, this we already established is positive two. This we already established, let me do this in the yellow color. This we already established is 1.4, and this part right over here is 0.36. So this is going to be equal to two plus 1.4 times 0.36 is 0.9, or 0.09. So this is going to be equal to, this is going to be equal to 2.09. So that is our approximation, and it should be, at least based on how I graphed it, a little bit higher than the actual value of the square root of 4.36. But we could write that up here."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So this is going to be equal to two plus 1.4 times 0.36 is 0.9, or 0.09. So this is going to be equal to, this is going to be equal to 2.09. So that is our approximation, and it should be, at least based on how I graphed it, a little bit higher than the actual value of the square root of 4.36. But we could write that up here. This is going to be approximately, let me just write it this way. The square root, I'll just write it down here. So we could say the square root of 4.36, which is the same thing as f of 4.36, this is approximately equal to 2.09."}, {"video_title": "Local linearization Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "But we could write that up here. This is going to be approximately, let me just write it this way. The square root, I'll just write it down here. So we could say the square root of 4.36, which is the same thing as f of 4.36, this is approximately equal to 2.09. Now, let's just say we happen to find a calculator, and just out of curiosity, let's see how good of an approximation this actually is. Let's get a calculator out. And so we want to do the square root of 4.36, and we get 2.088."}, {"video_title": "Particular solution given initial conditions for population Khan Academy.mp3", "Sentence": "In the last video, we established that if we say the rate of change of a population with respect to time is going to be proportional to the population, we were able to solve that differential equation and find a general solution which involves an exponential, that the population is going to be equal to some constant times e to some other constant times time. In the last video, we assumed time was days. So let's just apply this just to feel good that we can truly model population in this way. So let's use it with some concrete numbers. And once again, you've probably done that before. You probably started with the assumption that you can model with an exponential function and then you use some information, some conditions, to figure out what the constants are. We probably did this earlier in pre-calculus or algebra class."}, {"video_title": "Particular solution given initial conditions for population Khan Academy.mp3", "Sentence": "So let's use it with some concrete numbers. And once again, you've probably done that before. You probably started with the assumption that you can model with an exponential function and then you use some information, some conditions, to figure out what the constants are. We probably did this earlier in pre-calculus or algebra class. But let's just do it again just so we can feel that this thing right over here is useful. So let's give you some information. Let's say that at time equals zero, the population is equal to 100 insects or whatever we're measuring the population of."}, {"video_title": "Particular solution given initial conditions for population Khan Academy.mp3", "Sentence": "We probably did this earlier in pre-calculus or algebra class. But let's just do it again just so we can feel that this thing right over here is useful. So let's give you some information. Let's say that at time equals zero, the population is equal to 100 insects or whatever we're measuring the population of. And let's say that at time equals, let's say at time equals 50, the population, so after 50 days, the population is 200. So notice it doubled after 50 days. So given this information, can we solve for C and K?"}, {"video_title": "Particular solution given initial conditions for population Khan Academy.mp3", "Sentence": "Let's say that at time equals zero, the population is equal to 100 insects or whatever we're measuring the population of. And let's say that at time equals, let's say at time equals 50, the population, so after 50 days, the population is 200. So notice it doubled after 50 days. So given this information, can we solve for C and K? And I encourage you to pause the video and try to work through it on your own. So this first initial condition is pretty straightforward to use because when T is equal to zero, P is 100. So we could say, let's just use, so based on this first piece of information, we could say that 100 must be equal to C, C times E, times E, to the K times zero."}, {"video_title": "Particular solution given initial conditions for population Khan Academy.mp3", "Sentence": "So given this information, can we solve for C and K? And I encourage you to pause the video and try to work through it on your own. So this first initial condition is pretty straightforward to use because when T is equal to zero, P is 100. So we could say, let's just use, so based on this first piece of information, we could say that 100 must be equal to C, C times E, times E, to the K times zero. Well, that's just going to be E to the zero. Well, E to the zero is just one. So this is just the same thing as C times one."}, {"video_title": "Particular solution given initial conditions for population Khan Academy.mp3", "Sentence": "So we could say, let's just use, so based on this first piece of information, we could say that 100 must be equal to C, C times E, times E, to the K times zero. Well, that's just going to be E to the zero. Well, E to the zero is just one. So this is just the same thing as C times one. And just like that, we have figured out what C is. So now we can write, we can now write that the population is going to be equal to 100, E to the KT. And so you can see, expressed this way, our C is always going to be our initial population."}, {"video_title": "Particular solution given initial conditions for population Khan Academy.mp3", "Sentence": "So this is just the same thing as C times one. And just like that, we have figured out what C is. So now we can write, we can now write that the population is going to be equal to 100, E to the KT. And so you can see, expressed this way, our C is always going to be our initial population. So E to the KT, E to the KT. And now we can use the second piece of information. So our population is 200."}, {"video_title": "Particular solution given initial conditions for population Khan Academy.mp3", "Sentence": "And so you can see, expressed this way, our C is always going to be our initial population. So E to the KT, E to the KT. And now we can use the second piece of information. So our population is 200. Let's write that down. So our population is 200 when time is equal to 50, after 50 days. So 200 is equal to 100E, 100E to the K times 50, right?"}, {"video_title": "Particular solution given initial conditions for population Khan Academy.mp3", "Sentence": "So our population is 200. Let's write that down. So our population is 200 when time is equal to 50, after 50 days. So 200 is equal to 100E, 100E to the K times 50, right? T is now 50. So we can, let me just write that, times K times 50. Now we can divide both sides by 100, and we will get two is equal to E to the 50K, E to the 50K."}, {"video_title": "Particular solution given initial conditions for population Khan Academy.mp3", "Sentence": "So 200 is equal to 100E, 100E to the K times 50, right? T is now 50. So we can, let me just write that, times K times 50. Now we can divide both sides by 100, and we will get two is equal to E to the 50K, E to the 50K. Then we can take the natural log of both sides, natural log on the left-hand side, we get the natural log of two. And on the right-hand side, the natural log of E to the 50K, well, that's just going to be, that's the power that you need to raise E to to get E to the 50K. Well, that's just going to be 50K."}, {"video_title": "Particular solution given initial conditions for population Khan Academy.mp3", "Sentence": "Now we can divide both sides by 100, and we will get two is equal to E to the 50K, E to the 50K. Then we can take the natural log of both sides, natural log on the left-hand side, we get the natural log of two. And on the right-hand side, the natural log of E to the 50K, well, that's just going to be, that's the power that you need to raise E to to get E to the 50K. Well, that's just going to be 50K. That's just going to be 50K. All I did is took the natural log of both sides. Notice that this equation that I've just written expresses the same thing."}, {"video_title": "Particular solution given initial conditions for population Khan Academy.mp3", "Sentence": "Well, that's just going to be 50K. That's just going to be 50K. All I did is took the natural log of both sides. Notice that this equation that I've just written expresses the same thing. Natural log of two is equal to 50K, that means E to the 50K is equal to two, which is exactly what we had written there. And now we can solve for K, divide both sides by 50. And we are left with K is equal to the natural log of two, natural log of two over 50, over 50."}, {"video_title": "Particular solution given initial conditions for population Khan Academy.mp3", "Sentence": "Notice that this equation that I've just written expresses the same thing. Natural log of two is equal to 50K, that means E to the 50K is equal to two, which is exactly what we had written there. And now we can solve for K, divide both sides by 50. And we are left with K is equal to the natural log of two, natural log of two over 50, over 50. And we're done. We can now write the particular solution that meets these conditions. So we can now write that our population, and I can even write our population as a function of time, is going to be equal to, is going to be equal to 100, 100 times E to the, now K is natural log of two over 50, so I'll write that."}, {"video_title": "Functions with same limit at infinity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So if we were to make the general statement that the limit of some function f of x as x approaches infinity is equal to three, what I wanna do in this video is show some examples of that and to show that we can keep creating more and more examples, really an infinite number of examples where that is going to be true. So for example, we could look at this graph over here. And in other videos, we'll think about why this is the case, but just think about what happens when you have very, very large x's. When you have very, very large x's, the plus five doesn't matter as much, and so it gets closer and closer to three x squared over x squared, which is equal to three. And you could see that right over here. It's graphed in this green color. And you can see, even when x is equal to 10, we're getting awfully close to three right over there."}, {"video_title": "Functions with same limit at infinity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "When you have very, very large x's, the plus five doesn't matter as much, and so it gets closer and closer to three x squared over x squared, which is equal to three. And you could see that right over here. It's graphed in this green color. And you can see, even when x is equal to 10, we're getting awfully close to three right over there. Let me zoom out a little bit so you see our axis. So that is three. Actually, let me draw a dotted line at the asymptote."}, {"video_title": "Functions with same limit at infinity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And you can see, even when x is equal to 10, we're getting awfully close to three right over there. Let me zoom out a little bit so you see our axis. So that is three. Actually, let me draw a dotted line at the asymptote. That is y is equal to three, and so you see the function's getting closer and closer as x approaches infinity. But that's not the only function that could do that. As I keep saying, there's an infinite number of functions that could do that."}, {"video_title": "Functions with same limit at infinity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Actually, let me draw a dotted line at the asymptote. That is y is equal to three, and so you see the function's getting closer and closer as x approaches infinity. But that's not the only function that could do that. As I keep saying, there's an infinite number of functions that could do that. You could have this somewhat wild function that involves natural logs. That too, as x approaches infinity, it is getting closer and closer to three. It might be getting closer to three at a slightly slower rate than the one in green, but we're talking about infinity."}, {"video_title": "Functions with same limit at infinity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "As I keep saying, there's an infinite number of functions that could do that. You could have this somewhat wild function that involves natural logs. That too, as x approaches infinity, it is getting closer and closer to three. It might be getting closer to three at a slightly slower rate than the one in green, but we're talking about infinity. As x approaches infinity, this thing is approaching three. And as we've talked about in other videos, you could even have things that keep oscillating around the asymptote as long as they're getting closer and closer and closer to it as x gets larger and larger and larger. So, for example, that function right over there."}, {"video_title": "Functions with same limit at infinity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "It might be getting closer to three at a slightly slower rate than the one in green, but we're talking about infinity. As x approaches infinity, this thing is approaching three. And as we've talked about in other videos, you could even have things that keep oscillating around the asymptote as long as they're getting closer and closer and closer to it as x gets larger and larger and larger. So, for example, that function right over there. Let me zoom in. So let's zoom in. Let's say when x is equal to 14, we can see that they're all approaching three."}, {"video_title": "Functions with same limit at infinity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So, for example, that function right over there. Let me zoom in. So let's zoom in. Let's say when x is equal to 14, we can see that they're all approaching three. The purple one is oscillating around it. The other two are approaching three from below. But as we get much larger, let me actually zoom out a ways, and then I'll zoom in."}, {"video_title": "Functions with same limit at infinity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say when x is equal to 14, we can see that they're all approaching three. The purple one is oscillating around it. The other two are approaching three from below. But as we get much larger, let me actually zoom out a ways, and then I'll zoom in. So let's get to really large values. So, actually, even 100 isn't even that large if we're thinking about infinity. Even a trillion wouldn't be that large if we're thinking about infinity."}, {"video_title": "Functions with same limit at infinity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "But as we get much larger, let me actually zoom out a ways, and then I'll zoom in. So let's get to really large values. So, actually, even 100 isn't even that large if we're thinking about infinity. Even a trillion wouldn't be that large if we're thinking about infinity. But let's go to 200. 200 is much larger than the numbers we've been looking at. And let me zoom in when x is equal to 200."}, {"video_title": "Functions with same limit at infinity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Even a trillion wouldn't be that large if we're thinking about infinity. But let's go to 200. 200 is much larger than the numbers we've been looking at. And let me zoom in when x is equal to 200. And you can see we have to zoom in an awfully lot, an awful lot, just to even see that the graphs still aren't quite stabilized around the asymptote, that they are a little bit different than the asymptote. I am really zoomed in. I mean, look at the scale."}, {"video_title": "Functions with same limit at infinity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And let me zoom in when x is equal to 200. And you can see we have to zoom in an awfully lot, an awful lot, just to even see that the graphs still aren't quite stabilized around the asymptote, that they are a little bit different than the asymptote. I am really zoomed in. I mean, look at the scale. This is, each of these are now a hundredth, each square. And so we've gotten much, much, much closer to the asymptote. In fact, the green function, we still can't tell the difference."}, {"video_title": "Functions with same limit at infinity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "I mean, look at the scale. This is, each of these are now a hundredth, each square. And so we've gotten much, much, much closer to the asymptote. In fact, the green function, we still can't tell the difference. You can see the calculation. This is up to three or four decimal places. We're getting awfully close to three now, but we aren't there."}, {"video_title": "Functions with same limit at infinity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "In fact, the green function, we still can't tell the difference. You can see the calculation. This is up to three or four decimal places. We're getting awfully close to three now, but we aren't there. So the green function's got there the fastest, is an argument. But the whole point of this is to emphasize the fact that there's an infinite number of functions for which you could make the statement that we made, that the limit of the function as x approaches infinity, in this case, we said that limit is going to be equal to three and I just picked three arbitrarily. This could be true for any, for any function."}, {"video_title": "Functions with same limit at infinity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "We're getting awfully close to three now, but we aren't there. So the green function's got there the fastest, is an argument. But the whole point of this is to emphasize the fact that there's an infinite number of functions for which you could make the statement that we made, that the limit of the function as x approaches infinity, in this case, we said that limit is going to be equal to three and I just picked three arbitrarily. This could be true for any, for any function. And I'm trying to, I didn't realize how much I had zoomed in. So let me now go back to the origin where we had our original expression. So there we have it."}, {"video_title": "Functions with same limit at infinity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "This could be true for any, for any function. And I'm trying to, I didn't realize how much I had zoomed in. So let me now go back to the origin where we had our original expression. So there we have it. Maybe I could zoom in this way. So there you have it. Limit of any of these as x approaches infinity is equal to three."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation AP Calculus AB Khan Academy.mp3", "Sentence": "So what I want to do is figure out the slope at x is equal to 1. So when x is equal to 1, as you can imagine, once we implicitly take the derivative of this, we're going to have that as a function of x and y. So it will be useful to know what y value we get to when our x is equal to 1. So let's figure that out right now. So when x is equal to 1, our relationship right over here becomes 1 squared, which is just 1, plus y minus 1 to the third power is equal to 28. Subtract 1 from both sides, you get y minus 1 to the third power is equal to 27. It looks like the numbers work out quite neatly for us."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation AP Calculus AB Khan Academy.mp3", "Sentence": "So let's figure that out right now. So when x is equal to 1, our relationship right over here becomes 1 squared, which is just 1, plus y minus 1 to the third power is equal to 28. Subtract 1 from both sides, you get y minus 1 to the third power is equal to 27. It looks like the numbers work out quite neatly for us. Take the cube root of both sides, you get y minus 1 is equal to 3. Add 1 to both sides, you get y is equal to 4. So we really want to figure out the slope at the point 1, 1, 4, which is right over here."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation AP Calculus AB Khan Academy.mp3", "Sentence": "It looks like the numbers work out quite neatly for us. Take the cube root of both sides, you get y minus 1 is equal to 3. Add 1 to both sides, you get y is equal to 4. So we really want to figure out the slope at the point 1, 1, 4, which is right over here. When x is 1, y is 4. So we want to figure out the slope of the tangent line right over there. So let's start doing some implicit differentiation."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation AP Calculus AB Khan Academy.mp3", "Sentence": "So we really want to figure out the slope at the point 1, 1, 4, which is right over here. When x is 1, y is 4. So we want to figure out the slope of the tangent line right over there. So let's start doing some implicit differentiation. So we're going to take the derivative of both sides of this relationship or this equation, depending on how you want to view it. And so let's skip down here, past the orange. So the derivative with respect to x of x squared is going to be 2x."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation AP Calculus AB Khan Academy.mp3", "Sentence": "So let's start doing some implicit differentiation. So we're going to take the derivative of both sides of this relationship or this equation, depending on how you want to view it. And so let's skip down here, past the orange. So the derivative with respect to x of x squared is going to be 2x. And the derivative with respect to x of something to the third power is going to be 3 times that something squared times the derivative of that something with respect to x. So what's the derivative of this with respect to x? Well, the derivative of y with respect to x is just dy dx."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation AP Calculus AB Khan Academy.mp3", "Sentence": "So the derivative with respect to x of x squared is going to be 2x. And the derivative with respect to x of something to the third power is going to be 3 times that something squared times the derivative of that something with respect to x. So what's the derivative of this with respect to x? Well, the derivative of y with respect to x is just dy dx. And then the derivative of x with respect to x is just 1. So we have minus 1. And on the right-hand side, we just get 0."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the derivative of y with respect to x is just dy dx. And then the derivative of x with respect to x is just 1. So we have minus 1. And on the right-hand side, we just get 0. Derivative of a constant is just equal to 0. Now we need to solve for dy dx. So we get 2x."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation AP Calculus AB Khan Academy.mp3", "Sentence": "And on the right-hand side, we just get 0. Derivative of a constant is just equal to 0. Now we need to solve for dy dx. So we get 2x. And so if we distribute this business times the dy dx and times the negative 1, when we multiply it times the dy dx, we get, and actually I'm going to write it over here, so we get plus 3 times y minus x squared times dy dx. And then when we multiply it times the negative 1, we get negative 3 times y minus x squared. And then, of course, all of that is going to be equal to 0."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation AP Calculus AB Khan Academy.mp3", "Sentence": "So we get 2x. And so if we distribute this business times the dy dx and times the negative 1, when we multiply it times the dy dx, we get, and actually I'm going to write it over here, so we get plus 3 times y minus x squared times dy dx. And then when we multiply it times the negative 1, we get negative 3 times y minus x squared. And then, of course, all of that is going to be equal to 0. Now all we have to do is take this and put it on the right-hand side. So we'll subtract it from both sides of this equation. So on the left-hand side, and actually all the stuff that's not a dy dx I'm going to write in green."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation AP Calculus AB Khan Academy.mp3", "Sentence": "And then, of course, all of that is going to be equal to 0. Now all we have to do is take this and put it on the right-hand side. So we'll subtract it from both sides of this equation. So on the left-hand side, and actually all the stuff that's not a dy dx I'm going to write in green. So on the left-hand side, we're just left with 3 times y minus x squared times dy dx. dy, the derivative of y with respect to x, is equal to, I'm just going to subtract this from both sides, is equal to negative 2x plus this. So I could write it as 3 times y minus x squared plus, sorry, minus 2x."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation AP Calculus AB Khan Academy.mp3", "Sentence": "So on the left-hand side, and actually all the stuff that's not a dy dx I'm going to write in green. So on the left-hand side, we're just left with 3 times y minus x squared times dy dx. dy, the derivative of y with respect to x, is equal to, I'm just going to subtract this from both sides, is equal to negative 2x plus this. So I could write it as 3 times y minus x squared plus, sorry, minus 2x. So we're adding this to both sides, and we're subtracting this from both sides, minus 2x. And then to solve for dy dx, we've done this multiple times already, to solve for the derivative of y with respect to x. The derivative of y with respect to x is going to be equal to 3 times y minus x squared minus 2x, all of that over this stuff, 3 times y minus x squared."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation AP Calculus AB Khan Academy.mp3", "Sentence": "So I could write it as 3 times y minus x squared plus, sorry, minus 2x. So we're adding this to both sides, and we're subtracting this from both sides, minus 2x. And then to solve for dy dx, we've done this multiple times already, to solve for the derivative of y with respect to x. The derivative of y with respect to x is going to be equal to 3 times y minus x squared minus 2x, all of that over this stuff, 3 times y minus x squared. And we can leave it just like that for now. So what is the derivative of y with respect to x? What is the slope of the tangent line when x is 1 and y is equal to 4?"}, {"video_title": "Worked example Evaluating derivative with implicit differentiation AP Calculus AB Khan Academy.mp3", "Sentence": "The derivative of y with respect to x is going to be equal to 3 times y minus x squared minus 2x, all of that over this stuff, 3 times y minus x squared. And we can leave it just like that for now. So what is the derivative of y with respect to x? What is the slope of the tangent line when x is 1 and y is equal to 4? Well, we just have to substitute x is equal to 1 and y equals 4 into this expression. So it's going to be equal to 3 times 4 minus 1 squared minus 2 times 1, all of that over 3 times 4 minus 1 squared, which is equal to 4 minus 1 is 3. You square it, you get 9."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation AP Calculus AB Khan Academy.mp3", "Sentence": "What is the slope of the tangent line when x is 1 and y is equal to 4? Well, we just have to substitute x is equal to 1 and y equals 4 into this expression. So it's going to be equal to 3 times 4 minus 1 squared minus 2 times 1, all of that over 3 times 4 minus 1 squared, which is equal to 4 minus 1 is 3. You square it, you get 9. 9 times 3 is 27. You get 27 minus 2 in the numerator, which is going to be equal to 25. And then in the denominator, you get 3 times 9, which is 27."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation AP Calculus AB Khan Academy.mp3", "Sentence": "You square it, you get 9. 9 times 3 is 27. You get 27 minus 2 in the numerator, which is going to be equal to 25. And then in the denominator, you get 3 times 9, which is 27. So the slope is 25 27ths. So it's almost 1, but not quite. And that's actually what it looks like on this graph."}, {"video_title": "Worked example Evaluating derivative with implicit differentiation AP Calculus AB Khan Academy.mp3", "Sentence": "And then in the denominator, you get 3 times 9, which is 27. So the slope is 25 27ths. So it's almost 1, but not quite. And that's actually what it looks like on this graph. And actually, just to make sure you know where I got this graph, this was from Wolfram Alpha. Wolfram Alpha. I should have told you that from the beginning."}, {"video_title": "Worked example finding relative extrema AP Calculus AB Khan Academy.mp3", "Sentence": "And what we want to do without having to graph g, we want to figure out at what x values does g have a relative maximum. And just to remind us what's going on at a relative maximum, so let me draw a hypothetical function right over here. So a relative maximum is going to happen, so you can visually inspect this, okay, that looks like a relative maximum, this is kind of top of a mountain or top of a hill. These all look like relative maxima. And what's in common? Well, the graph, the function, is going from increasing to decreasing at each of those points. It's going from increasing to decreasing."}, {"video_title": "Worked example finding relative extrema AP Calculus AB Khan Academy.mp3", "Sentence": "These all look like relative maxima. And what's in common? Well, the graph, the function, is going from increasing to decreasing at each of those points. It's going from increasing to decreasing. Increasing to decreasing at either of the points. Or you could say that the first derivative is going from positive to negative. So if you look at this interval right over here, g prime is greater than zero, and then over the next interval, when you're decreasing, g prime would be less than zero."}, {"video_title": "Worked example finding relative extrema AP Calculus AB Khan Academy.mp3", "Sentence": "It's going from increasing to decreasing. Increasing to decreasing at either of the points. Or you could say that the first derivative is going from positive to negative. So if you look at this interval right over here, g prime is greater than zero, and then over the next interval, when you're decreasing, g prime would be less than zero. So what we really need to think about is when does g prime, so let me see, relative, we care about relative maximum point, and so that's essentially asking when does g prime go when does g prime go from positive to negative? From, I wrote frore, from g prime greater than zero to g prime less than zero. And the values that we could look at, or the points, are our critical points."}, {"video_title": "Worked example finding relative extrema AP Calculus AB Khan Academy.mp3", "Sentence": "So if you look at this interval right over here, g prime is greater than zero, and then over the next interval, when you're decreasing, g prime would be less than zero. So what we really need to think about is when does g prime, so let me see, relative, we care about relative maximum point, and so that's essentially asking when does g prime go when does g prime go from positive to negative? From, I wrote frore, from g prime greater than zero to g prime less than zero. And the values that we could look at, or the points, are our critical points. And critical points are where g prime is either zero or it is undefined. So let's think about it. Where is g prime of x equal to zero?"}, {"video_title": "Worked example finding relative extrema AP Calculus AB Khan Academy.mp3", "Sentence": "And the values that we could look at, or the points, are our critical points. And critical points are where g prime is either zero or it is undefined. So let's think about it. Where is g prime of x equal to zero? G prime of x is equal to zero when, let's just take g prime of x. We're gonna leverage the power rule right here. Four x to the third power, four x to the third minus five x to the fourth, minus five x to the fourth is equal to zero."}, {"video_title": "Worked example finding relative extrema AP Calculus AB Khan Academy.mp3", "Sentence": "Where is g prime of x equal to zero? G prime of x is equal to zero when, let's just take g prime of x. We're gonna leverage the power rule right here. Four x to the third power, four x to the third minus five x to the fourth, minus five x to the fourth is equal to zero. Let's see, we can factor out an x to the third. We have x to the third times four minus five x is equal to zero. So this is going to happen when x is equal to zero."}, {"video_title": "Worked example finding relative extrema AP Calculus AB Khan Academy.mp3", "Sentence": "Four x to the third power, four x to the third minus five x to the fourth, minus five x to the fourth is equal to zero. Let's see, we can factor out an x to the third. We have x to the third times four minus five x is equal to zero. So this is going to happen when x is equal to zero. That I could, let me not skip steps. So this is going to happen when x to the third is equal to zero, or four minus five x is equal to zero. For x to the third equaling zero, well that's only gonna happen when x is equal to zero."}, {"video_title": "Worked example finding relative extrema AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to happen when x is equal to zero. That I could, let me not skip steps. So this is going to happen when x to the third is equal to zero, or four minus five x is equal to zero. For x to the third equaling zero, well that's only gonna happen when x is equal to zero. And four minus five x equaling zero, we'll add five x to both sides. You get four is equal to five x. Divide both sides by five, you get four fifths is equal to x. So here, these are the two places where our derivative is equal to zero."}, {"video_title": "Worked example finding relative extrema AP Calculus AB Khan Academy.mp3", "Sentence": "For x to the third equaling zero, well that's only gonna happen when x is equal to zero. And four minus five x equaling zero, we'll add five x to both sides. You get four is equal to five x. Divide both sides by five, you get four fifths is equal to x. So here, these are the two places where our derivative is equal to zero. Now are there any places where our derivative is undefined? Well, our function right over here is just a straight up polynomial. Our derivative is another polynomial."}, {"video_title": "Worked example finding relative extrema AP Calculus AB Khan Academy.mp3", "Sentence": "So here, these are the two places where our derivative is equal to zero. Now are there any places where our derivative is undefined? Well, our function right over here is just a straight up polynomial. Our derivative is another polynomial. It is defined for all real numbers. So these are our two critical, our critical points, or we could even say critical values. So let's think about what g prime is doing on either side of these critical values."}, {"video_title": "Worked example finding relative extrema AP Calculus AB Khan Academy.mp3", "Sentence": "Our derivative is another polynomial. It is defined for all real numbers. So these are our two critical, our critical points, or we could even say critical values. So let's think about what g prime is doing on either side of these critical values. And I'll draw a little number line here to help us visualize this. And so, so there we go, a little bit of a number line. And let's see, we care about zero, and we care about four fifths."}, {"video_title": "Worked example finding relative extrema AP Calculus AB Khan Academy.mp3", "Sentence": "So let's think about what g prime is doing on either side of these critical values. And I'll draw a little number line here to help us visualize this. And so, so there we go, a little bit of a number line. And let's see, we care about zero, and we care about four fifths. So let's say this is negative one, this is zero, this is one. And so we have one critical point at, let me do this in magenta. We have one critical point here at x equals zero."}, {"video_title": "Worked example finding relative extrema AP Calculus AB Khan Academy.mp3", "Sentence": "And let's see, we care about zero, and we care about four fifths. So let's say this is negative one, this is zero, this is one. And so we have one critical point at, let me do this in magenta. We have one critical point here at x equals zero. And then we have another critical point, I will do this at x equals four fifths. So four fifths is right around there. So that is four fifths."}, {"video_title": "Worked example finding relative extrema AP Calculus AB Khan Academy.mp3", "Sentence": "We have one critical point here at x equals zero. And then we have another critical point, I will do this at x equals four fifths. So four fifths is right around there. So that is four fifths. And let's just think about what g, what g prime is doing in these intervals. And these critical points are the only places where g prime might switch sides, switch signs. So let's first think about this, let me pick some colors I haven't used yet."}, {"video_title": "Worked example finding relative extrema AP Calculus AB Khan Academy.mp3", "Sentence": "So that is four fifths. And let's just think about what g, what g prime is doing in these intervals. And these critical points are the only places where g prime might switch sides, switch signs. So let's first think about this, let me pick some colors I haven't used yet. So let's think about the interval from negative infinity to zero. So this is the open interval from negative infinity to zero. And we could just plug in a value, we could, let's try negative one."}, {"video_title": "Worked example finding relative extrema AP Calculus AB Khan Academy.mp3", "Sentence": "So let's first think about this, let me pick some colors I haven't used yet. So let's think about the interval from negative infinity to zero. So this is the open interval from negative infinity to zero. And we could just plug in a value, we could, let's try negative one. Negative one is pretty straightforward to evaluate. So let's see, you have four, you're gonna have four times negative one to the third power so that's gonna be four times negative one, minus five times negative one to the fourth power. So that's just gonna be one."}, {"video_title": "Worked example finding relative extrema AP Calculus AB Khan Academy.mp3", "Sentence": "And we could just plug in a value, we could, let's try negative one. Negative one is pretty straightforward to evaluate. So let's see, you have four, you're gonna have four times negative one to the third power so that's gonna be four times negative one, minus five times negative one to the fourth power. So that's just gonna be one. So let's see, this is going to be negative four minus five, which is negative nine. So right over here, g prime is equal to negative nine. And so we know over this whole interval, since it's to the left of this critical point, we know that g prime of x is less than zero."}, {"video_title": "Worked example finding relative extrema AP Calculus AB Khan Academy.mp3", "Sentence": "So that's just gonna be one. So let's see, this is going to be negative four minus five, which is negative nine. So right over here, g prime is equal to negative nine. And so we know over this whole interval, since it's to the left of this critical point, we know that g prime of x is less than zero. And so our function itself is decreasing over this interval. And so we know we need to go from increasing to decreasing. So you can already say, well we can't go from increasing to decreasing at this critical point because we're already decreasing to the left of it."}, {"video_title": "Worked example finding relative extrema AP Calculus AB Khan Academy.mp3", "Sentence": "And so we know over this whole interval, since it's to the left of this critical point, we know that g prime of x is less than zero. And so our function itself is decreasing over this interval. And so we know we need to go from increasing to decreasing. So you can already say, well we can't go from increasing to decreasing at this critical point because we're already decreasing to the left of it. But anyway, let's just think about what's happening in the other intervals. So in the interval between zero and 4 5ths, so that interval right over there, so it's between zero and 4 5ths. Well let's just sample a number there."}, {"video_title": "Worked example finding relative extrema AP Calculus AB Khan Academy.mp3", "Sentence": "So you can already say, well we can't go from increasing to decreasing at this critical point because we're already decreasing to the left of it. But anyway, let's just think about what's happening in the other intervals. So in the interval between zero and 4 5ths, so that interval right over there, so it's between zero and 4 5ths. Well let's just sample a number there. Let's say the number, I don't know, 1 1 2 might be really straightforward. So we can evaluate g prime of 1 1 2. G prime of 1 1 2 is equal to four times 1 1 2 to the third power."}, {"video_title": "Worked example finding relative extrema AP Calculus AB Khan Academy.mp3", "Sentence": "Well let's just sample a number there. Let's say the number, I don't know, 1 1 2 might be really straightforward. So we can evaluate g prime of 1 1 2. G prime of 1 1 2 is equal to four times 1 1 2 to the third power. 1 1 2 to the third power is 1 8th. So it's 4 8ths, or it's just 1 1 2, minus five times 1 1 2 to the fourth. So that's 5 1 6ths."}, {"video_title": "Worked example finding relative extrema AP Calculus AB Khan Academy.mp3", "Sentence": "G prime of 1 1 2 is equal to four times 1 1 2 to the third power. 1 1 2 to the third power is 1 8th. So it's 4 8ths, or it's just 1 1 2, minus five times 1 1 2 to the fourth. So that's 5 1 6ths. Minus 5 1 6ths. And so this is equal to 8 1 6ths minus 5 1 6ths, which is equal to 3 1 6ths, but the important thing is it's equal to a positive value. So in this blue interval right over there, and actually let me put 4 5ths in a different color so we see that it's not part of that interval."}, {"video_title": "Worked example finding relative extrema AP Calculus AB Khan Academy.mp3", "Sentence": "So that's 5 1 6ths. Minus 5 1 6ths. And so this is equal to 8 1 6ths minus 5 1 6ths, which is equal to 3 1 6ths, but the important thing is it's equal to a positive value. So in this blue interval right over there, and actually let me put 4 5ths in a different color so we see that it's not part of that interval. So in this light blue interval right here between zero and 4 5ths, g prime of x is greater than zero, so we know our function is increasing. And so let's see what's happening to the right of this. And the easiest value to try out would just be 1."}, {"video_title": "Worked example finding relative extrema AP Calculus AB Khan Academy.mp3", "Sentence": "So in this blue interval right over there, and actually let me put 4 5ths in a different color so we see that it's not part of that interval. So in this light blue interval right here between zero and 4 5ths, g prime of x is greater than zero, so we know our function is increasing. And so let's see what's happening to the right of this. And the easiest value to try out would just be 1. So let's try out x equals 1. It's in that interval. So when x equals 1, I'll just write g prime of 1 is equal to 4 minus 5, which is equal to negative 1."}, {"video_title": "Worked example finding relative extrema AP Calculus AB Khan Academy.mp3", "Sentence": "And the easiest value to try out would just be 1. So let's try out x equals 1. It's in that interval. So when x equals 1, I'll just write g prime of 1 is equal to 4 minus 5, which is equal to negative 1. So g prime of x is less than zero. g prime of x is less than zero. So our function, so we could say g is increasing here, it is decreasing, oh sorry, let me be careful."}, {"video_title": "Worked example finding relative extrema AP Calculus AB Khan Academy.mp3", "Sentence": "So when x equals 1, I'll just write g prime of 1 is equal to 4 minus 5, which is equal to negative 1. So g prime of x is less than zero. g prime of x is less than zero. So our function, so we could say g is increasing here, it is decreasing, oh sorry, let me be careful. G is decreasing here, the function itself is decreasing because our derivative is negative, then our function is increasing here because our derivative is positive, and then our function is decreasing here. So at what critical point are we going from increasing to decreasing? Well we're doing that at x equals 4 5ths."}, {"video_title": "Worked example finding relative extrema AP Calculus AB Khan Academy.mp3", "Sentence": "So our function, so we could say g is increasing here, it is decreasing, oh sorry, let me be careful. G is decreasing here, the function itself is decreasing because our derivative is negative, then our function is increasing here because our derivative is positive, and then our function is decreasing here. So at what critical point are we going from increasing to decreasing? Well we're doing that at x equals 4 5ths. So we have a relative maximum at x equals 4 5ths. If they said, well where do we have a relative minimum point? Well that's going to happen at x equals zero."}, {"video_title": "Finding derivative with fundamental theorem of calculus chain rule AP\u00ae\ufe0e Calculus Khan Academy.mp3", "Sentence": "Let's say that we have the function capital F of x, which we're going to define as the definite integral from one to sine of x, so that's an interesting upper bound right over there, of two t minus one, and of course, dt. And what we are curious about is trying to figure out what is F prime of x going to be equal to? So pause this video and see if you can figure that out. All right, so some of you might have been a little bit challenged by this notion of, hey, instead of an x on this upper bound, I now have a sine of x. If it was just an x, I could have used the fundamental theorem of calculus. Just to review that, if I had a function, let me call it h of x, if I have h of x that was defined as the definite integral from one to x of two t minus one dt, we know from the fundamental theorem of calculus that h prime of x would be simply this inner function with the t replaced by the x. It would just be two x minus one."}, {"video_title": "Finding derivative with fundamental theorem of calculus chain rule AP\u00ae\ufe0e Calculus Khan Academy.mp3", "Sentence": "All right, so some of you might have been a little bit challenged by this notion of, hey, instead of an x on this upper bound, I now have a sine of x. If it was just an x, I could have used the fundamental theorem of calculus. Just to review that, if I had a function, let me call it h of x, if I have h of x that was defined as the definite integral from one to x of two t minus one dt, we know from the fundamental theorem of calculus that h prime of x would be simply this inner function with the t replaced by the x. It would just be two x minus one. Pretty straightforward. But this one isn't quite as straightforward. Instead of having an x up here, our upper bound is a sine of x."}, {"video_title": "Finding derivative with fundamental theorem of calculus chain rule AP\u00ae\ufe0e Calculus Khan Academy.mp3", "Sentence": "It would just be two x minus one. Pretty straightforward. But this one isn't quite as straightforward. Instead of having an x up here, our upper bound is a sine of x. So one way to think about it is if we were to define g of x as being equal to sine of x, equal to sine of x, our capital F of x can be expressed as capital F of x is the same thing as h of, h of, instead of an x, everywhere we see an x, we're replacing it with a sine of x. So it's h of g of x, g of x. You can see the g of x right over there."}, {"video_title": "Finding derivative with fundamental theorem of calculus chain rule AP\u00ae\ufe0e Calculus Khan Academy.mp3", "Sentence": "Instead of having an x up here, our upper bound is a sine of x. So one way to think about it is if we were to define g of x as being equal to sine of x, equal to sine of x, our capital F of x can be expressed as capital F of x is the same thing as h of, h of, instead of an x, everywhere we see an x, we're replacing it with a sine of x. So it's h of g of x, g of x. You can see the g of x right over there. So you replace x with g of x for where in this expression, you get h of g of x, and that is capital F of x. Now why am I doing all of that? Well, this might start making you think about the chain rule, because if this is true, then that means that capital F prime of x is going to be equal to h prime of g of x, h prime of g of x, times g prime of x."}, {"video_title": "Finding derivative with fundamental theorem of calculus chain rule AP\u00ae\ufe0e Calculus Khan Academy.mp3", "Sentence": "You can see the g of x right over there. So you replace x with g of x for where in this expression, you get h of g of x, and that is capital F of x. Now why am I doing all of that? Well, this might start making you think about the chain rule, because if this is true, then that means that capital F prime of x is going to be equal to h prime of g of x, h prime of g of x, times g prime of x. And so what would that be? Well, we already know what h prime of x is. So, let me do this in another color."}, {"video_title": "Finding derivative with fundamental theorem of calculus chain rule AP\u00ae\ufe0e Calculus Khan Academy.mp3", "Sentence": "Well, this might start making you think about the chain rule, because if this is true, then that means that capital F prime of x is going to be equal to h prime of g of x, h prime of g of x, times g prime of x. And so what would that be? Well, we already know what h prime of x is. So, let me do this in another color. This part right over here is going to be equal to, everywhere we see an x here, we'll replace with a g of x. So it's going to be two, two times sine of x, two sine of x, and then minus one, minus one. This is this right over here."}, {"video_title": "Finding derivative with fundamental theorem of calculus chain rule AP\u00ae\ufe0e Calculus Khan Academy.mp3", "Sentence": "So, let me do this in another color. This part right over here is going to be equal to, everywhere we see an x here, we'll replace with a g of x. So it's going to be two, two times sine of x, two sine of x, and then minus one, minus one. This is this right over here. And then what's g prime of x? G prime of x, well, g prime of x is just, of course, the derivative of sine of x is cosine of x, is cosine of x. So this part right over here is going to be cosine of x."}, {"video_title": "Negative definite integrals Integration and accumulation of change AP Calculus AB Khan Academy.mp3", "Sentence": "We've already thought about what a definite integral means. If I'm taking the definite integral from a to b of f of x dx, I can just view that as the area below my function f, so if this is my y axis, this is my x axis, and y is equal to f of x, so something like that, y is equal to f of x, and if this is a and if this is b, I could just view this expression as being equal to this area, but what if my function was not above the x axis? What if it was below the x axis? So these are going to be equivalent. Let's say, let me just draw that scenario, so let me draw a scenario where it's my x axis, that is my y axis, and let's say I have, let's say I have a function that looks like that, so that is y is equal to g of x, and let's say that this right over here is a, and this right over here is b, and let's say that this area right over here is equal to five. Well, if I were to ask you what is the definite integral from a to b of g of x dx, what do you think it is going to be? Well, you might be tempted to say, hey, well, it's just the area again between my curve and the x axis."}, {"video_title": "Negative definite integrals Integration and accumulation of change AP Calculus AB Khan Academy.mp3", "Sentence": "So these are going to be equivalent. Let's say, let me just draw that scenario, so let me draw a scenario where it's my x axis, that is my y axis, and let's say I have, let's say I have a function that looks like that, so that is y is equal to g of x, and let's say that this right over here is a, and this right over here is b, and let's say that this area right over here is equal to five. Well, if I were to ask you what is the definite integral from a to b of g of x dx, what do you think it is going to be? Well, you might be tempted to say, hey, well, it's just the area again between my curve and the x axis. You might be tempted to say, hey, this is just going to be equal to five, but you have to be very careful, because if you're looking at the area above your curve and below your x axis versus below your curve and above the x axis, this definite integral is actually going to be the negative of the area. Now, we'll see later on why this will work out nicely with a whole set of integration properties, but if you want to get some intuition for it, let's just think about velocity versus time graphs. So if I, in my horizontal axis, that is time."}, {"video_title": "Negative definite integrals Integration and accumulation of change AP Calculus AB Khan Academy.mp3", "Sentence": "Well, you might be tempted to say, hey, well, it's just the area again between my curve and the x axis. You might be tempted to say, hey, this is just going to be equal to five, but you have to be very careful, because if you're looking at the area above your curve and below your x axis versus below your curve and above the x axis, this definite integral is actually going to be the negative of the area. Now, we'll see later on why this will work out nicely with a whole set of integration properties, but if you want to get some intuition for it, let's just think about velocity versus time graphs. So if I, in my horizontal axis, that is time. My vertical axis, this is velocity, and velocity is going to be measured in meters per second. Time is going to be measured in seconds. Time is measured in seconds, and actually I'm gonna do two scenarios here."}, {"video_title": "Negative definite integrals Integration and accumulation of change AP Calculus AB Khan Academy.mp3", "Sentence": "So if I, in my horizontal axis, that is time. My vertical axis, this is velocity, and velocity is going to be measured in meters per second. Time is going to be measured in seconds. Time is measured in seconds, and actually I'm gonna do two scenarios here. So let's say that I have a first velocity time graph. Let's just call it v one of t, which is equal to three, and it would be three meters per second. So one, two, three."}, {"video_title": "Negative definite integrals Integration and accumulation of change AP Calculus AB Khan Academy.mp3", "Sentence": "Time is measured in seconds, and actually I'm gonna do two scenarios here. So let's say that I have a first velocity time graph. Let's just call it v one of t, which is equal to three, and it would be three meters per second. So one, two, three. So it would look like that. That is v one of t. And if I were to look at the definite integral going from time equals one to time equals five of v sub one of t dt, what would this be equal to? Well, here my function is above my t axis, so I'll just go from one to five, which will be around there, and I could just think about the area here, and this area is pretty easy to calculate."}, {"video_title": "Negative definite integrals Integration and accumulation of change AP Calculus AB Khan Academy.mp3", "Sentence": "So one, two, three. So it would look like that. That is v one of t. And if I were to look at the definite integral going from time equals one to time equals five of v sub one of t dt, what would this be equal to? Well, here my function is above my t axis, so I'll just go from one to five, which will be around there, and I could just think about the area here, and this area is pretty easy to calculate. It's going to be three meters per second times four seconds. That's my change in time. And so this is going to be 12 meters."}, {"video_title": "Negative definite integrals Integration and accumulation of change AP Calculus AB Khan Academy.mp3", "Sentence": "Well, here my function is above my t axis, so I'll just go from one to five, which will be around there, and I could just think about the area here, and this area is pretty easy to calculate. It's going to be three meters per second times four seconds. That's my change in time. And so this is going to be 12 meters. And so this is going to be equal to 12. And one way to conceptualize this is this gives us our change in position. If my velocity is three meters per second, and since it's positive, you can conceptualize that as it's going to the right at three meters per second, what is my change in position?"}, {"video_title": "Negative definite integrals Integration and accumulation of change AP Calculus AB Khan Academy.mp3", "Sentence": "And so this is going to be 12 meters. And so this is going to be equal to 12. And one way to conceptualize this is this gives us our change in position. If my velocity is three meters per second, and since it's positive, you can conceptualize that as it's going to the right at three meters per second, what is my change in position? Well, I would have gone 12 meters to the right. And you don't need calculus to figure that out. Three meters per second times four seconds would be 12 meters."}, {"video_title": "Negative definite integrals Integration and accumulation of change AP Calculus AB Khan Academy.mp3", "Sentence": "If my velocity is three meters per second, and since it's positive, you can conceptualize that as it's going to the right at three meters per second, what is my change in position? Well, I would have gone 12 meters to the right. And you don't need calculus to figure that out. Three meters per second times four seconds would be 12 meters. But what if it were the other way around? What if I had another velocity function? Let's call that v sub two of t that is equal to negative two meters per second."}, {"video_title": "Negative definite integrals Integration and accumulation of change AP Calculus AB Khan Academy.mp3", "Sentence": "Three meters per second times four seconds would be 12 meters. But what if it were the other way around? What if I had another velocity function? Let's call that v sub two of t that is equal to negative two meters per second. And it's just a constant negative two meters per second. So this is v sub two of t right over here. What would or what should the definite integral from one to five of v sub two of t be dt be equal to?"}, {"video_title": "Negative definite integrals Integration and accumulation of change AP Calculus AB Khan Academy.mp3", "Sentence": "Let's call that v sub two of t that is equal to negative two meters per second. And it's just a constant negative two meters per second. So this is v sub two of t right over here. What would or what should the definite integral from one to five of v sub two of t be dt be equal to? Well, it should be equal to my change in position. But if my velocity is negative, that means I'm moving to the left. That means my change in position should be to the left as opposed to to the right."}, {"video_title": "Negative definite integrals Integration and accumulation of change AP Calculus AB Khan Academy.mp3", "Sentence": "What would or what should the definite integral from one to five of v sub two of t be dt be equal to? Well, it should be equal to my change in position. But if my velocity is negative, that means I'm moving to the left. That means my change in position should be to the left as opposed to to the right. And so we can just look at this area right over here. Well, if you just look at it as the rectangle, it's gonna be two times four, which is equal to eight. But you have to be very careful."}, {"video_title": "Negative definite integrals Integration and accumulation of change AP Calculus AB Khan Academy.mp3", "Sentence": "That means my change in position should be to the left as opposed to to the right. And so we can just look at this area right over here. Well, if you just look at it as the rectangle, it's gonna be two times four, which is equal to eight. But you have to be very careful. Since it is below my horizontal axis and above my function, this is going to be negative. And this should make a lot of sense. If I'm going two meters per second to the left for four seconds, or another way to think about it, if I'm going negative two meters per second for four seconds then my change in position is going to be negative eight meters."}, {"video_title": "Negative definite integrals Integration and accumulation of change AP Calculus AB Khan Academy.mp3", "Sentence": "But you have to be very careful. Since it is below my horizontal axis and above my function, this is going to be negative. And this should make a lot of sense. If I'm going two meters per second to the left for four seconds, or another way to think about it, if I'm going negative two meters per second for four seconds then my change in position is going to be negative eight meters. I would have moved eight meters to the left if we say the convention is negative means to the left. So the big takeaway is if it's below your function and above the horizontal axis, the definite integral, and if your a is less than b, then your definite integral is going to be positive. If your a is less than b, but your function over that interval is below the horizontal axis, then your definite integral is going to be negative."}, {"video_title": "Limits at infinity of quotients (Part 1) Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And what I want to think about is what is the limit of f of x as x approaches infinity? And there are several ways that you could do this. You could actually try to plug in larger and larger numbers for x and see if it seems to be approaching some value. Or you could reason through this. And when I talk about reasoning through this, it's to think about the behavior of this numerator and denominator as x gets very, very, very large. And when I'm talking about that, what I'm saying is as x gets very, very large, let's just focus on the numerator, as x gets very, very large, this term right over here in the numerator, 4x to the 5th, is going to become much, much more significant than any of these other things. Something squaring gets large, but something being raised to the 5th power gets raised that much, much faster."}, {"video_title": "Limits at infinity of quotients (Part 1) Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Or you could reason through this. And when I talk about reasoning through this, it's to think about the behavior of this numerator and denominator as x gets very, very, very large. And when I'm talking about that, what I'm saying is as x gets very, very large, let's just focus on the numerator, as x gets very, very large, this term right over here in the numerator, 4x to the 5th, is going to become much, much more significant than any of these other things. Something squaring gets large, but something being raised to the 5th power gets raised that much, much faster. Similarly, in the denominator, this term right over here, the highest degree term, 6x to the 5th, is going to grow much, much, much faster than any of these other terms. Even though this has 100 as a coefficient or a negative 100 as a coefficient, when you take something to the 5th power, it's going to grow so much faster than x squared. So as x gets very, very, very large, this thing is going to approximate 4x to the 5th over 6x to the 5th for very large x."}, {"video_title": "Limits at infinity of quotients (Part 1) Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Something squaring gets large, but something being raised to the 5th power gets raised that much, much faster. Similarly, in the denominator, this term right over here, the highest degree term, 6x to the 5th, is going to grow much, much, much faster than any of these other terms. Even though this has 100 as a coefficient or a negative 100 as a coefficient, when you take something to the 5th power, it's going to grow so much faster than x squared. So as x gets very, very, very large, this thing is going to approximate 4x to the 5th over 6x to the 5th for very large x. Or we could say as x approaches infinity. Now what could this be simplified to? Well, you have x to the 5th divided by x to the 5th."}, {"video_title": "Limits at infinity of quotients (Part 1) Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So as x gets very, very, very large, this thing is going to approximate 4x to the 5th over 6x to the 5th for very large x. Or we could say as x approaches infinity. Now what could this be simplified to? Well, you have x to the 5th divided by x to the 5th. These are going to grow together. So you can think of them as canceling out, and so you are left with 2 thirds. So what you could say is the limit of f of x as x approaches infinity, as x gets larger and larger and larger, all of these other terms aren't going to matter that much, and so it's going to approach 2 thirds."}, {"video_title": "Limits at infinity of quotients (Part 1) Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Well, you have x to the 5th divided by x to the 5th. These are going to grow together. So you can think of them as canceling out, and so you are left with 2 thirds. So what you could say is the limit of f of x as x approaches infinity, as x gets larger and larger and larger, all of these other terms aren't going to matter that much, and so it's going to approach 2 thirds. Now let's look at the graph and see if that actually makes sense. What we're actually saying is that we have a horizontal asymptote at y is equal to 2 thirds. So let's look at the graph."}, {"video_title": "Limits at infinity of quotients (Part 1) Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So what you could say is the limit of f of x as x approaches infinity, as x gets larger and larger and larger, all of these other terms aren't going to matter that much, and so it's going to approach 2 thirds. Now let's look at the graph and see if that actually makes sense. What we're actually saying is that we have a horizontal asymptote at y is equal to 2 thirds. So let's look at the graph. So right here is the graph. Got it from Wolfram Alpha. We see indeed as x gets larger and larger and larger, f of x seems to be approaching this value that looks right at around 2 thirds."}, {"video_title": "Limits at infinity of quotients (Part 1) Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So let's look at the graph. So right here is the graph. Got it from Wolfram Alpha. We see indeed as x gets larger and larger and larger, f of x seems to be approaching this value that looks right at around 2 thirds. So it looks like we have a horizontal asymptote right over here. Let me draw that a little bit neater. We have a horizontal asymptote right at 2 thirds."}, {"video_title": "Limits at infinity of quotients (Part 1) Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "We see indeed as x gets larger and larger and larger, f of x seems to be approaching this value that looks right at around 2 thirds. So it looks like we have a horizontal asymptote right over here. Let me draw that a little bit neater. We have a horizontal asymptote right at 2 thirds. So let me draw it as neatly as I can. So this right over here is y is equal to 2 thirds. The limit as x gets really, really large, as it approaches infinity, y is getting closer and closer and closer to 2 thirds."}, {"video_title": "Limits at infinity of quotients (Part 1) Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "We have a horizontal asymptote right at 2 thirds. So let me draw it as neatly as I can. So this right over here is y is equal to 2 thirds. The limit as x gets really, really large, as it approaches infinity, y is getting closer and closer and closer to 2 thirds. When we just look at the graph here, it seems like the same thing is happening from the bottom direction when x approaches negative infinity. So we could say the limit of f of x as x approaches negative infinity, that also looks like it's 2 thirds. We can use the exact same logic."}, {"video_title": "Limits at infinity of quotients (Part 1) Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "The limit as x gets really, really large, as it approaches infinity, y is getting closer and closer and closer to 2 thirds. When we just look at the graph here, it seems like the same thing is happening from the bottom direction when x approaches negative infinity. So we could say the limit of f of x as x approaches negative infinity, that also looks like it's 2 thirds. We can use the exact same logic. When x becomes a very, very negative number, as it becomes further and further to the left on the number line, the only terms that are going to matter are going to be the 4x to the 5th and the 6x to the 5th. So this is true for very large x's. It's also true for very negative x's."}, {"video_title": "Limits at infinity of quotients (Part 1) Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "We can use the exact same logic. When x becomes a very, very negative number, as it becomes further and further to the left on the number line, the only terms that are going to matter are going to be the 4x to the 5th and the 6x to the 5th. So this is true for very large x's. It's also true for very negative x's. So we could also say as x approaches negative infinity, this is also true. Then the x to the 5th over the x to the 5th is going to cancel out. These are the dominant terms."}, {"video_title": "Limits at infinity of quotients (Part 1) Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "It's also true for very negative x's. So we could also say as x approaches negative infinity, this is also true. Then the x to the 5th over the x to the 5th is going to cancel out. These are the dominant terms. We're going to get it equaling 2 thirds. Once again, you see that in the graph here. We have a horizontal asymptote at y is equal to 2 thirds."}, {"video_title": "Limits at infinity of quotients (Part 1) Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "These are the dominant terms. We're going to get it equaling 2 thirds. Once again, you see that in the graph here. We have a horizontal asymptote at y is equal to 2 thirds. Whether we take the limit of f of x as x approaches infinity, we get 2 thirds. And the limit of f of x as x approaches negative infinity is 2 thirds. In general, whenever you do this, you just have to think about what terms are going to dominate the rest and focus on those."}, {"video_title": "Definite integrals intro Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And as we'll see, they are all related, and we'll see that more and more in future videos, and we'll also get a better appreciation for even where the notation of a definite integral comes from. So let me draw some functions here. And we're actually gonna start thinking about areas under curves. So let me draw coordinate axes here. So that's my y-axis. This is my x-axis. Actually, I'm gonna do two cases."}, {"video_title": "Definite integrals intro Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So let me draw coordinate axes here. So that's my y-axis. This is my x-axis. Actually, I'm gonna do two cases. So this is my y-axis. This is my x-axis. And let's say I have some function here."}, {"video_title": "Definite integrals intro Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Actually, I'm gonna do two cases. So this is my y-axis. This is my x-axis. And let's say I have some function here. So this is f of x, right over there. And let's say that this is x equals a. And let me draw a line going straight up like that."}, {"video_title": "Definite integrals intro Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And let's say I have some function here. So this is f of x, right over there. And let's say that this is x equals a. And let me draw a line going straight up like that. And let's say that this is x equals b, just like that. And what we want to do is concern ourselves with the area under the graph, under the graph of y is equal to f of x, and above the x-axis, and between these two bounds, between x equals a and x equals b. So this area right over here."}, {"video_title": "Definite integrals intro Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And let me draw a line going straight up like that. And let's say that this is x equals b, just like that. And what we want to do is concern ourselves with the area under the graph, under the graph of y is equal to f of x, and above the x-axis, and between these two bounds, between x equals a and x equals b. So this area right over here. And you can already get an appreciation. We're not used to finding areas where one of the boundaries, or as we'll see in the future, many of the boundaries could actually be curves. But that's one of the powers of the definite integral, and one of the powers of integral calculus."}, {"video_title": "Definite integrals intro Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So this area right over here. And you can already get an appreciation. We're not used to finding areas where one of the boundaries, or as we'll see in the future, many of the boundaries could actually be curves. But that's one of the powers of the definite integral, and one of the powers of integral calculus. And so the notation for this area right over here would be the definite integral. And so we're gonna have our lower bound at x equals a, so we'll write it there. We'll have our upper bound at x equals b, right over there."}, {"video_title": "Definite integrals intro Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "But that's one of the powers of the definite integral, and one of the powers of integral calculus. And so the notation for this area right over here would be the definite integral. And so we're gonna have our lower bound at x equals a, so we'll write it there. We'll have our upper bound at x equals b, right over there. We're taking the area under the curve of f of x, f of x, and then dx. Now in the future, we're going to, especially once we start looking at Riemann sums, we'll get a better understanding of where this notation comes from. This actually comes from Leibniz, one of the founders of calculus."}, {"video_title": "Definite integrals intro Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "We'll have our upper bound at x equals b, right over there. We're taking the area under the curve of f of x, f of x, and then dx. Now in the future, we're going to, especially once we start looking at Riemann sums, we'll get a better understanding of where this notation comes from. This actually comes from Leibniz, one of the founders of calculus. This is known as the summa symbol. But for the sake of this video, you just need to know what this represents. This right over here, this represents the area under f of x between x equals a and x equals b."}, {"video_title": "Applying the chain rule twice Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And what we're curious about is what is the derivative of this with respect to x? What is dy dx, which we could also write as y prime? Well, there's a couple of ways to think about it. This isn't a straightforward expression here, but you might notice that I have something being raised to the third power. In fact, if we look at the outside of this expression, we have some business in here as being raised to the third power. And so one way to tackle this is to apply the chain rule. So if we apply the chain rule, it's gonna be the derivative of the outside with respect to the inside, or the something to the third power, the derivative of the something to the third power with respect to that something."}, {"video_title": "Applying the chain rule twice Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "This isn't a straightforward expression here, but you might notice that I have something being raised to the third power. In fact, if we look at the outside of this expression, we have some business in here as being raised to the third power. And so one way to tackle this is to apply the chain rule. So if we apply the chain rule, it's gonna be the derivative of the outside with respect to the inside, or the something to the third power, the derivative of the something to the third power with respect to that something. So it's going to be three times that something squared times the derivative with respect to x of that something. In this case, the something is sine, let me write that in a blue color, it is sine of x squared. It is sine of x squared."}, {"video_title": "Applying the chain rule twice Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So if we apply the chain rule, it's gonna be the derivative of the outside with respect to the inside, or the something to the third power, the derivative of the something to the third power with respect to that something. So it's going to be three times that something squared times the derivative with respect to x of that something. In this case, the something is sine, let me write that in a blue color, it is sine of x squared. It is sine of x squared. If I, no matter what was inside of these orange parentheses, I would put it inside of the orange parentheses and these orange brackets right over here. We learned that in the chain rule. So let's see, we know this is just a matter, the first part of the expression is just a matter of algebraic simplification."}, {"video_title": "Applying the chain rule twice Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "It is sine of x squared. If I, no matter what was inside of these orange parentheses, I would put it inside of the orange parentheses and these orange brackets right over here. We learned that in the chain rule. So let's see, we know this is just a matter, the first part of the expression is just a matter of algebraic simplification. But the second part, we need to now take the derivative of sine of x squared. Well now, we would want to use the chain rule again. So I'm gonna take the derivative, it's sine of something."}, {"video_title": "Applying the chain rule twice Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let's see, we know this is just a matter, the first part of the expression is just a matter of algebraic simplification. But the second part, we need to now take the derivative of sine of x squared. Well now, we would want to use the chain rule again. So I'm gonna take the derivative, it's sine of something. So this is going to be, the derivative of this is going to be the sine of something with respect to something. So that is cosine of that something times the derivative with respect to x of the something. In this case, the something is our x squared."}, {"video_title": "Applying the chain rule twice Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So I'm gonna take the derivative, it's sine of something. So this is going to be, the derivative of this is going to be the sine of something with respect to something. So that is cosine of that something times the derivative with respect to x of the something. In this case, the something is our x squared. And of course, we have all of this out front, which is the three times sine of x squared, and I could write it like this, squared. Alright, so we're getting close. Now we just have to figure out the derivative with respect to x of x squared, and we've seen that many times before."}, {"video_title": "Applying the chain rule twice Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "In this case, the something is our x squared. And of course, we have all of this out front, which is the three times sine of x squared, and I could write it like this, squared. Alright, so we're getting close. Now we just have to figure out the derivative with respect to x of x squared, and we've seen that many times before. That, we just use the power rule, that's going to be two x. Two x. And so if we wanted to write the dy dx, let me get a little bit of a mini drum roll here, this shouldn't take us too long."}, {"video_title": "Applying the chain rule twice Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Now we just have to figure out the derivative with respect to x of x squared, and we've seen that many times before. That, we just use the power rule, that's going to be two x. Two x. And so if we wanted to write the dy dx, let me get a little bit of a mini drum roll here, this shouldn't take us too long. Dy dx, I'll multiply the three times the two x, which is going to be six x, so I covered those so far, times sine squared of x squared, times sine squared of x squared, times cosine of x squared. And we are done. So we're applying the chain rule multiple times."}, {"video_title": "Finding relative extrema (first derivative test) AP Calculus AB Khan Academy.mp3", "Sentence": "But then we saw that the other way around isn't necessarily true. x equal a being a critical point does not necessarily mean that the function takes on a minimum or maximum value at that point. So what we're going to try to do in this video is try to come up with some criteria, especially involving the derivative of the function around x equals a, to figure out if it is a minimum or a maximum point. So let's look at what we saw in the last video. We saw that this point right over here is where the function takes on a maximum value. So this critical point in particular was x-naught. What made it a critical point was that the derivative is zero."}, {"video_title": "Finding relative extrema (first derivative test) AP Calculus AB Khan Academy.mp3", "Sentence": "So let's look at what we saw in the last video. We saw that this point right over here is where the function takes on a maximum value. So this critical point in particular was x-naught. What made it a critical point was that the derivative is zero. You have a critical point where either the derivative is zero or the derivative is undefined. So this is a critical point. And let's explore what the derivative is doing as we approach that point."}, {"video_title": "Finding relative extrema (first derivative test) AP Calculus AB Khan Academy.mp3", "Sentence": "What made it a critical point was that the derivative is zero. You have a critical point where either the derivative is zero or the derivative is undefined. So this is a critical point. And let's explore what the derivative is doing as we approach that point. So in order for this to be a maximum point, the function is increasing as we approach it. The function is increasing is another way of saying that the slope is positive. The slope is changing, but it stays positive the whole time, which means that the function is increasing."}, {"video_title": "Finding relative extrema (first derivative test) AP Calculus AB Khan Academy.mp3", "Sentence": "And let's explore what the derivative is doing as we approach that point. So in order for this to be a maximum point, the function is increasing as we approach it. The function is increasing is another way of saying that the slope is positive. The slope is changing, but it stays positive the whole time, which means that the function is increasing. And the slope being positive is another way of saying that the derivative is greater than zero as we approach that point. Now what happens as we pass that point? Right at that point, the slope is zero."}, {"video_title": "Finding relative extrema (first derivative test) AP Calculus AB Khan Academy.mp3", "Sentence": "The slope is changing, but it stays positive the whole time, which means that the function is increasing. And the slope being positive is another way of saying that the derivative is greater than zero as we approach that point. Now what happens as we pass that point? Right at that point, the slope is zero. But then as we pass that point, what has to happen in order for that to be a maximum point? Well, the value of the function has to go down. If the value of the function is going down, that means the slope is negative."}, {"video_title": "Finding relative extrema (first derivative test) AP Calculus AB Khan Academy.mp3", "Sentence": "Right at that point, the slope is zero. But then as we pass that point, what has to happen in order for that to be a maximum point? Well, the value of the function has to go down. If the value of the function is going down, that means the slope is negative. And that's another way of saying that the derivative is negative. So that seems like a pretty good criteria for identifying whether a critical point is a maximum point. So let's say that we have critical point A."}, {"video_title": "Finding relative extrema (first derivative test) AP Calculus AB Khan Academy.mp3", "Sentence": "If the value of the function is going down, that means the slope is negative. And that's another way of saying that the derivative is negative. So that seems like a pretty good criteria for identifying whether a critical point is a maximum point. So let's say that we have critical point A. We are at a maximum point if f prime of x switches signs from positive to negative as we cross x equals a. That's exactly what happened right over here. Let's make sure it happened at our other maximum point right over here."}, {"video_title": "Finding relative extrema (first derivative test) AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say that we have critical point A. We are at a maximum point if f prime of x switches signs from positive to negative as we cross x equals a. That's exactly what happened right over here. Let's make sure it happened at our other maximum point right over here. So right over here, as we approach that point, the function is increasing. The function increasing means that the slope is positive. It's a different positive slope."}, {"video_title": "Finding relative extrema (first derivative test) AP Calculus AB Khan Academy.mp3", "Sentence": "Let's make sure it happened at our other maximum point right over here. So right over here, as we approach that point, the function is increasing. The function increasing means that the slope is positive. It's a different positive slope. The slope is changing. It's actually getting more and more and more positive. But it is definitely positive."}, {"video_title": "Finding relative extrema (first derivative test) AP Calculus AB Khan Academy.mp3", "Sentence": "It's a different positive slope. The slope is changing. It's actually getting more and more and more positive. But it is definitely positive. So it's positive going into that point, and then it becomes negative after we cross that point. The slope was undefined right at the point, but it did switch signs from positive to negative as we crossed that critical point. So these both meet our criteria for being a maximum point."}, {"video_title": "Finding relative extrema (first derivative test) AP Calculus AB Khan Academy.mp3", "Sentence": "But it is definitely positive. So it's positive going into that point, and then it becomes negative after we cross that point. The slope was undefined right at the point, but it did switch signs from positive to negative as we crossed that critical point. So these both meet our criteria for being a maximum point. So, so far, our criteria seems pretty good. Now let's make sure that somehow this point right over here, which we identified in the last video as a critical point, let's make, and I think we call this, let's see, this was x zero, this was x one, this was x two, this was x one, this was x two, so this is x three. Let's make sure that this doesn't somehow meet the criteria because we see visually that this is not a maximum point."}, {"video_title": "Finding relative extrema (first derivative test) AP Calculus AB Khan Academy.mp3", "Sentence": "So these both meet our criteria for being a maximum point. So, so far, our criteria seems pretty good. Now let's make sure that somehow this point right over here, which we identified in the last video as a critical point, let's make, and I think we call this, let's see, this was x zero, this was x one, this was x two, this was x one, this was x two, so this is x three. Let's make sure that this doesn't somehow meet the criteria because we see visually that this is not a maximum point. So as we approach this, our slope is negative, and then as we cross it, our slope is still negative. We're still decreasing. So we haven't switched signs."}, {"video_title": "Finding relative extrema (first derivative test) AP Calculus AB Khan Academy.mp3", "Sentence": "Let's make sure that this doesn't somehow meet the criteria because we see visually that this is not a maximum point. So as we approach this, our slope is negative, and then as we cross it, our slope is still negative. We're still decreasing. So we haven't switched signs. So this does not meet our criteria, which is good. Now let's come up with a criteria for a minimum point, and I think you could see where this is likely to go. Well, we identified in the last video that this right over here is a minimum point."}, {"video_title": "Finding relative extrema (first derivative test) AP Calculus AB Khan Academy.mp3", "Sentence": "So we haven't switched signs. So this does not meet our criteria, which is good. Now let's come up with a criteria for a minimum point, and I think you could see where this is likely to go. Well, we identified in the last video that this right over here is a minimum point. We can see that. It's a local minimum just by looking at it. And what's the slope doing as we approach it?"}, {"video_title": "Finding relative extrema (first derivative test) AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we identified in the last video that this right over here is a minimum point. We can see that. It's a local minimum just by looking at it. And what's the slope doing as we approach it? So the function is decreasing. The slope is negative as we approach it. f prime of x is less than zero as we approach that point."}, {"video_title": "Finding relative extrema (first derivative test) AP Calculus AB Khan Academy.mp3", "Sentence": "And what's the slope doing as we approach it? So the function is decreasing. The slope is negative as we approach it. f prime of x is less than zero as we approach that point. And then right after we cross it, this wouldn't be a minimum point if the function were to keep decreasing somehow. The function needs to increase now. So let me do that same green."}, {"video_title": "Finding relative extrema (first derivative test) AP Calculus AB Khan Academy.mp3", "Sentence": "f prime of x is less than zero as we approach that point. And then right after we cross it, this wouldn't be a minimum point if the function were to keep decreasing somehow. The function needs to increase now. So let me do that same green. So right after that, the function starts increasing again. f prime of x is greater than zero. So this seems like pretty good criteria for a minimum point."}, {"video_title": "Finding relative extrema (first derivative test) AP Calculus AB Khan Academy.mp3", "Sentence": "So let me do that same green. So right after that, the function starts increasing again. f prime of x is greater than zero. So this seems like pretty good criteria for a minimum point. f prime of x switches signs from negative to positive as we cross a. If we have some critical point a, the function takes on a minimum value at a if the derivative of our function switches signs from negative to positive as we cross a, from negative to positive. Now once again, this point right over here, this critical point, x sub 3, does not meet that criteria."}, {"video_title": "Trig functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say that we have y is equal to the secant of three pi over two minus x. And what we want to do is we want to figure out what dy dx is, the derivative of y with respect to x is at x equal pi over four. And like always, pause this video and see if you could figure it out. Well as you can see here, we have a composite function. So we're taking the secant not just of x, but you could view this as of another expression that I guess you could define, or as of another function. So for example, if we call this right over here u of x. So let's do that."}, {"video_title": "Trig functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well as you can see here, we have a composite function. So we're taking the secant not just of x, but you could view this as of another expression that I guess you could define, or as of another function. So for example, if we call this right over here u of x. So let's do that. So if we say u of x is equal to three pi over two minus x, we could also figure out u prime of x is going to be equal to derivative of three pi over two, that's just going to be zero, derivative of minus x, so that's just going to be minus one, and you could just view that as a power rule. It's one times negative one times x to the zero power, which is just one. So there you go."}, {"video_title": "Trig functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So let's do that. So if we say u of x is equal to three pi over two minus x, we could also figure out u prime of x is going to be equal to derivative of three pi over two, that's just going to be zero, derivative of minus x, so that's just going to be minus one, and you could just view that as a power rule. It's one times negative one times x to the zero power, which is just one. So there you go. So we could view this as the derivative of secant with respect to u of x, and when we take the derivative, the derivative of secant with respect to u of x times the derivative of u with respect to x. And you might say, well what about the derivative of secant? Well in other videos we actually prove it out, and you could actually re-derive it."}, {"video_title": "Trig functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So there you go. So we could view this as the derivative of secant with respect to u of x, and when we take the derivative, the derivative of secant with respect to u of x times the derivative of u with respect to x. And you might say, well what about the derivative of secant? Well in other videos we actually prove it out, and you could actually re-derive it. Secant is just one over cosine of x, so it comes straight out of the chain rule. So in other videos we prove that the derivative of the secant of x, is equal to sine of x over cosine of x over cosine of x squared. So if we're trying to find the derivative of y with respect to x, well it's going to be the derivative of secant with respect to u of x times the derivative of u with respect to x."}, {"video_title": "Trig functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well in other videos we actually prove it out, and you could actually re-derive it. Secant is just one over cosine of x, so it comes straight out of the chain rule. So in other videos we prove that the derivative of the secant of x, is equal to sine of x over cosine of x over cosine of x squared. So if we're trying to find the derivative of y with respect to x, well it's going to be the derivative of secant with respect to u of x times the derivative of u with respect to x. So let's do that. The derivative of secant with respect to u of x, well instead of seeing an x everywhere, you're gonna see a u of x everywhere. So this is going to be sine of u of x, sine of u of x, u of x, and I could, I don't have to write u of x, I could write three pi over two minus x, but I'll write u of x right over here just to really visualize what we're doing."}, {"video_title": "Trig functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So if we're trying to find the derivative of y with respect to x, well it's going to be the derivative of secant with respect to u of x times the derivative of u with respect to x. So let's do that. The derivative of secant with respect to u of x, well instead of seeing an x everywhere, you're gonna see a u of x everywhere. So this is going to be sine of u of x, sine of u of x, u of x, and I could, I don't have to write u of x, I could write three pi over two minus x, but I'll write u of x right over here just to really visualize what we're doing. So sine of u of x over, over cosine squared of u of x. Cosine squared, let me do those parentheses in the blue color just to make sure that you identify it with the trig function. So cosine squared of u of x, u of x. So that's the derivative of secant with respect to u of x, and then the chain rule tells us it's gonna be that times u prime, u prime of x."}, {"video_title": "Trig functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be sine of u of x, sine of u of x, u of x, and I could, I don't have to write u of x, I could write three pi over two minus x, but I'll write u of x right over here just to really visualize what we're doing. So sine of u of x over, over cosine squared of u of x. Cosine squared, let me do those parentheses in the blue color just to make sure that you identify it with the trig function. So cosine squared of u of x, u of x. So that's the derivative of secant with respect to u of x, and then the chain rule tells us it's gonna be that times u prime, u prime of x. So what is this going to be equal to? Well, I could just substitute back. This is going to be equal to, I will write it like this, sine of u of x, which is three pi over two minus x, and I'll fill that in in a second, over cosine of u of x squared times u prime of x. U of x is three pi over two minus x."}, {"video_title": "Trig functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So that's the derivative of secant with respect to u of x, and then the chain rule tells us it's gonna be that times u prime, u prime of x. So what is this going to be equal to? Well, I could just substitute back. This is going to be equal to, I will write it like this, sine of u of x, which is three pi over two minus x, and I'll fill that in in a second, over cosine of u of x squared times u prime of x. U of x is three pi over two minus x. Three pi over two minus x. And then u prime of x, we already figured out is negative one, so I could write times negative one. Well, yeah, let me just leave it out there for now."}, {"video_title": "Trig functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "This is going to be equal to, I will write it like this, sine of u of x, which is three pi over two minus x, and I'll fill that in in a second, over cosine of u of x squared times u prime of x. U of x is three pi over two minus x. Three pi over two minus x. And then u prime of x, we already figured out is negative one, so I could write times negative one. Well, yeah, let me just leave it out there for now. I could have just put a negative out front, but I really want you to be able to see what I'm doing here. So now we want to evaluate at x equals pi over four. So when that is equal to pi over four, pi over four."}, {"video_title": "Trig functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, yeah, let me just leave it out there for now. I could have just put a negative out front, but I really want you to be able to see what I'm doing here. So now we want to evaluate at x equals pi over four. So when that is equal to pi over four, pi over four. So let's see, this is going to be, this is going to be equal to sine of, what's three pi over two minus pi over four? I'll do that over here. So if you have a common denominator, that is six pi over four, the same thing as three pi over two, minus pi over four, sorry, minus pi over four is equal to five, five pi over four."}, {"video_title": "Trig functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So when that is equal to pi over four, pi over four. So let's see, this is going to be, this is going to be equal to sine of, what's three pi over two minus pi over four? I'll do that over here. So if you have a common denominator, that is six pi over four, the same thing as three pi over two, minus pi over four, sorry, minus pi over four is equal to five, five pi over four. So it's sine of five pi over four, five pi over four, over cosine squared of five pi over four, and then times negative one. I could just put that out here. Now what is sine of five pi over four and cosine squared of five pi over four?"}, {"video_title": "Trig functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So if you have a common denominator, that is six pi over four, the same thing as three pi over two, minus pi over four, sorry, minus pi over four is equal to five, five pi over four. So it's sine of five pi over four, five pi over four, over cosine squared of five pi over four, and then times negative one. I could just put that out here. Now what is sine of five pi over four and cosine squared of five pi over four? Well, I don't have that memorized, but let's actually draw a unit circle and we should be able to figure out what that is. So a unit circle, I'm gonna try to hand draw it as best as I can. Please forgive me that this circle does not look really like a circle."}, {"video_title": "Trig functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Now what is sine of five pi over four and cosine squared of five pi over four? Well, I don't have that memorized, but let's actually draw a unit circle and we should be able to figure out what that is. So a unit circle, I'm gonna try to hand draw it as best as I can. Please forgive me that this circle does not look really like a circle. All right. Okay, so let me just remember my angles. So I, in my brain, I sometimes convert into degrees."}, {"video_title": "Trig functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Please forgive me that this circle does not look really like a circle. All right. Okay, so let me just remember my angles. So I, in my brain, I sometimes convert into degrees. Pi over four is 45 degrees. This is pi over two. This is three pi over four."}, {"video_title": "Trig functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So I, in my brain, I sometimes convert into degrees. Pi over four is 45 degrees. This is pi over two. This is three pi over four. This is four pi over four. This is five pi over four. Lands you right over there."}, {"video_title": "Trig functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "This is three pi over four. This is four pi over four. This is five pi over four. Lands you right over there. So if you wanted to see where you intersect the unit circle, this is at the point, this is at the point where your x coordinate is negative square root of two over two, negative square root of two over two, and your y coordinate is negative square root of two over two. If you're wondering how I got that, I encourage you to review the unit circle and some of the standard angles around the unit circle. You'll see that in the trigonometry section of Khan Academy."}, {"video_title": "Trig functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Lands you right over there. So if you wanted to see where you intersect the unit circle, this is at the point, this is at the point where your x coordinate is negative square root of two over two, negative square root of two over two, and your y coordinate is negative square root of two over two. If you're wondering how I got that, I encourage you to review the unit circle and some of the standard angles around the unit circle. You'll see that in the trigonometry section of Khan Academy. But this is enough for us because the sine is the y coordinate. It's the y coordinate here. So negative square root of two over two."}, {"video_title": "Trig functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "You'll see that in the trigonometry section of Khan Academy. But this is enough for us because the sine is the y coordinate. It's the y coordinate here. So negative square root of two over two. This is negative square root of two over two. And then the cosine is the x coordinate, which is also negative square root of two over two, but it's gonna be that squared. Negative square root of two over two."}, {"video_title": "Trig functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So negative square root of two over two. This is negative square root of two over two. And then the cosine is the x coordinate, which is also negative square root of two over two, but it's gonna be that squared. Negative square root of two over two. We're squaring it. So if we square this, it's gonna become, this is going to become, it's gonna become positive. And then square root of two squared is two."}, {"video_title": "Trig functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Negative square root of two over two. We're squaring it. So if we square this, it's gonna become, this is going to become, it's gonna become positive. And then square root of two squared is two. And then two squared is four. So it's 1 1 2. So this is, the denominator is equal to 1 1 2."}, {"video_title": "Trig functions differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And then square root of two squared is two. And then two squared is four. So it's 1 1 2. So this is, the denominator is equal to 1 1 2. See, the numerator, this negative cancels out with that negative. And so we are left with, and we deserve a little bit of a drum roll, that we are left with square root of two over two, that's the numerator, divided by 1 1 2. Well, that's the same thing as multiplying by two."}, {"video_title": "Definite integral as the limit of a Riemann sum AP Calculus AB Khan Academy.mp3", "Sentence": "We've done several videos already where we're approximating the area under a curve by breaking up that area into rectangles and then finding the sum of the areas of those rectangles as an approximation. This was actually the first example that we looked at where each of the rectangles had an equal width. So we equally partitioned the interval between our two boundaries, between a and b. And the height of the rectangle was the function evaluated at the left endpoint of each rectangle. And we wanted to generalize it and write it in sigma notation. It looked something like this. And this was one case."}, {"video_title": "Definite integral as the limit of a Riemann sum AP Calculus AB Khan Academy.mp3", "Sentence": "And the height of the rectangle was the function evaluated at the left endpoint of each rectangle. And we wanted to generalize it and write it in sigma notation. It looked something like this. And this was one case. Later on we looked at a situation where you define the height by the function evaluated at the right endpoint or at the midpoint. And then we even constructed trapezoids. And these are all particular instances of Riemann sums."}, {"video_title": "Definite integral as the limit of a Riemann sum AP Calculus AB Khan Academy.mp3", "Sentence": "And this was one case. Later on we looked at a situation where you define the height by the function evaluated at the right endpoint or at the midpoint. And then we even constructed trapezoids. And these are all particular instances of Riemann sums. So this right over here is a Riemann sum. And when people talk about Riemann sums, they're talking about the more general notion. You don't have to just do it this way."}, {"video_title": "Definite integral as the limit of a Riemann sum AP Calculus AB Khan Academy.mp3", "Sentence": "And these are all particular instances of Riemann sums. So this right over here is a Riemann sum. And when people talk about Riemann sums, they're talking about the more general notion. You don't have to just do it this way. You could use trapezoids. You don't even have to have equally spaced partitions. I used equally spaced partitions because it made things a little bit conceptually simpler."}, {"video_title": "Definite integral as the limit of a Riemann sum AP Calculus AB Khan Academy.mp3", "Sentence": "You don't have to just do it this way. You could use trapezoids. You don't even have to have equally spaced partitions. I used equally spaced partitions because it made things a little bit conceptually simpler. And this right here is a picture of the person that Riemann sums was named after. This is Bernard Riemann. And he made many contributions to mathematics."}, {"video_title": "Definite integral as the limit of a Riemann sum AP Calculus AB Khan Academy.mp3", "Sentence": "I used equally spaced partitions because it made things a little bit conceptually simpler. And this right here is a picture of the person that Riemann sums was named after. This is Bernard Riemann. And he made many contributions to mathematics. But what he's most known for, at least if you're taking a first year calculus course, is the Riemann sum and how this is used to define the Riemann integral. Both Newton and Leibniz had come up with the idea of the integral when they had formulated calculus. But the Riemann integral is kind of the most mainstream formal or I would say rigorous definition of what an integral is."}, {"video_title": "Definite integral as the limit of a Riemann sum AP Calculus AB Khan Academy.mp3", "Sentence": "And he made many contributions to mathematics. But what he's most known for, at least if you're taking a first year calculus course, is the Riemann sum and how this is used to define the Riemann integral. Both Newton and Leibniz had come up with the idea of the integral when they had formulated calculus. But the Riemann integral is kind of the most mainstream formal or I would say rigorous definition of what an integral is. So as you can imagine, this is one instance of a Riemann sum. We have n right over here. The larger n is, the better an approximation it's going to be."}, {"video_title": "Definite integral as the limit of a Riemann sum AP Calculus AB Khan Academy.mp3", "Sentence": "But the Riemann integral is kind of the most mainstream formal or I would say rigorous definition of what an integral is. So as you can imagine, this is one instance of a Riemann sum. We have n right over here. The larger n is, the better an approximation it's going to be. So his definition of an integral, which is the actual area under the curve, or his definition of a definite integral, which is the actual area under a curve between a and b, is to take this Riemann sum, it doesn't have to be this one, take any Riemann sum, and take the limit as n approaches infinity. So just to be clear, what's happening when n approaches infinity? Let me draw another diagram here."}, {"video_title": "Definite integral as the limit of a Riemann sum AP Calculus AB Khan Academy.mp3", "Sentence": "The larger n is, the better an approximation it's going to be. So his definition of an integral, which is the actual area under the curve, or his definition of a definite integral, which is the actual area under a curve between a and b, is to take this Riemann sum, it doesn't have to be this one, take any Riemann sum, and take the limit as n approaches infinity. So just to be clear, what's happening when n approaches infinity? Let me draw another diagram here. So let's say that's my y-axis, this is my x-axis, this is my function. As n approaches infinity, so this is a, this is b, you're just going to have a ton of rectangles. You're just going to get a ton of rectangles over there and they're going to become better and better approximations for the actual area."}, {"video_title": "Definite integral as the limit of a Riemann sum AP Calculus AB Khan Academy.mp3", "Sentence": "Let me draw another diagram here. So let's say that's my y-axis, this is my x-axis, this is my function. As n approaches infinity, so this is a, this is b, you're just going to have a ton of rectangles. You're just going to get a ton of rectangles over there and they're going to become better and better approximations for the actual area. And the actual area under the curve is denoted by the integral from a to b of f of x times dx. And just to make it, you see where this is coming from, or how these notations are close, or at least in my brain, how they're connected, delta x was the distance for each of these, was the width for each of these sections. This right here is delta x, so that is a delta x."}, {"video_title": "Definite integral as the limit of a Riemann sum AP Calculus AB Khan Academy.mp3", "Sentence": "You're just going to get a ton of rectangles over there and they're going to become better and better approximations for the actual area. And the actual area under the curve is denoted by the integral from a to b of f of x times dx. And just to make it, you see where this is coming from, or how these notations are close, or at least in my brain, how they're connected, delta x was the distance for each of these, was the width for each of these sections. This right here is delta x, so that is a delta x. This is another delta x, this is another delta x. A reasonable way to conceptualize what dx is, or what a differential is, is what delta x approaches if it becomes infinitely small. So you can view this, you can conceptualize this, and it's not a very rigorous way of thinking about it, it's an infinitely small delta x."}, {"video_title": "Definite integral as the limit of a Riemann sum AP Calculus AB Khan Academy.mp3", "Sentence": "This right here is delta x, so that is a delta x. This is another delta x, this is another delta x. A reasonable way to conceptualize what dx is, or what a differential is, is what delta x approaches if it becomes infinitely small. So you can view this, you can conceptualize this, and it's not a very rigorous way of thinking about it, it's an infinitely small delta x. It's one way that you can conceptualize this. So once again, you have your function times a little small change in delta x, and you are summing, although you're summing an infinite number of these things from a to b. So I'm going to leave you there, just so you see the connection, you know the name for these things, and once again, this one over here, this isn't the only Riemann sum, in fact this is often called the left Riemann sum, if you're using it with rectangles, you can do a right Riemann sum, you could use the midpoint, you could use a trapezoid, but if you take the limit of any of those Riemann sums, as n approaches infinity, then that you get as a Riemann definition of the integral."}, {"video_title": "Integrating scaled version of function AP Calculus AB Khan Academy.mp3", "Sentence": "Now what I want to explore in this video, and it'll come up with kind of an answer that you probably could have guessed on your own, but at least get an intuition for it, is I want to start thinking about the area under the curve that's a scaled version of f of x. Let's say it's y is equal to c times f of x. Y is equal to some number times f of x, so it's scaling f of x. And so I want this to be kind of some arbitrary number, but just to help me visualize, you know, I have to draw something. So I'm just gonna kind of in my head, let's just pretend like the c is a three, just for visualization purposes. So it's gonna be three times this all, instead of one, instead of this far right over here, it's gonna be about this far far right over here. Instead of this far right over here, it's gonna be that, and another right over there. And then instead of, it's gonna be about there, and then instead of it being like that, it's going to be one, two, and then three right around there."}, {"video_title": "Integrating scaled version of function AP Calculus AB Khan Academy.mp3", "Sentence": "So I'm just gonna kind of in my head, let's just pretend like the c is a three, just for visualization purposes. So it's gonna be three times this all, instead of one, instead of this far right over here, it's gonna be about this far far right over here. Instead of this far right over here, it's gonna be that, and another right over there. And then instead of, it's gonna be about there, and then instead of it being like that, it's going to be one, two, and then three right around there. So I'm starting to get a sense of what this curve is going to look like. It's going to, a scaled version of f of x, and for at least what I'm drawing, it's pretty close to three times f of x, but just to give you an idea. It's gonna look something like, and over here, let's see, at this distance, do a second one, a third one, it's gonna be up here."}, {"video_title": "Integrating scaled version of function AP Calculus AB Khan Academy.mp3", "Sentence": "And then instead of, it's gonna be about there, and then instead of it being like that, it's going to be one, two, and then three right around there. So I'm starting to get a sense of what this curve is going to look like. It's going to, a scaled version of f of x, and for at least what I'm drawing, it's pretty close to three times f of x, but just to give you an idea. It's gonna look something like, and over here, let's see, at this distance, do a second one, a third one, it's gonna be up here. It's gonna look something, something like this. It's gonna look something like that. So this is a scaled version."}, {"video_title": "Integrating scaled version of function AP Calculus AB Khan Academy.mp3", "Sentence": "It's gonna look something like, and over here, let's see, at this distance, do a second one, a third one, it's gonna be up here. It's gonna look something, something like this. It's gonna look something like that. So this is a scaled version. And the scale I did right here, I assumed a positive, I assumed a positive c greater than zero, but this is just for visualization purposes. Now, what do we think the area under this curve is going to be between a and b? So what do we think this, what do we think, what do we think this area, this area right over here is going to be?"}, {"video_title": "Integrating scaled version of function AP Calculus AB Khan Academy.mp3", "Sentence": "So this is a scaled version. And the scale I did right here, I assumed a positive, I assumed a positive c greater than zero, but this is just for visualization purposes. Now, what do we think the area under this curve is going to be between a and b? So what do we think this, what do we think, what do we think this area, this area right over here is going to be? And we already know how we can denote it. That area right over there is equal to the definite integral from a to b of the function we're integrating is c f of x dx. I guess to make the question a little bit clearer, how does this relate to this?"}, {"video_title": "Integrating scaled version of function AP Calculus AB Khan Academy.mp3", "Sentence": "So what do we think this, what do we think, what do we think this area, this area right over here is going to be? And we already know how we can denote it. That area right over there is equal to the definite integral from a to b of the function we're integrating is c f of x dx. I guess to make the question a little bit clearer, how does this relate to this? How does this green area relate to this yellow area? Well, one way to think about it is we just scaled the vertical dimension up by c. So one way that you could reason it is, well, if I'm finding the area of something, if I have the area of a rectangle, and I have the vertical dimension is, let's say the vertical dimension is, I don't want to use those same letters over and over again. Well, I'll just, well, let's say the vertical dimension is alpha and the horizontal dimension is beta."}, {"video_title": "Integrating scaled version of function AP Calculus AB Khan Academy.mp3", "Sentence": "I guess to make the question a little bit clearer, how does this relate to this? How does this green area relate to this yellow area? Well, one way to think about it is we just scaled the vertical dimension up by c. So one way that you could reason it is, well, if I'm finding the area of something, if I have the area of a rectangle, and I have the vertical dimension is, let's say the vertical dimension is, I don't want to use those same letters over and over again. Well, I'll just, well, let's say the vertical dimension is alpha and the horizontal dimension is beta. We know that the area, we know that the area is going to be alpha times beta. Now, if I scale up the vertical dimension by c, so instead of alpha, this is c times alpha, and this is the width is beta. If I scale up the vertical dimension by c, so this is, I have essentially, this is now c times alpha, what's the area going to be?"}, {"video_title": "Integrating scaled version of function AP Calculus AB Khan Academy.mp3", "Sentence": "Well, I'll just, well, let's say the vertical dimension is alpha and the horizontal dimension is beta. We know that the area, we know that the area is going to be alpha times beta. Now, if I scale up the vertical dimension by c, so instead of alpha, this is c times alpha, and this is the width is beta. If I scale up the vertical dimension by c, so this is, I have essentially, this is now c times alpha, what's the area going to be? Well, it's going to be c alpha times beta. Or another way to think of it, I have just taken, when I scale one of the dimensions by c, I take my old area and I scale up my old area by c. And that's what we're doing. We're scaling up the vertical dimension by c. When you multiply c times f of x, f of x is giving us the vertical height."}, {"video_title": "Integrating scaled version of function AP Calculus AB Khan Academy.mp3", "Sentence": "If I scale up the vertical dimension by c, so this is, I have essentially, this is now c times alpha, what's the area going to be? Well, it's going to be c alpha times beta. Or another way to think of it, I have just taken, when I scale one of the dimensions by c, I take my old area and I scale up my old area by c. And that's what we're doing. We're scaling up the vertical dimension by c. When you multiply c times f of x, f of x is giving us the vertical height. Now, obviously, that changes as our x changes, but when you think back to the Riemann sums, the f of x is what gave us the height of our rectangles. We're now scaling up the height, or scaling, I should say, because we might be scaling down, depending on the c. We're scaling it, we're scaling one dimension by c. If you scale one dimension by c, you're going to scale the area by c. So this right over here, the integral, let me just rewrite it, the integral from a to b of c f of x dx, that's just going to be the scaled, we're just going to take the area of f of x, so let me do that in that same color. We're going to take the area under the curve f of x from a to b f of x dx, and we're just going to scale it up, we're going to scale it up by this c. We're just going to scale it up by this c. So you might say, okay, maybe I could have felt that that was, you know, if it was a c inside the integral, now I can take the c out of the integral."}, {"video_title": "Integrating scaled version of function AP Calculus AB Khan Academy.mp3", "Sentence": "We're scaling up the vertical dimension by c. When you multiply c times f of x, f of x is giving us the vertical height. Now, obviously, that changes as our x changes, but when you think back to the Riemann sums, the f of x is what gave us the height of our rectangles. We're now scaling up the height, or scaling, I should say, because we might be scaling down, depending on the c. We're scaling it, we're scaling one dimension by c. If you scale one dimension by c, you're going to scale the area by c. So this right over here, the integral, let me just rewrite it, the integral from a to b of c f of x dx, that's just going to be the scaled, we're just going to take the area of f of x, so let me do that in that same color. We're going to take the area under the curve f of x from a to b f of x dx, and we're just going to scale it up, we're going to scale it up by this c. We're just going to scale it up by this c. So you might say, okay, maybe I could have felt that that was, you know, if it was a c inside the integral, now I can take the c out of the integral. Once again, this is not a rigorous proof based on the definition of the definite integral, but it hopefully gives you a little bit of intuition why you can do this. If you scale up the function, you're essentially scaling up the vertical dimension, so the area under this is going to just be a scaled up version of the area under the original function f of x. And once again, really, really, really useful property, really, really useful property of definite integrals that's going to help us solve a bunch of definite integrals and kind of clarify what we're even doing with them."}, {"video_title": "Analyzing related rates problems expressions AP Calculus AB Khan Academy.mp3", "Sentence": "What is the rate of change of the area, a of t, so our area's going to be a function of t, what is the rate of change of the area of the triangle at that instant? And so what we're going to do in this exercise, instead of going straight and trying to solve it, what we need to do here is to identify the various units of different expressions, and then try to think about what information is given and what's not, and then that will actually equip us to actually solve this rate of change problem. So let's just do this first part. Let's match each expression with its units, and like always, pause the video and see if you can do it on your own. All right, so the first one is b prime of t. So this is the rate of change of which the base is changing with respect to time. So if we think about it, b of t, that is the base, that is going to be in meters. So this is going to be in meters."}, {"video_title": "Analyzing related rates problems expressions AP Calculus AB Khan Academy.mp3", "Sentence": "Let's match each expression with its units, and like always, pause the video and see if you can do it on your own. All right, so the first one is b prime of t. So this is the rate of change of which the base is changing with respect to time. So if we think about it, b of t, that is the base, that is going to be in meters. So this is going to be in meters. If we say b prime of t, this is going to be how much our base is changing with respect to time. So this is going to be meters per, and they give us right over here, they say it's decreasing at a rate of 13 meters per hour. So the units here are meters per hour."}, {"video_title": "Analyzing related rates problems expressions AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be in meters. If we say b prime of t, this is going to be how much our base is changing with respect to time. So this is going to be meters per, and they give us right over here, they say it's decreasing at a rate of 13 meters per hour. So the units here are meters per hour. And so b prime of t, that is going to be in meters per hour. A at time t sub zero. Remember, a is the area of our triangle, and we're measuring everything in meters, as you can tell from the information they've given us."}, {"video_title": "Analyzing related rates problems expressions AP Calculus AB Khan Academy.mp3", "Sentence": "So the units here are meters per hour. And so b prime of t, that is going to be in meters per hour. A at time t sub zero. Remember, a is the area of our triangle, and we're measuring everything in meters, as you can tell from the information they've given us. And so area is going to be in square units, and so it's going to be in square meters. Now the height at time t sub zero. Well, both the base and the height, those are lengths, they're gonna be measured in meters."}, {"video_title": "Analyzing related rates problems expressions AP Calculus AB Khan Academy.mp3", "Sentence": "Remember, a is the area of our triangle, and we're measuring everything in meters, as you can tell from the information they've given us. And so area is going to be in square units, and so it's going to be in square meters. Now the height at time t sub zero. Well, both the base and the height, those are lengths, they're gonna be measured in meters. And so our height at time t sub zero is going to be in meters. And then here we have the rate of change of our area with respect to time. So our area, we already know, is in meters squared."}, {"video_title": "Analyzing related rates problems expressions AP Calculus AB Khan Academy.mp3", "Sentence": "Well, both the base and the height, those are lengths, they're gonna be measured in meters. And so our height at time t sub zero is going to be in meters. And then here we have the rate of change of our area with respect to time. So our area, we already know, is in meters squared. But we wanna know, this here, this is going to be the rate of change of our area with respect to time. So it's going to be an amount of area per unit time, and time here, we're using hours, as you can see from some of the information they've given us. So this is going to be area per unit time, or meters squared per hour."}, {"video_title": "Analyzing related rates problems expressions AP Calculus AB Khan Academy.mp3", "Sentence": "So our area, we already know, is in meters squared. But we wanna know, this here, this is going to be the rate of change of our area with respect to time. So it's going to be an amount of area per unit time, and time here, we're using hours, as you can see from some of the information they've given us. So this is going to be area per unit time, or meters squared per hour. So it's going to be right over here. So it's area per unit time, and the length we're using in this is meters, and time is hours. All right."}, {"video_title": "Analyzing related rates problems expressions AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be area per unit time, or meters squared per hour. So it's going to be right over here. So it's area per unit time, and the length we're using in this is meters, and time is hours. All right. Now they say, match each expression with its given value. So what is the base of the triangle at time t sub zero? Do they give that to us?"}, {"video_title": "Analyzing related rates problems expressions AP Calculus AB Khan Academy.mp3", "Sentence": "All right. Now they say, match each expression with its given value. So what is the base of the triangle at time t sub zero? Do they give that to us? Well, let's see. They say at a certain time, at a certain instant, t sub zero, the base, I'm gonna underline this in a different color, the base at a certain instant, t sub zero, the base is five meters. So they say the base at time t sub zero, the base is a function of time, but they tell us that it is five meters."}, {"video_title": "Analyzing related rates problems expressions AP Calculus AB Khan Academy.mp3", "Sentence": "Do they give that to us? Well, let's see. They say at a certain time, at a certain instant, t sub zero, the base, I'm gonna underline this in a different color, the base at a certain instant, t sub zero, the base is five meters. So they say the base at time t sub zero, the base is a function of time, but they tell us that it is five meters. So this is five meters right over here. Now what about the rate of change of the base with respect to time? Do they tell us that?"}, {"video_title": "Analyzing related rates problems expressions AP Calculus AB Khan Academy.mp3", "Sentence": "So they say the base at time t sub zero, the base is a function of time, but they tell us that it is five meters. So this is five meters right over here. Now what about the rate of change of the base with respect to time? Do they tell us that? Well, look right over here. That's actually the first piece of information they gave us. So the base, B of t of a triangle, is decreasing at a rate of 13 meters per hour."}, {"video_title": "Analyzing related rates problems expressions AP Calculus AB Khan Academy.mp3", "Sentence": "Do they tell us that? Well, look right over here. That's actually the first piece of information they gave us. So the base, B of t of a triangle, is decreasing at a rate of 13 meters per hour. So the rate of change of the base, that is B prime of t, which is equal to dB dt, and they tell us that that is, it's decreasing at a rate of 13 meters per hour. So that would be negative 13 meters per hour. And so the rate of change of the base with respect to time is going to be negative 13."}, {"video_title": "Analyzing related rates problems expressions AP Calculus AB Khan Academy.mp3", "Sentence": "So the base, B of t of a triangle, is decreasing at a rate of 13 meters per hour. So the rate of change of the base, that is B prime of t, which is equal to dB dt, and they tell us that that is, it's decreasing at a rate of 13 meters per hour. So that would be negative 13 meters per hour. And so the rate of change of the base with respect to time is going to be negative 13. They gave us that. Now A prime of t, this is the rate of change of the area at time t sub zero. Did they give us this?"}, {"video_title": "Analyzing related rates problems expressions AP Calculus AB Khan Academy.mp3", "Sentence": "And so the rate of change of the base with respect to time is going to be negative 13. They gave us that. Now A prime of t, this is the rate of change of the area at time t sub zero. Did they give us this? Well, they ask us that. What is the rate of change of the area, A of t of the triangle at that instant? So this is what we actually need to figure out, but they haven't given it to us, otherwise there's no problem to solve."}, {"video_title": "Analyzing related rates problems expressions AP Calculus AB Khan Academy.mp3", "Sentence": "Did they give us this? Well, they ask us that. What is the rate of change of the area, A of t of the triangle at that instant? So this is what we actually need to figure out, but they haven't given it to us, otherwise there's no problem to solve. So this one right over here is not given. This is what we are trying to solve for. And then finally we have the first derivative of the height with respect to time."}, {"video_title": "Analyzing related rates problems expressions AP Calculus AB Khan Academy.mp3", "Sentence": "So this is what we actually need to figure out, but they haven't given it to us, otherwise there's no problem to solve. So this one right over here is not given. This is what we are trying to solve for. And then finally we have the first derivative of the height with respect to time. So you could view this as dh dt. What is this going to be? Do they give it to us?"}, {"video_title": "Analyzing related rates problems expressions AP Calculus AB Khan Academy.mp3", "Sentence": "And then finally we have the first derivative of the height with respect to time. So you could view this as dh dt. What is this going to be? Do they give it to us? Well, look right over here. They say the height of the triangle is increasing at a rate of six meters per hour. So if they're saying h of t is increasing, they're telling us the rate of change of h of t with respect to time, so that's h prime of t, and they're telling us that it is increasing at six meters per hour."}, {"video_title": "Analyzing related rates problems expressions AP Calculus AB Khan Academy.mp3", "Sentence": "Do they give it to us? Well, look right over here. They say the height of the triangle is increasing at a rate of six meters per hour. So if they're saying h of t is increasing, they're telling us the rate of change of h of t with respect to time, so that's h prime of t, and they're telling us that it is increasing at six meters per hour. So it's gonna be positive six meters per hour. So they did indeed give us that. Now why is all of this a useful exercise to go through?"}, {"video_title": "Analyzing related rates problems expressions AP Calculus AB Khan Academy.mp3", "Sentence": "So if they're saying h of t is increasing, they're telling us the rate of change of h of t with respect to time, so that's h prime of t, and they're telling us that it is increasing at six meters per hour. So it's gonna be positive six meters per hour. So they did indeed give us that. Now why is all of this a useful exercise to go through? Well, now we are really ready to solve the question because in general, if we're talking about any triangle, we know that area is equal to 1 1 2 base times height. Now in this situation, area and our base and our height, they're all going to be functions of t. So we could write a of t is equal to 1 1 2 times b of t times h of t. And if we wanna find the rate of change of our area at that instant, and the instant that they're talking about is at time t sub zero, well then what we would wanna do is take the derivative of both sides with respect to t. So the derivative on the left-hand side with respect to t would be a prime of t, and then on the right-hand side, it would be 1 1 2 times, and we would actually use a combination of, well, it's really just the product rule right over here, the derivative of the first function with respect to t, so it's b prime of t times the second function. This is just the product rule here, plus the first function, b of t, times the derivative of the second function with respect to time."}, {"video_title": "Analyzing related rates problems expressions AP Calculus AB Khan Academy.mp3", "Sentence": "Now why is all of this a useful exercise to go through? Well, now we are really ready to solve the question because in general, if we're talking about any triangle, we know that area is equal to 1 1 2 base times height. Now in this situation, area and our base and our height, they're all going to be functions of t. So we could write a of t is equal to 1 1 2 times b of t times h of t. And if we wanna find the rate of change of our area at that instant, and the instant that they're talking about is at time t sub zero, well then what we would wanna do is take the derivative of both sides with respect to t. So the derivative on the left-hand side with respect to t would be a prime of t, and then on the right-hand side, it would be 1 1 2 times, and we would actually use a combination of, well, it's really just the product rule right over here, the derivative of the first function with respect to t, so it's b prime of t times the second function. This is just the product rule here, plus the first function, b of t, times the derivative of the second function with respect to time. And we need to figure out not just the general expression, they want us to know what the rate of change of the area, so a prime of t at that instant, at t sub zero. So what we wanna figure out, we wanna figure out a prime at time t sub zero. Well, that's just going to be equal to 1 1 2 times b prime of t sub zero times h of t sub zero plus b of t sub zero times h prime of t sub zero."}, {"video_title": "Analyzing related rates problems expressions AP Calculus AB Khan Academy.mp3", "Sentence": "This is just the product rule here, plus the first function, b of t, times the derivative of the second function with respect to time. And we need to figure out not just the general expression, they want us to know what the rate of change of the area, so a prime of t at that instant, at t sub zero. So what we wanna figure out, we wanna figure out a prime at time t sub zero. Well, that's just going to be equal to 1 1 2 times b prime of t sub zero times h of t sub zero plus b of t sub zero times h prime of t sub zero. Now, this might seem daunting, except they've given us a lot of this information. What is b prime of t sub zero? Well, they tell us the rate of change of b with respect to time, and it seems like it's just gonna stay at negative 13 meters per hour, so they gave us this."}, {"video_title": "Analyzing related rates problems expressions AP Calculus AB Khan Academy.mp3", "Sentence": "Well, that's just going to be equal to 1 1 2 times b prime of t sub zero times h of t sub zero plus b of t sub zero times h prime of t sub zero. Now, this might seem daunting, except they've given us a lot of this information. What is b prime of t sub zero? Well, they tell us the rate of change of b with respect to time, and it seems like it's just gonna stay at negative 13 meters per hour, so they gave us this. And h, what is the height at time t sub zero? Well, they tell us right over here. At a certain instant, the base is five meters and the height is one meter, so they give us both b and h at t sub zero, so they gave us this, they gave us this, and what is the rate of change of the height at time t sub zero?"}, {"video_title": "Analyzing related rates problems expressions AP Calculus AB Khan Academy.mp3", "Sentence": "Well, they tell us the rate of change of b with respect to time, and it seems like it's just gonna stay at negative 13 meters per hour, so they gave us this. And h, what is the height at time t sub zero? Well, they tell us right over here. At a certain instant, the base is five meters and the height is one meter, so they give us both b and h at t sub zero, so they gave us this, they gave us this, and what is the rate of change of the height at time t sub zero? Well, they tell us the height of the triangle is increasing at a rate of six meters per hour. So they tell us that as well. All of that stuff is given, and so you just have to plug it in to figure out what is the rate of change of the area at t sub zero, at that instant."}, {"video_title": "Worked example Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So it tells us that if we assume that we have some f of x, if we have some f of x that is positive, positive, continuous, continuous, continuous, and decreasing, and decreasing, and decreasing on some interval on, so starting at k and including k, all the way to infinity, then we can make one of two statements. We could say either that if the improper integral from k to infinity of f of x dx is convergent, is convergent, then, then the sum, the infinite series from n is equal to k to infinity of f of n is also convergent, is also convergent, convergent. And this is actually the case that we saw when we looked at one over n squared, but I'll look at that in a second. But the second claim that we could make, or the second deduction that we might be able to make using the integral test, is if it's the other way around. That if the integral from k to infinity, the improper integral of f of x dx is divergent, divergent, then the same thing is true for the corresponding infinite series. Then this infinite series right over here is also going to be, is also divergent. And as I already mentioned, in the last video, we already saw this in the case of f of x is equal to one over x squared."}, {"video_title": "Worked example Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "But the second claim that we could make, or the second deduction that we might be able to make using the integral test, is if it's the other way around. That if the integral from k to infinity, the improper integral of f of x dx is divergent, divergent, then the same thing is true for the corresponding infinite series. Then this infinite series right over here is also going to be, is also divergent. And as I already mentioned, in the last video, we already saw this in the case of f of x is equal to one over x squared. We saw that since, since the integral from one to infinity of one over x squared, over one over x squared dx is convergent, in fact it equals one, it equals one, because of that, we were able to say that the sum from n equals, the sum from n is equal to one to infinity of one over n squared is also convergent, also convergent. And now we can see an example where we go the other way. For example, we know that this integral, let me write the integral down, let's start with this integral."}, {"video_title": "Worked example Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And as I already mentioned, in the last video, we already saw this in the case of f of x is equal to one over x squared. We saw that since, since the integral from one to infinity of one over x squared, over one over x squared dx is convergent, in fact it equals one, it equals one, because of that, we were able to say that the sum from n equals, the sum from n is equal to one to infinity of one over n squared is also convergent, also convergent. And now we can see an example where we go the other way. For example, we know that this integral, let me write the integral down, let's start with this integral. From one to infinity, not of f of x is equal to one over x squared, but let's say that f of x is equal to one over x. Actually, let me just write that down. Let's just start with f of x is equal to one over x."}, {"video_title": "Worked example Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "For example, we know that this integral, let me write the integral down, let's start with this integral. From one to infinity, not of f of x is equal to one over x squared, but let's say that f of x is equal to one over x. Actually, let me just write that down. Let's just start with f of x is equal to one over x. It definitely meets our conditions that it is positive, and let's say we're gonna consider it over the interval, over the interval from one to infinity. So it meets this first constraint. Over this interval, one over x is positive, it is continuous, and it is decreasing."}, {"video_title": "Worked example Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let's just start with f of x is equal to one over x. It definitely meets our conditions that it is positive, and let's say we're gonna consider it over the interval, over the interval from one to infinity. So it meets this first constraint. Over this interval, one over x is positive, it is continuous, and it is decreasing. As x increases, f of x decreases. So the integral test should apply. So let's see what this, what the improper integral from one to infinity of this would be."}, {"video_title": "Worked example Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Over this interval, one over x is positive, it is continuous, and it is decreasing. As x increases, f of x decreases. So the integral test should apply. So let's see what this, what the improper integral from one to infinity of this would be. So if we take, if we go from one to infinity of one over x, of one over x dx, this is equal to, we could write this as the limit as t approaches infinity of the definite integral from one to t of one over x dx, which is equal to the limit as t approaches infinity of, take the antiderivative, is going to be of the natural log of x, the natural log of x going from one to t, one to t. We could do the, well, it's really the absolute value of x, but we're dealing with positive x's here, so it's just gonna be the natural log of x, which is going to be the limit as t approaches infinity of the natural log of t, or I could even say the natural log of the absolute value of t, which is just gonna be the natural log of t, because it's positive t's, minus the natural log of one, minus the natural log of the absolute value of one. Well, the natural log of one is zero, so it's going to be the natural log of t, the limit as that approaches infinity, but the limit as that approaches infinity is just gonna be unbounded. This is gonna go to infinity."}, {"video_title": "Worked example Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's see what this, what the improper integral from one to infinity of this would be. So if we take, if we go from one to infinity of one over x, of one over x dx, this is equal to, we could write this as the limit as t approaches infinity of the definite integral from one to t of one over x dx, which is equal to the limit as t approaches infinity of, take the antiderivative, is going to be of the natural log of x, the natural log of x going from one to t, one to t. We could do the, well, it's really the absolute value of x, but we're dealing with positive x's here, so it's just gonna be the natural log of x, which is going to be the limit as t approaches infinity of the natural log of t, or I could even say the natural log of the absolute value of t, which is just gonna be the natural log of t, because it's positive t's, minus the natural log of one, minus the natural log of the absolute value of one. Well, the natural log of one is zero, so it's going to be the natural log of t, the limit as that approaches infinity, but the limit as that approaches infinity is just gonna be unbounded. This is gonna go to infinity. This right over here is divergent. So this right over here is divergent. So this is divergent."}, {"video_title": "Worked example Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "This is gonna go to infinity. This right over here is divergent. So this right over here is divergent. So this is divergent. And because this is divergent, we can then say, we can then say, by the integral test, we can then say, the integral test, once again, our function over this interval, positive continuous decreasing, we saw that this improper integral right over here is divergent, and then by the second point of the integral test, we can say, therefore, and I haven't rigorously proved it yet, but hopefully I gave you a good intuitive justification in the previous video, that the integral, that the infinite series from n equals one to infinity of one over n, which is the harmonic series, that this is also, this is also, this is also divergent. So we've already shown that the harmonic series is divergent using that very beautiful, elegant proof by O'Ram, I think I'm probably mispronouncing his name, that used the comparison test, but just like this, we have used the integral test to show that it is also divergent. And once again, let's remember what the whole motivation of the integral test is."}, {"video_title": "Worked example Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So this is divergent. And because this is divergent, we can then say, we can then say, by the integral test, we can then say, the integral test, once again, our function over this interval, positive continuous decreasing, we saw that this improper integral right over here is divergent, and then by the second point of the integral test, we can say, therefore, and I haven't rigorously proved it yet, but hopefully I gave you a good intuitive justification in the previous video, that the integral, that the infinite series from n equals one to infinity of one over n, which is the harmonic series, that this is also, this is also, this is also divergent. So we've already shown that the harmonic series is divergent using that very beautiful, elegant proof by O'Ram, I think I'm probably mispronouncing his name, that used the comparison test, but just like this, we have used the integral test to show that it is also divergent. And once again, let's remember what the whole motivation of the integral test is. Let me draw f of x is equal to one over x. So f of x is equal to one over x would look like, do my best attempt here. So let's say that's one, two, three, that is one, two, and so see when x is one, f of x is one, when x is two, f of x is 1 1.5, 1 1.3, that's 1 1.5 here, it'll be two over here, so it looks like this."}, {"video_title": "Worked example Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And once again, let's remember what the whole motivation of the integral test is. Let me draw f of x is equal to one over x. So f of x is equal to one over x would look like, do my best attempt here. So let's say that's one, two, three, that is one, two, and so see when x is one, f of x is one, when x is two, f of x is 1 1.5, 1 1.3, that's 1 1.5 here, it'll be two over here, so it looks like this. So this is f of x is equal to one over n, and once again we see that over the interval we care about from one to infinity, it's definitely positive, continuous, and decreasing. And if we look at this sum right over here, we could view this sum as, let's do that, let me write it down. So the sum, the sum from n equals one to infinity of one over n is equal to one plus 1 1.5, plus 1 1.3, and of course we keep going on and on and on and on."}, {"video_title": "Worked example Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's say that's one, two, three, that is one, two, and so see when x is one, f of x is one, when x is two, f of x is 1 1.5, 1 1.3, that's 1 1.5 here, it'll be two over here, so it looks like this. So this is f of x is equal to one over n, and once again we see that over the interval we care about from one to infinity, it's definitely positive, continuous, and decreasing. And if we look at this sum right over here, we could view this sum as, let's do that, let me write it down. So the sum, the sum from n equals one to infinity of one over n is equal to one plus 1 1.5, plus 1 1.3, and of course we keep going on and on and on and on. In this case, since we want to show it's divergent, we could say, hey look, this is an overestimate of this area here, let me be clear. So we have this area, we have this area in green, which is what the improper integral is denoting. So that right over there is the improper integral from one to infinity of one over x dx."}, {"video_title": "Worked example Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So the sum, the sum from n equals one to infinity of one over n is equal to one plus 1 1.5, plus 1 1.3, and of course we keep going on and on and on and on. In this case, since we want to show it's divergent, we could say, hey look, this is an overestimate of this area here, let me be clear. So we have this area, we have this area in green, which is what the improper integral is denoting. So that right over there is the improper integral from one to infinity of one over x dx. Now you could view this as an overestimate of that area. So this first, this one right over here, you could say that this is this one, height times one width, so that's that block right over there, that's the area of that, is going to be equal to one. Then this over here, 1 1.5, you could view that as the area of the next block, of the next block."}, {"video_title": "Worked example Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So that right over there is the improper integral from one to infinity of one over x dx. Now you could view this as an overestimate of that area. So this first, this one right over here, you could say that this is this one, height times one width, so that's that block right over there, that's the area of that, is going to be equal to one. Then this over here, 1 1.5, you could view that as the area of the next block, of the next block. And so you could kind of view this as a left-sided Riemann sum, I guess is one way to think about it. And so this is going to be 1 1.5, yeah, left-sided Riemann sum, so this is gonna be 1 1.5, and then the 1 3rd is going to be this one, is going to be this one. And notice, they're all, the actual area we care about, that the improper integral, it's all contained in these blocks."}, {"video_title": "Worked example Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Then this over here, 1 1.5, you could view that as the area of the next block, of the next block. And so you could kind of view this as a left-sided Riemann sum, I guess is one way to think about it. And so this is going to be 1 1.5, yeah, left-sided Riemann sum, so this is gonna be 1 1.5, and then the 1 3rd is going to be this one, is going to be this one. And notice, they're all, the actual area we care about, that the improper integral, it's all contained in these blocks. So this is going to be an over, is going to be larger than this, but we've already seen that this is unbounded towards infinity, this is divergent. So if this is larger than this, and this is divergent, this goes to infinity, then this must also go to infinity. So that's exactly where the integral test is coming from."}, {"video_title": "Worked example Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And like always, I encourage you to pause the video and see if you can work through this on your own. So let's look at this first statement. So this first statement says both the limit of g of x as x approaches six from the right-hand side and the limit as x approaches six from the left-hand side of g of x exist. All right, so let's first think about the limit of g of x as x approaches six from the right-hand side, as we approach six from values greater than six. So if we look over here, we could say, okay, when x is equal to nine, g of nine is right over there. G of eight is right over here. G of seven is right over here."}, {"video_title": "Worked example Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "All right, so let's first think about the limit of g of x as x approaches six from the right-hand side, as we approach six from values greater than six. So if we look over here, we could say, okay, when x is equal to nine, g of nine is right over there. G of eight is right over here. G of seven is right over here. It looks like it's between negative three and negative four. G of 6.5 looks like it's a little bit, it's still between negative three and negative four, but it's closer to negative three. G of 6.1 is even closer to negative three."}, {"video_title": "Worked example Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "G of seven is right over here. It looks like it's between negative three and negative four. G of 6.5 looks like it's a little bit, it's still between negative three and negative four, but it's closer to negative three. G of 6.1 is even closer to negative three. G of 6.01 is even closer to negative three. So it looks like the limit from the right-hand side does exist. So it looks like this one exists."}, {"video_title": "Worked example Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "G of 6.1 is even closer to negative three. G of 6.01 is even closer to negative three. So it looks like the limit from the right-hand side does exist. So it looks like this one exists. Now let's see, and I'm just looking at it graphically, and that's all they can expect you to do in an exercise like this. Now let's think about the limit as x approaches six from the left-hand side. So I could start anywhere, but let's say when x is equal to three, g of three is a little more than one."}, {"video_title": "Worked example Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So it looks like this one exists. Now let's see, and I'm just looking at it graphically, and that's all they can expect you to do in an exercise like this. Now let's think about the limit as x approaches six from the left-hand side. So I could start anywhere, but let's say when x is equal to three, g of three is a little more than one. G of four looks like it's a little bit less than two. G of five looks like it's close to three. G of 5.5 looks like it's between five and six."}, {"video_title": "Worked example Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So I could start anywhere, but let's say when x is equal to three, g of three is a little more than one. G of four looks like it's a little bit less than two. G of five looks like it's close to three. G of 5.5 looks like it's between five and six. G of 5.75 looks like it's approaching nine. And as we get closer and closer, as x gets closer and closer to six from below, from values to the left of six, it looks like we're unbounded. We are approaching infinity."}, {"video_title": "Worked example Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "G of 5.5 looks like it's between five and six. G of 5.75 looks like it's approaching nine. And as we get closer and closer, as x gets closer and closer to six from below, from values to the left of six, it looks like we're unbounded. We are approaching infinity. And so technically, we would say this limit does not exist. So this one does not exist. So I won't check this one off."}, {"video_title": "Worked example Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "We are approaching infinity. And so technically, we would say this limit does not exist. So this one does not exist. So I won't check this one off. Some people will say the limit is approaching infinity, but that technically is, infinity is not a value that you can say it is approaching in the classical, formal definition of a limit. So for these purposes, we would just say this does not exist. Now let's see."}, {"video_title": "Worked example Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So I won't check this one off. Some people will say the limit is approaching infinity, but that technically is, infinity is not a value that you can say it is approaching in the classical, formal definition of a limit. So for these purposes, we would just say this does not exist. Now let's see. They say the limit as x approaches six of g of x exists. Well, the only way that the limit exists is if both the left and the right limits exist, and they approach the same thing. Well, we don't even, our limit as x approaches six from the negative side, or from the left-hand side, I guess I could say, does not even exist."}, {"video_title": "Worked example Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's see. They say the limit as x approaches six of g of x exists. Well, the only way that the limit exists is if both the left and the right limits exist, and they approach the same thing. Well, we don't even, our limit as x approaches six from the negative side, or from the left-hand side, I guess I could say, does not even exist. So this cannot be true. So that's not gonna be true. The first one's not gonna be true."}, {"video_title": "Worked example Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we don't even, our limit as x approaches six from the negative side, or from the left-hand side, I guess I could say, does not even exist. So this cannot be true. So that's not gonna be true. The first one's not gonna be true. G is defined at x equals six. So at x equals six, it doesn't look like g is defined. And looking at this graph, I can't tell you what g of six should be."}, {"video_title": "Worked example Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "The first one's not gonna be true. G is defined at x equals six. So at x equals six, it doesn't look like g is defined. And looking at this graph, I can't tell you what g of six should be. We have an open circle over here, so g of six is not equal to negative three, and this goes up to infinity, and we have a vertical asymptote actually drawn right over here at x equals six. So g is not defined at x equals six. So I'll rule that one out."}, {"video_title": "Worked example Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And looking at this graph, I can't tell you what g of six should be. We have an open circle over here, so g of six is not equal to negative three, and this goes up to infinity, and we have a vertical asymptote actually drawn right over here at x equals six. So g is not defined at x equals six. So I'll rule that one out. G is continuous at x equals six. Well, you can see that it goes up to infinity, then it jumps down, back down here, then continues. So just, when you just think about it in common sense language, it looks very discontinuous."}, {"video_title": "Worked example Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So I'll rule that one out. G is continuous at x equals six. Well, you can see that it goes up to infinity, then it jumps down, back down here, then continues. So just, when you just think about it in common sense language, it looks very discontinuous. And if you wanna think about it more formally, in order for something to be continuous, the limit needs to exist at that value. The function needs to be defined at that value, and the value of the function needs to be equal to the value of the limit. And neither of these, the first two conditions aren't true, and so these can't even equal each other, because neither of these exist."}, {"video_title": "Worked example Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So just, when you just think about it in common sense language, it looks very discontinuous. And if you wanna think about it more formally, in order for something to be continuous, the limit needs to exist at that value. The function needs to be defined at that value, and the value of the function needs to be equal to the value of the limit. And neither of these, the first two conditions aren't true, and so these can't even equal each other, because neither of these exist. So this is not continuous at x equals six, and so the only thing I could check here is none of the above. Let's do another one of these. So the first statement, both the right hand and the left hand limit exist as x approaches three."}, {"video_title": "Worked example Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And neither of these, the first two conditions aren't true, and so these can't even equal each other, because neither of these exist. So this is not continuous at x equals six, and so the only thing I could check here is none of the above. Let's do another one of these. So the first statement, both the right hand and the left hand limit exist as x approaches three. So let's think about it. So x equals three is where we have this little discontinuity here, this jump discontinuity. So let's approach, let's go from the positive, from values larger than three."}, {"video_title": "Worked example Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So the first statement, both the right hand and the left hand limit exist as x approaches three. So let's think about it. So x equals three is where we have this little discontinuity here, this jump discontinuity. So let's approach, let's go from the positive, from values larger than three. So when x is equal to five, g of five is a little bit more negative than negative three. G of four is between negative two and negative three. G of 3.5 is getting a little bit closer to negative two."}, {"video_title": "Worked example Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So let's approach, let's go from the positive, from values larger than three. So when x is equal to five, g of five is a little bit more negative than negative three. G of four is between negative two and negative three. G of 3.5 is getting a little bit closer to negative two. G of 3.1, it's getting even closer, closer to negative two. G of 3.01 is even closer to negative two. So it looks like this limit right over here, oh, I'm circling the wrong one."}, {"video_title": "Worked example Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "G of 3.5 is getting a little bit closer to negative two. G of 3.1, it's getting even closer, closer to negative two. G of 3.01 is even closer to negative two. So it looks like this limit right over here, oh, I'm circling the wrong one. It looks like this limit exists, and in fact, it looks like it is approaching negative two. So this right over here is equal to negative two, the limit of g of x as x approaches three from the right hand side. And now let's think about it from the left hand side."}, {"video_title": "Worked example Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So it looks like this limit right over here, oh, I'm circling the wrong one. It looks like this limit exists, and in fact, it looks like it is approaching negative two. So this right over here is equal to negative two, the limit of g of x as x approaches three from the right hand side. And now let's think about it from the left hand side. So we can start, I can start here. G of one looks like it's a little bit greater than negative one. G of two, it's less than one."}, {"video_title": "Worked example Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And now let's think about it from the left hand side. So we can start, I can start here. G of one looks like it's a little bit greater than negative one. G of two, it's less than one. G of 2.5 is between one and two. G of 2.9, looks like it's a little bit less than two. G of 2.99 is getting even closer to two."}, {"video_title": "Worked example Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "G of two, it's less than one. G of 2.5 is between one and two. G of 2.9, looks like it's a little bit less than two. G of 2.99 is getting even closer to two. G of 2.999999 would be even closer to, so it looks like this thing right over here is approaching two. So both of these limits, the limit from the right and the limit from the left, exist. The limit of g of x as x approaches three exists."}, {"video_title": "Worked example Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "G of 2.99 is getting even closer to two. G of 2.999999 would be even closer to, so it looks like this thing right over here is approaching two. So both of these limits, the limit from the right and the limit from the left, exist. The limit of g of x as x approaches three exists. So these are the one-sided limit. This is the actual limit. Now in order for this to exist, both the right and left-handed limits need to exist, and they need to approach the same value."}, {"video_title": "Worked example Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "The limit of g of x as x approaches three exists. So these are the one-sided limit. This is the actual limit. Now in order for this to exist, both the right and left-handed limits need to exist, and they need to approach the same value. Well this first statement, we saw that both of these exist, but they aren't approaching the same value. From the left, we are, sorry, from the right, we are approaching negative two, and from the left, we are approaching two. So this limit does not exist."}, {"video_title": "Worked example Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Now in order for this to exist, both the right and left-handed limits need to exist, and they need to approach the same value. Well this first statement, we saw that both of these exist, but they aren't approaching the same value. From the left, we are, sorry, from the right, we are approaching negative two, and from the left, we are approaching two. So this limit does not exist. So I will not check that out, or I will not check that box. G is defined at x equals three. Well when x equals three, we see a solid dot right over there."}, {"video_title": "Worked example Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So this limit does not exist. So I will not check that out, or I will not check that box. G is defined at x equals three. Well when x equals three, we see a solid dot right over there. And so it is indeed defined. It is indeed defined there. G is continuous at x equals three."}, {"video_title": "Worked example Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Well when x equals three, we see a solid dot right over there. And so it is indeed defined. It is indeed defined there. G is continuous at x equals three. Well in order for g to be continuous at x equals three, the limit must exist there. It must be defined there, and the value of the function there needs to be equal to the value of the limit. Well the function is defined there, but the limit doesn't exist there."}, {"video_title": "Worked example Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "G is continuous at x equals three. Well in order for g to be continuous at x equals three, the limit must exist there. It must be defined there, and the value of the function there needs to be equal to the value of the limit. Well the function is defined there, but the limit doesn't exist there. So it cannot be continuous. It cannot be continuous there. So I would cross that out."}, {"video_title": "Optimization sum of squares Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We are being asked, what is the smallest, this is a little typo here, what is the smallest possible sum of squares of two numbers if their product is negative 16? So let's say that these two numbers are x and y, x and y. So how could we define the sum of the squares of the two numbers? So I'll just call that the sum of the squares, s for sum of the squares, and it would just be equal to x squared plus y squared. And this is what we wanna minimize. We want to minimize, minimize s. Now, right now, s is expressed as a function of x and y. We don't know how to minimize with respect to two variables, so we have to get this in terms of only one variable, and lucky for us, they give us another piece of information."}, {"video_title": "Optimization sum of squares Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So I'll just call that the sum of the squares, s for sum of the squares, and it would just be equal to x squared plus y squared. And this is what we wanna minimize. We want to minimize, minimize s. Now, right now, s is expressed as a function of x and y. We don't know how to minimize with respect to two variables, so we have to get this in terms of only one variable, and lucky for us, they give us another piece of information. Their product is negative 16. So x times y is equal to negative 16. So let's say we wanted this expression right over here only in terms of x."}, {"video_title": "Optimization sum of squares Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We don't know how to minimize with respect to two variables, so we have to get this in terms of only one variable, and lucky for us, they give us another piece of information. Their product is negative 16. So x times y is equal to negative 16. So let's say we wanted this expression right over here only in terms of x. Well, then we could solve, we can figure out what y is in terms of x, and then substitute. So let's do that right over here. If we divide both sides by x, we get y is equal to negative 16 over x."}, {"video_title": "Optimization sum of squares Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say we wanted this expression right over here only in terms of x. Well, then we could solve, we can figure out what y is in terms of x, and then substitute. So let's do that right over here. If we divide both sides by x, we get y is equal to negative 16 over x. And so let's replace our y in this expression with negative 16 over x. So then we would get our sum of squares as a function of x is going to be equal to x squared plus y squared. Y is negative 16 over x."}, {"video_title": "Optimization sum of squares Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "If we divide both sides by x, we get y is equal to negative 16 over x. And so let's replace our y in this expression with negative 16 over x. So then we would get our sum of squares as a function of x is going to be equal to x squared plus y squared. Y is negative 16 over x. Negative 16 over x. And then that's what we will now square. So this is equal to x squared plus, what is this?"}, {"video_title": "Optimization sum of squares Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Y is negative 16 over x. Negative 16 over x. And then that's what we will now square. So this is equal to x squared plus, what is this? 256, 256 over x squared, or we could write that as 256, 256 x to the negative 2 power. That is the sum of our squares that we now want to minimize. Well, to minimize this, we would wanna look at the critical points of this, which is where the derivative is either zero or undefined, and see whether those critical points are possibly a minimum or a maximum point."}, {"video_title": "Optimization sum of squares Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So this is equal to x squared plus, what is this? 256, 256 over x squared, or we could write that as 256, 256 x to the negative 2 power. That is the sum of our squares that we now want to minimize. Well, to minimize this, we would wanna look at the critical points of this, which is where the derivative is either zero or undefined, and see whether those critical points are possibly a minimum or a maximum point. They don't have to be, but those are the ones, if we have a minimum or a maximum point, they're going to be one of the critical points. So let's take the derivative. So the derivative, s prime, let me do this in a different color, s prime of x, I'll do it right over here, actually."}, {"video_title": "Optimization sum of squares Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, to minimize this, we would wanna look at the critical points of this, which is where the derivative is either zero or undefined, and see whether those critical points are possibly a minimum or a maximum point. They don't have to be, but those are the ones, if we have a minimum or a maximum point, they're going to be one of the critical points. So let's take the derivative. So the derivative, s prime, let me do this in a different color, s prime of x, I'll do it right over here, actually. The derivative, s prime of x, with respect to x, is going to be equal to 2x times negative 2 times 5, 2x plus 256 times negative 2, so that's minus 512 x to the negative 3 power. X to the negative 3 power. Now, this is going to be undefined, this is going to be undefined when x is equal to zero, but if x is equal to zero, then y is undefined, so this whole thing breaks down."}, {"video_title": "Optimization sum of squares Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So the derivative, s prime, let me do this in a different color, s prime of x, I'll do it right over here, actually. The derivative, s prime of x, with respect to x, is going to be equal to 2x times negative 2 times 5, 2x plus 256 times negative 2, so that's minus 512 x to the negative 3 power. X to the negative 3 power. Now, this is going to be undefined, this is going to be undefined when x is equal to zero, but if x is equal to zero, then y is undefined, so this whole thing breaks down. So that isn't a useful critical point, x equals zero. So let's think about any other ones. Well, it's defined everywhere else, so let's think about where the derivative is equal to zero."}, {"video_title": "Optimization sum of squares Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Now, this is going to be undefined, this is going to be undefined when x is equal to zero, but if x is equal to zero, then y is undefined, so this whole thing breaks down. So that isn't a useful critical point, x equals zero. So let's think about any other ones. Well, it's defined everywhere else, so let's think about where the derivative is equal to zero. So when does this thing equal zero? So when does 2x minus 512 x to the negative 3 equal zero? Well, we can add 512 x to the negative 3 to both sides, so you get 2x is equal to 512 x to the negative 3rd power."}, {"video_title": "Optimization sum of squares Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, it's defined everywhere else, so let's think about where the derivative is equal to zero. So when does this thing equal zero? So when does 2x minus 512 x to the negative 3 equal zero? Well, we can add 512 x to the negative 3 to both sides, so you get 2x is equal to 512 x to the negative 3rd power. We can multiply both sides times x to the 3rd power, multiply both sides times x to the 3rd, so all of the x's go away on the right-hand side. So you get 2x to the 4th is equal to 512. We can divide both sides by 2, and you get x to the 4th power is equal to 256."}, {"video_title": "Optimization sum of squares Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we can add 512 x to the negative 3 to both sides, so you get 2x is equal to 512 x to the negative 3rd power. We can multiply both sides times x to the 3rd power, multiply both sides times x to the 3rd, so all of the x's go away on the right-hand side. So you get 2x to the 4th is equal to 512. We can divide both sides by 2, and you get x to the 4th power is equal to 256. And so what is the 4th root of 256? Well, we could take the square root of both sides just to help us here. So let's see, if we take, so it's going to be x squared is going to be equal to, 256 is 16 squared, so this is 16, this is going to be x squared is equal to 16, or x is equal to 4."}, {"video_title": "Optimization sum of squares Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We can divide both sides by 2, and you get x to the 4th power is equal to 256. And so what is the 4th root of 256? Well, we could take the square root of both sides just to help us here. So let's see, if we take, so it's going to be x squared is going to be equal to, 256 is 16 squared, so this is 16, this is going to be x squared is equal to 16, or x is equal to 4. Now, that's our only critical point we have, so that's probably the x value that minimizes our sum of squares right over here, but let's make sure it's a minimum value, and to do that, we can just do our second derivative test. So let's figure out, let's take the second derivative, s prime prime of x, and figure out if we are concave upwards or downwards when x is equal to 4. So s prime prime of x is going to be equal to 2, and then we're going to have negative 3 times negative 512, so I'll just write that as plus 3 times 512, that's going to be 1536, is that right?"}, {"video_title": "Optimization sum of squares Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let's see, if we take, so it's going to be x squared is going to be equal to, 256 is 16 squared, so this is 16, this is going to be x squared is equal to 16, or x is equal to 4. Now, that's our only critical point we have, so that's probably the x value that minimizes our sum of squares right over here, but let's make sure it's a minimum value, and to do that, we can just do our second derivative test. So let's figure out, let's take the second derivative, s prime prime of x, and figure out if we are concave upwards or downwards when x is equal to 4. So s prime prime of x is going to be equal to 2, and then we're going to have negative 3 times negative 512, so I'll just write that as plus 3 times 512, that's going to be 1536, is that right? Yeah, 3 times 500 is 1500, 3 times 12 is 36, x to the negative 4 power. So this thing is going to be, this thing right over here is actually going to be positive for any x, this thing is going to be positive for any x. x to the negative 4, even if it's a negative x value, that's going to be positive, everything else is positive, this thing is always positive. So we are always in a concave upwards situation."}, {"video_title": "Optimization sum of squares Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So s prime prime of x is going to be equal to 2, and then we're going to have negative 3 times negative 512, so I'll just write that as plus 3 times 512, that's going to be 1536, is that right? Yeah, 3 times 500 is 1500, 3 times 12 is 36, x to the negative 4 power. So this thing is going to be, this thing right over here is actually going to be positive for any x, this thing is going to be positive for any x. x to the negative 4, even if it's a negative x value, that's going to be positive, everything else is positive, this thing is always positive. So we are always in a concave upwards situation. Concave upwards means that our graph might look something like that. Actually, I don't want to draw a little squiggle, it might look something like that, and you see the reason why the second derivative implies concave upwards, a second derivative positive means that our derivative is constantly increasing, so the derivative is constantly increasing, it's negative, less negative, even less negative, let me do that in a different color. You see it's negative, less negative, even less negative, zero, positive, more positive, so it's increasing over the entire place."}, {"video_title": "Optimization sum of squares Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So we are always in a concave upwards situation. Concave upwards means that our graph might look something like that. Actually, I don't want to draw a little squiggle, it might look something like that, and you see the reason why the second derivative implies concave upwards, a second derivative positive means that our derivative is constantly increasing, so the derivative is constantly increasing, it's negative, less negative, even less negative, let me do that in a different color. You see it's negative, less negative, even less negative, zero, positive, more positive, so it's increasing over the entire place. So if you have a critical point where the derivative is equal to zero, so the slope is equal to zero, and it's concave upwards, you see pretty clearly that we have minimized the function. So what is y going to be equal to? We actually don't even have to figure out what y has to be equal to in order to minimize the sum of squares, we could just put it back into this, but just for fun, we see that y would be negative 16 over x, so y would be equal to negative 4, and we could just figure out now what our sum of squares is."}, {"video_title": "Optimization sum of squares Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "You see it's negative, less negative, even less negative, zero, positive, more positive, so it's increasing over the entire place. So if you have a critical point where the derivative is equal to zero, so the slope is equal to zero, and it's concave upwards, you see pretty clearly that we have minimized the function. So what is y going to be equal to? We actually don't even have to figure out what y has to be equal to in order to minimize the sum of squares, we could just put it back into this, but just for fun, we see that y would be negative 16 over x, so y would be equal to negative 4, and we could just figure out now what our sum of squares is. Our minimum sum of squares is going to be equal to, it's going to be equal to four squared, which is 16, plus negative four squared, plus another 16, which is equal to 32. Now I know some of you might be thinking, hey, I could have done this without calculus. I could have just tried out numbers whose product is negative 16, and I probably would have tried out four and negative four in not too much time, and then I would have been able to maybe figure out that it's lower than if I did two and negative eight, or negative two and eight, or one and 16, and that's true."}, {"video_title": "Optimization sum of squares Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We actually don't even have to figure out what y has to be equal to in order to minimize the sum of squares, we could just put it back into this, but just for fun, we see that y would be negative 16 over x, so y would be equal to negative 4, and we could just figure out now what our sum of squares is. Our minimum sum of squares is going to be equal to, it's going to be equal to four squared, which is 16, plus negative four squared, plus another 16, which is equal to 32. Now I know some of you might be thinking, hey, I could have done this without calculus. I could have just tried out numbers whose product is negative 16, and I probably would have tried out four and negative four in not too much time, and then I would have been able to maybe figure out that it's lower than if I did two and negative eight, or negative two and eight, or one and 16, and that's true. You probably would have been able to do that, but you still wouldn't have been able to feel good that that was a minimum value, because you wouldn't have tried out 4.01 or 4.0011. In fact, you couldn't have tried out all of the possible values. Remember, we didn't say that this is only integers."}, {"video_title": "Optimization sum of squares Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "I could have just tried out numbers whose product is negative 16, and I probably would have tried out four and negative four in not too much time, and then I would have been able to maybe figure out that it's lower than if I did two and negative eight, or negative two and eight, or one and 16, and that's true. You probably would have been able to do that, but you still wouldn't have been able to feel good that that was a minimum value, because you wouldn't have tried out 4.01 or 4.0011. In fact, you couldn't have tried out all of the possible values. Remember, we didn't say that this is only integers. It just happened to be that our values worked out to be, just worked out to be integers in this situation. You can imagine what would happen if the problem wasn't if their product is negative 16, but what if their product is negative 17, or what if their product is negative 16.5, or what if their product was pi squared? Then you wouldn't be able to try everything else out, and you would have to resort to doing what we did in this video."}, {"video_title": "Introduction to infinite limits Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "This is y is equal to one over x. And we explored what's the limit as x approaches zero in either of those scenarios. And in this left scenario, we saw as x becomes less and less negative, as it approaches zero from the left-hand side, our, the value of one over x squared is unbounded in the positive direction. And the same thing happens as we approach x from the right. As we become less and less positive, but we are still positive, the value of one over x squared becomes unbounded in the positive direction. So in that video, we just said, hey, one could say that this limit is unbounded. But what we're going to do in this video is introduce new notation."}, {"video_title": "Introduction to infinite limits Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And the same thing happens as we approach x from the right. As we become less and less positive, but we are still positive, the value of one over x squared becomes unbounded in the positive direction. So in that video, we just said, hey, one could say that this limit is unbounded. But what we're going to do in this video is introduce new notation. Instead of just saying it's unbounded, we could say, hey, from both the left and the right, it looks like we're going to positive infinity. So we can introduce this notation of saying, hey, this is going to infinity, which you will sometimes see used. Some people would call this unbounded."}, {"video_title": "Introduction to infinite limits Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "But what we're going to do in this video is introduce new notation. Instead of just saying it's unbounded, we could say, hey, from both the left and the right, it looks like we're going to positive infinity. So we can introduce this notation of saying, hey, this is going to infinity, which you will sometimes see used. Some people would call this unbounded. Some people say it does not exist because it's not approaching some finite value. While some people will use this notation of the limit going to infinity. But what about this scenario?"}, {"video_title": "Introduction to infinite limits Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Some people would call this unbounded. Some people say it does not exist because it's not approaching some finite value. While some people will use this notation of the limit going to infinity. But what about this scenario? Can we use our new notation here? Well, when we approach zero from the left, it looks like we're unbounded in the negative direction. And when we approach zero from the right, we're unbounded in the positive direction."}, {"video_title": "Introduction to infinite limits Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "But what about this scenario? Can we use our new notation here? Well, when we approach zero from the left, it looks like we're unbounded in the negative direction. And when we approach zero from the right, we're unbounded in the positive direction. So here, you still could not say that the limit is approaching infinity because from the right, it's approaching infinity, but from the left, it's approaching negative infinity. So you would still say that this does not exist. You could do one-sided limits here, which if you're not familiar with, I encourage you to review it on Khan Academy."}, {"video_title": "Introduction to infinite limits Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And when we approach zero from the right, we're unbounded in the positive direction. So here, you still could not say that the limit is approaching infinity because from the right, it's approaching infinity, but from the left, it's approaching negative infinity. So you would still say that this does not exist. You could do one-sided limits here, which if you're not familiar with, I encourage you to review it on Khan Academy. If you said the limit of one over x as x approaches zero from the left-hand side from values less than zero, well, then you would look at this right over here and say, well, look, it looks like we're going unbounded in the negative direction. So you would say this is equal to negative infinity. And of course, if you said the limit as x approaches zero from the right of one over x, well, here, you're unbounded in the positive direction, so that's going to be equal to positive infinity."}, {"video_title": "Introduction to infinite limits Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "You could do one-sided limits here, which if you're not familiar with, I encourage you to review it on Khan Academy. If you said the limit of one over x as x approaches zero from the left-hand side from values less than zero, well, then you would look at this right over here and say, well, look, it looks like we're going unbounded in the negative direction. So you would say this is equal to negative infinity. And of course, if you said the limit as x approaches zero from the right of one over x, well, here, you're unbounded in the positive direction, so that's going to be equal to positive infinity. Let's do an example problem from Khan Academy based on this idea and this notation. So here it says, consider graphs A, B, and C. The dashed lines represent asymptotes. Which of the graphs agree with this statement that the limit as x approaches one of h of x is equal to infinity?"}, {"video_title": "Introduction to infinite limits Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And of course, if you said the limit as x approaches zero from the right of one over x, well, here, you're unbounded in the positive direction, so that's going to be equal to positive infinity. Let's do an example problem from Khan Academy based on this idea and this notation. So here it says, consider graphs A, B, and C. The dashed lines represent asymptotes. Which of the graphs agree with this statement that the limit as x approaches one of h of x is equal to infinity? Pause this video and see if you can figure it out. All right, let's go through each of these. So we wanna think about what happens at x equals one, so that's right over here on graph A."}, {"video_title": "Introduction to infinite limits Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Which of the graphs agree with this statement that the limit as x approaches one of h of x is equal to infinity? Pause this video and see if you can figure it out. All right, let's go through each of these. So we wanna think about what happens at x equals one, so that's right over here on graph A. So as we approach x equals one, so let me write this. So the limit, let me do this for the different graphs. So for graph A, the limit as x approaches one from the left, that looks like it's unbounded in the positive direction, that equals infinity."}, {"video_title": "Introduction to infinite limits Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So we wanna think about what happens at x equals one, so that's right over here on graph A. So as we approach x equals one, so let me write this. So the limit, let me do this for the different graphs. So for graph A, the limit as x approaches one from the left, that looks like it's unbounded in the positive direction, that equals infinity. And the limit as x approaches one from the right, well, that looks like it's going to negative infinity. That equals negative infinity. And since these are going in two different directions, you wouldn't be able to say that the limit as x approaches one from both directions is equal to infinity, so I would rule this one out."}, {"video_title": "Introduction to infinite limits Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So for graph A, the limit as x approaches one from the left, that looks like it's unbounded in the positive direction, that equals infinity. And the limit as x approaches one from the right, well, that looks like it's going to negative infinity. That equals negative infinity. And since these are going in two different directions, you wouldn't be able to say that the limit as x approaches one from both directions is equal to infinity, so I would rule this one out. Now let's look at choice B. What's the limit as x approaches one from the left? And of course, these are of h of x. Gotta write that down."}, {"video_title": "Introduction to infinite limits Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And since these are going in two different directions, you wouldn't be able to say that the limit as x approaches one from both directions is equal to infinity, so I would rule this one out. Now let's look at choice B. What's the limit as x approaches one from the left? And of course, these are of h of x. Gotta write that down. So of h of x, right over here. Well, as we approach from the left, we are going to, looks like we're going to positive infinity and it looks like the limit of h of x as we approach one from the right is also going to positive infinity. And so since we're approaching, you could say the same direction of infinity, you could say this for B."}, {"video_title": "Definite integral of radical function AP Calculus AB Khan Academy.mp3", "Sentence": "So we want to evaluate the definite integral from negative one to eight of 12 times the cube root of x dx. Let's see, this is going to be the same thing as the definite integral from negative one to eight of 12 times, the cube root is the same thing as saying x to the 1 3rd power dx. And so now, if we want to take the antiderivative of the stuff on the inside, we're just going to do, essentially the power rule, you could use this as the power rule of integrals, or it's the reverse of the power rule for derivatives, where we increase this exponent by one, and then we divide by that increased exponent. So this is going to be equal to 12 times x to the 1 3rd plus one, let me do it in another color just so we can keep track of it. X to the 1 3rd plus one, and then we're going to divide by 1 3rd plus one. And so what's 1 3rd plus one? Well, that's 4 3rds, 1 3rd plus 3 3rds, that's 4 3rds."}, {"video_title": "Definite integral of radical function AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be equal to 12 times x to the 1 3rd plus one, let me do it in another color just so we can keep track of it. X to the 1 3rd plus one, and then we're going to divide by 1 3rd plus one. And so what's 1 3rd plus one? Well, that's 4 3rds, 1 3rd plus 3 3rds, that's 4 3rds. So I could write it this way, I could write this x to the 4 3rds divided by 4 3rds. And this is going to be, and I'm going to evaluate this at the bounds. So I'm going to evaluate this at, and I'll do this in different colors, I'm going to evaluate it at eight, and I'm going to evaluate it at negative one, and I'm going to subtract it, evaluate it at negative one from this expression evaluated at eight."}, {"video_title": "Definite integral of radical function AP Calculus AB Khan Academy.mp3", "Sentence": "Well, that's 4 3rds, 1 3rd plus 3 3rds, that's 4 3rds. So I could write it this way, I could write this x to the 4 3rds divided by 4 3rds. And this is going to be, and I'm going to evaluate this at the bounds. So I'm going to evaluate this at, and I'll do this in different colors, I'm going to evaluate it at eight, and I'm going to evaluate it at negative one, and I'm going to subtract it, evaluate it at negative one from this expression evaluated at eight. And so what is this going to be equal to? Well, actually, let me simplify a little bit more. What is 12 divided by 4 3rds?"}, {"video_title": "Definite integral of radical function AP Calculus AB Khan Academy.mp3", "Sentence": "So I'm going to evaluate this at, and I'll do this in different colors, I'm going to evaluate it at eight, and I'm going to evaluate it at negative one, and I'm going to subtract it, evaluate it at negative one from this expression evaluated at eight. And so what is this going to be equal to? Well, actually, let me simplify a little bit more. What is 12 divided by 4 3rds? So 12, I'll do it right, well, I'll do it right over here. 12 over 4 3rds is equal to 12 times three over four, which we could use 12 over one times three over four. 12 divided by four is three, so this is going to be equal to nine."}, {"video_title": "Definite integral of radical function AP Calculus AB Khan Academy.mp3", "Sentence": "What is 12 divided by 4 3rds? So 12, I'll do it right, well, I'll do it right over here. 12 over 4 3rds is equal to 12 times three over four, which we could use 12 over one times three over four. 12 divided by four is three, so this is going to be equal to nine. 3 4ths of 12 is nine. So this, we could rewrite this, we could write this as nine x to the 4 3rd power. So if we evaluate it at eight, this is going to be nine times eight to the 4 3rds power, and from that, we're going to subtract, it evaluated at negative one."}, {"video_title": "Definite integral of radical function AP Calculus AB Khan Academy.mp3", "Sentence": "12 divided by four is three, so this is going to be equal to nine. 3 4ths of 12 is nine. So this, we could rewrite this, we could write this as nine x to the 4 3rd power. So if we evaluate it at eight, this is going to be nine times eight to the 4 3rds power, and from that, we're going to subtract, it evaluated at negative one. So this is going to be nine times negative one to the 4 3rd power. So what is eight to the 4 3rd power? I'll do it over here."}, {"video_title": "Definite integral of radical function AP Calculus AB Khan Academy.mp3", "Sentence": "So if we evaluate it at eight, this is going to be nine times eight to the 4 3rds power, and from that, we're going to subtract, it evaluated at negative one. So this is going to be nine times negative one to the 4 3rd power. So what is eight to the 4 3rd power? I'll do it over here. So eight to the 4 3rds is equal to eight to the 1 3rd to the 4th power. These are just exponent properties here. Eight to the 1 3rd, the cube root of eight, or eight to the 1 3rd power, that's two, because two to the 3rd power is eight."}, {"video_title": "Definite integral of radical function AP Calculus AB Khan Academy.mp3", "Sentence": "I'll do it over here. So eight to the 4 3rds is equal to eight to the 1 3rd to the 4th power. These are just exponent properties here. Eight to the 1 3rd, the cube root of eight, or eight to the 1 3rd power, that's two, because two to the 3rd power is eight. And two to the 4th power, well, two to the 4th power is equal to 16. So eight to the 4 3rds is 16. And what's negative one to the 4 3rds?"}, {"video_title": "Definite integral of radical function AP Calculus AB Khan Academy.mp3", "Sentence": "Eight to the 1 3rd, the cube root of eight, or eight to the 1 3rd power, that's two, because two to the 3rd power is eight. And two to the 4th power, well, two to the 4th power is equal to 16. So eight to the 4 3rds is 16. And what's negative one to the 4 3rds? Well, same idea. Negative one to the 4 3rds is equal to negative one, there's several ways you could do it. You could say negative one to the 4th, and then the cube root of that, or the cube root of negative one, and then raise that to the 4th power, either way."}, {"video_title": "Definite integral of radical function AP Calculus AB Khan Academy.mp3", "Sentence": "And what's negative one to the 4 3rds? Well, same idea. Negative one to the 4 3rds is equal to negative one, there's several ways you could do it. You could say negative one to the 4th, and then the cube root of that, or the cube root of negative one, and then raise that to the 4th power, either way. So let's do it the first way. Negative one to the 4th, and then take the cube root of that. Well, negative one to the 4th is just one, and then one to the 1 3rd power, well, that's just going to be equal to one."}, {"video_title": "Definite integral of radical function AP Calculus AB Khan Academy.mp3", "Sentence": "You could say negative one to the 4th, and then the cube root of that, or the cube root of negative one, and then raise that to the 4th power, either way. So let's do it the first way. Negative one to the 4th, and then take the cube root of that. Well, negative one to the 4th is just one, and then one to the 1 3rd power, well, that's just going to be equal to one. So what we have here in blue, that's just equal to one. So we have nine times 16 minus nine times one. Well, that's just going to be nine times 15."}, {"video_title": "Definite integral of radical function AP Calculus AB Khan Academy.mp3", "Sentence": "Well, negative one to the 4th is just one, and then one to the 1 3rd power, well, that's just going to be equal to one. So what we have here in blue, that's just equal to one. So we have nine times 16 minus nine times one. Well, that's just going to be nine times 15. We have 16 nines, and then we're gonna take away a nine. So that's gonna be nine times 15. So what is that?"}, {"video_title": "Derivatives expressed as limits Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "Let's see if we can find the limit as h approaches zero of five log of two plus h minus five log of two, all of that over h. And I'll give you a little bit of a hint, because I know you're about to pause the video and try to work through it. Think of your derivative properties, especially the derivative of logarithmic functions, especially logarithmic functions, in this case, with base 10. If someone just writes log without the base, you can just assume that that is a 10 right over there. So pause the video and see if you can work through it. All right, so the key here is to remember that if I have f of x, let me do it over here, I'll do it over here, f of x, and I want to find f prime of, let's say f prime of some number, let's say a, this is going to be equal to the limit as x, or sorry, as h approaches zero is one way of thinking about it, as h approaches zero of f of a plus h minus f of a, all of that over h. So this looks pretty close to that limit definition, except we have these fives here. But lucky for us, we can factor out those fives, and we could factor them out out front here, but if you just have a scalar times the expression, we know from our limit properties that we can actually take those out of the limit themselves. So let's do that."}, {"video_title": "Derivatives expressed as limits Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "So pause the video and see if you can work through it. All right, so the key here is to remember that if I have f of x, let me do it over here, I'll do it over here, f of x, and I want to find f prime of, let's say f prime of some number, let's say a, this is going to be equal to the limit as x, or sorry, as h approaches zero is one way of thinking about it, as h approaches zero of f of a plus h minus f of a, all of that over h. So this looks pretty close to that limit definition, except we have these fives here. But lucky for us, we can factor out those fives, and we could factor them out out front here, but if you just have a scalar times the expression, we know from our limit properties that we can actually take those out of the limit themselves. So let's do that. Let's take both of these fives and factor them out. And so this whole thing is going to simplify to five times the limit as h approaches zero of log of two plus h minus, minus log of two, all of that over h. Minus log of two, all of that over h. Now, you might recognize what we have in yellow here. Let's think about it."}, {"video_title": "Derivatives expressed as limits Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "So let's do that. Let's take both of these fives and factor them out. And so this whole thing is going to simplify to five times the limit as h approaches zero of log of two plus h minus, minus log of two, all of that over h. Minus log of two, all of that over h. Now, you might recognize what we have in yellow here. Let's think about it. What this is, if we had f of x is equal to log of x, and we wanted to know what f prime of, well actually, let's say f prime of two is, well, this would be the limit as h approaches zero of log of two plus h, two plus h, minus log of two, minus log of two, all of that over h. So this is really just a, what we see here, this by definition, this right over here is f prime of two. If f of x is log of x, this is f prime of two. F prime of two."}, {"video_title": "Derivatives expressed as limits Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "Let's think about it. What this is, if we had f of x is equal to log of x, and we wanted to know what f prime of, well actually, let's say f prime of two is, well, this would be the limit as h approaches zero of log of two plus h, two plus h, minus log of two, minus log of two, all of that over h. So this is really just a, what we see here, this by definition, this right over here is f prime of two. If f of x is log of x, this is f prime of two. F prime of two. So can we figure that out? If f of x is log of x, what is f prime of x? F prime of x, we don't need to use the limit definition."}, {"video_title": "Derivatives expressed as limits Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "F prime of two. So can we figure that out? If f of x is log of x, what is f prime of x? F prime of x, we don't need to use the limit definition. In fact, the limit definition is quite hard to evaluate this limit, but we know how to take the derivative of logarithmic functions. So f prime of x is going to be equal to one over the natural log of our base, our base here, we already talked about that, that is 10. So one over natural log of 10 times, times x."}, {"video_title": "Derivatives expressed as limits Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "F prime of x, we don't need to use the limit definition. In fact, the limit definition is quite hard to evaluate this limit, but we know how to take the derivative of logarithmic functions. So f prime of x is going to be equal to one over the natural log of our base, our base here, we already talked about that, that is 10. So one over natural log of 10 times, times x. If this was a natural log, well then this would be one over natural log of e times x, natural log of e is just one, so that's where you get the one over x, but if you have any other base, you put the natural log of that base right over here in the denominator. So what is f prime of two? F prime of two is one over the natural log of 10 times two."}, {"video_title": "Derivatives expressed as limits Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "So one over natural log of 10 times, times x. If this was a natural log, well then this would be one over natural log of e times x, natural log of e is just one, so that's where you get the one over x, but if you have any other base, you put the natural log of that base right over here in the denominator. So what is f prime of two? F prime of two is one over the natural log of 10 times two. So this whole thing has simplified, this whole thing is equal to five times this business. So I could actually just write it as, it's equal to five over, five over the natural log of 10, natural log of 10 times two. I could have written it as two natural log of 10s."}, {"video_title": "Derivatives expressed as limits Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "F prime of two is one over the natural log of 10 times two. So this whole thing has simplified, this whole thing is equal to five times this business. So I could actually just write it as, it's equal to five over, five over the natural log of 10, natural log of 10 times two. I could have written it as two natural log of 10s. The key here for this type of exercise, you might be like, oh let me see if I can evaluate this limit, but you're like, well, this looks a lot like the derivative of a logarithmic function, especially the derivative when x is equal to two, if we could just factor these fives out. So you factor out the five, you say, hey, this is the derivative of log of x when x is equal to two. And so we know how to take the derivative of log of x."}, {"video_title": "Exponential functions differentiation intro Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So we've already seen that the derivative with respect to x of e to the x is equal to e to the x, which is a pretty amazing thing. One of the many things that makes e somewhat special, that when you have an exponential with your base right over here as e, the derivative of it, the slope at any point, is equal to the value of that actual function. But now let's see, let's start exploring when we have other bases. Can we somehow figure out what is the derivative, what is the derivative with respect to x when we have a to the x, where a could be any number? Is there some way to figure this out? And maybe using our knowledge that the derivative of e to the x is e to the x. Well, can we somehow use a little bit of algebra and exponent properties to rewrite this so it does look like something with e as a base?"}, {"video_title": "Exponential functions differentiation intro Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Can we somehow figure out what is the derivative, what is the derivative with respect to x when we have a to the x, where a could be any number? Is there some way to figure this out? And maybe using our knowledge that the derivative of e to the x is e to the x. Well, can we somehow use a little bit of algebra and exponent properties to rewrite this so it does look like something with e as a base? Well, you could view a, you could view a as being equal to e, let me write it this way, well, I'll write it, a as being equal to e to the natural log of a. Now, I want you to, if this isn't obvious to you, I really want you to think about it. What is the natural log of a?"}, {"video_title": "Exponential functions differentiation intro Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, can we somehow use a little bit of algebra and exponent properties to rewrite this so it does look like something with e as a base? Well, you could view a, you could view a as being equal to e, let me write it this way, well, I'll write it, a as being equal to e to the natural log of a. Now, I want you to, if this isn't obvious to you, I really want you to think about it. What is the natural log of a? The natural log, the natural log of a is the power you need to raise e to to get to a. So if you actually raise e to that power, if you raise e to the power you need to raise e to to get to a, well, then you're just going to get to a. So really think about this."}, {"video_title": "Exponential functions differentiation intro Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "What is the natural log of a? The natural log, the natural log of a is the power you need to raise e to to get to a. So if you actually raise e to that power, if you raise e to the power you need to raise e to to get to a, well, then you're just going to get to a. So really think about this. Don't just accept this as a leap of faith. It should make sense to you, and it just comes out of really what a logarithm is. And so we can replace a with this whole expression here."}, {"video_title": "Exponential functions differentiation intro Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So really think about this. Don't just accept this as a leap of faith. It should make sense to you, and it just comes out of really what a logarithm is. And so we can replace a with this whole expression here. We can, if a is the same thing as e to the natural log of a, well, then this is going to be, then this is going to be equal to the derivative with respect to x of e, e to the natural log, I keep writing la, to the natural log of a, and then we're gonna raise that to the xth power. We're gonna raise that to the xth power. And now this, just using our exponent properties, this is going to be equal to the derivative with respect to x of, and I'll just keep color coding it."}, {"video_title": "Exponential functions differentiation intro Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And so we can replace a with this whole expression here. We can, if a is the same thing as e to the natural log of a, well, then this is going to be, then this is going to be equal to the derivative with respect to x of e, e to the natural log, I keep writing la, to the natural log of a, and then we're gonna raise that to the xth power. We're gonna raise that to the xth power. And now this, just using our exponent properties, this is going to be equal to the derivative with respect to x of, and I'll just keep color coding it. If I raise something to an exponent and then raise that to an exponent, that's the same thing as raising our original base to the product of those exponents. That's just a basic exponent property. So that's going to be the same thing as e to the natural log of a, natural log of a, times x power."}, {"video_title": "Exponential functions differentiation intro Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And now this, just using our exponent properties, this is going to be equal to the derivative with respect to x of, and I'll just keep color coding it. If I raise something to an exponent and then raise that to an exponent, that's the same thing as raising our original base to the product of those exponents. That's just a basic exponent property. So that's going to be the same thing as e to the natural log of a, natural log of a, times x power. Times x power. And now we can use the chain rule to evaluate this derivative. So what we will do is we will first take the derivative of the outside function, so e to the natural log of a times x, with respect to the inside function, with respect to natural log of a times x."}, {"video_title": "Exponential functions differentiation intro Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So that's going to be the same thing as e to the natural log of a, natural log of a, times x power. Times x power. And now we can use the chain rule to evaluate this derivative. So what we will do is we will first take the derivative of the outside function, so e to the natural log of a times x, with respect to the inside function, with respect to natural log of a times x. And so this is going to be equal to e to the natural log of a times x. And then we take the derivative of that inside function with respect to x. Well, natural log of a might not immediately jump out to you but that's just going to be a number."}, {"video_title": "Exponential functions differentiation intro Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So what we will do is we will first take the derivative of the outside function, so e to the natural log of a times x, with respect to the inside function, with respect to natural log of a times x. And so this is going to be equal to e to the natural log of a times x. And then we take the derivative of that inside function with respect to x. Well, natural log of a might not immediately jump out to you but that's just going to be a number. So that's just going to be, so times, if it was the derivative of three x, it would just be three. If it's the derivative of natural log of a times x, it's just going to be natural log, natural log of a. And so this is going to give us the natural log of a times e to the natural log of a, and I'm going to write it like this, natural log of a to the x power."}, {"video_title": "Exponential functions differentiation intro Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, natural log of a might not immediately jump out to you but that's just going to be a number. So that's just going to be, so times, if it was the derivative of three x, it would just be three. If it's the derivative of natural log of a times x, it's just going to be natural log, natural log of a. And so this is going to give us the natural log of a times e to the natural log of a, and I'm going to write it like this, natural log of a to the x power. Well, we've already seen this, let me, this right over here is just a. So it all simplifies, it all simplifies to the natural log of a times, times a to the x, which is a pretty neat result. So if you're taking the derivative of e to the x, it's just going to be e to the x."}, {"video_title": "Exponential functions differentiation intro Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And so this is going to give us the natural log of a times e to the natural log of a, and I'm going to write it like this, natural log of a to the x power. Well, we've already seen this, let me, this right over here is just a. So it all simplifies, it all simplifies to the natural log of a times, times a to the x, which is a pretty neat result. So if you're taking the derivative of e to the x, it's just going to be e to the x. If you're taking the derivative of the natural log, if you're taking the derivative of a to the x, it's just going to be the natural log of a times a to the x. And so we can now use this result to actually take the derivatives of, of these types of expressions with bases other than e. So if I want to find the derivative with respect to x of eight times three to the x power, well, what's that going to be? Well, that's just going to be eight times, and then the derivative of this right over here is going to be, based on what we just saw, it's going to be the natural log of our base, natural log of three times three to the x, times three to the x."}, {"video_title": "Worked example Product rule with mixed implicit & explicit AP Calculus AB Khan Academy.mp3", "Sentence": "Let g be the function g of x is equal to one over x. Let capital F be a function defined as the product of those other two functions. What is capital F prime of negative one? Well, we can just apply the product rule here. Let me just rewrite, let me just essentially state the product rule. Capital F prime of x is going to be equal to, since capital F of x is the product of these two functions, we apply the product rule, this is going to be f prime of x times g of x plus f of x times g prime of x. If we want to evaluate this at negative one, capital F prime at negative one is equal to f prime of negative one times g of negative one plus function f evaluated at negative one times the derivative of g evaluated at negative one."}, {"video_title": "Worked example Product rule with mixed implicit & explicit AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we can just apply the product rule here. Let me just rewrite, let me just essentially state the product rule. Capital F prime of x is going to be equal to, since capital F of x is the product of these two functions, we apply the product rule, this is going to be f prime of x times g of x plus f of x times g prime of x. If we want to evaluate this at negative one, capital F prime at negative one is equal to f prime of negative one times g of negative one plus function f evaluated at negative one times the derivative of g evaluated at negative one. Now let's see if we can figure these things out. Do they tell us this anywhere? Can we figure this out, f prime of negative one?"}, {"video_title": "Worked example Product rule with mixed implicit & explicit AP Calculus AB Khan Academy.mp3", "Sentence": "If we want to evaluate this at negative one, capital F prime at negative one is equal to f prime of negative one times g of negative one plus function f evaluated at negative one times the derivative of g evaluated at negative one. Now let's see if we can figure these things out. Do they tell us this anywhere? Can we figure this out, f prime of negative one? Well, they tell us right over here, f prime of negative one is equal to five. This is equal to five. Now, let's actually stick with f. What is f of negative one?"}, {"video_title": "Worked example Product rule with mixed implicit & explicit AP Calculus AB Khan Academy.mp3", "Sentence": "Can we figure this out, f prime of negative one? Well, they tell us right over here, f prime of negative one is equal to five. This is equal to five. Now, let's actually stick with f. What is f of negative one? Well, they tell that to us right over here. f of negative one is equal to three. So f of negative one is equal to three."}, {"video_title": "Worked example Product rule with mixed implicit & explicit AP Calculus AB Khan Academy.mp3", "Sentence": "Now, let's actually stick with f. What is f of negative one? Well, they tell that to us right over here. f of negative one is equal to three. So f of negative one is equal to three. Now, g of negative one and g prime of negative one, they don't give it to us explicitly here, but we can figure it out. We know that, well, if g of x is equal to this, g of negative one is equal to one over negative one, which is equal to negative one. So this is equal to negative one."}, {"video_title": "Worked example Product rule with mixed implicit & explicit AP Calculus AB Khan Academy.mp3", "Sentence": "So f of negative one is equal to three. Now, g of negative one and g prime of negative one, they don't give it to us explicitly here, but we can figure it out. We know that, well, if g of x is equal to this, g of negative one is equal to one over negative one, which is equal to negative one. So this is equal to negative one. And then last but not least, if we want to find g prime of negative one, we just have to take the derivative of this. So g prime of x, actually, let me just rewrite g of x. g of x, one over x is just the same thing as x to the negative one. So we're gonna use a power rule to figure out g prime of x is equal to, bring that exponent out front, negative one times x to the, and then decrement the exponent negative two power."}, {"video_title": "Worked example Product rule with mixed implicit & explicit AP Calculus AB Khan Academy.mp3", "Sentence": "So this is equal to negative one. And then last but not least, if we want to find g prime of negative one, we just have to take the derivative of this. So g prime of x, actually, let me just rewrite g of x. g of x, one over x is just the same thing as x to the negative one. So we're gonna use a power rule to figure out g prime of x is equal to, bring that exponent out front, negative one times x to the, and then decrement the exponent negative two power. So g prime of negative one, of negative one is equal to negative one times negative one to the negative two power. And that's just the same thing as negative one over negative one squared. This is one, so this is just all going to evaluate to negative one."}, {"video_title": "Worked example Product rule with mixed implicit & explicit AP Calculus AB Khan Academy.mp3", "Sentence": "So we're gonna use a power rule to figure out g prime of x is equal to, bring that exponent out front, negative one times x to the, and then decrement the exponent negative two power. So g prime of negative one, of negative one is equal to negative one times negative one to the negative two power. And that's just the same thing as negative one over negative one squared. This is one, so this is just all going to evaluate to negative one. So this is negative one. And so we have five times negative one, which is negative five, plus three times negative one, which is negative three, which is equal to negative eight. So f prime of negative one is equal to negative eight."}, {"video_title": "Finding decreasing interval given the function Calculus Khan Academy.mp3", "Sentence": "Pause the video and try to figure that out. All right, now let's do this together. So we know that a function is decreasing when its derivative is negative. Or another way to say it, it's going to be decreasing when f prime of x is less than zero. So what is f prime of x? Well, we could use the derivative rules and derivative properties we know. We apply the power rule to x to the sixth."}, {"video_title": "Finding decreasing interval given the function Calculus Khan Academy.mp3", "Sentence": "Or another way to say it, it's going to be decreasing when f prime of x is less than zero. So what is f prime of x? Well, we could use the derivative rules and derivative properties we know. We apply the power rule to x to the sixth. We bring the six out front or multiply the one coefficient here times six to get six x to the fifth, decrement that exponent, minus, bring the five times the three, minus 15 x to the, we decrement the five, so x to the fourth. And we need to figure out over what intervals is this going to be less than zero? And now let's see how we can simplify this a little bit."}, {"video_title": "Finding decreasing interval given the function Calculus Khan Academy.mp3", "Sentence": "We apply the power rule to x to the sixth. We bring the six out front or multiply the one coefficient here times six to get six x to the fifth, decrement that exponent, minus, bring the five times the three, minus 15 x to the, we decrement the five, so x to the fourth. And we need to figure out over what intervals is this going to be less than zero? And now let's see how we can simplify this a little bit. Both of these terms are divisible by x to the fourth and they're both divisible by three. So let's factor out a three x to the fourth times, you factor out a three x to the fourth here, you're left with a two x. And then over here you have minus five has to be less than zero."}, {"video_title": "Finding decreasing interval given the function Calculus Khan Academy.mp3", "Sentence": "And now let's see how we can simplify this a little bit. Both of these terms are divisible by x to the fourth and they're both divisible by three. So let's factor out a three x to the fourth times, you factor out a three x to the fourth here, you're left with a two x. And then over here you have minus five has to be less than zero. Any interval where this is true, we are going to be decreasing. Now, how do we get this to be less than zero? Well, if I take the product of two things and it's less than zero, that means that they have to have different signs, either one's positive and the other is negative or one's negative and the other's positive."}, {"video_title": "Finding decreasing interval given the function Calculus Khan Academy.mp3", "Sentence": "And then over here you have minus five has to be less than zero. Any interval where this is true, we are going to be decreasing. Now, how do we get this to be less than zero? Well, if I take the product of two things and it's less than zero, that means that they have to have different signs, either one's positive and the other is negative or one's negative and the other's positive. So we have two situations. So we could say either, either three x to the fourth is greater than zero and, and two x minus five is less than zero. So that's one situation."}, {"video_title": "Finding decreasing interval given the function Calculus Khan Academy.mp3", "Sentence": "Well, if I take the product of two things and it's less than zero, that means that they have to have different signs, either one's positive and the other is negative or one's negative and the other's positive. So we have two situations. So we could say either, either three x to the fourth is greater than zero and, and two x minus five is less than zero. So that's one situation. I'll do this in a different color. Or, and I'll do this one in a different color, three x to the fourth is less than zero and two x minus five is greater than zero. Actually, let me stay on the second case first."}, {"video_title": "Finding decreasing interval given the function Calculus Khan Academy.mp3", "Sentence": "So that's one situation. I'll do this in a different color. Or, and I'll do this one in a different color, three x to the fourth is less than zero and two x minus five is greater than zero. Actually, let me stay on the second case first. Are there any situations where three x to the fourth can be less than zero? You take any number, you take it to the fourth power, even if it's a negative, it's going to become a positive. So you can't get a negative expression right over here."}, {"video_title": "Finding decreasing interval given the function Calculus Khan Academy.mp3", "Sentence": "Actually, let me stay on the second case first. Are there any situations where three x to the fourth can be less than zero? You take any number, you take it to the fourth power, even if it's a negative, it's going to become a positive. So you can't get a negative expression right over here. So actually, the second condition is impossible to obtain. You can't get any situation for any x where three x to the fourth is less than zero. So we can rule this one out."}, {"video_title": "Finding decreasing interval given the function Calculus Khan Academy.mp3", "Sentence": "So you can't get a negative expression right over here. So actually, the second condition is impossible to obtain. You can't get any situation for any x where three x to the fourth is less than zero. So we can rule this one out. And so this is our best hope. So under what conditions is three x to the fourth greater than zero? Well, if you divide both sides by three, you get x to the fourth is greater than zero."}, {"video_title": "Finding decreasing interval given the function Calculus Khan Academy.mp3", "Sentence": "So we can rule this one out. And so this is our best hope. So under what conditions is three x to the fourth greater than zero? Well, if you divide both sides by three, you get x to the fourth is greater than zero. And if you think about it, this is gonna be true for any x value that is not equal to zero. Even if you have a negative value there, if you have a negative one, you take it to the fourth power, it becomes a positive one. Only zero will be equal to zero when you take it to the fourth power."}, {"video_title": "Finding decreasing interval given the function Calculus Khan Academy.mp3", "Sentence": "Well, if you divide both sides by three, you get x to the fourth is greater than zero. And if you think about it, this is gonna be true for any x value that is not equal to zero. Even if you have a negative value there, if you have a negative one, you take it to the fourth power, it becomes a positive one. Only zero will be equal to zero when you take it to the fourth power. So one way we could say this is going to be true for any non-zero x, or we could just say x does not equal zero. And this is a little bit more straightforward. We add five to both sides, we get two x is less than five."}, {"video_title": "Finding decreasing interval given the function Calculus Khan Academy.mp3", "Sentence": "Only zero will be equal to zero when you take it to the fourth power. So one way we could say this is going to be true for any non-zero x, or we could just say x does not equal zero. And this is a little bit more straightforward. We add five to both sides, we get two x is less than five. Divide both sides by two, you get x is less than 5 1\u20442. So it might be tempting to say, all right, the intervals that matter are all the x's less than 5 1\u20442, but x cannot be equal to zero. Now, is that the entire interval where our function is decreasing?"}, {"video_title": "Finding decreasing interval given the function Calculus Khan Academy.mp3", "Sentence": "We add five to both sides, we get two x is less than five. Divide both sides by two, you get x is less than 5 1\u20442. So it might be tempting to say, all right, the intervals that matter are all the x's less than 5 1\u20442, but x cannot be equal to zero. Now, is that the entire interval where our function is decreasing? Well, let's think about what happens at zero itself. We're decreasing over the interval from negative infinity all the way up to zero, and we're also decreasing from zero to 5 1\u20442. And so if we're decreasing right to the left of zero, we're decreasing right to the right of zero, we're actually going to be decreasing at zero at, we're also going to be decreasing at zero as well."}, {"video_title": "Finding decreasing interval given the function Calculus Khan Academy.mp3", "Sentence": "Now, is that the entire interval where our function is decreasing? Well, let's think about what happens at zero itself. We're decreasing over the interval from negative infinity all the way up to zero, and we're also decreasing from zero to 5 1\u20442. And so if we're decreasing right to the left of zero, we're decreasing right to the right of zero, we're actually going to be decreasing at zero at, we're also going to be decreasing at zero as well. So there's something interesting here. Even though the derivative at x equals zero is going to be equal to zero, we are still decreasing there. And so the interval that we care about, the interval over which we're decreasing is just x is less than 5 1\u20442."}, {"video_title": "Finding decreasing interval given the function Calculus Khan Academy.mp3", "Sentence": "And so if we're decreasing right to the left of zero, we're decreasing right to the right of zero, we're actually going to be decreasing at zero at, we're also going to be decreasing at zero as well. So there's something interesting here. Even though the derivative at x equals zero is going to be equal to zero, we are still decreasing there. And so the interval that we care about, the interval over which we're decreasing is just x is less than 5 1\u20442. And we can see that by graphing the function. I graphed it on Desmos. And you can see here that the function is decreasing from negative infinity, it's decreasing at a slower and slower rate, we get to zero, still decreasing to the left of zero, and then it continues to decrease to the right of zero."}, {"video_title": "Finding decreasing interval given the function Calculus Khan Academy.mp3", "Sentence": "And so the interval that we care about, the interval over which we're decreasing is just x is less than 5 1\u20442. And we can see that by graphing the function. I graphed it on Desmos. And you can see here that the function is decreasing from negative infinity, it's decreasing at a slower and slower rate, we get to zero, still decreasing to the left of zero, and then it continues to decrease to the right of zero. So any x value to the left of zero, the value of the function is going to be larger than f of zero. And x to the right of zero, the value of the function is going to be less than the function at zero. So it's actually decreasing through zero, even though the slope of the tangent line at zero is zero, even though it's non-negative."}, {"video_title": "Calculating integral disc around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "In the last video, we had set up the definite integral to evaluate the volume of this upside down gumdrop truffle looking thing. So now in this video, we can actually evaluate the definite integral. So what we need to do is really just expand out this expression, the square root of y plus 1 plus 2. So let's do that. So this is going to be equal to pi times the definite integral from y is equal to negative 1 to y is equal to 3. If you expand this out, you get square root of y plus 1 squared, which is just going to be y plus 1. And then you're going to have 2 times the product of both of these terms."}, {"video_title": "Calculating integral disc around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "So let's do that. So this is going to be equal to pi times the definite integral from y is equal to negative 1 to y is equal to 3. If you expand this out, you get square root of y plus 1 squared, which is just going to be y plus 1. And then you're going to have 2 times the product of both of these terms. 2 times 2 times square root of y plus 1 is going to be plus 4 times the square root of y plus 1. And then you have 2 squared, so plus 4. So you have this whole thing times dy."}, {"video_title": "Calculating integral disc around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "And then you're going to have 2 times the product of both of these terms. 2 times 2 times square root of y plus 1 is going to be plus 4 times the square root of y plus 1. And then you have 2 squared, so plus 4. So you have this whole thing times dy. We can simplify a little bit. You have a 1 plus a 4. We can add the 1 to the 4 and get a 5."}, {"video_title": "Calculating integral disc around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "So you have this whole thing times dy. We can simplify a little bit. You have a 1 plus a 4. We can add the 1 to the 4 and get a 5. And now we're ready to take the antiderivative. This is going to be equal to pi times, let's take the antiderivative of all of this business, and I'll color code it. The antiderivative of y is just y squared over 2."}, {"video_title": "Calculating integral disc around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "We can add the 1 to the 4 and get a 5. And now we're ready to take the antiderivative. This is going to be equal to pi times, let's take the antiderivative of all of this business, and I'll color code it. The antiderivative of y is just y squared over 2. The antiderivative of 4 times the square root of y plus 1, you just really have to think of it as 4 times y plus 1 to the 1 half power. We could use u substitution explicitly, but you probably are pretty practiced and this can do this in your head. y plus 1 raised to the 1 half power."}, {"video_title": "Calculating integral disc around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "The antiderivative of y is just y squared over 2. The antiderivative of 4 times the square root of y plus 1, you just really have to think of it as 4 times y plus 1 to the 1 half power. We could use u substitution explicitly, but you probably are pretty practiced and this can do this in your head. y plus 1 raised to the 1 half power. Derivative of y plus 1 is just 1, which is essentially out here. So you can really just, if you did u substitution, you'd say u is equal to y plus 1. But this antiderivative is going to be equal to, well, if you increment this exponent, you get 3 halves multiplied by the reciprocal 2 thirds."}, {"video_title": "Calculating integral disc around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "y plus 1 raised to the 1 half power. Derivative of y plus 1 is just 1, which is essentially out here. So you can really just, if you did u substitution, you'd say u is equal to y plus 1. But this antiderivative is going to be equal to, well, if you increment this exponent, you get 3 halves multiplied by the reciprocal 2 thirds. 2 thirds times 4 is 8 thirds. So it's plus 8 thirds times y plus 1 to the 3 halves. And you can verify, if you take the derivative here, you will get this expression right over here."}, {"video_title": "Calculating integral disc around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "But this antiderivative is going to be equal to, well, if you increment this exponent, you get 3 halves multiplied by the reciprocal 2 thirds. 2 thirds times 4 is 8 thirds. So it's plus 8 thirds times y plus 1 to the 3 halves. And you can verify, if you take the derivative here, you will get this expression right over here. 3 halves times 8 thirds is 4. Decrement it, you have y plus 1 to the 1 half power. And then finally, you have this 5."}, {"video_title": "Calculating integral disc around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "And you can verify, if you take the derivative here, you will get this expression right over here. 3 halves times 8 thirds is 4. Decrement it, you have y plus 1 to the 1 half power. And then finally, you have this 5. The antiderivative of 5 is just 5y. And we are going to evaluate it from at 3 and at negative 1. y is equals 3 and y equals negative 1. So this is going to be equal to pi."}, {"video_title": "Calculating integral disc around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "And then finally, you have this 5. The antiderivative of 5 is just 5y. And we are going to evaluate it from at 3 and at negative 1. y is equals 3 and y equals negative 1. So this is going to be equal to pi. So let's evaluate all this business at 3. So 3 squared over 2 is 9 halves. 3 plus 1 is 4 to the 3 halves."}, {"video_title": "Calculating integral disc around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be equal to pi. So let's evaluate all this business at 3. So 3 squared over 2 is 9 halves. 3 plus 1 is 4 to the 3 halves. Well, that's, let's see, square root of 4 is 2 to the third power is 8. 8 times 8 thirds is 64 over 3. So plus 64 over 3."}, {"video_title": "Calculating integral disc around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "3 plus 1 is 4 to the 3 halves. Well, that's, let's see, square root of 4 is 2 to the third power is 8. 8 times 8 thirds is 64 over 3. So plus 64 over 3. You have 5 times 3. Well, that's going to be 15. Plus 15."}, {"video_title": "Calculating integral disc around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "So plus 64 over 3. You have 5 times 3. Well, that's going to be 15. Plus 15. And from that, we're going to subtract all this business evaluated at negative 1. So you have negative 1 squared over 2. Well, that's just 1 half."}, {"video_title": "Calculating integral disc around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "Plus 15. And from that, we're going to subtract all this business evaluated at negative 1. So you have negative 1 squared over 2. Well, that's just 1 half. Negative 1 plus 1 is 0 to the 3 half power. That's going to be 0 times 8 thirds. This is all going to be 0, so we don't have to even write it."}, {"video_title": "Calculating integral disc around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "Well, that's just 1 half. Negative 1 plus 1 is 0 to the 3 half power. That's going to be 0 times 8 thirds. This is all going to be 0, so we don't have to even write it. And then finally, you have negative 1 times 5. Well, this is going to be negative 5. And we are in the home stretch."}, {"video_title": "Calculating integral disc around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "This is all going to be 0, so we don't have to even write it. And then finally, you have negative 1 times 5. Well, this is going to be negative 5. And we are in the home stretch. We really just have to do a little bit of arithmetic, add some hairy fractions right over here. So let's do it. So this whole thing is going to simplify to pi times."}, {"video_title": "Calculating integral disc around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "And we are in the home stretch. We really just have to do a little bit of arithmetic, add some hairy fractions right over here. So let's do it. So this whole thing is going to simplify to pi times. And it looks like, let's see, our least common multiple of all of these denominators is going to be 6. So let's put everything over a denominator of 6. So 9 halves is the same thing as 27 over 6."}, {"video_title": "Calculating integral disc around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "So this whole thing is going to simplify to pi times. And it looks like, let's see, our least common multiple of all of these denominators is going to be 6. So let's put everything over a denominator of 6. So 9 halves is the same thing as 27 over 6. 64 thirds is the same thing as 128 over 6. 15 is the same thing as 90 over 6. 1 half is the same thing as 3 over 6."}, {"video_title": "Calculating integral disc around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "So 9 halves is the same thing as 27 over 6. 64 thirds is the same thing as 128 over 6. 15 is the same thing as 90 over 6. 1 half is the same thing as 3 over 6. So this is, we distribute the negative signs. This is negative 3 6. And negative times negative is positive."}, {"video_title": "Calculating integral disc around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "1 half is the same thing as 3 over 6. So this is, we distribute the negative signs. This is negative 3 6. And negative times negative is positive. 5 is the same thing as 30 over 6. So plus 30 over 6. And so this is going to give us, our denominator is going to be over 6."}, {"video_title": "Calculating integral disc around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "And negative times negative is positive. 5 is the same thing as 30 over 6. So plus 30 over 6. And so this is going to give us, our denominator is going to be over 6. We're going to multiply something times pi. We have this pi over here. And then we just have to figure out what the numerator is."}, {"video_title": "Calculating integral disc around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "And so this is going to give us, our denominator is going to be over 6. We're going to multiply something times pi. We have this pi over here. And then we just have to figure out what the numerator is. So let's see if I can do this in my head. So 27 plus 128 is going to be, let me see, that's going to be 155. Is that right?"}, {"video_title": "Calculating integral disc around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "And then we just have to figure out what the numerator is. So let's see if I can do this in my head. So 27 plus 128 is going to be, let me see, that's going to be 155. Is that right? 155. Let's see, if we get to 48 plus another 7. Yeah, 155 plus 90 gets us to 245."}, {"video_title": "Calculating integral disc around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "Is that right? 155. Let's see, if we get to 48 plus another 7. Yeah, 155 plus 90 gets us to 245. Is that right? Yeah, plus 90 gets us to 245. Minus 3, you subtract 3 from that, you get to 242."}, {"video_title": "Calculating integral disc around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "Yeah, 155 plus 90 gets us to 245. Is that right? Yeah, plus 90 gets us to 245. Minus 3, you subtract 3 from that, you get to 242. And then you add 30 to that, you get to 272. So we're left with 272 pi over 6. But then we can, let's see, 272 and 6 are both divisible by 2."}, {"video_title": "Calculating integral disc around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "Minus 3, you subtract 3 from that, you get to 242. And then you add 30 to that, you get to 272. So we're left with 272 pi over 6. But then we can, let's see, 272 and 6 are both divisible by 2. So this is equal to, see, 272 divided by 2 is going to be 136 pi over, and if you divide this denominator right over here, by 2 over 3. Is that right? Yeah, 136 pi over 3."}, {"video_title": "Calculating integral disc around vertical line AP Calculus AB Khan Academy.mp3", "Sentence": "But then we can, let's see, 272 and 6 are both divisible by 2. So this is equal to, see, 272 divided by 2 is going to be 136 pi over, and if you divide this denominator right over here, by 2 over 3. Is that right? Yeah, 136 pi over 3. And 136 is not divisible by 3. So we have it as simplified as we can. This right over here is the volume of our little upside down gumdrop looking thing."}, {"video_title": "Fancy algebra to find a limit and make a function continuous Differential Calculus Khan Academy.mp3", "Sentence": "So if we know that f is continuous at x equals 5, that means that the limit, the limit as x approaches 5 of f of x is equal to f of 5. This is the definition of continuity. And they tell us that f of 5, when x equals 5, the value of the function is equal to c. So this must be equal to c. So what we really need to do is figure out what the limit of f of x as x approaches 5 actually is. Now if we just try to substitute 5 into the expression right up here, in the numerator you have 5 plus 4 is 9, the square root of that is positive 3, the principal root is positive 3, 3 minus 3 is 0, so you get a 0 in the numerator, and then you get 5 minus 5 in the denominator, so you get 0 in the denominator. So you get this indeterminate form of 0 over 0. And in the future we will see that we do have a tool that allows us, or gives us an option to attempt to find limits when we get this indeterminate form. It's called L'Hopital's Rule."}, {"video_title": "Fancy algebra to find a limit and make a function continuous Differential Calculus Khan Academy.mp3", "Sentence": "Now if we just try to substitute 5 into the expression right up here, in the numerator you have 5 plus 4 is 9, the square root of that is positive 3, the principal root is positive 3, 3 minus 3 is 0, so you get a 0 in the numerator, and then you get 5 minus 5 in the denominator, so you get 0 in the denominator. So you get this indeterminate form of 0 over 0. And in the future we will see that we do have a tool that allows us, or gives us an option to attempt to find limits when we get this indeterminate form. It's called L'Hopital's Rule. But we can actually tackle this with a little bit of fancy algebra. And to do that I'm going to try to get this radical out of the numerator. So let's rewrite it."}, {"video_title": "Fancy algebra to find a limit and make a function continuous Differential Calculus Khan Academy.mp3", "Sentence": "It's called L'Hopital's Rule. But we can actually tackle this with a little bit of fancy algebra. And to do that I'm going to try to get this radical out of the numerator. So let's rewrite it. So we have the square root of x plus 4 minus 3 over x minus 5. And any time you see a radical plus or minus something else, to get rid of the radical what you can do is multiply by the radical doing the, or if you have a radical minus 3 you multiply by the radical plus 3. So in this situation you just multiply the numerator by square root of x plus 4 plus 3 over the square root of x plus 4 plus 3."}, {"video_title": "Fancy algebra to find a limit and make a function continuous Differential Calculus Khan Academy.mp3", "Sentence": "So let's rewrite it. So we have the square root of x plus 4 minus 3 over x minus 5. And any time you see a radical plus or minus something else, to get rid of the radical what you can do is multiply by the radical doing the, or if you have a radical minus 3 you multiply by the radical plus 3. So in this situation you just multiply the numerator by square root of x plus 4 plus 3 over the square root of x plus 4 plus 3. We obviously have to multiply the numerator and the denominator by the same thing so that we actually don't change the value of the expression. If this right over here had a plus 3 then we would do a minus 3 here. This is a technique that we learn in algebra or sometimes in pre-calculus class to rationalize usually denominators but to rationalize numerators or denominators."}, {"video_title": "Fancy algebra to find a limit and make a function continuous Differential Calculus Khan Academy.mp3", "Sentence": "So in this situation you just multiply the numerator by square root of x plus 4 plus 3 over the square root of x plus 4 plus 3. We obviously have to multiply the numerator and the denominator by the same thing so that we actually don't change the value of the expression. If this right over here had a plus 3 then we would do a minus 3 here. This is a technique that we learn in algebra or sometimes in pre-calculus class to rationalize usually denominators but to rationalize numerators or denominators. It's also a very similar technique that we use oftentimes to get rid of complex numbers usually in denominators. But if you multiply this out, and I encourage you to do it, you notice this has the pattern that you learned in algebra class. This is the difference of squares."}, {"video_title": "Fancy algebra to find a limit and make a function continuous Differential Calculus Khan Academy.mp3", "Sentence": "This is a technique that we learn in algebra or sometimes in pre-calculus class to rationalize usually denominators but to rationalize numerators or denominators. It's also a very similar technique that we use oftentimes to get rid of complex numbers usually in denominators. But if you multiply this out, and I encourage you to do it, you notice this has the pattern that you learned in algebra class. This is the difference of squares. Something minus something times something plus something. So the first term is going to be the first something squared. So square root of x plus 4 squared is x plus 4."}, {"video_title": "Fancy algebra to find a limit and make a function continuous Differential Calculus Khan Academy.mp3", "Sentence": "This is the difference of squares. Something minus something times something plus something. So the first term is going to be the first something squared. So square root of x plus 4 squared is x plus 4. The second term is going to be the second something or you're going to subtract the second something squared. So you're going to have minus 3 squared, so minus 9. In the denominator you're of course going to have x minus 5 times the square root of x plus 4 plus 3."}, {"video_title": "Fancy algebra to find a limit and make a function continuous Differential Calculus Khan Academy.mp3", "Sentence": "So square root of x plus 4 squared is x plus 4. The second term is going to be the second something or you're going to subtract the second something squared. So you're going to have minus 3 squared, so minus 9. In the denominator you're of course going to have x minus 5 times the square root of x plus 4 plus 3. So this has, I guess you could say simplified to, although not arguably any simpler, but at least we've gotten our radical. We're really just playing around with it algebraically to see if we can then substitute x equals 5 or if we can somehow simplify it to figure out what the limit is. When you simplify the numerator up here, you get x plus 4 minus 9."}, {"video_title": "Fancy algebra to find a limit and make a function continuous Differential Calculus Khan Academy.mp3", "Sentence": "In the denominator you're of course going to have x minus 5 times the square root of x plus 4 plus 3. So this has, I guess you could say simplified to, although not arguably any simpler, but at least we've gotten our radical. We're really just playing around with it algebraically to see if we can then substitute x equals 5 or if we can somehow simplify it to figure out what the limit is. When you simplify the numerator up here, you get x plus 4 minus 9. Well that's x minus 5 over x minus 5 times the square root of x plus 4 plus 3. And now it pops out at you. Both the numerator and the denominator are now divisible by x minus 5."}, {"video_title": "Fancy algebra to find a limit and make a function continuous Differential Calculus Khan Academy.mp3", "Sentence": "When you simplify the numerator up here, you get x plus 4 minus 9. Well that's x minus 5 over x minus 5 times the square root of x plus 4 plus 3. And now it pops out at you. Both the numerator and the denominator are now divisible by x minus 5. So you can have a completely identical expression if you say that this is the same thing. You can divide the numerator and the denominator by x minus 5 if you assume x does not equal 5. So this is going to be the same thing as 1 over square root of x plus 4 plus 3 for x does not equal 5."}, {"video_title": "Fancy algebra to find a limit and make a function continuous Differential Calculus Khan Academy.mp3", "Sentence": "Both the numerator and the denominator are now divisible by x minus 5. So you can have a completely identical expression if you say that this is the same thing. You can divide the numerator and the denominator by x minus 5 if you assume x does not equal 5. So this is going to be the same thing as 1 over square root of x plus 4 plus 3 for x does not equal 5. Which is fine because in the first part of this function definition, this is for the case for x does not equal 5. So we could actually replace this, and this is a simpler expression, with 1 over square root of x plus 4 plus 3. And so now when we take the limit as x approaches 5, we're going to get closer and closer to 5."}, {"video_title": "Fancy algebra to find a limit and make a function continuous Differential Calculus Khan Academy.mp3", "Sentence": "So this is going to be the same thing as 1 over square root of x plus 4 plus 3 for x does not equal 5. Which is fine because in the first part of this function definition, this is for the case for x does not equal 5. So we could actually replace this, and this is a simpler expression, with 1 over square root of x plus 4 plus 3. And so now when we take the limit as x approaches 5, we're going to get closer and closer to 5. We're going to get x values closer and closer to 5, but not quite at 5. We can use this expression right over here. So the limit of f of x as x approaches 5 is going to be the same thing as the limit of 1 over the square root of x plus 4 plus 3 as x approaches 5."}, {"video_title": "Fancy algebra to find a limit and make a function continuous Differential Calculus Khan Academy.mp3", "Sentence": "And so now when we take the limit as x approaches 5, we're going to get closer and closer to 5. We're going to get x values closer and closer to 5, but not quite at 5. We can use this expression right over here. So the limit of f of x as x approaches 5 is going to be the same thing as the limit of 1 over the square root of x plus 4 plus 3 as x approaches 5. And now we can substitute a 5 in here. It's going to be 1 over 5 plus 4 is 9, the principal root of that is 3, 3 plus 3 is 6. So if c is equal to 1 6, then the limit of our function as x approaches 5 is going to be equal to f of 5, and we are continuous at x equals 5."}, {"video_title": "Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3", "Sentence": "So this first one is the definite integral from negative six to negative two of f of x dx. Pause this video and see if you can figure this one out from this graph. All right, we're going from x equals negative six to x equals negative two, and the definite integral is going to be the area below our graph and above the x-axis. So it's going to be this area right over here. And how do we figure that out? Well, this is a semicircle, and we know how to find the area of a circle if we know its radius, and this circle has radius two, has a radius of two. No matter what direction we go in from the center, it has a radius of two."}, {"video_title": "Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3", "Sentence": "So it's going to be this area right over here. And how do we figure that out? Well, this is a semicircle, and we know how to find the area of a circle if we know its radius, and this circle has radius two, has a radius of two. No matter what direction we go in from the center, it has a radius of two. And so the area of a circle is pi r squared, so it'd be pi times our radius, which is two squared, but this is a semicircle, so I'm gonna divide by two. It's only half the area of the full circle. So this is going to be four pi over two, which is equal to two pi."}, {"video_title": "Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3", "Sentence": "No matter what direction we go in from the center, it has a radius of two. And so the area of a circle is pi r squared, so it'd be pi times our radius, which is two squared, but this is a semicircle, so I'm gonna divide by two. It's only half the area of the full circle. So this is going to be four pi over two, which is equal to two pi. All right, let's do another one. So here we have the definite integral from negative two to one of f of x dx. Pause the video and see if you can figure that out."}, {"video_title": "Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be four pi over two, which is equal to two pi. All right, let's do another one. So here we have the definite integral from negative two to one of f of x dx. Pause the video and see if you can figure that out. All right, let's do it together. So we're going from negative two to one, and so we have to be a little bit careful here. So the definite integral, you could view it as the area below the function and above the x-axis."}, {"video_title": "Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3", "Sentence": "Pause the video and see if you can figure that out. All right, let's do it together. So we're going from negative two to one, and so we have to be a little bit careful here. So the definite integral, you could view it as the area below the function and above the x-axis. But here, the function is below the x-axis. And so what we can do is we can figure out this area, just knowing what we know about geometry, and then we have to realize that this is going to be a negative value for the definite integral because our function is below the x-axis. So what's the area here?"}, {"video_title": "Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3", "Sentence": "So the definite integral, you could view it as the area below the function and above the x-axis. But here, the function is below the x-axis. And so what we can do is we can figure out this area, just knowing what we know about geometry, and then we have to realize that this is going to be a negative value for the definite integral because our function is below the x-axis. So what's the area here? Well, there's a couple of ways to think about it. We could split it up into a few shapes. So you could just view it as a trapezoid, or you could just split it up into a rectangle and two triangles."}, {"video_title": "Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3", "Sentence": "So what's the area here? Well, there's a couple of ways to think about it. We could split it up into a few shapes. So you could just view it as a trapezoid, or you could just split it up into a rectangle and two triangles. So if you split it up like this, this triangle right over here has an area of one times two times 1 1\u20442. So this has an area of one. This rectangle right over here has an area of two times one."}, {"video_title": "Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3", "Sentence": "So you could just view it as a trapezoid, or you could just split it up into a rectangle and two triangles. So if you split it up like this, this triangle right over here has an area of one times two times 1 1\u20442. So this has an area of one. This rectangle right over here has an area of two times one. So it has an area of two. And then this triangle right over here is the same area as the first one. It's going to have a base of one, a height of two."}, {"video_title": "Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3", "Sentence": "This rectangle right over here has an area of two times one. So it has an area of two. And then this triangle right over here is the same area as the first one. It's going to have a base of one, a height of two. So it's one times two times 1 1\u20442. Remember, the area of a triangle is 1 1\u20442 base times height, so it's one. So if you add up those areas, one plus two plus one is four."}, {"video_title": "Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3", "Sentence": "It's going to have a base of one, a height of two. So it's one times two times 1 1\u20442. Remember, the area of a triangle is 1 1\u20442 base times height, so it's one. So if you add up those areas, one plus two plus one is four. And so you might be tempted to say, oh, is this going to be equal to four? But remember, our function is below the x-axis here. And so this is going to be a negative four."}, {"video_title": "Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3", "Sentence": "So if you add up those areas, one plus two plus one is four. And so you might be tempted to say, oh, is this going to be equal to four? But remember, our function is below the x-axis here. And so this is going to be a negative four. All right, let's do another one. So now we're gonna go from one to four of f of x dx. So pause the video and see if you can figure that out."}, {"video_title": "Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3", "Sentence": "And so this is going to be a negative four. All right, let's do another one. So now we're gonna go from one to four of f of x dx. So pause the video and see if you can figure that out. So we're gonna go from here to here. And so it's gonna be this area right over there. So how do we figure that out?"}, {"video_title": "Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3", "Sentence": "So pause the video and see if you can figure that out. So we're gonna go from here to here. And so it's gonna be this area right over there. So how do we figure that out? Well, just the formula for the area of a triangle, base times height times 1 1\u20442. So, or you could say 1 1\u20442 times our base, which is a length of, see, we have a base of three right over here, we go from one to four. So 1 1\u20442 times three times our height, which is one, two, three, four, times four."}, {"video_title": "Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3", "Sentence": "So how do we figure that out? Well, just the formula for the area of a triangle, base times height times 1 1\u20442. So, or you could say 1 1\u20442 times our base, which is a length of, see, we have a base of three right over here, we go from one to four. So 1 1\u20442 times three times our height, which is one, two, three, four, times four. Well, this is just going to get us six. All right, last but not least, if we are going from four to six of f of x dx, so that's going to be this area right over here. But we have to be careful."}, {"video_title": "Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3", "Sentence": "So 1 1\u20442 times three times our height, which is one, two, three, four, times four. Well, this is just going to get us six. All right, last but not least, if we are going from four to six of f of x dx, so that's going to be this area right over here. But we have to be careful. Our function is below the x-axis. So we'll figure out this area, and then it's going to be negative. So this is a half of a circle of radius one."}, {"video_title": "Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3", "Sentence": "But we have to be careful. Our function is below the x-axis. So we'll figure out this area, and then it's going to be negative. So this is a half of a circle of radius one. And so the area of a circle is pi times r squared, so it's pi times one squared. That would be the area if we went all the way around like that. But this is only half of the circle, so divided by two."}, {"video_title": "Finding definite integrals using area formulas AP Calculus AB Khan Academy.mp3", "Sentence": "So this is a half of a circle of radius one. And so the area of a circle is pi times r squared, so it's pi times one squared. That would be the area if we went all the way around like that. But this is only half of the circle, so divided by two. And since this area is above the function and below the x-axis, it's going to be negative. So this is going to be equal to negative pi over two. And we are done."}, {"video_title": "Second derivatives (parametric functions) Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "And in both cases, it's going to be in terms of t. So let's get to it. So first let's find the first derivative of y with respect to x. First derivative of y with respect to x. And we've seen this before in other videos where this is going to be the derivative of y with respect to t over the derivative of x with respect to t. And so this is going to be equal to, well what is the derivative of y with respect to t? Dy dt is equal to, well let's see, the derivative of e to the three t with respect to three t is just e to the three t. And then the derivative of three t with respect to t is going to be three. So I could say times three like that or I could put that three out front. And then the derivative of negative one, well negative, a constant doesn't change no matter what you do to your t. So that's just going to be zero."}, {"video_title": "Second derivatives (parametric functions) Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "And we've seen this before in other videos where this is going to be the derivative of y with respect to t over the derivative of x with respect to t. And so this is going to be equal to, well what is the derivative of y with respect to t? Dy dt is equal to, well let's see, the derivative of e to the three t with respect to three t is just e to the three t. And then the derivative of three t with respect to t is going to be three. So I could say times three like that or I could put that three out front. And then the derivative of negative one, well negative, a constant doesn't change no matter what you do to your t. So that's just going to be zero. So that's dy dt. So it's going to be equal to three e to the three t. All of that over, well what's the derivative of x with respect to t? Derivative of x with respect to t is equal to, well we're gonna have the three out front."}, {"video_title": "Second derivatives (parametric functions) Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "And then the derivative of negative one, well negative, a constant doesn't change no matter what you do to your t. So that's just going to be zero. So that's dy dt. So it's going to be equal to three e to the three t. All of that over, well what's the derivative of x with respect to t? Derivative of x with respect to t is equal to, well we're gonna have the three out front. And so the derivative of e to the two t with respect to two t is going to be e to the two t. And then we're going to take the derivative of two t with respect to t, which is just two. So this is going to be six e to the two t. Six e to the two t. And let's see, we could simplify this a little bit. I'll now go to a neutral color."}, {"video_title": "Second derivatives (parametric functions) Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "Derivative of x with respect to t is equal to, well we're gonna have the three out front. And so the derivative of e to the two t with respect to two t is going to be e to the two t. And then we're going to take the derivative of two t with respect to t, which is just two. So this is going to be six e to the two t. Six e to the two t. And let's see, we could simplify this a little bit. I'll now go to a neutral color. This is equal to, so e to the, so this is going to be 1 1 2. That's three over six. E to the three t minus two t. Three t minus two t. And I'm just using exponent properties right over here."}, {"video_title": "Second derivatives (parametric functions) Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "I'll now go to a neutral color. This is equal to, so e to the, so this is going to be 1 1 2. That's three over six. E to the three t minus two t. Three t minus two t. And I'm just using exponent properties right over here. But three, if I have three t's and I take away two of those t's, I'm just gonna have a t. So this is just going to simplify to a t right over here. So now that we know, we've now figured out the first derivative of y with respect to x in terms of t. Now how do we find the second derivative? How do we find the second derivative of y with respect to x now?"}, {"video_title": "Second derivatives (parametric functions) Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "E to the three t minus two t. Three t minus two t. And I'm just using exponent properties right over here. But three, if I have three t's and I take away two of those t's, I'm just gonna have a t. So this is just going to simplify to a t right over here. So now that we know, we've now figured out the first derivative of y with respect to x in terms of t. Now how do we find the second derivative? How do we find the second derivative of y with respect to x now? And I'll give you a hint. We're going to use this same idea. If you want to find the rate of change of something with respect to x, you find the rate of change of that something with respect to t and divide it by the rate of change of x with respect to t. So what this is going to be, we want to find the derivative of the first derivative with respect to t. So let me write this down."}, {"video_title": "Second derivatives (parametric functions) Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "How do we find the second derivative of y with respect to x now? And I'll give you a hint. We're going to use this same idea. If you want to find the rate of change of something with respect to x, you find the rate of change of that something with respect to t and divide it by the rate of change of x with respect to t. So what this is going to be, we want to find the derivative of the first derivative with respect to t. So let me write this down. So we want to find, we want to take the derivative with respect to t in the numerator of the first derivative, which I will put in blue now, of dy dx, all of that over, all of that over, the all of that over dx, dx dt. Now I want you to, if it doesn't jump out at you why this is the same thing that we did before, I encourage you to pause the video and think about it. Think about what we did over here the first time."}, {"video_title": "Second derivatives (parametric functions) Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "If you want to find the rate of change of something with respect to x, you find the rate of change of that something with respect to t and divide it by the rate of change of x with respect to t. So what this is going to be, we want to find the derivative of the first derivative with respect to t. So let me write this down. So we want to find, we want to take the derivative with respect to t in the numerator of the first derivative, which I will put in blue now, of dy dx, all of that over, all of that over, the all of that over dx, dx dt. Now I want you to, if it doesn't jump out at you why this is the same thing that we did before, I encourage you to pause the video and think about it. Think about what we did over here the first time. When we wanted to find the derivative of y with respect to x, we found the derivative of y with respect to t and then divided that by the derivative of x with respect to t. Here we want to find the derivative, we want to find the second derivative of y with respect to x. Actually let me just write it down out here a little bit clearer. What we really want to do is we want to find the derivative with respect, let me write it this way."}, {"video_title": "Second derivatives (parametric functions) Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "Think about what we did over here the first time. When we wanted to find the derivative of y with respect to x, we found the derivative of y with respect to t and then divided that by the derivative of x with respect to t. Here we want to find the derivative, we want to find the second derivative of y with respect to x. Actually let me just write it down out here a little bit clearer. What we really want to do is we want to find the derivative with respect, let me write it this way. When we wanted to find the derivative with respect to x of y, that was equal to derivative of y with respect to t over the derivative of x with respect to t. Now we want to find the derivative with respect to x of the first derivative with respect to x. Everywhere we saw a y here, replace that with the first derivative. This is going to be equal to, in the numerator, the derivative with respect to t of dy dx."}, {"video_title": "Second derivatives (parametric functions) Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "What we really want to do is we want to find the derivative with respect, let me write it this way. When we wanted to find the derivative with respect to x of y, that was equal to derivative of y with respect to t over the derivative of x with respect to t. Now we want to find the derivative with respect to x of the first derivative with respect to x. Everywhere we saw a y here, replace that with the first derivative. This is going to be equal to, in the numerator, the derivative with respect to t of dy dx. Notice this was the derivative with respect to t of y. In fact, let me write it that way just so you can see it. If I clear this out, if I clear that out, we're gonna get, this is the derivative with respect to t of y. Hopefully you see, before we had a y there, now we have a dy dx, dx dt."}, {"video_title": "Second derivatives (parametric functions) Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "This is going to be equal to, in the numerator, the derivative with respect to t of dy dx. Notice this was the derivative with respect to t of y. In fact, let me write it that way just so you can see it. If I clear this out, if I clear that out, we're gonna get, this is the derivative with respect to t of y. Hopefully you see, before we had a y there, now we have a dy dx, dx dt. Now this might seem really daunting and complicated except for the fact that these are actually fairly straight things to evaluate. Taking the derivative with respect to t of the first derivative, well that's just taking the derivative with respect to t of this. This is pretty easy."}, {"video_title": "Second derivatives (parametric functions) Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "If I clear this out, if I clear that out, we're gonna get, this is the derivative with respect to t of y. Hopefully you see, before we had a y there, now we have a dy dx, dx dt. Now this might seem really daunting and complicated except for the fact that these are actually fairly straight things to evaluate. Taking the derivative with respect to t of the first derivative, well that's just taking the derivative with respect to t of this. This is pretty easy. The derivative was just going to be 1 1 2. The derivative with respect to t of e to the t is just e to the t. That's going to be over the derivative of x with respect to t, which we already saw is 6e to the 2t. 6e to the 2t."}, {"video_title": "Second derivatives (parametric functions) Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "This is pretty easy. The derivative was just going to be 1 1 2. The derivative with respect to t of e to the t is just e to the t. That's going to be over the derivative of x with respect to t, which we already saw is 6e to the 2t. 6e to the 2t. We can write this. We can write this as, let's see, 1 1 2 divided by 6 is 1 over 12. Then e to the t minus 2t, which is equal to, we could write this as 1 12 e to the negative t, or we could write this as 1 over 12 e to the t. We're done."}, {"video_title": "Analyzing functions for discontinuities (discontinuity example) AP Calculus AB Khan Academy.mp3", "Sentence": "For any x is larger than two, well then f of x is going to be x squared times the natural log of x. And what we want to do is, we want to find the limit of f of x as x approaches two. And what's interesting about the value two is that that's essentially the boundary between these two intervals. If we wanted to evaluate it at two, we would fall into this first interval, f of two. Well, two is less than or equal to two, and it's greater than zero. So f of two, f of two would be pretty straightforward, that would just be natural log of two. But that's not necessarily what the limit is going to be."}, {"video_title": "Analyzing functions for discontinuities (discontinuity example) AP Calculus AB Khan Academy.mp3", "Sentence": "If we wanted to evaluate it at two, we would fall into this first interval, f of two. Well, two is less than or equal to two, and it's greater than zero. So f of two, f of two would be pretty straightforward, that would just be natural log of two. But that's not necessarily what the limit is going to be. For to figure out what the limit is going to be, we should think about, well what's the limit as we approach from the left? What's the limit as we approach from the right? And do those exist?"}, {"video_title": "Analyzing functions for discontinuities (discontinuity example) AP Calculus AB Khan Academy.mp3", "Sentence": "But that's not necessarily what the limit is going to be. For to figure out what the limit is going to be, we should think about, well what's the limit as we approach from the left? What's the limit as we approach from the right? And do those exist? And if they do exist, are they the same thing? And if they are the same thing, well then we have a well-defined limit. So let's do that."}, {"video_title": "Analyzing functions for discontinuities (discontinuity example) AP Calculus AB Khan Academy.mp3", "Sentence": "And do those exist? And if they do exist, are they the same thing? And if they are the same thing, well then we have a well-defined limit. So let's do that. Let's first think about the limit, the limit of f of x as we approach two from the left, from values lower than two. Well, this is gonna be the case where we're gonna be operating in this interval right over here. We're operating from values less than two, and we're going to be approaching two from the left."}, {"video_title": "Analyzing functions for discontinuities (discontinuity example) AP Calculus AB Khan Academy.mp3", "Sentence": "So let's do that. Let's first think about the limit, the limit of f of x as we approach two from the left, from values lower than two. Well, this is gonna be the case where we're gonna be operating in this interval right over here. We're operating from values less than two, and we're going to be approaching two from the left. And so we'll fall under this clause. And so since this clause or case is continuous over the interval in which we're operating, and for sure between, or for all values greater than zero and less than or equal to two, this limit is going to be equal to just this clause evaluated at two, because it's continuous over the interval. So this is just going to be the natural log of two."}, {"video_title": "Analyzing functions for discontinuities (discontinuity example) AP Calculus AB Khan Academy.mp3", "Sentence": "We're operating from values less than two, and we're going to be approaching two from the left. And so we'll fall under this clause. And so since this clause or case is continuous over the interval in which we're operating, and for sure between, or for all values greater than zero and less than or equal to two, this limit is going to be equal to just this clause evaluated at two, because it's continuous over the interval. So this is just going to be the natural log of two. All right, so now let's think about the limit from the right-hand side, from values greater than two. So the limit, the limit of f of x as x approaches two from the right-hand side. Well, even though two falls into this clause, as soon as we go anything greater than two, we fall in this clause."}, {"video_title": "Analyzing functions for discontinuities (discontinuity example) AP Calculus AB Khan Academy.mp3", "Sentence": "So this is just going to be the natural log of two. All right, so now let's think about the limit from the right-hand side, from values greater than two. So the limit, the limit of f of x as x approaches two from the right-hand side. Well, even though two falls into this clause, as soon as we go anything greater than two, we fall in this clause. So we're gonna be approaching two essentially using this case. And once again, this case here is continuous for all x values not only greater than two, actually greater than or equal to two. And so for this one over here, we can make the same argument that this limit is going to be this clause evaluated at two."}, {"video_title": "Analyzing functions for discontinuities (discontinuity example) AP Calculus AB Khan Academy.mp3", "Sentence": "Well, even though two falls into this clause, as soon as we go anything greater than two, we fall in this clause. So we're gonna be approaching two essentially using this case. And once again, this case here is continuous for all x values not only greater than two, actually greater than or equal to two. And so for this one over here, we can make the same argument that this limit is going to be this clause evaluated at two. Because once again, if we were just evaluated the function at two, it falls under this clause. But if we're approaching from the right, well, if we're approaching from the right, those are x values greater than two, so this clause is what's at play. So we'll evaluate this clause at two."}, {"video_title": "Analyzing functions for discontinuities (discontinuity example) AP Calculus AB Khan Academy.mp3", "Sentence": "And so for this one over here, we can make the same argument that this limit is going to be this clause evaluated at two. Because once again, if we were just evaluated the function at two, it falls under this clause. But if we're approaching from the right, well, if we're approaching from the right, those are x values greater than two, so this clause is what's at play. So we'll evaluate this clause at two. So because it is continuous. So this is going to be two squared times the natural log of two. And so this is equal to four times the natural log of two."}, {"video_title": "Analyzing functions for discontinuities (discontinuity example) AP Calculus AB Khan Academy.mp3", "Sentence": "So we'll evaluate this clause at two. So because it is continuous. So this is going to be two squared times the natural log of two. And so this is equal to four times the natural log of two. Four times the natural log of two. So the right-hand limit does exist. The left-hand limit does exist."}, {"video_title": "Analyzing functions for discontinuities (discontinuity example) AP Calculus AB Khan Academy.mp3", "Sentence": "And so this is equal to four times the natural log of two. Four times the natural log of two. So the right-hand limit does exist. The left-hand limit does exist. But the thing that might jump out at you is that these are two different values. We approach a different value from the left as we do from the right. If you were to graph this, you would see a jump in the actual graph."}, {"video_title": "Analyzing functions for discontinuities (discontinuity example) AP Calculus AB Khan Academy.mp3", "Sentence": "The left-hand limit does exist. But the thing that might jump out at you is that these are two different values. We approach a different value from the left as we do from the right. If you were to graph this, you would see a jump in the actual graph. You would see a discontinuity occurring there. And so for this one in particular, you have that jump discontinuity. This limit would not exist because the left-hand limit and the right-hand limit go to two different values."}, {"video_title": "Chain rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And you're going to use it any time you take the derivative of anything even reasonably complex. And it's called the chain rule. And when you're first exposed to it, it can seem a little daunting and a little bit convoluted. But as you see more and more examples, it'll start to make sense. And hopefully, it'll even start to seem a little bit simple and intuitive over time. So let's say that I had a function. Let's say I have a function h of x."}, {"video_title": "Chain rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "But as you see more and more examples, it'll start to make sense. And hopefully, it'll even start to seem a little bit simple and intuitive over time. So let's say that I had a function. Let's say I have a function h of x. And it is equal to, just for example, let's say it's equal to sine of x. Let's say it's equal to sine of x squared. Now, I could have written it like this, sine squared of x."}, {"video_title": "Chain rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say I have a function h of x. And it is equal to, just for example, let's say it's equal to sine of x. Let's say it's equal to sine of x squared. Now, I could have written it like this, sine squared of x. But it'll be a little bit clearer using that type of notation. So let me make it. So I have h of x."}, {"video_title": "Chain rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Now, I could have written it like this, sine squared of x. But it'll be a little bit clearer using that type of notation. So let me make it. So I have h of x. And what I'm curious about is what is h prime of x? So I want to know h prime of x, which another way of writing it is the derivative of h with respect to x. These are just different notations."}, {"video_title": "Chain rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So I have h of x. And what I'm curious about is what is h prime of x? So I want to know h prime of x, which another way of writing it is the derivative of h with respect to x. These are just different notations. And to do this, I'm going to use the chain rule. And the chain rule comes into play anytime your function can be used as a composition of more than one function. And that might not seem obvious right now."}, {"video_title": "Chain rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "These are just different notations. And to do this, I'm going to use the chain rule. And the chain rule comes into play anytime your function can be used as a composition of more than one function. And that might not seem obvious right now. But it will hopefully maybe by the end of this video or the next one. Now, what I want to do is a little bit of a thought experiment. If I were to ask you, what is the derivative with respect to x, if I were to supply the derivative operator to x squared with respect to x, what do I get?"}, {"video_title": "Chain rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And that might not seem obvious right now. But it will hopefully maybe by the end of this video or the next one. Now, what I want to do is a little bit of a thought experiment. If I were to ask you, what is the derivative with respect to x, if I were to supply the derivative operator to x squared with respect to x, what do I get? Well, this gives me 2x. We've seen that many, many, many, many times. Now, what if I were to take the derivative with respect to a of a squared?"}, {"video_title": "Chain rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "If I were to ask you, what is the derivative with respect to x, if I were to supply the derivative operator to x squared with respect to x, what do I get? Well, this gives me 2x. We've seen that many, many, many, many times. Now, what if I were to take the derivative with respect to a of a squared? Well, it's the exact same thing. I just swapped an a for the x's. This is still going to be equal to 2a."}, {"video_title": "Chain rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Now, what if I were to take the derivative with respect to a of a squared? Well, it's the exact same thing. I just swapped an a for the x's. This is still going to be equal to 2a. Now, I will do something that might be a little bit more bizarre. What if I were to take the derivative with respect to sine of x with respect to sine of x of sine of x squared? Well, wherever I had the x's up here or the a's over here, I just replaced it with a sine of x."}, {"video_title": "Chain rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "This is still going to be equal to 2a. Now, I will do something that might be a little bit more bizarre. What if I were to take the derivative with respect to sine of x with respect to sine of x of sine of x squared? Well, wherever I had the x's up here or the a's over here, I just replaced it with a sine of x. So this is just going to be 2 times the thing that I had. So whatever I'm taking the derivative with respect to. Here it was with respect to x."}, {"video_title": "Chain rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, wherever I had the x's up here or the a's over here, I just replaced it with a sine of x. So this is just going to be 2 times the thing that I had. So whatever I'm taking the derivative with respect to. Here it was with respect to x. Here it was with respect to a. Here it's with respect to sine of x. So it's going to be 2 times sine of x."}, {"video_title": "Chain rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Here it was with respect to x. Here it was with respect to a. Here it's with respect to sine of x. So it's going to be 2 times sine of x. Now, so the chain rule tells us that this derivative is going to be the derivative of our whole function with respect or the derivative of this outer function, x squared, the derivative of x squared, the derivative of this outer function with respect to sine of x. So that's going to be 2 sine of x. So we could view it as the derivative of the outer function with respect to the inner, 2 sine of x."}, {"video_title": "Chain rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So it's going to be 2 times sine of x. Now, so the chain rule tells us that this derivative is going to be the derivative of our whole function with respect or the derivative of this outer function, x squared, the derivative of x squared, the derivative of this outer function with respect to sine of x. So that's going to be 2 sine of x. So we could view it as the derivative of the outer function with respect to the inner, 2 sine of x. We could just treat sine of x like it's kind of an x. And it would have been just 2x. But instead, it's a sine of x."}, {"video_title": "Chain rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So we could view it as the derivative of the outer function with respect to the inner, 2 sine of x. We could just treat sine of x like it's kind of an x. And it would have been just 2x. But instead, it's a sine of x. So we say 2 sine of x times the derivative of sine of x with respect to x. Well, that's more straightforward, a little bit more intuitive. The derivative of sine of x with respect to x, we've seen multiple times, is cosine of x."}, {"video_title": "Chain rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "But instead, it's a sine of x. So we say 2 sine of x times the derivative of sine of x with respect to x. Well, that's more straightforward, a little bit more intuitive. The derivative of sine of x with respect to x, we've seen multiple times, is cosine of x. So times cosine of x. And so there we've applied the chain rule. It was the derivative of the outer function with respect to the inner."}, {"video_title": "Chain rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "The derivative of sine of x with respect to x, we've seen multiple times, is cosine of x. So times cosine of x. And so there we've applied the chain rule. It was the derivative of the outer function with respect to the inner. So derivative of sine of x squared with respect to sine of x is 2 sine of x. And then we multiply that times the derivative of sine of x with respect to x. Times the derivative of sine of x with respect to x."}, {"video_title": "Chain rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "It was the derivative of the outer function with respect to the inner. So derivative of sine of x squared with respect to sine of x is 2 sine of x. And then we multiply that times the derivative of sine of x with respect to x. Times the derivative of sine of x with respect to x. So let me make it clear. This right over here is the derivative. We're taking the derivative of sine of x squared."}, {"video_title": "Chain rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Times the derivative of sine of x with respect to x. So let me make it clear. This right over here is the derivative. We're taking the derivative of sine of x squared. So let me make it clear. That's what we're taking the derivative of with respect to sine of x. With respect to sine of x."}, {"video_title": "Chain rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "We're taking the derivative of sine of x squared. So let me make it clear. That's what we're taking the derivative of with respect to sine of x. With respect to sine of x. And then we're multiplying that times the derivative of sine of x with respect to x. And this is where it might start making a little bit of intuition. You can't really treat these differentials, this d whatever, this dx, this d sine of x, as a number."}, {"video_title": "Chain rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "With respect to sine of x. And then we're multiplying that times the derivative of sine of x with respect to x. And this is where it might start making a little bit of intuition. You can't really treat these differentials, this d whatever, this dx, this d sine of x, as a number. And the notation makes it look like a fraction, because intuitively that's what we're doing. But if you were to treat them like fractions, then you could think about canceling that and that. And once again, this isn't a rigorous thing to do, but it can help with the intuition."}, {"video_title": "Chain rule Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "You can't really treat these differentials, this d whatever, this dx, this d sine of x, as a number. And the notation makes it look like a fraction, because intuitively that's what we're doing. But if you were to treat them like fractions, then you could think about canceling that and that. And once again, this isn't a rigorous thing to do, but it can help with the intuition. And then what you're left with is the derivative of this whole sine of x squared with respect to x. So you're left with the derivative of essentially our original function, sine of x squared with respect to x. Which is exactly what dh dx is."}, {"video_title": "Antiderivatives and indefinite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "We know how to take derivatives of functions. If I apply the derivative operator to x squared, I get 2x. Now, if I also apply the derivative operator to x squared plus 1, I also get 2x. If I apply the derivative operator to x squared plus pi, I also get 2x. The derivative of x squared is 2x. The derivative with respect to x of pi of a constant is just 0. The derivative with respect to x of 1 is just a constant."}, {"video_title": "Antiderivatives and indefinite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "If I apply the derivative operator to x squared plus pi, I also get 2x. The derivative of x squared is 2x. The derivative with respect to x of pi of a constant is just 0. The derivative with respect to x of 1 is just a constant. It's just 0. So once again, this is just going to be equal to 2x. In general, the derivative with respect to x of x squared plus any constant is going to be equal to 2x."}, {"video_title": "Antiderivatives and indefinite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "The derivative with respect to x of 1 is just a constant. It's just 0. So once again, this is just going to be equal to 2x. In general, the derivative with respect to x of x squared plus any constant is going to be equal to 2x. The derivative of x squared with respect to x is 2x. The derivative of a constant with respect to x, a constant does not change with respect to x, so it's just equal to 0. You apply the derivative operator to any of these expressions, and you get 2x."}, {"video_title": "Antiderivatives and indefinite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "In general, the derivative with respect to x of x squared plus any constant is going to be equal to 2x. The derivative of x squared with respect to x is 2x. The derivative of a constant with respect to x, a constant does not change with respect to x, so it's just equal to 0. You apply the derivative operator to any of these expressions, and you get 2x. Now let's go the other way around. Let's think about the antiderivative. One way to think about it is we're doing the opposite of the derivative operator."}, {"video_title": "Antiderivatives and indefinite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "You apply the derivative operator to any of these expressions, and you get 2x. Now let's go the other way around. Let's think about the antiderivative. One way to think about it is we're doing the opposite of the derivative operator. In the derivative operator, you get an expression, and you find its derivative. Now what we want to do is, given some expression, we want to find what it could be the derivative of. If someone were to ask you, what is 2x the derivative of?"}, {"video_title": "Antiderivatives and indefinite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "One way to think about it is we're doing the opposite of the derivative operator. In the derivative operator, you get an expression, and you find its derivative. Now what we want to do is, given some expression, we want to find what it could be the derivative of. If someone were to ask you, what is 2x the derivative of? They're essentially asking you for the antiderivative. You could say, well, 2x is the derivative of x squared. You could also say 2x is the derivative of x squared plus 1."}, {"video_title": "Antiderivatives and indefinite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "If someone were to ask you, what is 2x the derivative of? They're essentially asking you for the antiderivative. You could say, well, 2x is the derivative of x squared. You could also say 2x is the derivative of x squared plus 1. You could also say that 2x is the derivative of x squared plus pi. I think you get the general idea. If you wanted to write it in the most general sense, you would write that 2x is the derivative of x squared plus some constant."}, {"video_title": "Antiderivatives and indefinite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "You could also say 2x is the derivative of x squared plus 1. You could also say that 2x is the derivative of x squared plus pi. I think you get the general idea. If you wanted to write it in the most general sense, you would write that 2x is the derivative of x squared plus some constant. This is what you would consider the antiderivative of 2x. That's all nice, but this is kind of clumsy to have to write a sentence like this, so let's come up with some notation for the antiderivative. The convention here is to use a strange-looking notation."}, {"video_title": "Antiderivatives and indefinite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "If you wanted to write it in the most general sense, you would write that 2x is the derivative of x squared plus some constant. This is what you would consider the antiderivative of 2x. That's all nice, but this is kind of clumsy to have to write a sentence like this, so let's come up with some notation for the antiderivative. The convention here is to use a strange-looking notation. It's to use a big, elongated, S-looking thing like that, and a dx around the function that you're trying to take the antiderivative of. In this case, it would look something like this. This is just saying this is equal to the antiderivative of 2x, and the antiderivative of 2x, we have already seen, is x squared plus c. Now, you might be saying, why do we use this type of crazy notation?"}, {"video_title": "Antiderivatives and indefinite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "The convention here is to use a strange-looking notation. It's to use a big, elongated, S-looking thing like that, and a dx around the function that you're trying to take the antiderivative of. In this case, it would look something like this. This is just saying this is equal to the antiderivative of 2x, and the antiderivative of 2x, we have already seen, is x squared plus c. Now, you might be saying, why do we use this type of crazy notation? It will become more obvious when we study the definite integral and areas under curves, and taking sums of rectangles in order to approximate the area of the curve. Here, it really should just be viewed as a notation for antiderivative. This notation right over here, this whole expression, is called the indefinite integral of 2x, which is another way of just saying the antiderivative of 2x."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Not a great approximation. If you have a first degree polynomial, you can at least get the slope right at that point. If you get to a second degree polynomial, you can get something that hugs the function a little bit longer. If you go to a third degree polynomial, maybe something that hugs the function even a little bit longer than that. But all of that was focused on approximating the function around x is equal to 0. And that's why we call it the Maclaurin series or the Taylor series at x is equal to 0. What I want to do now is expand it a little bit, generalize it a little bit, and focus on the Taylor expansion at x equals anything."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "If you go to a third degree polynomial, maybe something that hugs the function even a little bit longer than that. But all of that was focused on approximating the function around x is equal to 0. And that's why we call it the Maclaurin series or the Taylor series at x is equal to 0. What I want to do now is expand it a little bit, generalize it a little bit, and focus on the Taylor expansion at x equals anything. So let's say we want to approximate this function when x, so this is our x-axis, when x is equal to c. So we can do the exact same thing. We could say, look, our first approximation is that our polynomial at c should be equal to, or actually even our polynomial could just be, if it's just going to be a constant, if it's going to be a constant, it should at least equal to whatever the function equals at c. So it should just equal f of c. f of c is a constant. It's that value right over there."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "What I want to do now is expand it a little bit, generalize it a little bit, and focus on the Taylor expansion at x equals anything. So let's say we want to approximate this function when x, so this is our x-axis, when x is equal to c. So we can do the exact same thing. We could say, look, our first approximation is that our polynomial at c should be equal to, or actually even our polynomial could just be, if it's just going to be a constant, if it's going to be a constant, it should at least equal to whatever the function equals at c. So it should just equal f of c. f of c is a constant. It's that value right over there. We're assuming that c is given. And then you would have, this would just be a horizontal line that goes through f of c. That's p of x is equal to f of c. Not a great approximation, but then we can try to go for this having this constraint matched plus having the derivative matched. So what this constraint gave us, just as a reminder, this gave us the fact that at least p of c, the approximation at c, our polynomial at c, at least is going to be equal to f of c. If you put c over here, it doesn't change what's on the right-hand side, because this is just going to be a constant."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "It's that value right over there. We're assuming that c is given. And then you would have, this would just be a horizontal line that goes through f of c. That's p of x is equal to f of c. Not a great approximation, but then we can try to go for this having this constraint matched plus having the derivative matched. So what this constraint gave us, just as a reminder, this gave us the fact that at least p of c, the approximation at c, our polynomial at c, at least is going to be equal to f of c. If you put c over here, it doesn't change what's on the right-hand side, because this is just going to be a constant. Now let's get the constraint one more step. What if we want a situation where this is true and we want the derivative of our polynomial to be the same thing as the derivative of our function when either of them are at c? So for this situation, what if we set up our polynomial?"}, {"video_title": "Taylor & Maclaurin polynomials intro (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So what this constraint gave us, just as a reminder, this gave us the fact that at least p of c, the approximation at c, our polynomial at c, at least is going to be equal to f of c. If you put c over here, it doesn't change what's on the right-hand side, because this is just going to be a constant. Now let's get the constraint one more step. What if we want a situation where this is true and we want the derivative of our polynomial to be the same thing as the derivative of our function when either of them are at c? So for this situation, what if we set up our polynomial? And you'll see a complete parallel to what we did in earlier videos. We're just going to shift it a little bit for the fact that we're not at 0. So now let's define p of x to be equal to f of c plus f prime of c. So whatever the slope is at this point of the function, whatever the function slope is, times, and you're going to see something slightly different over here, x minus c. And let's think about why we put what this minus c is doing."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So for this situation, what if we set up our polynomial? And you'll see a complete parallel to what we did in earlier videos. We're just going to shift it a little bit for the fact that we're not at 0. So now let's define p of x to be equal to f of c plus f prime of c. So whatever the slope is at this point of the function, whatever the function slope is, times, and you're going to see something slightly different over here, x minus c. And let's think about why we put what this minus c is doing. What this minus c is doing. So let's test, first of all, that we didn't mess up our previous constraint. So let's evaluate this at c. So now we know that p of c, and I'm using this exact example."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So now let's define p of x to be equal to f of c plus f prime of c. So whatever the slope is at this point of the function, whatever the function slope is, times, and you're going to see something slightly different over here, x minus c. And let's think about why we put what this minus c is doing. What this minus c is doing. So let's test, first of all, that we didn't mess up our previous constraint. So let's evaluate this at c. So now we know that p of c, and I'm using this exact example. So p of c, let me just add a new color. Let me try it out. So p, that's not a new color, p of c. p of c is going to be equal to f of c plus f prime of c times c minus c. Wherever you see an x, you put a c in there."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's evaluate this at c. So now we know that p of c, and I'm using this exact example. So p of c, let me just add a new color. Let me try it out. So p, that's not a new color, p of c. p of c is going to be equal to f of c plus f prime of c times c minus c. Wherever you see an x, you put a c in there. c minus c. Well, this term right over here is going to be 0. And so this whole term right over here is going to be 0. And so you're just left with p of c is equal to f of c. You're just left with that constraint right over there."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So p, that's not a new color, p of c. p of c is going to be equal to f of c plus f prime of c times c minus c. Wherever you see an x, you put a c in there. c minus c. Well, this term right over here is going to be 0. And so this whole term right over here is going to be 0. And so you're just left with p of c is equal to f of c. You're just left with that constraint right over there. And the only reason why we were able to blank out this second term right over here is because we had f prime of c times x minus c. The x minus c makes all of the terms after this irrelevant. We can go now verify that this is now true. So let's try p prime of x is going to be the derivative of this, which is just 0, because this is going to be a constant, plus the derivative of this right over here."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so you're just left with p of c is equal to f of c. You're just left with that constraint right over there. And the only reason why we were able to blank out this second term right over here is because we had f prime of c times x minus c. The x minus c makes all of the terms after this irrelevant. We can go now verify that this is now true. So let's try p prime of x is going to be the derivative of this, which is just 0, because this is going to be a constant, plus the derivative of this right over here. And what's that going to be? Well, that's going to be, you can expand this out to be f prime of c times x minus f prime of c times c, which would just be constant. So if you take the derivative of this thing right here, you're just going to be left with an f prime of c. So the derivative of our polynomial is now constant."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's try p prime of x is going to be the derivative of this, which is just 0, because this is going to be a constant, plus the derivative of this right over here. And what's that going to be? Well, that's going to be, you can expand this out to be f prime of c times x minus f prime of c times c, which would just be constant. So if you take the derivative of this thing right here, you're just going to be left with an f prime of c. So the derivative of our polynomial is now constant. So obviously, if you were to evaluate this at c, p prime at c, you're going to get f prime of c. So once again, it meets the second constraint. And now when you have both of these terms, maybe our approximation will look something like this. It'll at least have the right slope as f of x."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So if you take the derivative of this thing right here, you're just going to be left with an f prime of c. So the derivative of our polynomial is now constant. So obviously, if you were to evaluate this at c, p prime at c, you're going to get f prime of c. So once again, it meets the second constraint. And now when you have both of these terms, maybe our approximation will look something like this. It'll at least have the right slope as f of x. Our approximation is getting a little bit better. And if we keep doing this, and we're using the exact same logic that we used when we did it around 0, when we did the Maclaurin expansion, you get the Taylor expansion, the general Taylor expansion for the approximation of f of x around c to be the polynomial. So the polynomial p of x is going to be equal to, and I'll just expand it out."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "It'll at least have the right slope as f of x. Our approximation is getting a little bit better. And if we keep doing this, and we're using the exact same logic that we used when we did it around 0, when we did the Maclaurin expansion, you get the Taylor expansion, the general Taylor expansion for the approximation of f of x around c to be the polynomial. So the polynomial p of x is going to be equal to, and I'll just expand it out. And this is very similar to what we saw before. f of c plus f prime of c times x minus c. You might even guess what the next few terms are going to be. It's the exact same logic."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So the polynomial p of x is going to be equal to, and I'll just expand it out. And this is very similar to what we saw before. f of c plus f prime of c times x minus c. You might even guess what the next few terms are going to be. It's the exact same logic. Watch the videos on Maclaurin series where I go a few more terms into it. It becomes a little bit more complicated taking the second and third derivatives and all of the rest, just because you have to kind of expand out these binomials. But it's the exact same logic."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "It's the exact same logic. Watch the videos on Maclaurin series where I go a few more terms into it. It becomes a little bit more complicated taking the second and third derivatives and all of the rest, just because you have to kind of expand out these binomials. But it's the exact same logic. So then you have plus your second degree term, f prime prime of c divided by 2 factorial. And this is just like what we saw in the Maclaurin expansion. And just to be clear, you could say that there's a 1 factorial down here."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "But it's the exact same logic. So then you have plus your second degree term, f prime prime of c divided by 2 factorial. And this is just like what we saw in the Maclaurin expansion. And just to be clear, you could say that there's a 1 factorial down here. I didn't take the trouble to write it, because it doesn't change the value. And then that times x minus c squared plus the third derivative of the function evaluated at c over 3 factorial times x minus c to the third power. And I think you get the general idea."}, {"video_title": "Taylor & Maclaurin polynomials intro (part 2) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And just to be clear, you could say that there's a 1 factorial down here. I didn't take the trouble to write it, because it doesn't change the value. And then that times x minus c squared plus the third derivative of the function evaluated at c over 3 factorial times x minus c to the third power. And I think you get the general idea. You can keep adding more and more and more terms like this. Unfortunately, it makes it a little bit harder, especially if you're willing to do the work, it's not so bad. But adding this x, instead of having just x here, instead of just having an x squared here, having an x minus c squared, having an x minus c to the third, this makes the analytical math a little bit hairier, a little bit more difficult."}, {"video_title": "Worked example Motion problems with derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "A particle moves along the x-axis. The function x of t gives the particle's position at any time t is greater than or equal to zero, and they give us x of t right over here. What is the particle's velocity v of t at t is equal to two? So pause this video, see if you can figure that out. Well, the key thing to realize is that your velocity is a function of time, is the derivative of position, and so this is going to be equal to, we just take the derivative with respect to t up here. So derivative of t to the third with respect to t is three t squared. If that's unfamiliar, I encourage you to review the power rule."}, {"video_title": "Worked example Motion problems with derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So pause this video, see if you can figure that out. Well, the key thing to realize is that your velocity is a function of time, is the derivative of position, and so this is going to be equal to, we just take the derivative with respect to t up here. So derivative of t to the third with respect to t is three t squared. If that's unfamiliar, I encourage you to review the power rule. The derivative of negative four t squared with respect to t is negative eight t, and derivative of three t with respect to t is plus three. Derivative of a constant doesn't change with respect to time, so that's just zero. And so here we have velocity as a function of time, and so if we wanna know our velocity at time t equals two, we just substitute two wherever we see the t's, so it's gonna be three times four, three times two squared, so it's 12 minus eight times two, minus 16, plus three, which is equal to negative one."}, {"video_title": "Worked example Motion problems with derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "If that's unfamiliar, I encourage you to review the power rule. The derivative of negative four t squared with respect to t is negative eight t, and derivative of three t with respect to t is plus three. Derivative of a constant doesn't change with respect to time, so that's just zero. And so here we have velocity as a function of time, and so if we wanna know our velocity at time t equals two, we just substitute two wherever we see the t's, so it's gonna be three times four, three times two squared, so it's 12 minus eight times two, minus 16, plus three, which is equal to negative one. And you might say, well, negative one by itself doesn't sound like a velocity. Well, if they gave us units, if they told us that x was in meters and that t was in seconds, well, then x would be, well, I already said, would be in meters, and velocity would be negative one meters per second. You might also be saying, well, what does the negative mean?"}, {"video_title": "Worked example Motion problems with derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And so here we have velocity as a function of time, and so if we wanna know our velocity at time t equals two, we just substitute two wherever we see the t's, so it's gonna be three times four, three times two squared, so it's 12 minus eight times two, minus 16, plus three, which is equal to negative one. And you might say, well, negative one by itself doesn't sound like a velocity. Well, if they gave us units, if they told us that x was in meters and that t was in seconds, well, then x would be, well, I already said, would be in meters, and velocity would be negative one meters per second. You might also be saying, well, what does the negative mean? Well, that means that we are moving to the left. Remember, we're moving along the x-axis, so if our velocity is negative, that means that x is decreasing or we are moving to the left. What is the particle's acceleration, a of t, at t equals three?"}, {"video_title": "Worked example Motion problems with derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "You might also be saying, well, what does the negative mean? Well, that means that we are moving to the left. Remember, we're moving along the x-axis, so if our velocity is negative, that means that x is decreasing or we are moving to the left. What is the particle's acceleration, a of t, at t equals three? So pause this video again and see if you can do that. Well, here, the realization is that acceleration is a function of time. It's just the derivative of velocity, which is the second derivative of our position, which is just going to be equal to the derivative of this right over here."}, {"video_title": "Worked example Motion problems with derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "What is the particle's acceleration, a of t, at t equals three? So pause this video again and see if you can do that. Well, here, the realization is that acceleration is a function of time. It's just the derivative of velocity, which is the second derivative of our position, which is just going to be equal to the derivative of this right over here. And so I'm just going to get derivative of three t squared with respect to t is six t, derivative of negative eight t with respect to t is minus eight and derivative of constants is zero. So it's just going to be six t minus eight. So our acceleration at time t equals three is going to be six times three, which is 18 minus eight."}, {"video_title": "Worked example Motion problems with derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "It's just the derivative of velocity, which is the second derivative of our position, which is just going to be equal to the derivative of this right over here. And so I'm just going to get derivative of three t squared with respect to t is six t, derivative of negative eight t with respect to t is minus eight and derivative of constants is zero. So it's just going to be six t minus eight. So our acceleration at time t equals three is going to be six times three, which is 18 minus eight. So minus eight, which is going to be equal to positive 10. All right, now they ask us, what is the direction of the particle's motion at t equals two? Well, I already talked about this, but pause this video and see if you can answer that yourself."}, {"video_title": "Worked example Motion problems with derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So our acceleration at time t equals three is going to be six times three, which is 18 minus eight. So minus eight, which is going to be equal to positive 10. All right, now they ask us, what is the direction of the particle's motion at t equals two? Well, I already talked about this, but pause this video and see if you can answer that yourself. Well, we've already looked at the sign right over here. The fact that we have a negative sign on our velocity means we are moving towards the left. So I'll fill that in right over there."}, {"video_title": "Worked example Motion problems with derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, I already talked about this, but pause this video and see if you can answer that yourself. Well, we've already looked at the sign right over here. The fact that we have a negative sign on our velocity means we are moving towards the left. So I'll fill that in right over there. At t equals three, is the particle's speed increasing, decreasing, or neither? So pause this video and try to answer that. All right, now we have to be very careful here."}, {"video_title": "Worked example Motion problems with derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So I'll fill that in right over there. At t equals three, is the particle's speed increasing, decreasing, or neither? So pause this video and try to answer that. All right, now we have to be very careful here. If it says is the particle's velocity increasing, decreasing, or neither, then we would just have to look at the acceleration. We see that the acceleration is positive, and so we know that the velocity is increasing. But here they're not saying velocity, they're saying speed."}, {"video_title": "Worked example Motion problems with derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "All right, now we have to be very careful here. If it says is the particle's velocity increasing, decreasing, or neither, then we would just have to look at the acceleration. We see that the acceleration is positive, and so we know that the velocity is increasing. But here they're not saying velocity, they're saying speed. And just as a reminder, speed is the magnitude of velocity. So for example, at time t equals two, our velocity is negative one. If the units were meters in second, it would be negative one meters per second."}, {"video_title": "Worked example Motion problems with derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "But here they're not saying velocity, they're saying speed. And just as a reminder, speed is the magnitude of velocity. So for example, at time t equals two, our velocity is negative one. If the units were meters in second, it would be negative one meters per second. But our speed would just be one meter per second. Speed, you're not talking about the direction, so you would not have that sign there. And so in order to figure out if the speed is increasing or decreasing or neither, if the acceleration is positive and the velocity is positive, that means the magnitude of your velocity is increasing."}, {"video_title": "Worked example Motion problems with derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "If the units were meters in second, it would be negative one meters per second. But our speed would just be one meter per second. Speed, you're not talking about the direction, so you would not have that sign there. And so in order to figure out if the speed is increasing or decreasing or neither, if the acceleration is positive and the velocity is positive, that means the magnitude of your velocity is increasing. So that means your speed is increasing. If your velocity is negative and your acceleration is also negative, that also means that your speed is increasing. But if your velocity and acceleration have different signs, well that means that your speed is decreasing."}, {"video_title": "Worked example Motion problems with derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And so in order to figure out if the speed is increasing or decreasing or neither, if the acceleration is positive and the velocity is positive, that means the magnitude of your velocity is increasing. So that means your speed is increasing. If your velocity is negative and your acceleration is also negative, that also means that your speed is increasing. But if your velocity and acceleration have different signs, well that means that your speed is decreasing. The magnitude of your velocity would be coming less. So let's look at our velocity at time t equals three. Our velocity at time three, we just go back right over here."}, {"video_title": "Worked example Motion problems with derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "But if your velocity and acceleration have different signs, well that means that your speed is decreasing. The magnitude of your velocity would be coming less. So let's look at our velocity at time t equals three. Our velocity at time three, we just go back right over here. It's gonna be three times nine, which is 27, three times three squared, minus 24 plus three, plus three. So this is going to be equal to six. So our velocity and acceleration are both, you could say, in the same direction."}, {"video_title": "Worked example Motion problems with derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Our velocity at time three, we just go back right over here. It's gonna be three times nine, which is 27, three times three squared, minus 24 plus three, plus three. So this is going to be equal to six. So our velocity and acceleration are both, you could say, in the same direction. They are both positive. And so our velocity is only going to become more positive, or the magnitude of our velocity is only going to increase. So our speed is increasing."}, {"video_title": "Power rule (with rewriting the expression) AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say we need to take the derivative with respect to x of one over x. What is that going to be equal to? Pause this video and try to figure it out. So at first you might say, how does the power rule apply here? The power rule, just to remind ourselves, it tells us that if we're taking the derivative of x to the n with respect to x, so if we're taking the derivative of that, that that's going to be equal to, we take the exponent, bring it out front, and we've proven it in other videos, but this is going to be n times x to the, and then we decrement the exponent, so n minus one. But this does not look like that, and the key is to appreciate that one over x is the same thing as x to the negative one. So this is going to be the derivative with respect to x of x to the negative one."}, {"video_title": "Power rule (with rewriting the expression) AP Calculus AB Khan Academy.mp3", "Sentence": "So at first you might say, how does the power rule apply here? The power rule, just to remind ourselves, it tells us that if we're taking the derivative of x to the n with respect to x, so if we're taking the derivative of that, that that's going to be equal to, we take the exponent, bring it out front, and we've proven it in other videos, but this is going to be n times x to the, and then we decrement the exponent, so n minus one. But this does not look like that, and the key is to appreciate that one over x is the same thing as x to the negative one. So this is going to be the derivative with respect to x of x to the negative one. And now this looks a lot more like what you might be used to where this is going to be equal to, you take our exponent, bring it out front, so that's negative one, times x to the negative one minus one, negative one minus one, and so this is going to be equal to negative x to the negative two, and we're done. Let's do another example. Let's say that we're told that f of x is equal to the cube root of x, and we wanna figure out what f prime of x is equal to."}, {"video_title": "Power rule (with rewriting the expression) AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be the derivative with respect to x of x to the negative one. And now this looks a lot more like what you might be used to where this is going to be equal to, you take our exponent, bring it out front, so that's negative one, times x to the negative one minus one, negative one minus one, and so this is going to be equal to negative x to the negative two, and we're done. Let's do another example. Let's say that we're told that f of x is equal to the cube root of x, and we wanna figure out what f prime of x is equal to. Pause the video and see if you can figure it out again. Well, once again, you might say, hey, how do I take the derivative of something like this, especially if my goal, or if I'm thinking that maybe the power rule might be useful? And the idea is to rewrite this as an exponent."}, {"video_title": "Power rule (with rewriting the expression) AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say that we're told that f of x is equal to the cube root of x, and we wanna figure out what f prime of x is equal to. Pause the video and see if you can figure it out again. Well, once again, you might say, hey, how do I take the derivative of something like this, especially if my goal, or if I'm thinking that maybe the power rule might be useful? And the idea is to rewrite this as an exponent. If you could rewrite the cube root as x to the 1 3rd power. And so the derivative, you take the 1 3rd, bring it out front, so it's 1 3rd, x to the 1 3rd minus one power. And so this is going to be 1 3rd times x to the 1 3rd minus one is negative 2 3rds, negative 2 3rd power, and we are done."}, {"video_title": "Power rule (with rewriting the expression) AP Calculus AB Khan Academy.mp3", "Sentence": "And the idea is to rewrite this as an exponent. If you could rewrite the cube root as x to the 1 3rd power. And so the derivative, you take the 1 3rd, bring it out front, so it's 1 3rd, x to the 1 3rd minus one power. And so this is going to be 1 3rd times x to the 1 3rd minus one is negative 2 3rds, negative 2 3rd power, and we are done. And hopefully through these examples, you're seeing that the power rule is incredibly powerful. You can tackle a far broader range of derivatives than you might have initially thought. Let's do another example, and I'll make this one really nice and hairy."}, {"video_title": "Power rule (with rewriting the expression) AP Calculus AB Khan Academy.mp3", "Sentence": "And so this is going to be 1 3rd times x to the 1 3rd minus one is negative 2 3rds, negative 2 3rd power, and we are done. And hopefully through these examples, you're seeing that the power rule is incredibly powerful. You can tackle a far broader range of derivatives than you might have initially thought. Let's do another example, and I'll make this one really nice and hairy. Let's say we wanna figure out the derivative with respect to x of the cube root of x squared. What is this going to be? And actually, let's just not figure out what the derivative is."}, {"video_title": "Power rule (with rewriting the expression) AP Calculus AB Khan Academy.mp3", "Sentence": "Let's do another example, and I'll make this one really nice and hairy. Let's say we wanna figure out the derivative with respect to x of the cube root of x squared. What is this going to be? And actually, let's just not figure out what the derivative is. Let's figure out the derivative at x equals eight. Pause this video again and see if you can figure that out. Well, what we're gonna do is first just figure out what this is, and then we're going to evaluate it at x equals eight."}, {"video_title": "Power rule (with rewriting the expression) AP Calculus AB Khan Academy.mp3", "Sentence": "And actually, let's just not figure out what the derivative is. Let's figure out the derivative at x equals eight. Pause this video again and see if you can figure that out. Well, what we're gonna do is first just figure out what this is, and then we're going to evaluate it at x equals eight. And the key thing to appreciate is this is the same thing, and we're just gonna do what we did up here, as the derivative with respect to x. Instead of saying the cube root of x squared, we could say this is x squared to the 1 3rd power, which is the same thing as the derivative with respect to x of, well, x squared, if I raise something to an exponent and then raise that to an exponent, I could just take the product of the exponents. And so this is just going to be x to the two times 1 3rd power, or to the 2 3rds power."}, {"video_title": "Power rule (with rewriting the expression) AP Calculus AB Khan Academy.mp3", "Sentence": "Well, what we're gonna do is first just figure out what this is, and then we're going to evaluate it at x equals eight. And the key thing to appreciate is this is the same thing, and we're just gonna do what we did up here, as the derivative with respect to x. Instead of saying the cube root of x squared, we could say this is x squared to the 1 3rd power, which is the same thing as the derivative with respect to x of, well, x squared, if I raise something to an exponent and then raise that to an exponent, I could just take the product of the exponents. And so this is just going to be x to the two times 1 3rd power, or to the 2 3rds power. And now this is just going to be equal to, I'll do it right over here, bring the 2 3rds out front, 2 3rds times x to the, what's 2 3rds minus one? Well, that's 2 3rds minus 3 3rd, or it would be negative 1 3rd power. And we wanna know what happens at x equals eight, so let's just evaluate that."}, {"video_title": "Power rule (with rewriting the expression) AP Calculus AB Khan Academy.mp3", "Sentence": "And so this is just going to be x to the two times 1 3rd power, or to the 2 3rds power. And now this is just going to be equal to, I'll do it right over here, bring the 2 3rds out front, 2 3rds times x to the, what's 2 3rds minus one? Well, that's 2 3rds minus 3 3rd, or it would be negative 1 3rd power. And we wanna know what happens at x equals eight, so let's just evaluate that. That's going to be 2 3rds times x is equal to eight to the negative 1 3rd power. Well, what's eight to the 1 3rd power? Eight to the 1 3rd power is going to be equal to two, and so eight to the negative 1 3rd power is 1 1.5."}, {"video_title": "Power rule (with rewriting the expression) AP Calculus AB Khan Academy.mp3", "Sentence": "And we wanna know what happens at x equals eight, so let's just evaluate that. That's going to be 2 3rds times x is equal to eight to the negative 1 3rd power. Well, what's eight to the 1 3rd power? Eight to the 1 3rd power is going to be equal to two, and so eight to the negative 1 3rd power is 1 1.5. Actually, let me just do that step by step. So this is going to be equal to 2 3rds times, we could do it this way, one over eight to the 1 3rd power. And so this is just one over two, 2 3rds times 1 1.5."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "A rectangular storage container with an open top needs to have a volume of 10 cubic meters. The length of its base is twice the width. Material for the base costs $10 per square meter. Material for the sides costs $6 per square meter. Find the cost of the material for the cheapest container. So let's draw this open storage container, this open rectangular storage container. So it's going to have an open top."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Material for the sides costs $6 per square meter. Find the cost of the material for the cheapest container. So let's draw this open storage container, this open rectangular storage container. So it's going to have an open top. So let me draw its open top as good as I can. So it's going to have an open top. That's the top of my container."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So it's going to have an open top. So let me draw its open top as good as I can. So it's going to have an open top. That's the top of my container. And then let me draw the sides just like that. So it might look something like that. And then I could draw, and since it's open top, I can see through."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "That's the top of my container. And then let me draw the sides just like that. So it might look something like that. And then I could draw, and since it's open top, I can see through. I can see the inside of the container as well. So the container would look something like that. And so what do they tell us?"}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And then I could draw, and since it's open top, I can see through. I can see the inside of the container as well. So the container would look something like that. And so what do they tell us? They tell us that the volume needs to be 10 cubic meters. So let me write that down. The volume needs to be equal to 10 meters cubed."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And so what do they tell us? They tell us that the volume needs to be 10 cubic meters. So let me write that down. The volume needs to be equal to 10 meters cubed. The length of its base is twice the width. So the length, let's call this the width x. So the length is going to be twice that."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "The volume needs to be equal to 10 meters cubed. The length of its base is twice the width. So the length, let's call this the width x. So the length is going to be twice that. It's going to be 2x. That's what they tell us right over here. They tell us material for the base costs $10 per square meter."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So the length is going to be twice that. It's going to be 2x. That's what they tell us right over here. They tell us material for the base costs $10 per square meter. So the base, so this area right over here, if I was transparent, I could continue to draw it down here. But this right over here, that material costs $10 per square meter. Let me label that."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "They tell us material for the base costs $10 per square meter. So the base, so this area right over here, if I was transparent, I could continue to draw it down here. But this right over here, that material costs $10 per square meter. Let me label that. $10 per square meter. And then they say material for the sides costs $6 per square meter. So the material over here costs $6 per meter squared."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Let me label that. $10 per square meter. And then they say material for the sides costs $6 per square meter. So the material over here costs $6 per meter squared. So let's see if we can come up with a value or how much this box would cost to make as a function of x. But x only gives us the dimensions of the base. We also need a dimension for height."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So the material over here costs $6 per meter squared. So let's see if we can come up with a value or how much this box would cost to make as a function of x. But x only gives us the dimensions of the base. We also need a dimension for height. So it'll be a function of x and height for now. So let's write h as the height right over here. So what is the cost of this container going to be?"}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We also need a dimension for height. So it'll be a function of x and height for now. So let's write h as the height right over here. So what is the cost of this container going to be? So the cost is going to be equal to the cost of the base. Well, the cost of the base is going to be $10 times, I'll just write 10. It's just going to be 10 times the area of the base."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So what is the cost of this container going to be? So the cost is going to be equal to the cost of the base. Well, the cost of the base is going to be $10 times, I'll just write 10. It's just going to be 10 times the area of the base. Well, what's the area of the base? Well, it's going to be the width times the length. So 10 times x times 2x."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "It's just going to be 10 times the area of the base. Well, what's the area of the base? Well, it's going to be the width times the length. So 10 times x times 2x. That is the cost of base. And now what's going to be the cost of the sides? Well, the different sides are going to have different dimensions."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So 10 times x times 2x. That is the cost of base. And now what's going to be the cost of the sides? Well, the different sides are going to have different dimensions. You have this side over here and this side over here, which have the same dimension. They both have an area of x times h. You have x times h. And then our material is $6 per square meter. So 6 times x times h would be the cost of one of these side panels."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the different sides are going to have different dimensions. You have this side over here and this side over here, which have the same dimension. They both have an area of x times h. You have x times h. And then our material is $6 per square meter. So 6 times x times h would be the cost of one of these side panels. But so for two of them, we have to multiply by 2. So plus 2 times 6 times h. And then we have these two side panels. We have this side panel right over here."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So 6 times x times h would be the cost of one of these side panels. But so for two of them, we have to multiply by 2. So plus 2 times 6 times h. And then we have these two side panels. We have this side panel right over here. And we have this side panel right over here. The area of each of them is going to be 2x times h. So it's going to be 2x times h. The cost of the material is going to be 6. So the cost of one of the panels is going to be $6 per square meters times 2xh meters squared."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We have this side panel right over here. And we have this side panel right over here. The area of each of them is going to be 2x times h. So it's going to be 2x times h. The cost of the material is going to be 6. So the cost of one of the panels is going to be $6 per square meters times 2xh meters squared. But we have two of these panels, one panel and two panels. So we have to multiply by 2. And so we will get, so this is right over here."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So the cost of one of the panels is going to be $6 per square meters times 2xh meters squared. But we have two of these panels, one panel and two panels. So we have to multiply by 2. And so we will get, so this is right over here. This is the cost of the sides. And so let's see if we can simplify this. And I'll write it all in a neutral color."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And so we will get, so this is right over here. This is the cost of the sides. And so let's see if we can simplify this. And I'll write it all in a neutral color. So this is going to be equal to 10. Let's see, 10 times 2 is 20. x times x is x squared. And you have 2 times 6 times xh."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And I'll write it all in a neutral color. So this is going to be equal to 10. Let's see, 10 times 2 is 20. x times x is x squared. And you have 2 times 6 times xh. So this is going to be plus 12xh. And then this is going to be 2 times 6, which is 12 times 2 is 24xh. Plus 24xh."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And you have 2 times 6 times xh. So this is going to be plus 12xh. And then this is going to be 2 times 6, which is 12 times 2 is 24xh. Plus 24xh. So this is going to be equal to 20x squared plus 36xh. So this is going to be my cost. But I'm not ready to optimize it yet."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Plus 24xh. So this is going to be equal to 20x squared plus 36xh. So this is going to be my cost. But I'm not ready to optimize it yet. We don't know how to optimize with respect to two variables. We only know how to optimize with respect to one variable. And maybe I'll say, let's optimize with respect to x."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "But I'm not ready to optimize it yet. We don't know how to optimize with respect to two variables. We only know how to optimize with respect to one variable. And maybe I'll say, let's optimize with respect to x. But if we want to optimize with respect to x, we have to express h as a function of x. So how can we do that? How can we express h as a function of x?"}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And maybe I'll say, let's optimize with respect to x. But if we want to optimize with respect to x, we have to express h as a function of x. So how can we do that? How can we express h as a function of x? Well, we know that the volume has to be 10 cubic meters. So we know that x, the width times the length times 2x times the height times h needs to be equal to 10. Or another way of saying that, this tells us that 2x squared times h needs to be equal to 10."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "How can we express h as a function of x? Well, we know that the volume has to be 10 cubic meters. So we know that x, the width times the length times 2x times the height times h needs to be equal to 10. Or another way of saying that, this tells us that 2x squared times h needs to be equal to 10. And so if we want h as a function of x, we just divide both sides by 2x squared. And we get h is equal to 10 over 2x squared. Or we could say that h is equal to 5 over x squared."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Or another way of saying that, this tells us that 2x squared times h needs to be equal to 10. And so if we want h as a function of x, we just divide both sides by 2x squared. And we get h is equal to 10 over 2x squared. Or we could say that h is equal to 5 over x squared. And then we can substitute back right over here. h is equal to 5 over x squared. So all of this business is going to be equal to 20 times x squared plus 36 times x times 5 over x squared."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Or we could say that h is equal to 5 over x squared. And then we can substitute back right over here. h is equal to 5 over x squared. So all of this business is going to be equal to 20 times x squared plus 36 times x times 5 over x squared. So our cost as a function of x is going to be 20x squared. 36 times 5. Let's see, 30 times 5 is 150."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So all of this business is going to be equal to 20 times x squared plus 36 times x times 5 over x squared. So our cost as a function of x is going to be 20x squared. 36 times 5. Let's see, 30 times 5 is 150. Plus another 30 is going to be 180. So it's going to be plus 180 times x times x to the negative 2, 180x to the negative 1 power. So we finally have cost as a function of x."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Let's see, 30 times 5 is 150. Plus another 30 is going to be 180. So it's going to be plus 180 times x times x to the negative 2, 180x to the negative 1 power. So we finally have cost as a function of x. Now we're ready to optimize. To optimize, we just have to figure out what are the critical points here and whether those critical points are minimum or maximum value. So let's see what we can do."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So we finally have cost as a function of x. Now we're ready to optimize. To optimize, we just have to figure out what are the critical points here and whether those critical points are minimum or maximum value. So let's see what we can do. So to find a critical point, we take the derivative, figure out where the derivative is undefined equal to 0. And those are our candidate critical points. And then from the critical points we find, they might be minimum or maximum values."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let's see what we can do. So to find a critical point, we take the derivative, figure out where the derivative is undefined equal to 0. And those are our candidate critical points. And then from the critical points we find, they might be minimum or maximum values. So the derivative of c of our cost with respect to x is going to be equal to 40 times x minus 180 times x to the negative 2 power. Now this seems, well, it's defined for all x except for x equaling 0. But x equaling 0 is not interesting to us as a critical point because then we're going to have a degenerate."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And then from the critical points we find, they might be minimum or maximum values. So the derivative of c of our cost with respect to x is going to be equal to 40 times x minus 180 times x to the negative 2 power. Now this seems, well, it's defined for all x except for x equaling 0. But x equaling 0 is not interesting to us as a critical point because then we're going to have a degenerate. This is going to have no base at all. So we don't want to worry about that critical point. We would have no volume at all."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "But x equaling 0 is not interesting to us as a critical point because then we're going to have a degenerate. This is going to have no base at all. So we don't want to worry about that critical point. We would have no volume at all. So it would not work out. And actually, if x equals 0, then our height is undefined as well. So let's see."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We would have no volume at all. So it would not work out. And actually, if x equals 0, then our height is undefined as well. So let's see. So this was defined for everything else, for anything other than x equals 0. So let's see when this derivative is equal to 0 in our search for potential critical points. So when does?"}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let's see. So this was defined for everything else, for anything other than x equals 0. So let's see when this derivative is equal to 0 in our search for potential critical points. So when does? I'll do it over here. When does 40x minus 180x to the negative 2 equal 0? Well, we could add the 180x to the negative 2 to both sides."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So when does? I'll do it over here. When does 40x minus 180x to the negative 2 equal 0? Well, we could add the 180x to the negative 2 to both sides. We get 40x is equal to 180. And I could write it as 180 over x squared. Now let's see."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we could add the 180x to the negative 2 to both sides. We get 40x is equal to 180. And I could write it as 180 over x squared. Now let's see. We could multiply both sides of this equation by x squared. And we would get 40x to the third is equal to 180. Divide both sides by 40."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's see. We could multiply both sides of this equation by x squared. And we would get 40x to the third is equal to 180. Divide both sides by 40. You get x to the third is equal to 180 over 40, which is the same thing as 18 over 4, which is the same thing as 9 over 2. And so if we want to solve for x, we get that x is equal to a critical point. We get a critical point at x is equal to 9 halves to the 1 third power, the cube root of 9 halves."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Divide both sides by 40. You get x to the third is equal to 180 over 40, which is the same thing as 18 over 4, which is the same thing as 9 over 2. And so if we want to solve for x, we get that x is equal to a critical point. We get a critical point at x is equal to 9 halves to the 1 third power, the cube root of 9 halves. So let's see. Let's get an approximate value for what that is. So if we take 9 halves, 9 divided by 2, I guess we could call that 4.5."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We get a critical point at x is equal to 9 halves to the 1 third power, the cube root of 9 halves. So let's see. Let's get an approximate value for what that is. So if we take 9 halves, 9 divided by 2, I guess we could call that 4.5. And we want to raise it to the 1 third power. We get 1.65. So it's approximately equal to 1.65 as our critical point."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So if we take 9 halves, 9 divided by 2, I guess we could call that 4.5. And we want to raise it to the 1 third power. We get 1.65. So it's approximately equal to 1.65 as our critical point. Now the way the problem is asked, we're only getting one legitimate critical point here. So that's probably going to be the x at which we achieve a minimum value. But let's use our second derivative test just in case to make sure that we're definitely concave upwards over here, in which case that this will definitely be the x value at which we achieve a minimum value."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So it's approximately equal to 1.65 as our critical point. Now the way the problem is asked, we're only getting one legitimate critical point here. So that's probably going to be the x at which we achieve a minimum value. But let's use our second derivative test just in case to make sure that we're definitely concave upwards over here, in which case that this will definitely be the x value at which we achieve a minimum value. So the second derivative of our cost function is just the derivative of this, which is going to be equal to 40 minus 180 times negative 2, which is negative 360. So it's going to be plus 360 over x to the third. The derivative of this is negative 2 times negative 180, which is positive 360, x to the negative 3 power, which is exactly this right over here."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "But let's use our second derivative test just in case to make sure that we're definitely concave upwards over here, in which case that this will definitely be the x value at which we achieve a minimum value. So the second derivative of our cost function is just the derivative of this, which is going to be equal to 40 minus 180 times negative 2, which is negative 360. So it's going to be plus 360 over x to the third. The derivative of this is negative 2 times negative 180, which is positive 360, x to the negative 3 power, which is exactly this right over here. So when x is equal to 1.65, this is going to be positive. This is going to be positive. So let me write this down."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "The derivative of this is negative 2 times negative 180, which is positive 360, x to the negative 3 power, which is exactly this right over here. So when x is equal to 1.65, this is going to be positive. This is going to be positive. So let me write this down. c prime prime of 1.65 is definitely greater than 0. So we're definitely concave upwards when x is 1.65, concave upwards, which means that our graph is going to look something like this. And so where the derivative equal to 0, which is right over there, we are at a minimum point."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let me write this down. c prime prime of 1.65 is definitely greater than 0. So we're definitely concave upwards when x is 1.65, concave upwards, which means that our graph is going to look something like this. And so where the derivative equal to 0, which is right over there, we are at a minimum point. We are minimizing our cost. And so if we go back to the question, the only thing that we have to do now, we know the x value that minimizes our cost. We now have to find the cost of the material for the cheapest container."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And so where the derivative equal to 0, which is right over there, we are at a minimum point. We are minimizing our cost. And so if we go back to the question, the only thing that we have to do now, we know the x value that minimizes our cost. We now have to find the cost of the material for the cheapest container. So we just have to figure out what our cost is. And we already know what our cost is as a function of x. So we just have to put 1.65 into this equation, evaluate the function at 1.65."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We now have to find the cost of the material for the cheapest container. So we just have to figure out what our cost is. And we already know what our cost is as a function of x. So we just have to put 1.65 into this equation, evaluate the function at 1.65. So let's do that. We get our cost is going to be equal to 20 times 1.65. I should say approximately equal to, because I'm using an approximation of this original value."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So we just have to put 1.65 into this equation, evaluate the function at 1.65. So let's do that. We get our cost is going to be equal to 20 times 1.65. I should say approximately equal to, because I'm using an approximation of this original value. 1.65 squared plus 180. I could say divided by 1.65. That's the same thing as multiplying by 1.65 to the negative 1."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "I should say approximately equal to, because I'm using an approximation of this original value. 1.65 squared plus 180. I could say divided by 1.65. That's the same thing as multiplying by 1.65 to the negative 1. So divided by 1.65, which is equal to $163.5. So it's approximately. So the cost when, let me do this in a new color."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "That's the same thing as multiplying by 1.65 to the negative 1. So divided by 1.65, which is equal to $163.5. So it's approximately. So the cost when, let me do this in a new color. We deserve a drum roll now. The cost when x is 1.65 is approximately equal to $163.54, which is quite an expensive box. So this is fairly expensive material here, although it's a fairly large box."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So the cost when, let me do this in a new color. We deserve a drum roll now. The cost when x is 1.65 is approximately equal to $163.54, which is quite an expensive box. So this is fairly expensive material here, although it's a fairly large box. 1.65 meters in width. It's going to be twice that in length. And then you could figure out what its height is going to be, although it's not going to be too tall."}, {"video_title": "Optimization cost of materials Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So this is fairly expensive material here, although it's a fairly large box. 1.65 meters in width. It's going to be twice that in length. And then you could figure out what its height is going to be, although it's not going to be too tall. 5 divided by 1.65 squared. I don't know. It'll be roughly a little under 2 meters tall."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "Here we've graphed the function y is equal to the square root of x. And we're going to create a solid of revolution, but we're not going to do it by rotating this around the x or the y-axis. Instead, we're going to rotate it around another somewhat arbitrary line. And in this case, I will rotate it around the line y is equal to 1. So let's say that this right over here is the line y equals 1. So this is what we're going to rotate it around, y equals 1. So the first thing we want to do is visualize what we're doing."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "And in this case, I will rotate it around the line y is equal to 1. So let's say that this right over here is the line y equals 1. So this is what we're going to rotate it around, y equals 1. So the first thing we want to do is visualize what we're doing. And we actually care about the interval. So let's say that the interval is between this point right over here, where the two points intersect. And let's say between that and x is equal to 4."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "So the first thing we want to do is visualize what we're doing. And we actually care about the interval. So let's say that the interval is between this point right over here, where the two points intersect. And let's say between that and x is equal to 4. So let's say that this right over here, so this is x equals 4. So it's going to be just right over here. So this is the interval that we're rotating."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "And let's say between that and x is equal to 4. So let's say that this right over here, so this is x equals 4. So it's going to be just right over here. So this is the interval that we're rotating. We're going to rotate it around y equals 1, not around the x-axis. So what would our figure look like? Well, we're going to rotate it around like this."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "So this is the interval that we're rotating. We're going to rotate it around y equals 1, not around the x-axis. So what would our figure look like? Well, we're going to rotate it around like this. We're going to rotate it around something like this. And so your figure is going to look, I guess you could call it, it'll almost look like a cone head if you view it on the side, or something like a bullet, but not quite looking like a bullet. But it would be a shape that looks something like that."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we're going to rotate it around like this. We're going to rotate it around something like this. And so your figure is going to look, I guess you could call it, it'll almost look like a cone head if you view it on the side, or something like a bullet, but not quite looking like a bullet. But it would be a shape that looks something like that. So hopefully we can visualize this. We've done this several times already, attempting to visualize these shapes. But let's think about how we could actually figure out the volume of this solid of revolution."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "But it would be a shape that looks something like that. So hopefully we can visualize this. We've done this several times already, attempting to visualize these shapes. But let's think about how we could actually figure out the volume of this solid of revolution. Well, let's just think about it disk by disk. So let's construct a disk right over here. And we've done this many, many, many times already, where we essentially want the volume of each of these disks, and then we're going to take this for each of these x's in our interval, and then we're going to take the sum of the volumes of all of the disks and we're going to find the volume."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "But let's think about how we could actually figure out the volume of this solid of revolution. Well, let's just think about it disk by disk. So let's construct a disk right over here. And we've done this many, many, many times already, where we essentially want the volume of each of these disks, and then we're going to take this for each of these x's in our interval, and then we're going to take the sum of the volumes of all of the disks and we're going to find the volume. We're going to stack them all up, or I guess put them next to each other in this case, and find the volume of our entire figure. So to figure out the volume of each disk, we just have to figure out the area of its face, and I'll do that in purple. We just have to figure out the area of the face and multiply it by the depth."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "And we've done this many, many, many times already, where we essentially want the volume of each of these disks, and then we're going to take this for each of these x's in our interval, and then we're going to take the sum of the volumes of all of the disks and we're going to find the volume. We're going to stack them all up, or I guess put them next to each other in this case, and find the volume of our entire figure. So to figure out the volume of each disk, we just have to figure out the area of its face, and I'll do that in purple. We just have to figure out the area of the face and multiply it by the depth. So what's the area of this face? Well, it's going to be pi times the radius of the face squared. Now what's the radius of the face here?"}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "We just have to figure out the area of the face and multiply it by the depth. So what's the area of this face? Well, it's going to be pi times the radius of the face squared. Now what's the radius of the face here? Well, it's not just square root of x. Square root of x would tell us the distance between the x-axis and our function. It's now square root of x minus 1."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "Now what's the radius of the face here? Well, it's not just square root of x. Square root of x would tell us the distance between the x-axis and our function. It's now square root of x minus 1. This length right over here is square root of x minus 1 for any given x in our interval. So it's going to be equal to square root of... Let me do that same orange color just to make it clear where I'm getting it from. It's going to be equal to square root of x minus 1."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "It's now square root of x minus 1. This length right over here is square root of x minus 1 for any given x in our interval. So it's going to be equal to square root of... Let me do that same orange color just to make it clear where I'm getting it from. It's going to be equal to square root of x minus 1. It's essentially our function minus what we are rotating around. That gives us the radius, and so this will give us the area of each of our faces, and then we just multiply that times the depth. The depth, of course, is dx."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "It's going to be equal to square root of x minus 1. It's essentially our function minus what we are rotating around. That gives us the radius, and so this will give us the area of each of our faces, and then we just multiply that times the depth. The depth, of course, is dx. We've done that multiple times, so times dx. This is the thing that we want to sum up. This is the thing that we want to sum up over our interval."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "The depth, of course, is dx. We've done that multiple times, so times dx. This is the thing that we want to sum up. This is the thing that we want to sum up over our interval. Our interval, let's see, this point right over here where the square root of x is equal to 1, that's just going to be x equals 1. This is just x equals 1 over here. We said that we would do this all the way until x equals 4."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "This is the thing that we want to sum up over our interval. Our interval, let's see, this point right over here where the square root of x is equal to 1, that's just going to be x equals 1. This is just x equals 1 over here. We said that we would do this all the way until x equals 4. This was kind of our end of our interval. This is until x is equal to... Let me be careful. This is the x-axis right over here."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "We said that we would do this all the way until x equals 4. This was kind of our end of our interval. This is until x is equal to... Let me be careful. This is the x-axis right over here. This is the x-axis right over here, all the way until x equals 4. You get confused. This is the x-axis."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "This is the x-axis right over here. This is the x-axis right over here, all the way until x equals 4. You get confused. This is the x-axis. We're going from x equals 1 to x equals 4, and we're essentially taking this area, one way to think about it, between our square root of x and 1, and we're rotating it around 1. We're using these little disks right over here, so the interval between 1 and 4. This is going to give us the volume of our solid of revolution."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "This is the x-axis. We're going from x equals 1 to x equals 4, and we're essentially taking this area, one way to think about it, between our square root of x and 1, and we're rotating it around 1. We're using these little disks right over here, so the interval between 1 and 4. This is going to give us the volume of our solid of revolution. Now we just have to evaluate this definite integral. Let's give a shot at it. This is going to be equal to the integral from 1 to 4, and we can factor out, or we can put the pi outside the integral sign, and then we can expand this out."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "This is going to give us the volume of our solid of revolution. Now we just have to evaluate this definite integral. Let's give a shot at it. This is going to be equal to the integral from 1 to 4, and we can factor out, or we can put the pi outside the integral sign, and then we can expand this out. Square root of x squared. All we're going to do is expand this binomial. Square root of x minus 1 times square root of x minus 1."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "This is going to be equal to the integral from 1 to 4, and we can factor out, or we can put the pi outside the integral sign, and then we can expand this out. Square root of x squared. All we're going to do is expand this binomial. Square root of x minus 1 times square root of x minus 1. Square root of x times square root of x is x. Square root of x times negative 1 is negative square root of x. Negative 1 times square root of x is another negative square root of x."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "Square root of x minus 1 times square root of x minus 1. Square root of x times square root of x is x. Square root of x times negative 1 is negative square root of x. Negative 1 times square root of x is another negative square root of x. The negative 1 times negative 1 is equal to positive 1. This part right over here will simplify to x minus 2 square roots of x. This is minus 2 square roots of x."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "Negative 1 times square root of x is another negative square root of x. The negative 1 times negative 1 is equal to positive 1. This part right over here will simplify to x minus 2 square roots of x. This is minus 2 square roots of x. Minus 2 square roots of x, and then you have plus 1. Then all of that times dx. This is going to be equal to, let's put our pi out there, the antiderivative of this business."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "This is minus 2 square roots of x. Minus 2 square roots of x, and then you have plus 1. Then all of that times dx. This is going to be equal to, let's put our pi out there, the antiderivative of this business. The antiderivative of x is x squared over 2. The antiderivative, so minus 2 times the antiderivative of square root of x. Square root of x is just x to the 1 half power, so we increment the power by 1."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "This is going to be equal to, let's put our pi out there, the antiderivative of this business. The antiderivative of x is x squared over 2. The antiderivative, so minus 2 times the antiderivative of square root of x. Square root of x is just x to the 1 half power, so we increment the power by 1. It's going to be x to the 3 half power times 2 thirds. We could say it's going to be times 2 thirds x to the 3 halves. Let me make it clear."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "Square root of x is just x to the 1 half power, so we increment the power by 1. It's going to be x to the 3 half power times 2 thirds. We could say it's going to be times 2 thirds x to the 3 halves. Let me make it clear. This negative 2 right here is this negative 2. This expression right over here is the antiderivative of square root of x. You can verify this."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "Let me make it clear. This negative 2 right here is this negative 2. This expression right over here is the antiderivative of square root of x. You can verify this. The antiderivative of this 3 halves times 2 thirds is 1. Decrement the 3 halves, you get x to the 1 half. Now let's take the antiderivative of 1."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "You can verify this. The antiderivative of this 3 halves times 2 thirds is 1. Decrement the 3 halves, you get x to the 1 half. Now let's take the antiderivative of 1. That's just going to be equal to x. We're going to evaluate this from 1 to 4. We're in the home stretch."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's take the antiderivative of 1. That's just going to be equal to x. We're going to evaluate this from 1 to 4. We're in the home stretch. This is going to be equal to pi times, first let's do it at 4. See if 4 squared, I'll write it all out, is 1 squared over 2 minus 4 to the 3, let me do this part first, minus 4 thirds, 4 thirds, and that was 4 to the 3 halves. 4 to the 1 half is 2, and then you raise that to the 3rd power, you get 8."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "We're in the home stretch. This is going to be equal to pi times, first let's do it at 4. See if 4 squared, I'll write it all out, is 1 squared over 2 minus 4 to the 3, let me do this part first, minus 4 thirds, 4 thirds, and that was 4 to the 3 halves. 4 to the 1 half is 2, and then you raise that to the 3rd power, you get 8. Times 8 plus 4, and then you subtract out all of this stuff evaluated at 1. 1 squared over 2, that's just going to be minus 1 half. Then we're subtracting this now."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "4 to the 1 half is 2, and then you raise that to the 3rd power, you get 8. Times 8 plus 4, and then you subtract out all of this stuff evaluated at 1. 1 squared over 2, that's just going to be minus 1 half. Then we're subtracting this now. Actually, I don't want to skip too many steps. Let's do, it's going to be this. This is when we evaluated at 4, and then we're going to subtract when we evaluate this whole thing at 1."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "Then we're subtracting this now. Actually, I don't want to skip too many steps. Let's do, it's going to be this. This is when we evaluated at 4, and then we're going to subtract when we evaluate this whole thing at 1. When we evaluate it at 1, we do this in this green color. That's not the green color I thought I was going to do. When you evaluate it at 1, you get 1 squared over 2, which is 1 half, and then you get 1 to the 3 halves, which is just 1."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "This is when we evaluated at 4, and then we're going to subtract when we evaluate this whole thing at 1. When we evaluate it at 1, we do this in this green color. That's not the green color I thought I was going to do. When you evaluate it at 1, you get 1 squared over 2, which is 1 half, and then you get 1 to the 3 halves, which is just 1. This becomes minus 4 thirds, and then you have plus 1. Let's simplify this, and I'll do it all in the same color now. This is going to be equal to pi times 4 squared over 2, that is 16 over 2, which is equal to 8."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "When you evaluate it at 1, you get 1 squared over 2, which is 1 half, and then you get 1 to the 3 halves, which is just 1. This becomes minus 4 thirds, and then you have plus 1. Let's simplify this, and I'll do it all in the same color now. This is going to be equal to pi times 4 squared over 2, that is 16 over 2, which is equal to 8. Then you have 4 times 8, which is 32 over 3, so minus 32 over 3 plus 4. Then you have minus 1 half, we're just distributing the negative, plus 4 thirds, plus 4 thirds, and then you have minus 1. Now we just have to add up a bunch of fractions to simplify this thing."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "This is going to be equal to pi times 4 squared over 2, that is 16 over 2, which is equal to 8. Then you have 4 times 8, which is 32 over 3, so minus 32 over 3 plus 4. Then you have minus 1 half, we're just distributing the negative, plus 4 thirds, plus 4 thirds, and then you have minus 1. Now we just have to add up a bunch of fractions to simplify this thing. What do we get? This is going to be equal to pi times, see our least common multiple looks like 6. We're going to put everything over a denominator of 6."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "Now we just have to add up a bunch of fractions to simplify this thing. What do we get? This is going to be equal to pi times, see our least common multiple looks like 6. We're going to put everything over a denominator of 6. 8 is the same thing as 48 over 6. 32 over 3 is the same thing as 64 over 6, so minus 64 over 6. 4 is the same thing as 24 over 6."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "We're going to put everything over a denominator of 6. 8 is the same thing as 48 over 6. 32 over 3 is the same thing as 64 over 6, so minus 64 over 6. 4 is the same thing as 24 over 6. 1 half is the same thing as 3 over 6, so this is negative 3 over 6, the same thing as negative 1 half. 4 thirds is the same thing as 8 over 6, and then negative 1 is the same thing as negative 6 over 6. A little bit of arithmetic here, and let's see what we get."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "4 is the same thing as 24 over 6. 1 half is the same thing as 3 over 6, so this is negative 3 over 6, the same thing as negative 1 half. 4 thirds is the same thing as 8 over 6, and then negative 1 is the same thing as negative 6 over 6. A little bit of arithmetic here, and let's see what we get. If we take 48 and we subtract 64 from 48, we get negative 16, is that right? Yes, we get negative 16. Then if we add 24 to negative 16, you get positive 8."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "A little bit of arithmetic here, and let's see what we get. If we take 48 and we subtract 64 from 48, we get negative 16, is that right? Yes, we get negative 16. Then if we add 24 to negative 16, you get positive 8. Positive 8 minus 3 is 5. 5 plus 8 is 13. 13 minus 6 is 7."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "Then if we add 24 to negative 16, you get positive 8. Positive 8 minus 3 is 5. 5 plus 8 is 13. 13 minus 6 is 7. This whole numerator simplifies to 7, so we get 7 pi over 6 as our volume. Let me just verify I did that right. This is going to be negative 16."}, {"video_title": "Disc method rotation around horizontal line AP Calculus AB Khan Academy.mp3", "Sentence": "13 minus 6 is 7. This whole numerator simplifies to 7, so we get 7 pi over 6 as our volume. Let me just verify I did that right. This is going to be negative 16. We get to positive 8. Positive 8 plus positive 8 is 16. 16 minus 9 is, yes, it is indeed 7."}, {"video_title": "Integration by parts \u00c2\u00bax____dx AP Calculus BC Khan Academy.mp3", "Sentence": "Let's see if we can take the antiderivative of x squared times e to the x dx. Now, the key is to recognize when you can at least attempt to use integration by parts. It might be a little bit obvious because this video is about integration by parts. But the clue that integration by parts may be applicable is to say, look, I've got a function that's the product of two other functions, in this case x squared and e to the x. And integration by parts can be useful as if I can take the derivative of one of them and it becomes simpler. And if I take the antiderivative of the other one, it becomes no more complicated. So in this case, if I were to take the antiderivative of x squared, it does become simpler and it becomes 2x."}, {"video_title": "Integration by parts \u00c2\u00bax____dx AP Calculus BC Khan Academy.mp3", "Sentence": "But the clue that integration by parts may be applicable is to say, look, I've got a function that's the product of two other functions, in this case x squared and e to the x. And integration by parts can be useful as if I can take the derivative of one of them and it becomes simpler. And if I take the antiderivative of the other one, it becomes no more complicated. So in this case, if I were to take the antiderivative of x squared, it does become simpler and it becomes 2x. And if I take the antiderivative of e to the x, it doesn't become any more complicated. So let's assign f of x to be equal to x squared. And we want that one to be the one where if I take the derivative, it becomes simpler because I'm going to have to take the derivative of f of x right over here in the integration by parts formula."}, {"video_title": "Integration by parts \u00c2\u00bax____dx AP Calculus BC Khan Academy.mp3", "Sentence": "So in this case, if I were to take the antiderivative of x squared, it does become simpler and it becomes 2x. And if I take the antiderivative of e to the x, it doesn't become any more complicated. So let's assign f of x to be equal to x squared. And we want that one to be the one where if I take the derivative, it becomes simpler because I'm going to have to take the derivative of f of x right over here in the integration by parts formula. And let's assign g prime of x to be equal to e to the x because later I'm going to have to take its antiderivative and the antiderivative of e to the x is still just e to the x. So let me write this down. So we are saying that f of x is equal to x squared, in which case f prime of x is going to be equal to 2x."}, {"video_title": "Integration by parts \u00c2\u00bax____dx AP Calculus BC Khan Academy.mp3", "Sentence": "And we want that one to be the one where if I take the derivative, it becomes simpler because I'm going to have to take the derivative of f of x right over here in the integration by parts formula. And let's assign g prime of x to be equal to e to the x because later I'm going to have to take its antiderivative and the antiderivative of e to the x is still just e to the x. So let me write this down. So we are saying that f of x is equal to x squared, in which case f prime of x is going to be equal to 2x. And I'm not worrying about the constants right now. We'll just add a constant at the end to make sure that our antiderivative is in the most general form. And then g prime of x is equal to e to the x, which means its antiderivative, g of x, is still equal to e to the x."}, {"video_title": "Integration by parts \u00c2\u00bax____dx AP Calculus BC Khan Academy.mp3", "Sentence": "So we are saying that f of x is equal to x squared, in which case f prime of x is going to be equal to 2x. And I'm not worrying about the constants right now. We'll just add a constant at the end to make sure that our antiderivative is in the most general form. And then g prime of x is equal to e to the x, which means its antiderivative, g of x, is still equal to e to the x. And now we're ready to apply the right-hand side right over here. So all this thing right over here is going to be equal to f of x, which is x squared, and let me put it right underneath, x squared times g of x, which is e to the x, minus the antiderivative of f prime of x, where f prime of x is 2x, times g of x. g of x is e to the x dx. So you might say, hey, Sal, we're left with another antiderivative, another indefinite integral right over here."}, {"video_title": "Integration by parts \u00c2\u00bax____dx AP Calculus BC Khan Academy.mp3", "Sentence": "And then g prime of x is equal to e to the x, which means its antiderivative, g of x, is still equal to e to the x. And now we're ready to apply the right-hand side right over here. So all this thing right over here is going to be equal to f of x, which is x squared, and let me put it right underneath, x squared times g of x, which is e to the x, minus the antiderivative of f prime of x, where f prime of x is 2x, times g of x. g of x is e to the x dx. So you might say, hey, Sal, we're left with another antiderivative, another indefinite integral right over here. How do we solve this one? And, as you might guess, the key might be integration by parts again. And we're making progress."}, {"video_title": "Integration by parts \u00c2\u00bax____dx AP Calculus BC Khan Academy.mp3", "Sentence": "So you might say, hey, Sal, we're left with another antiderivative, another indefinite integral right over here. How do we solve this one? And, as you might guess, the key might be integration by parts again. And we're making progress. This right over here is a simpler expression than this. Notice we were able to reduce the degree of this x squared. It now is just a 2x."}, {"video_title": "Integration by parts \u00c2\u00bax____dx AP Calculus BC Khan Academy.mp3", "Sentence": "And we're making progress. This right over here is a simpler expression than this. Notice we were able to reduce the degree of this x squared. It now is just a 2x. So let's actually, and what we can do to simplify this a little bit, since 2 is just a scalar, it's a constant, it's multiplying the function, we can take that out of the integral sign. So let's take it this way. So let me rewrite it this way."}, {"video_title": "Integration by parts \u00c2\u00bax____dx AP Calculus BC Khan Academy.mp3", "Sentence": "It now is just a 2x. So let's actually, and what we can do to simplify this a little bit, since 2 is just a scalar, it's a constant, it's multiplying the function, we can take that out of the integral sign. So let's take it this way. So let me rewrite it this way. We can only do that with a constant that's multiplying the function. So let me put the 2 right out here. And so now what we're concerned about is finding the integral."}, {"video_title": "Integration by parts \u00c2\u00bax____dx AP Calculus BC Khan Academy.mp3", "Sentence": "So let me rewrite it this way. We can only do that with a constant that's multiplying the function. So let me put the 2 right out here. And so now what we're concerned about is finding the integral. Let me write it right over here. The integral of xe to the x dx. And now this is another integration by parts problem."}, {"video_title": "Integration by parts \u00c2\u00bax____dx AP Calculus BC Khan Academy.mp3", "Sentence": "And so now what we're concerned about is finding the integral. Let me write it right over here. The integral of xe to the x dx. And now this is another integration by parts problem. And so let's again apply the same principles of integration by parts. When I take its derivative, it's going to get simpler. Well, x is going to get simpler when I take its derivative."}, {"video_title": "Integration by parts \u00c2\u00bax____dx AP Calculus BC Khan Academy.mp3", "Sentence": "And now this is another integration by parts problem. And so let's again apply the same principles of integration by parts. When I take its derivative, it's going to get simpler. Well, x is going to get simpler when I take its derivative. So now for the purposes of integration by parts, let's redefine f of x to be equal to just x. And then we can still have g prime of x equaling e to the x. And so in this case, let me write it all down."}, {"video_title": "Integration by parts \u00c2\u00bax____dx AP Calculus BC Khan Academy.mp3", "Sentence": "Well, x is going to get simpler when I take its derivative. So now for the purposes of integration by parts, let's redefine f of x to be equal to just x. And then we can still have g prime of x equaling e to the x. And so in this case, let me write it all down. f of x is equal to x. f prime of x is equal to 1. g prime of x is equal to e to the x. g of x is equal to, just the antiderivative of this, is equal to e to the x. So let's apply integration by parts again. So this is going to be equal to f of x times g of x."}, {"video_title": "Integration by parts \u00c2\u00bax____dx AP Calculus BC Khan Academy.mp3", "Sentence": "And so in this case, let me write it all down. f of x is equal to x. f prime of x is equal to 1. g prime of x is equal to e to the x. g of x is equal to, just the antiderivative of this, is equal to e to the x. So let's apply integration by parts again. So this is going to be equal to f of x times g of x. Now f of x is x. g of x is e to the x minus the antiderivative of f prime of x. Well, that's just 1. Times g of x, e to the x."}, {"video_title": "Integration by parts \u00c2\u00bax____dx AP Calculus BC Khan Academy.mp3", "Sentence": "So this is going to be equal to f of x times g of x. Now f of x is x. g of x is e to the x minus the antiderivative of f prime of x. Well, that's just 1. Times g of x, e to the x. It's just 1 times e to the x. 1 times e to the x dx. And remember, all I'm doing right now, you might have lost track of things."}, {"video_title": "Integration by parts \u00c2\u00bax____dx AP Calculus BC Khan Academy.mp3", "Sentence": "Times g of x, e to the x. It's just 1 times e to the x. 1 times e to the x dx. And remember, all I'm doing right now, you might have lost track of things. I'm just focused on this antiderivative. That antiderivative is that antiderivative there. If we can figure out what it is, we can then substitute back into our original expression."}, {"video_title": "Integration by parts \u00c2\u00bax____dx AP Calculus BC Khan Academy.mp3", "Sentence": "And remember, all I'm doing right now, you might have lost track of things. I'm just focused on this antiderivative. That antiderivative is that antiderivative there. If we can figure out what it is, we can then substitute back into our original expression. Now, you might appreciate integration by parts. What does this right over here simplify to? What is the antiderivative of 1 times e to the x dx?"}, {"video_title": "Integration by parts \u00c2\u00bax____dx AP Calculus BC Khan Academy.mp3", "Sentence": "If we can figure out what it is, we can then substitute back into our original expression. Now, you might appreciate integration by parts. What does this right over here simplify to? What is the antiderivative of 1 times e to the x dx? Or what is the antiderivative of 1 times e to the x? Well, it's just the antiderivative of e to the x, which is just e to the x. So this simplifies to x times e to the x minus the antiderivative of e to the x, which is just e to the x."}, {"video_title": "Integration by parts \u00c2\u00bax____dx AP Calculus BC Khan Academy.mp3", "Sentence": "What is the antiderivative of 1 times e to the x dx? Or what is the antiderivative of 1 times e to the x? Well, it's just the antiderivative of e to the x, which is just e to the x. So this simplifies to x times e to the x minus the antiderivative of e to the x, which is just e to the x. So minus e to the x. And then we can take this and substitute it back. This is the antiderivative of this."}, {"video_title": "Integration by parts \u00c2\u00bax____dx AP Calculus BC Khan Academy.mp3", "Sentence": "So this simplifies to x times e to the x minus the antiderivative of e to the x, which is just e to the x. So minus e to the x. And then we can take this and substitute it back. This is the antiderivative of this. So we can substitute it back up here to figure out the antiderivative of our original expression. So the antiderivative of our original expression, we're getting really close, is going to be equal to different colors so we can keep track of things. It's going to be equal to x squared times e to the x minus 2 times all of this business."}, {"video_title": "Integration by parts \u00c2\u00bax____dx AP Calculus BC Khan Academy.mp3", "Sentence": "This is the antiderivative of this. So we can substitute it back up here to figure out the antiderivative of our original expression. So the antiderivative of our original expression, we're getting really close, is going to be equal to different colors so we can keep track of things. It's going to be equal to x squared times e to the x minus 2 times all of this business. So minus 2 times, well, this antiderivative we just figured out is this. Minus 2 times x e to the x minus e to the x. And if we want now is a good time to put our plus c. And of course, we can simplify this."}, {"video_title": "Integration by parts \u00c2\u00bax____dx AP Calculus BC Khan Academy.mp3", "Sentence": "It's going to be equal to x squared times e to the x minus 2 times all of this business. So minus 2 times, well, this antiderivative we just figured out is this. Minus 2 times x e to the x minus e to the x. And if we want now is a good time to put our plus c. And of course, we can simplify this. This is equal to x squared. I like to keep the same colors. This is equal to x squared e to the x."}, {"video_title": "Integration by parts \u00c2\u00bax____dx AP Calculus BC Khan Academy.mp3", "Sentence": "And if we want now is a good time to put our plus c. And of course, we can simplify this. This is equal to x squared. I like to keep the same colors. This is equal to x squared e to the x. Distribute the negative 2. You get minus 2x e to the x plus 2 e to the x. And then finally, plus c. And we're done."}, {"video_title": "Direct comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's think of two series. So let's say that I have this magenta series here. It's an infinite series from n equals one to infinity of a sub n. We're speaking in generalities here. And let's say I have another one. And that's the series b sub n from n equals one to infinity. And we know some things about these series. The first thing we know is that all of the terms in these series are non-negative."}, {"video_title": "Direct comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And let's say I have another one. And that's the series b sub n from n equals one to infinity. And we know some things about these series. The first thing we know is that all of the terms in these series are non-negative. So a sub n and b sub n are greater than or equal to zero, which tells us that these are either gonna diverge to positive infinity or they're going to converge to some finite value. They're not going to oscillate because you're not gonna have negative values here. You can't go to negative infinity because you don't have negative values here."}, {"video_title": "Direct comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "The first thing we know is that all of the terms in these series are non-negative. So a sub n and b sub n are greater than or equal to zero, which tells us that these are either gonna diverge to positive infinity or they're going to converge to some finite value. They're not going to oscillate because you're not gonna have negative values here. You can't go to negative infinity because you don't have negative values here. Now, let's say we also know that each of the corresponding terms in the first series are less than or equal to the corresponding term in the second series, less than or equal to b sub n. And once again, this is true for all the n's that we care about. So n equals one, two, three, all the way on and on and on. So the comparison test tells us that because this series, all of the corresponding terms of this series are less than the corresponding terms here, but they're greater than zero, that if this series converges, the one that's larger, if this one converges, well, then the one that is smaller than it, or I guess one way to think about it is kind of bounded by this one, must also converge."}, {"video_title": "Direct comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "You can't go to negative infinity because you don't have negative values here. Now, let's say we also know that each of the corresponding terms in the first series are less than or equal to the corresponding term in the second series, less than or equal to b sub n. And once again, this is true for all the n's that we care about. So n equals one, two, three, all the way on and on and on. So the comparison test tells us that because this series, all of the corresponding terms of this series are less than the corresponding terms here, but they're greater than zero, that if this series converges, the one that's larger, if this one converges, well, then the one that is smaller than it, or I guess one way to think about it is kind of bounded by this one, must also converge. And I'm not doing a formal proof here, but hopefully that gives you a little bit of intuition. So the comparison test tells us if, if, I guess what in my brain, the larger series, the one whose corresponding terms are at least as large as the ones here, if this one converges, if this one doesn't go unbounded towards infinity, it goes to, it sums to some finite value, then that tells us that the one that is, in some ways, I guess you could say smaller, must also converge, must also. So this one, this one right over here must also converge."}, {"video_title": "Direct comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So the comparison test tells us that because this series, all of the corresponding terms of this series are less than the corresponding terms here, but they're greater than zero, that if this series converges, the one that's larger, if this one converges, well, then the one that is smaller than it, or I guess one way to think about it is kind of bounded by this one, must also converge. And I'm not doing a formal proof here, but hopefully that gives you a little bit of intuition. So the comparison test tells us if, if, I guess what in my brain, the larger series, the one whose corresponding terms are at least as large as the ones here, if this one converges, if this one doesn't go unbounded towards infinity, it goes to, it sums to some finite value, then that tells us that the one that is, in some ways, I guess you could say smaller, must also converge, must also. So this one, this one right over here must also converge. So that must also converge. So why is that useful? And we'll see this in future videos."}, {"video_title": "Direct comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So this one, this one right over here must also converge. So that must also converge. So why is that useful? And we'll see this in future videos. Well, if you find, if you're looking, say you have your a sub n, and you're like, hey, gee, I wish I could prove that it converges, I kind of have a gut feeling it converges, the comparison test tells us, well, just find another series that is always, whose corresponding terms are at least as large as the corresponding terms here, and if you can prove that one converges, then you're good with this one. And of course, it would only apply to the case where your original series, each of the terms, each of the terms are non-negative. Now what if we went the other way around?"}, {"video_title": "Direct comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And we'll see this in future videos. Well, if you find, if you're looking, say you have your a sub n, and you're like, hey, gee, I wish I could prove that it converges, I kind of have a gut feeling it converges, the comparison test tells us, well, just find another series that is always, whose corresponding terms are at least as large as the corresponding terms here, and if you can prove that one converges, then you're good with this one. And of course, it would only apply to the case where your original series, each of the terms, each of the terms are non-negative. Now what if we went the other way around? What if you could prove that the magenta series, the smaller one, and I guess I could kind of put them in quotes, you know, this one right over here is the smaller, the smaller, I guess, each of its corresponding terms are smaller, what if you could prove this one diverges? Well, if this one diverges, it's going to go unbounded to infinity, it's not going to go to negative infinity, all the terms are positive, it's not gonna oscillate, it's not gonna diverge because it oscillates between two values, once again, if it's oscillating between values, that would, the only way you could do that is if you had negative terms here, so this would kind of be unbounded towards infinity. Well, if this one is unbounded, each of these corresponding terms are larger, so this one must also be unbounded, so let's write that down."}, {"video_title": "Direct comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Now what if we went the other way around? What if you could prove that the magenta series, the smaller one, and I guess I could kind of put them in quotes, you know, this one right over here is the smaller, the smaller, I guess, each of its corresponding terms are smaller, what if you could prove this one diverges? Well, if this one diverges, it's going to go unbounded to infinity, it's not going to go to negative infinity, all the terms are positive, it's not gonna oscillate, it's not gonna diverge because it oscillates between two values, once again, if it's oscillating between values, that would, the only way you could do that is if you had negative terms here, so this would kind of be unbounded towards infinity. Well, if this one is unbounded, each of these corresponding terms are larger, so this one must also be unbounded, so let's write that down. So the comparison test tells us if our smaller, smaller series diverges, if this one diverges, then the larger one must also diverge. Then the larger one must also diverge. And so once again, if you wanted to prove that this thing right over here is going to diverge, and if you have a, once again, and you know that all the b sub n's are greater than or equal to zero, you wanna prove it diverges, well, maybe you could try to find another series where each of the corresponding terms are less than the corresponding terms here, and you could prove this one diverges, then you would be all set."}, {"video_title": "Integrating sums of functions Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "This is the graph of y is equal to f of x. This is the graph of y is equal to g of x. And we already know something, or we know ways to represent the area under the curve y equals f of x between these two points, x is equal to a and x equal to b. So this area right over here, this area right over here, between the curve and the x axis, between x equals a and x equals b, we know we can write that as the definite integral from a to b of f of x dx. And we can do the same thing over here. We could call this area, let me pick a color that I have not used. Well, this is a slightly different green."}, {"video_title": "Integrating sums of functions Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So this area right over here, this area right over here, between the curve and the x axis, between x equals a and x equals b, we know we can write that as the definite integral from a to b of f of x dx. And we can do the same thing over here. We could call this area, let me pick a color that I have not used. Well, this is a slightly different green. I could call this area right over here, the area under the curve y is equal to g of x, and above the positive x axis between x equals a and x equals b, we could call that the definite integral from a to b of g of x dx. Now, given these two things, let's actually think about the area under the curve of the function created by the sum of these two functions. So what do I mean by that?"}, {"video_title": "Integrating sums of functions Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Well, this is a slightly different green. I could call this area right over here, the area under the curve y is equal to g of x, and above the positive x axis between x equals a and x equals b, we could call that the definite integral from a to b of g of x dx. Now, given these two things, let's actually think about the area under the curve of the function created by the sum of these two functions. So what do I mean by that? So let me, this is actually a fun thing to do. Let me start again. That's exactly what we have over here."}, {"video_title": "Integrating sums of functions Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So what do I mean by that? So let me, this is actually a fun thing to do. Let me start again. That's exactly what we have over here. This is the graph of y is equal to f of x. But what I want to do is I want to approximate the graph of y is equal to, so my goal is to graph y is equal to f of x, f of x plus g of x, plus g of x, plus g of x. So for any given x, it's going to be f of x, so that's the f of x, and I'm going to add the g of x to it."}, {"video_title": "Integrating sums of functions Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "That's exactly what we have over here. This is the graph of y is equal to f of x. But what I want to do is I want to approximate the graph of y is equal to, so my goal is to graph y is equal to f of x, f of x plus g of x, plus g of x, plus g of x. So for any given x, it's going to be f of x, so that's the f of x, and I'm going to add the g of x to it. So what would that look like? So that's going to look like, let's see, at when x is zero, g of x looks like it's about that length right there. I'm obviously approximating it, so I'm going to have to add that length right over here, so it'd probably be right around there."}, {"video_title": "Integrating sums of functions Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So for any given x, it's going to be f of x, so that's the f of x, and I'm going to add the g of x to it. So what would that look like? So that's going to look like, let's see, at when x is zero, g of x looks like it's about that length right there. I'm obviously approximating it, so I'm going to have to add that length right over here, so it'd probably be right around there. At x equals a, it's a little bit more, but now my f of x curve has gone more, has increased, but if I take that same distance above it, if I add the g of x there, it gets me right about there. Once again, I'm just eyeballing it, trying to get an approximation. Give you an intuition, actually, for what f of x plus g of x is."}, {"video_title": "Integrating sums of functions Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "I'm obviously approximating it, so I'm going to have to add that length right over here, so it'd probably be right around there. At x equals a, it's a little bit more, but now my f of x curve has gone more, has increased, but if I take that same distance above it, if I add the g of x there, it gets me right about there. Once again, I'm just eyeballing it, trying to get an approximation. Give you an intuition, actually, for what f of x plus g of x is. I'm just trying to add g of x for a given x. Now let's see, if I'm a little bit, let's say that I'm between a and b, g of x is about that distance right over there. So if I wanted to put that same distance right over here, it gets me right about, it gets me right about there."}, {"video_title": "Integrating sums of functions Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Give you an intuition, actually, for what f of x plus g of x is. I'm just trying to add g of x for a given x. Now let's see, if I'm a little bit, let's say that I'm between a and b, g of x is about that distance right over there. So if I wanted to put that same distance right over here, it gets me right about, it gets me right about there. And then when x is equal to b, g of x is about that big, so I have to add that length, which is about, which looks something like that. That actually looks like a little bit too much. Maybe something like that."}, {"video_title": "Integrating sums of functions Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So if I wanted to put that same distance right over here, it gets me right about, it gets me right about there. And then when x is equal to b, g of x is about that big, so I have to add that length, which is about, which looks something like that. That actually looks like a little bit too much. Maybe something like that. So if I were to add the two, if I were to add the two, I get a curve that looks something like this. That looks something like this, and maybe it just keeps on going higher and higher. So this is the curve, or it's a pretty good approximation of the curve of f of x plus g of x."}, {"video_title": "Integrating sums of functions Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Maybe something like that. So if I were to add the two, if I were to add the two, I get a curve that looks something like this. That looks something like this, and maybe it just keeps on going higher and higher. So this is the curve, or it's a pretty good approximation of the curve of f of x plus g of x. Now an interesting question is, is what would be, well, we know how we can represent this area. So the area under the curve, f of x plus g of x, above the positive x-axis, between x equals a and x equals b, we know we can represent that as, let me see, I have not used pink yet. So this area right over here, we know that that could be represented as the definite integral from a to b of f of x plus g of x dx."}, {"video_title": "Integrating sums of functions Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So this is the curve, or it's a pretty good approximation of the curve of f of x plus g of x. Now an interesting question is, is what would be, well, we know how we can represent this area. So the area under the curve, f of x plus g of x, above the positive x-axis, between x equals a and x equals b, we know we can represent that as, let me see, I have not used pink yet. So this area right over here, we know that that could be represented as the definite integral from a to b of f of x plus g of x dx. Now the question is, how does this thing relate to d, or how does this area relate to these areas right over here? Well, the important thing to realize is this area that we have in yellow, that's going to be this area right over here. That one's pretty clear."}, {"video_title": "Integrating sums of functions Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So this area right over here, we know that that could be represented as the definite integral from a to b of f of x plus g of x dx. Now the question is, how does this thing relate to d, or how does this area relate to these areas right over here? Well, the important thing to realize is this area that we have in yellow, that's going to be this area right over here. That one's pretty clear. But how does this area in green relate to this area there? And to think about that, we just have to think about what does an integral mean? Remember, what does it represent?"}, {"video_title": "Integrating sums of functions Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "That one's pretty clear. But how does this area in green relate to this area there? And to think about that, we just have to think about what does an integral mean? Remember, what does it represent? We've already thought about these like these really small rectangles. We're taking the sum of an infinitely, or the limit as we get an infinite number of these infinitely thin rectangles. But when we're thinking about Riemann sums, we're thinking about, okay, we have some change in x, and then you multiply it times essentially the height, which is going to be the value of the function at that point."}, {"video_title": "Integrating sums of functions Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Remember, what does it represent? We've already thought about these like these really small rectangles. We're taking the sum of an infinitely, or the limit as we get an infinite number of these infinitely thin rectangles. But when we're thinking about Riemann sums, we're thinking about, okay, we have some change in x, and then you multiply it times essentially the height, which is going to be the value of the function at that point. Well, over here, you could have the same change in x, you could have the exact same change in x, and what is the height right over here? Well, that's going to be this exact height right over here. We saw that when we constructed it."}, {"video_title": "Integrating sums of functions Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "But when we're thinking about Riemann sums, we're thinking about, okay, we have some change in x, and then you multiply it times essentially the height, which is going to be the value of the function at that point. Well, over here, you could have the same change in x, you could have the exact same change in x, and what is the height right over here? Well, that's going to be this exact height right over here. We saw that when we constructed it. This is going to be the g of x at that value. So even though the rectangles look like they're kind of shifted around a little bit, and they're actually all shifted up by the f of x, the heights of these rectangles that I'm drawing right over here are exactly the same thing as the heights of the rectangles that I'm drawing over here. They, once again, they are all just shifted up and down by this f of x function, but these are the exact same rectangles, or they have the exact same heights."}, {"video_title": "Integrating sums of functions Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "We saw that when we constructed it. This is going to be the g of x at that value. So even though the rectangles look like they're kind of shifted around a little bit, and they're actually all shifted up by the f of x, the heights of these rectangles that I'm drawing right over here are exactly the same thing as the heights of the rectangles that I'm drawing over here. They, once again, they are all just shifted up and down by this f of x function, but these are the exact same rectangles, or they have the exact same heights. And the limit as you get more and more of these by making them thinner and thinner is going to be the same as the limit as you get more and more of these as you get thinner and thinner. And so this area right over here, and I'm obviously not doing a rigorous proof, I'm giving you the intuition for it, is the exact same thing as this area right over here. So the area under this curve, the definite integral from a to b of f of x plus g of x dx, is just going to be the sum of these two, is just going to be the sum of these two definite integrals."}, {"video_title": "Integrating sums of functions Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "They, once again, they are all just shifted up and down by this f of x function, but these are the exact same rectangles, or they have the exact same heights. And the limit as you get more and more of these by making them thinner and thinner is going to be the same as the limit as you get more and more of these as you get thinner and thinner. And so this area right over here, and I'm obviously not doing a rigorous proof, I'm giving you the intuition for it, is the exact same thing as this area right over here. So the area under this curve, the definite integral from a to b of f of x plus g of x dx, is just going to be the sum of these two, is just going to be the sum of these two definite integrals. And you might say, oh, this is obvious, or maybe it's not so obvious, but when is this actually useful? Well, as you later actually learn to evaluate these integrals, you'll see that one of the most powerful ideas is being able to decompose them in this way, to say, okay, if I'm taking the definite integral from zero to one of x squared plus sine of x, which you may or may not have learned to do so far, you can at least start to break this down. You can say, okay, well, this is going to be the same thing as the integral from zero to one of x squared dx, plus the integral from zero to one of sine of x dx."}, {"video_title": "Power series intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "And the most common one that you will see in your mathematical careers is the power series. Is the power series. And I'm about to write a general case of the power series. So I can imagine a function, f of x, being defined as the infinite sum, so going from n equals zero to infinity, of a sub n, so a sub n is just going to be the coefficient on each term, times our variable x minus some constant c. You can almost imagine this is shifting our function to the nth power. So if I were to expand this out, I have my first term's coefficient, a sub zero, times x minus c to the zeroth power, plus a sub one times x minus c to the first power. This one, of course, will simplify to just a sub zero. This would simplify to a sub one times x minus c, plus a sub two times x minus c squared."}, {"video_title": "Power series intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "So I can imagine a function, f of x, being defined as the infinite sum, so going from n equals zero to infinity, of a sub n, so a sub n is just going to be the coefficient on each term, times our variable x minus some constant c. You can almost imagine this is shifting our function to the nth power. So if I were to expand this out, I have my first term's coefficient, a sub zero, times x minus c to the zeroth power, plus a sub one times x minus c to the first power. This one, of course, will simplify to just a sub zero. This would simplify to a sub one times x minus c, plus a sub two times x minus c squared. And I could just keep going on and on and on. And when you see this, you might say, well, aren't our geometric series, don't those look like a special case of a power series? If our common ratio was an x instead of an r in that case, or if our common ratio was a variable, I guess I could say, and you would be right."}, {"video_title": "Power series intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "This would simplify to a sub one times x minus c, plus a sub two times x minus c squared. And I could just keep going on and on and on. And when you see this, you might say, well, aren't our geometric series, don't those look like a special case of a power series? If our common ratio was an x instead of an r in that case, or if our common ratio was a variable, I guess I could say, and you would be right. That absolutely would be the case. So a geometric series. So let's just think about defining a function in terms of a geometric series."}, {"video_title": "Power series intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "If our common ratio was an x instead of an r in that case, or if our common ratio was a variable, I guess I could say, and you would be right. That absolutely would be the case. So a geometric series. So let's just think about defining a function in terms of a geometric series. And of course, we don't have to use x all the time as the independent variable, but this is kind of the most typical convention. I guess we could also use r as an independent variable if we wanted as well. But let's imagine a function g of x."}, {"video_title": "Power series intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's just think about defining a function in terms of a geometric series. And of course, we don't have to use x all the time as the independent variable, but this is kind of the most typical convention. I guess we could also use r as an independent variable if we wanted as well. But let's imagine a function g of x. We could have g of r if we wanted, but let's do g of x is equal to the sum from n equals zero to infinity of a times x to the n. So this is kind of a typical geometric series here. And what's the difference between this and this? Well, the difference is, is here, for every term, we're gonna have the same coefficient a, while over here we have a sub n. We're multiplying by a different thing."}, {"video_title": "Power series intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "But let's imagine a function g of x. We could have g of r if we wanted, but let's do g of x is equal to the sum from n equals zero to infinity of a times x to the n. So this is kind of a typical geometric series here. And what's the difference between this and this? Well, the difference is, is here, for every term, we're gonna have the same coefficient a, while over here we have a sub n. We're multiplying by a different thing. Every time up here, we're multiplying by the same thing over here. And in this case, this particular geometric series I just made, instead of having x minus c to the n, we have just x to the n. So you could say, well, this is a special case when c is equal to zero. And we can expand it out."}, {"video_title": "Power series intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, the difference is, is here, for every term, we're gonna have the same coefficient a, while over here we have a sub n. We're multiplying by a different thing. Every time up here, we're multiplying by the same thing over here. And in this case, this particular geometric series I just made, instead of having x minus c to the n, we have just x to the n. So you could say, well, this is a special case when c is equal to zero. And we can expand it out. This is a times x to the zero, which is just gonna be a, plus a times x to the first, plus a times x squared, and we just go on and on and on forever. Now, what's exciting about this is we know, we know under what, we know that this, under certain conditions, this will actually give us a finite value. This will actually converge."}, {"video_title": "Power series intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "And we can expand it out. This is a times x to the zero, which is just gonna be a, plus a times x to the first, plus a times x squared, and we just go on and on and on forever. Now, what's exciting about this is we know, we know under what, we know that this, under certain conditions, this will actually give us a finite value. This will actually converge. This will actually, I guess, give us a sensical answer. So under what conditions does that happen? Well, this converges as if each of these terms get smaller and smaller and smaller, and each of these terms get smaller and smaller and smaller if the absolute value of our common ratio, the absolute value of our common ratio is less than one."}, {"video_title": "Power series intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "This will actually converge. This will actually, I guess, give us a sensical answer. So under what conditions does that happen? Well, this converges as if each of these terms get smaller and smaller and smaller, and each of these terms get smaller and smaller and smaller if the absolute value of our common ratio, the absolute value of our common ratio is less than one. So let me write that down. So this converges, converges if the absolute value of our common ratio is less than one. Or another way of thinking about it, this is another way of saying, this is another way of saying is that x is in the interval between, it's less than one and it is greater than negative one."}, {"video_title": "Power series intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, this converges as if each of these terms get smaller and smaller and smaller, and each of these terms get smaller and smaller and smaller if the absolute value of our common ratio, the absolute value of our common ratio is less than one. So let me write that down. So this converges, converges if the absolute value of our common ratio is less than one. Or another way of thinking about it, this is another way of saying, this is another way of saying is that x is in the interval between, it's less than one and it is greater than negative one. And this term right over here, now x is a variable. X can vary between those values. We're defined a function in terms of x."}, {"video_title": "Power series intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "Or another way of thinking about it, this is another way of saying, this is another way of saying is that x is in the interval between, it's less than one and it is greater than negative one. And this term right over here, now x is a variable. X can vary between those values. We're defined a function in terms of x. We call this the interval of convergence. Interval, interval of convergence. And so we know that if x is in this interval, this is going to give us a finite sum, and we know what that finite sum is."}, {"video_title": "Power series intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "We're defined a function in terms of x. We call this the interval of convergence. Interval, interval of convergence. And so we know that if x is in this interval, this is going to give us a finite sum, and we know what that finite sum is. It's going to be, it's going to be equal to, if it converges, so if it converges, this is going to be equal to our first term, which is just a, this simplifies to a right over here, over one minus our common ratio. What's our common ratio? Well our common ratio in this example is x."}, {"video_title": "Power series intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so we know that if x is in this interval, this is going to give us a finite sum, and we know what that finite sum is. It's going to be, it's going to be equal to, if it converges, so if it converges, this is going to be equal to our first term, which is just a, this simplifies to a right over here, over one minus our common ratio. What's our common ratio? Well our common ratio in this example is x. Going from one term to the next, we're just multiplying by x. We're just multiplying by x right over there. Now this is pretty neat, because we're going to be able to use this fact to put more traditionally defined functions into this form, and then try to expand them out using a geometric series."}, {"video_title": "Power series intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well our common ratio in this example is x. Going from one term to the next, we're just multiplying by x. We're just multiplying by x right over there. Now this is pretty neat, because we're going to be able to use this fact to put more traditionally defined functions into this form, and then try to expand them out using a geometric series. And this whole idea of using power series, or in this special case, geometric series to represent functions, has all sorts of applications in engineering and finance, using a finite number of terms of these, of these series. You can kind of approximate the functions in a way that's simpler for the human brain to understand, or maybe a simpler way to manipulate in some way. But what's interesting here, is instead of just going from the sum to, instead of going from this expanded out version to this kind of finite value, we're now going to start being able to take something in this form, and expand it out into a geometric series."}, {"video_title": "Power series intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "Now this is pretty neat, because we're going to be able to use this fact to put more traditionally defined functions into this form, and then try to expand them out using a geometric series. And this whole idea of using power series, or in this special case, geometric series to represent functions, has all sorts of applications in engineering and finance, using a finite number of terms of these, of these series. You can kind of approximate the functions in a way that's simpler for the human brain to understand, or maybe a simpler way to manipulate in some way. But what's interesting here, is instead of just going from the sum to, instead of going from this expanded out version to this kind of finite value, we're now going to start being able to take something in this form, and expand it out into a geometric series. But we have to be careful to make sure that we're only doing it over the interval of convergence. This is only going to be true over the interval of convergence. Now one other term you might see in your mathematical career, is radius, radius of convergence, convergence."}, {"video_title": "Power series intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "But what's interesting here, is instead of just going from the sum to, instead of going from this expanded out version to this kind of finite value, we're now going to start being able to take something in this form, and expand it out into a geometric series. But we have to be careful to make sure that we're only doing it over the interval of convergence. This is only going to be true over the interval of convergence. Now one other term you might see in your mathematical career, is radius, radius of convergence, convergence. And this is how far, up to what value, but not including this value. So as long as our, as long as our x value stays no, stays less than a certain amount from our c value, then this thing will converge. Now in this case, our c value is zero."}, {"video_title": "Power series intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "Now one other term you might see in your mathematical career, is radius, radius of convergence, convergence. And this is how far, up to what value, but not including this value. So as long as our, as long as our x value stays no, stays less than a certain amount from our c value, then this thing will converge. Now in this case, our c value is zero. So our, so we could ask ourselves a question of, what is, how, as long as x stays within some value of zero, this thing is going to converge. Well you see it right over here. As long as x stays within one of zero, so as long as it can get, it can't go all the way to one, but as long as it stays less than one, or as long as it stays greater than negative one."}, {"video_title": "Power series intro Series AP Calculus BC Khan Academy.mp3", "Sentence": "Now in this case, our c value is zero. So our, so we could ask ourselves a question of, what is, how, as long as x stays within some value of zero, this thing is going to converge. Well you see it right over here. As long as x stays within one of zero, so as long as it can get, it can't go all the way to one, but as long as it stays less than one, or as long as it stays greater than negative one. So it can stray no more, or I guess it should, it has to, it can stray anything less than one away from zero, either in the positive direction or the negative direction, then this thing will still converge. So we could say that our radius of convergence, our radius of convergence is equal to one. Another way to think about it, our interval of convergence, we're going from negative one to one, not including those two boundaries, so our interval is two, so our radius of convergence is half of that."}, {"video_title": "Types of discontinuities Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So let's first review the classification of discontinuities. So here on the left, you see that this curve looks just like y equals x squared until we get to x equals three, and instead of it being three squared, at this point, you have this opening, and instead, the function at three is defined at four, but then it keeps going, and it looks just like y equals x squared. This is known as a point or a removable discontinuity, and it's called that for obvious reasons. You're discontinuous at that point. You might imagine of defining or redefining the function at that point, so it is continuous, so the discontinuity is removable, but then how does this relate to our definition of continuity? Well, let's remind ourselves our definition of continuity. We say f is continuous, continuous, if and only if, or let me write f continuous at x equals c, if and only if the limit as x approaches c of f of x is equal to the actual value of the function when x is equal to c. So why does this one fail?"}, {"video_title": "Types of discontinuities Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "You're discontinuous at that point. You might imagine of defining or redefining the function at that point, so it is continuous, so the discontinuity is removable, but then how does this relate to our definition of continuity? Well, let's remind ourselves our definition of continuity. We say f is continuous, continuous, if and only if, or let me write f continuous at x equals c, if and only if the limit as x approaches c of f of x is equal to the actual value of the function when x is equal to c. So why does this one fail? Well, the two-sided limit actually exists. You could find, if we say c in this case is three, the limit as x approaches three of f of x, it looks like, if you graphically inspect this, and I actually know this is the graph of y equals x squared except at that discontinuity right over there, this is equal to nine, but the issue is is the way this graph has been depicted, this is not the same thing as the value of the function. This function, f of three, the way it's been graphed, f of three is equal to four."}, {"video_title": "Types of discontinuities Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "We say f is continuous, continuous, if and only if, or let me write f continuous at x equals c, if and only if the limit as x approaches c of f of x is equal to the actual value of the function when x is equal to c. So why does this one fail? Well, the two-sided limit actually exists. You could find, if we say c in this case is three, the limit as x approaches three of f of x, it looks like, if you graphically inspect this, and I actually know this is the graph of y equals x squared except at that discontinuity right over there, this is equal to nine, but the issue is is the way this graph has been depicted, this is not the same thing as the value of the function. This function, f of three, the way it's been graphed, f of three is equal to four. So this is a situation where this two-sided limit exists, but it's not equal to the value of that function. You might see other circumstances where the function isn't even defined there, so that isn't even there, and so once again, the limit might exist, but the function might not be defined there, so in either case, you aren't going to meet this criteria for continuity, and so that's how a point or removable discontinuity, why it is discontinuous, with regards to our limit definition of continuity. So now let's look at this second example."}, {"video_title": "Types of discontinuities Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "This function, f of three, the way it's been graphed, f of three is equal to four. So this is a situation where this two-sided limit exists, but it's not equal to the value of that function. You might see other circumstances where the function isn't even defined there, so that isn't even there, and so once again, the limit might exist, but the function might not be defined there, so in either case, you aren't going to meet this criteria for continuity, and so that's how a point or removable discontinuity, why it is discontinuous, with regards to our limit definition of continuity. So now let's look at this second example. If we looked at our intuitive continuity test, if we were just trying to trace this thing, we see that once we get to x equals two, I have to pick up my pencil to keep tracing it, and so that's a pretty good sign that we are discontinuous. We see that over here as well. If I'm tracing this function, I gotta pick up my pencil to, I can't go through that point or I have to jump down here and then keep going right over there, so in either case, I have to pick up my pencil, and so intuitively, it is discontinuous, but this particular type of discontinuity, where I'm making a jump from one point and then I'm making a jump down here to continue, it is intuitively called a jump discontinuity."}, {"video_title": "Types of discontinuities Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So now let's look at this second example. If we looked at our intuitive continuity test, if we were just trying to trace this thing, we see that once we get to x equals two, I have to pick up my pencil to keep tracing it, and so that's a pretty good sign that we are discontinuous. We see that over here as well. If I'm tracing this function, I gotta pick up my pencil to, I can't go through that point or I have to jump down here and then keep going right over there, so in either case, I have to pick up my pencil, and so intuitively, it is discontinuous, but this particular type of discontinuity, where I'm making a jump from one point and then I'm making a jump down here to continue, it is intuitively called a jump discontinuity. Discontinuity, and this is, of course, a point or removable discontinuity. And so how does this relate to limits? Well, here, the left and right-handed limits exist, but they're not the same thing, so you don't have a two-sided limit."}, {"video_title": "Types of discontinuities Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "If I'm tracing this function, I gotta pick up my pencil to, I can't go through that point or I have to jump down here and then keep going right over there, so in either case, I have to pick up my pencil, and so intuitively, it is discontinuous, but this particular type of discontinuity, where I'm making a jump from one point and then I'm making a jump down here to continue, it is intuitively called a jump discontinuity. Discontinuity, and this is, of course, a point or removable discontinuity. And so how does this relate to limits? Well, here, the left and right-handed limits exist, but they're not the same thing, so you don't have a two-sided limit. So for example, for this one in particular, but for all the x values up to and including x equals two, this is the graph of y equals x squared, and then for x greater than two, it's the graph of square root of x. So in this scenario, if you were to take the limit of f of x as x approaches two from the left, from the left, this is going to be equal to four. You're approaching this value, and that actually is the value of the function, but if you were to take the limit as x approaches two from the right of f of x, what is that going to be equal to?"}, {"video_title": "Types of discontinuities Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Well, here, the left and right-handed limits exist, but they're not the same thing, so you don't have a two-sided limit. So for example, for this one in particular, but for all the x values up to and including x equals two, this is the graph of y equals x squared, and then for x greater than two, it's the graph of square root of x. So in this scenario, if you were to take the limit of f of x as x approaches two from the left, from the left, this is going to be equal to four. You're approaching this value, and that actually is the value of the function, but if you were to take the limit as x approaches two from the right of f of x, what is that going to be equal to? Well, we're approaching from the right, this is actually the square root of x, so it's approaching the square root of two. You can't, you wouldn't know it's the square root of two just by looking at this. I know that just because when I went onto Desmos and defined the function, that's the function that I used."}, {"video_title": "Types of discontinuities Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "You're approaching this value, and that actually is the value of the function, but if you were to take the limit as x approaches two from the right of f of x, what is that going to be equal to? Well, we're approaching from the right, this is actually the square root of x, so it's approaching the square root of two. You can't, you wouldn't know it's the square root of two just by looking at this. I know that just because when I went onto Desmos and defined the function, that's the function that I used. But it's clear, even visually, that you're approaching two different values when you approach from the left than when you approach from the right. So even though the one-sided limits exist, they're not approaching the same thing, so the two-sided limit doesn't exist, and if the two-sided limit doesn't exist, it for sure cannot be equal to the value of the function there, even if the function is defined. So that's why the jump discontinuity is failing this test."}, {"video_title": "Types of discontinuities Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "I know that just because when I went onto Desmos and defined the function, that's the function that I used. But it's clear, even visually, that you're approaching two different values when you approach from the left than when you approach from the right. So even though the one-sided limits exist, they're not approaching the same thing, so the two-sided limit doesn't exist, and if the two-sided limit doesn't exist, it for sure cannot be equal to the value of the function there, even if the function is defined. So that's why the jump discontinuity is failing this test. And once again, it's intuitive. You're seeing that, hey, I gotta jump, I gotta pick up my pencil. These two things are not connected to each other."}, {"video_title": "Types of discontinuities Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So that's why the jump discontinuity is failing this test. And once again, it's intuitive. You're seeing that, hey, I gotta jump, I gotta pick up my pencil. These two things are not connected to each other. Finally, what you see here is, when you learned precalculus, often known as an asymptotic discontinuity, asymptotic, asymptotic discontinuity, discontinuity, and intuitively, you have an asymptote here. If you, it's a vertical asymptote at x equals two. If I were to try to trace the graph from the left, I'd be, I would just keep on going."}, {"video_title": "Types of discontinuities Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "These two things are not connected to each other. Finally, what you see here is, when you learned precalculus, often known as an asymptotic discontinuity, asymptotic, asymptotic discontinuity, discontinuity, and intuitively, you have an asymptote here. If you, it's a vertical asymptote at x equals two. If I were to try to trace the graph from the left, I'd be, I would just keep on going. In fact, I would be doing it forever because it's, it would be, it would be infinitely, it would be unbounded as I get closer and closer to x equals two from the left. And if I get to, try to get to x equals two from the right, once again, I get unbounded up. But even if I could, and when I say it's unbounded, it goes to infinity, so it's actually impossible in a mortal's lifespan to try to trace the whole thing, but you get the sense that, hey, there's no way that I could draw from here to here without picking up my pencil."}, {"video_title": "Types of discontinuities Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "If I were to try to trace the graph from the left, I'd be, I would just keep on going. In fact, I would be doing it forever because it's, it would be, it would be infinitely, it would be unbounded as I get closer and closer to x equals two from the left. And if I get to, try to get to x equals two from the right, once again, I get unbounded up. But even if I could, and when I say it's unbounded, it goes to infinity, so it's actually impossible in a mortal's lifespan to try to trace the whole thing, but you get the sense that, hey, there's no way that I could draw from here to here without picking up my pencil. And if you wanna relate it to our notion of limits, it's that both the left and right-handed limits are unbounded, so they officially don't exist. So if they don't exist, then we can't meet these conditions. So if I were to say the limit as x approaches two from the left-hand side of f of x, we can see that it goes unbounded in the negative direction."}, {"video_title": "Types of discontinuities Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "But even if I could, and when I say it's unbounded, it goes to infinity, so it's actually impossible in a mortal's lifespan to try to trace the whole thing, but you get the sense that, hey, there's no way that I could draw from here to here without picking up my pencil. And if you wanna relate it to our notion of limits, it's that both the left and right-handed limits are unbounded, so they officially don't exist. So if they don't exist, then we can't meet these conditions. So if I were to say the limit as x approaches two from the left-hand side of f of x, we can see that it goes unbounded in the negative direction. You might sometimes see someone write something like this, negative infinity, but that's a little hand-wavy with the math. The more correct way to say it, it's just unbounded. Unbounded."}, {"video_title": "Types of discontinuities Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So if I were to say the limit as x approaches two from the left-hand side of f of x, we can see that it goes unbounded in the negative direction. You might sometimes see someone write something like this, negative infinity, but that's a little hand-wavy with the math. The more correct way to say it, it's just unbounded. Unbounded. And likewise, if we thought about the limit as x approaches two from the right of f of x, it is now unbounded towards positive infinity. So this, once again, this is also unbounded. And because it's unbounded and this limit does not exist, it can't meet these conditions, and so we are going to be discontinuous."}, {"video_title": "Formal definition of limits Part 2 building the idea AP Calculus AB Khan Academy.mp3", "Sentence": "Let's try to come up with a mathematically rigorous definition for what this statement means, the statement that the limit of f of x as x approaches c is equal to L. So let's say that this means that you can get f of x as close to L as you want. I'll put that in quotes right over here because it's kind of a little loosey-goosey is how close is that. But as close as you want by getting x sufficiently close to c. So another way of saying this is if you tell me, hey, I want to get my f of x to be within 0.5 of this limit, then you're telling me if this limit is actually true, you should be able to hand me a value around c that if x is within that range, that f of x is definitely going to be as close to L as I desire. So let me draw that out to make it a little bit clearer. And I'm going to zoom in. I'll draw another diagram. So let's say that this right over here is my y-axis."}, {"video_title": "Formal definition of limits Part 2 building the idea AP Calculus AB Khan Academy.mp3", "Sentence": "So let me draw that out to make it a little bit clearer. And I'm going to zoom in. I'll draw another diagram. So let's say that this right over here is my y-axis. And I'm going to zoom in. I'm going to draw a slightly different function just so we can really focus on what's going on around here, the ranges around c and the ranges around L. So that's x. This right over here is y."}, {"video_title": "Formal definition of limits Part 2 building the idea AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say that this right over here is my y-axis. And I'm going to zoom in. I'm going to draw a slightly different function just so we can really focus on what's going on around here, the ranges around c and the ranges around L. So that's x. This right over here is y. Let's say that this is c. And let's just zoom in on our function. So let's say our function is doing something like, let's say it does something like, let's see, I don't want it to be defined at c, at least just for the, it could be. You can always find a limit even where it is defined."}, {"video_title": "Formal definition of limits Part 2 building the idea AP Calculus AB Khan Academy.mp3", "Sentence": "This right over here is y. Let's say that this is c. And let's just zoom in on our function. So let's say our function is doing something like, let's say it does something like, let's see, I don't want it to be defined at c, at least just for the, it could be. You can always find a limit even where it is defined. But let's say our function looks something like that. And it can even have a little kink in it the way I drew it. So it looks something like this."}, {"video_title": "Formal definition of limits Part 2 building the idea AP Calculus AB Khan Academy.mp3", "Sentence": "You can always find a limit even where it is defined. But let's say our function looks something like that. And it can even have a little kink in it the way I drew it. So it looks something like this. It's undefined. Let me draw it a little bit different. So it is undefined when x is equal to c. So this is the point where there is a hole."}, {"video_title": "Formal definition of limits Part 2 building the idea AP Calculus AB Khan Academy.mp3", "Sentence": "So it looks something like this. It's undefined. Let me draw it a little bit different. So it is undefined when x is equal to c. So this is the point where there is a hole. It is undefined when x is equal to c. And it even has a little kink in it just like that. And what we want to do is prove that the limit of f of x, and let me make it clear, this is the graph of y is equal to f of x. We want to get an idea for what this definition is saying if we're claiming that the limit of f of x as x approaches c is L. So conceptually, we get the gist already."}, {"video_title": "Formal definition of limits Part 2 building the idea AP Calculus AB Khan Academy.mp3", "Sentence": "So it is undefined when x is equal to c. So this is the point where there is a hole. It is undefined when x is equal to c. And it even has a little kink in it just like that. And what we want to do is prove that the limit of f of x, and let me make it clear, this is the graph of y is equal to f of x. We want to get an idea for what this definition is saying if we're claiming that the limit of f of x as x approaches c is L. So conceptually, we get the gist already. We already get the gist that this right over here is L. But what is this definition saying? What's saying that you can get f of x as close to L as you want. So if you tell someone, I want to get f of x within a certain range of L, then if this limit is actually true, if the limit of f of x as x approaches c really is equal to L, then they should be able to find a range around c that as long as x is around that range, your f of x is going to be in the range that you want."}, {"video_title": "Formal definition of limits Part 2 building the idea AP Calculus AB Khan Academy.mp3", "Sentence": "We want to get an idea for what this definition is saying if we're claiming that the limit of f of x as x approaches c is L. So conceptually, we get the gist already. We already get the gist that this right over here is L. But what is this definition saying? What's saying that you can get f of x as close to L as you want. So if you tell someone, I want to get f of x within a certain range of L, then if this limit is actually true, if the limit of f of x as x approaches c really is equal to L, then they should be able to find a range around c that as long as x is around that range, your f of x is going to be in the range that you want. So let me actually go through that exercise. It really is a little bit like a game. So someone comes up to you and says, well, OK, I don't necessarily believe that you're claiming that the limit of f of x as x approaches c is equal to L. I'm not really sure if that's the case."}, {"video_title": "Formal definition of limits Part 2 building the idea AP Calculus AB Khan Academy.mp3", "Sentence": "So if you tell someone, I want to get f of x within a certain range of L, then if this limit is actually true, if the limit of f of x as x approaches c really is equal to L, then they should be able to find a range around c that as long as x is around that range, your f of x is going to be in the range that you want. So let me actually go through that exercise. It really is a little bit like a game. So someone comes up to you and says, well, OK, I don't necessarily believe that you're claiming that the limit of f of x as x approaches c is equal to L. I'm not really sure if that's the case. But I agree with this definition. So I want to get f of x within 0.5 of L. So this right over here would be L plus 0.5. And this right over here is L minus 0.5."}, {"video_title": "Formal definition of limits Part 2 building the idea AP Calculus AB Khan Academy.mp3", "Sentence": "So someone comes up to you and says, well, OK, I don't necessarily believe that you're claiming that the limit of f of x as x approaches c is equal to L. I'm not really sure if that's the case. But I agree with this definition. So I want to get f of x within 0.5 of L. So this right over here would be L plus 0.5. And this right over here is L minus 0.5. And then you say, fine, I'm going to give you a range around c that if you take any x within that range, your f of x is always going to fall in this range that you care about. And so you look at this. And obviously, we haven't explicitly defined this function."}, {"video_title": "Formal definition of limits Part 2 building the idea AP Calculus AB Khan Academy.mp3", "Sentence": "And this right over here is L minus 0.5. And then you say, fine, I'm going to give you a range around c that if you take any x within that range, your f of x is always going to fall in this range that you care about. And so you look at this. And obviously, we haven't explicitly defined this function. But you can even eyeball it the way this function is defined. It won't be that easy for all functions. But you look at it like this."}, {"video_title": "Formal definition of limits Part 2 building the idea AP Calculus AB Khan Academy.mp3", "Sentence": "And obviously, we haven't explicitly defined this function. But you can even eyeball it the way this function is defined. It won't be that easy for all functions. But you look at it like this. And you say that this value, just the way it's drawn right over here, let's say that this is c minus 0.25. And let's say that this value right over here is c plus 0.25. And so you tell them, look, as long as you get x within 0.25 of c, so as long as your x's are sitting someplace over here, the corresponding f of x is going to sit in the range that you care about."}, {"video_title": "Formal definition of limits Part 2 building the idea AP Calculus AB Khan Academy.mp3", "Sentence": "But you look at it like this. And you say that this value, just the way it's drawn right over here, let's say that this is c minus 0.25. And let's say that this value right over here is c plus 0.25. And so you tell them, look, as long as you get x within 0.25 of c, so as long as your x's are sitting someplace over here, the corresponding f of x is going to sit in the range that you care about. And you say, OK, fine. You won that round. Let me make it even tighter."}, {"video_title": "Formal definition of limits Part 2 building the idea AP Calculus AB Khan Academy.mp3", "Sentence": "And so you tell them, look, as long as you get x within 0.25 of c, so as long as your x's are sitting someplace over here, the corresponding f of x is going to sit in the range that you care about. And you say, OK, fine. You won that round. Let me make it even tighter. Maybe instead of saying within 0.5, I want to get within 0.05. And then you'd have to do this exercise again and find another range. And in order for this to be true, you have to be able to do this for any range that they give you."}, {"video_title": "Formal definition of limits Part 2 building the idea AP Calculus AB Khan Academy.mp3", "Sentence": "Let me make it even tighter. Maybe instead of saying within 0.5, I want to get within 0.05. And then you'd have to do this exercise again and find another range. And in order for this to be true, you have to be able to do this for any range that they give you. For any range around L that they give you, you have to be able to get f of x within that range by finding a range around c that as long as x is that range around c, f of x is going to sit within that range. I'll let you think about that a little bit. There's a lot to think about."}, {"video_title": "Formal definition of limits Part 2 building the idea AP Calculus AB Khan Academy.mp3", "Sentence": "And in order for this to be true, you have to be able to do this for any range that they give you. For any range around L that they give you, you have to be able to get f of x within that range by finding a range around c that as long as x is that range around c, f of x is going to sit within that range. I'll let you think about that a little bit. There's a lot to think about. But hopefully this made sense. We did it for the particular example of someone hands you the 0.5. I want f of x within 0.5 of L. And you say, well, as long as x is within 0.25 of c, you're going to match it."}, {"video_title": "Formal definition of limits Part 2 building the idea AP Calculus AB Khan Academy.mp3", "Sentence": "There's a lot to think about. But hopefully this made sense. We did it for the particular example of someone hands you the 0.5. I want f of x within 0.5 of L. And you say, well, as long as x is within 0.25 of c, you're going to match it. You need to be able to do that for any range they give you around L. And then this limit will definitely be true. So in the next video, we will now generalize that. And that will really bring us to the famous epsilon delta definition of limits."}, {"video_title": "Product rule example.mp3", "Sentence": "So when you look at this, you might say, well I know how to find the derivative with respect of e to the x. That's in fact just e to the x. And let me write this down. We know a few things. We know the derivative with respect to x of e to the x. E to the x is e to the x. We know how to find the derivative of cosine of x. The derivative with respect to x of cosine of x is equal to negative sine of x."}, {"video_title": "Product rule example.mp3", "Sentence": "We know a few things. We know the derivative with respect to x of e to the x. E to the x is e to the x. We know how to find the derivative of cosine of x. The derivative with respect to x of cosine of x is equal to negative sine of x. But how do we find the derivative of their product? Well as you can imagine, this might involve the product rule. And let me just write down the product rule generally first."}, {"video_title": "Product rule example.mp3", "Sentence": "The derivative with respect to x of cosine of x is equal to negative sine of x. But how do we find the derivative of their product? Well as you can imagine, this might involve the product rule. And let me just write down the product rule generally first. So if we would take the derivative with respect to x of the first expression in terms of x. So this is, we could call this u of x u of x times another expression that involves x. So u times v of x."}, {"video_title": "Product rule example.mp3", "Sentence": "And let me just write down the product rule generally first. So if we would take the derivative with respect to x of the first expression in terms of x. So this is, we could call this u of x u of x times another expression that involves x. So u times v of x. This is going to be equal to, and I'm color coding it so we can really keep track of things. This is going to be equal to the derivative of the first expression. So I could write that as u prime of x times just the second expression, not the derivative of it, just the second expression."}, {"video_title": "Product rule example.mp3", "Sentence": "So u times v of x. This is going to be equal to, and I'm color coding it so we can really keep track of things. This is going to be equal to the derivative of the first expression. So I could write that as u prime of x times just the second expression, not the derivative of it, just the second expression. So times v of x. V of x. And then we have plus, plus the first expression, not its derivative, just the first expression, u of x times the derivative of the second expression. Times the derivative of the second expression."}, {"video_title": "Product rule example.mp3", "Sentence": "So I could write that as u prime of x times just the second expression, not the derivative of it, just the second expression. So times v of x. V of x. And then we have plus, plus the first expression, not its derivative, just the first expression, u of x times the derivative of the second expression. Times the derivative of the second expression. So the way you remember it is you have these two things here. You're going to end up with two different terms. And each of them, you're going to take the derivative of one of them but not the other one."}, {"video_title": "Product rule example.mp3", "Sentence": "Times the derivative of the second expression. So the way you remember it is you have these two things here. You're going to end up with two different terms. And each of them, you're going to take the derivative of one of them but not the other one. And then the other one, you'll take the derivative of the other one but not the first one. So derivative of u times v is u prime times v plus u times v prime. Now when you just look at it like that, it seems a little bit abstract and that might even be a little confusing but that's why we have a tangible example here."}, {"video_title": "Product rule example.mp3", "Sentence": "And each of them, you're going to take the derivative of one of them but not the other one. And then the other one, you'll take the derivative of the other one but not the first one. So derivative of u times v is u prime times v plus u times v prime. Now when you just look at it like that, it seems a little bit abstract and that might even be a little confusing but that's why we have a tangible example here. And I color coded it intentionally so we can say that u of x is equal to e to the x and v of x is equal to cosine of x. So v of x is equal to cosine of x. And if u of x is equal to e to the x, we know that the derivative of that with respect to x is still e to the x."}, {"video_title": "Product rule example.mp3", "Sentence": "Now when you just look at it like that, it seems a little bit abstract and that might even be a little confusing but that's why we have a tangible example here. And I color coded it intentionally so we can say that u of x is equal to e to the x and v of x is equal to cosine of x. So v of x is equal to cosine of x. And if u of x is equal to e to the x, we know that the derivative of that with respect to x is still e to the x. That's one of the most magical things in mathematics. One of the things that makes e so special. So u prime of x is still equal to e to the x."}, {"video_title": "Product rule example.mp3", "Sentence": "And if u of x is equal to e to the x, we know that the derivative of that with respect to x is still e to the x. That's one of the most magical things in mathematics. One of the things that makes e so special. So u prime of x is still equal to e to the x. And v prime of x, we know is negative sine of x. And so what's this going to be equal to? This is going to be equal to the derivative of the first expression."}, {"video_title": "Product rule example.mp3", "Sentence": "So u prime of x is still equal to e to the x. And v prime of x, we know is negative sine of x. And so what's this going to be equal to? This is going to be equal to the derivative of the first expression. So the derivative of e to the x which is just e to the x times the second expression not taking its derivative so times cosine of x plus the first expression not taking its derivative so e to the x times the derivative of the second expression. So times the derivative of cosine of x which is negative sine, negative sine of x. And it might be a little bit confusing because e to the x is its own derivative, but this right over here, you can view this as this was the derivative of e to the x, which happens to be e to the x."}, {"video_title": "Product rule example.mp3", "Sentence": "This is going to be equal to the derivative of the first expression. So the derivative of e to the x which is just e to the x times the second expression not taking its derivative so times cosine of x plus the first expression not taking its derivative so e to the x times the derivative of the second expression. So times the derivative of cosine of x which is negative sine, negative sine of x. And it might be a little bit confusing because e to the x is its own derivative, but this right over here, you can view this as this was the derivative of e to the x, which happens to be e to the x. That's what's exciting about that expression or that function. And then this is just e to the x without taking its derivative. They're of course the same thing."}, {"video_title": "Product rule example.mp3", "Sentence": "And it might be a little bit confusing because e to the x is its own derivative, but this right over here, you can view this as this was the derivative of e to the x, which happens to be e to the x. That's what's exciting about that expression or that function. And then this is just e to the x without taking its derivative. They're of course the same thing. But anyway, now we can just simplify it. This is going to be equal to, this is going to be equal to, I could write this either as e to the x times cosine of x, times cosine of x, minus e to the x, e to the x times sine of x, times sine of x. Or if you want, you could factor out an e to the x."}, {"video_title": "Product rule example.mp3", "Sentence": "They're of course the same thing. But anyway, now we can just simplify it. This is going to be equal to, this is going to be equal to, I could write this either as e to the x times cosine of x, times cosine of x, minus e to the x, e to the x times sine of x, times sine of x. Or if you want, you could factor out an e to the x. This is the same thing as e to the x times cosine of x minus sine of x, cosine of x minus sine of x. So hopefully this makes the product rule a little bit more tangible. And once you have this in your tool belt, there's a whole broader class of functions and expressions that we can start to differentiate."}, {"video_title": "Product rule proof Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So if I have the function f of x, and if I wanted to take the derivative of it, by definition, by definition, the derivative of f of x is the limit as h approaches zero of f of x plus h minus f of x, all of that over, all of that over h. If we want to think of it visually, this is the slope of the tangent line and all of that. But now I want to do something a little bit more interesting. I want to find the derivative with respect to x, not just of f of x, but the product of two functions, f of x times g of x. And if I can come up with a simple thing for this, that essentially is the product rule. Well, if we just apply the definition of a derivative, that means I'm going to take the limit as h approaches zero. In the denominator, I'm going to have an h. In the denominator, I'm going to write a big, it's going to be a big rational expression. In the denominator, I'm going to have an h. And then I'm going to evaluate this thing at x plus h. So that's going to be f of x plus h, g of x plus h. And from that, I'm going to subtract this thing, evaluate it at f of x."}, {"video_title": "Product rule proof Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And if I can come up with a simple thing for this, that essentially is the product rule. Well, if we just apply the definition of a derivative, that means I'm going to take the limit as h approaches zero. In the denominator, I'm going to have an h. In the denominator, I'm going to write a big, it's going to be a big rational expression. In the denominator, I'm going to have an h. And then I'm going to evaluate this thing at x plus h. So that's going to be f of x plus h, g of x plus h. And from that, I'm going to subtract this thing, evaluate it at f of x. Or, sorry, this thing evaluate it at x. So that's going to be f of x times g of x. And I'm going to put a big, awkward space here, and you're going to see why in a second."}, {"video_title": "Product rule proof Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "In the denominator, I'm going to have an h. And then I'm going to evaluate this thing at x plus h. So that's going to be f of x plus h, g of x plus h. And from that, I'm going to subtract this thing, evaluate it at f of x. Or, sorry, this thing evaluate it at x. So that's going to be f of x times g of x. And I'm going to put a big, awkward space here, and you're going to see why in a second. So if I just, if I evaluate this at x, this is going to be minus f of x, g of x. All I did so far is I just applied the definition of the derivative. Instead of applying it to f of x, I applied it to f of x times g of x."}, {"video_title": "Product rule proof Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And I'm going to put a big, awkward space here, and you're going to see why in a second. So if I just, if I evaluate this at x, this is going to be minus f of x, g of x. All I did so far is I just applied the definition of the derivative. Instead of applying it to f of x, I applied it to f of x times g of x. So you have f of x plus h, g of x plus h, minus f of x, g of x, all of that over h, limit as h approaches zero. Now why did I put this big, awkward space here? Because just the way I've written it right now, it doesn't seem easy to algebraically manipulate."}, {"video_title": "Product rule proof Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Instead of applying it to f of x, I applied it to f of x times g of x. So you have f of x plus h, g of x plus h, minus f of x, g of x, all of that over h, limit as h approaches zero. Now why did I put this big, awkward space here? Because just the way I've written it right now, it doesn't seem easy to algebraically manipulate. I don't know how to evaluate this limit. There doesn't seem to be anything obvious to do. And what I'm about to show you, I guess you could view it as a little bit of a trick."}, {"video_title": "Product rule proof Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Because just the way I've written it right now, it doesn't seem easy to algebraically manipulate. I don't know how to evaluate this limit. There doesn't seem to be anything obvious to do. And what I'm about to show you, I guess you could view it as a little bit of a trick. I can't claim that I would have figured it out on my own, maybe eventually if I were spending hours on it. And I'm assuming somebody was fumbling with it long enough they said, oh wait, wait, look, if I just add and subtract the same term here, I can begin to algebraically manipulate it and get it to what we all know is the classic product rule. So what do I add and subtract here?"}, {"video_title": "Product rule proof Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And what I'm about to show you, I guess you could view it as a little bit of a trick. I can't claim that I would have figured it out on my own, maybe eventually if I were spending hours on it. And I'm assuming somebody was fumbling with it long enough they said, oh wait, wait, look, if I just add and subtract the same term here, I can begin to algebraically manipulate it and get it to what we all know is the classic product rule. So what do I add and subtract here? Well, let me give you a clue. So if we have plus, actually let me change this, minus f of x plus h, g of x. I can't just subtract, if I subtract it, I've got to add it too, so I don't change the value of this expression. So plus f of x plus h, g of x."}, {"video_title": "Product rule proof Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So what do I add and subtract here? Well, let me give you a clue. So if we have plus, actually let me change this, minus f of x plus h, g of x. I can't just subtract, if I subtract it, I've got to add it too, so I don't change the value of this expression. So plus f of x plus h, g of x. Now I haven't changed the value, I just added and subtracted the same thing. But now this thing can be manipulated in interesting algebraic ways to get us to what we all love about the product rule. And at any point you get inspired, I encourage you to pause this video."}, {"video_title": "Product rule proof Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So plus f of x plus h, g of x. Now I haven't changed the value, I just added and subtracted the same thing. But now this thing can be manipulated in interesting algebraic ways to get us to what we all love about the product rule. And at any point you get inspired, I encourage you to pause this video. Well, to keep going, let's just keep exploring this expression. So all of this is going to be equal to, it's all going to be equal to the limit as h approaches zero. So the first thing I'm going to do is I'm going to look at this part, this part of the expression."}, {"video_title": "Product rule proof Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And at any point you get inspired, I encourage you to pause this video. Well, to keep going, let's just keep exploring this expression. So all of this is going to be equal to, it's all going to be equal to the limit as h approaches zero. So the first thing I'm going to do is I'm going to look at this part, this part of the expression. And in particular, let's see, I am going to factor out an f of x plus h. So if you factor out an f of x plus h, this part right over here is going to be f of x plus h, f of x plus h, times, you're going to be left with g of x plus h, g of, it's a slightly shader, different shade of green, g of x plus h, that's that there, minus g of x, minus g of x, whoops, I forgot the parentheses. Oh, it's a different color. I got a new software program and it's making it hard for me to change colors."}, {"video_title": "Product rule proof Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So the first thing I'm going to do is I'm going to look at this part, this part of the expression. And in particular, let's see, I am going to factor out an f of x plus h. So if you factor out an f of x plus h, this part right over here is going to be f of x plus h, f of x plus h, times, you're going to be left with g of x plus h, g of, it's a slightly shader, different shade of green, g of x plus h, that's that there, minus g of x, minus g of x, whoops, I forgot the parentheses. Oh, it's a different color. I got a new software program and it's making it hard for me to change colors. My apologies, this is not a straightforward proof and the least I could do is change colors more smoothly. All right, g of x plus h, minus g of x, that's that one right over there, and then all of that over this h, all of that over h, so that's this part here, and then this part over here, this part over here, and actually it's still over h, so let me actually circle it like this. So, this part over here, I can write as, so then we're going to have plus, actually here let me, let me factor out a g of x here."}, {"video_title": "Product rule proof Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "I got a new software program and it's making it hard for me to change colors. My apologies, this is not a straightforward proof and the least I could do is change colors more smoothly. All right, g of x plus h, minus g of x, that's that one right over there, and then all of that over this h, all of that over h, so that's this part here, and then this part over here, this part over here, and actually it's still over h, so let me actually circle it like this. So, this part over here, I can write as, so then we're going to have plus, actually here let me, let me factor out a g of x here. So, plus g of x, plus g of x, times this f of x plus h, times f of x plus h, minus this f of x, minus that f of x, all of that over h, all of that over h. Now, we know from our limit properties, the limit of all of this business, well, that's just going to be the same thing as the limit of this as h approaches zero, plus the limit of this as h approaches zero, and then the limit of the product is going to be the same thing as the product of the limits. So, if I use both of those limit properties, I can rewrite this whole thing as the limit, let me give myself some real estate, the limit as h approaches zero of f of x plus h, of f of x plus h, times, times the limit as h approaches zero of all of this business, g of x plus h, minus g of x, minus g of x, all of that over h, I think you might see where this is going, very exciting, all right, plus, plus the limit, let me write that a little bit more clearly, plus the limit as h approaches zero of g of x, our nice brown colored g of x, times, now we're going to have a product here, the limit, the limit as h approaches zero of f of x plus h, of f of x plus h, minus f of x, minus f of x, all of that, all of that over h, and let me put the parentheses where they're appropriate, so that, that, that, that, and all I did here, the limit, the limit of this sum, that's going to be the sum of the limits, that's going to be the limit of this, plus the limit of that, and then the limit of the products is going to be the same thing as the product of the limits, so I just use those limit properties here, but now let's evaluate them. What's the limit, and I'll do it in different colors, what's this thing right over here?"}, {"video_title": "Product rule proof Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So, this part over here, I can write as, so then we're going to have plus, actually here let me, let me factor out a g of x here. So, plus g of x, plus g of x, times this f of x plus h, times f of x plus h, minus this f of x, minus that f of x, all of that over h, all of that over h. Now, we know from our limit properties, the limit of all of this business, well, that's just going to be the same thing as the limit of this as h approaches zero, plus the limit of this as h approaches zero, and then the limit of the product is going to be the same thing as the product of the limits. So, if I use both of those limit properties, I can rewrite this whole thing as the limit, let me give myself some real estate, the limit as h approaches zero of f of x plus h, of f of x plus h, times, times the limit as h approaches zero of all of this business, g of x plus h, minus g of x, minus g of x, all of that over h, I think you might see where this is going, very exciting, all right, plus, plus the limit, let me write that a little bit more clearly, plus the limit as h approaches zero of g of x, our nice brown colored g of x, times, now we're going to have a product here, the limit, the limit as h approaches zero of f of x plus h, of f of x plus h, minus f of x, minus f of x, all of that, all of that over h, and let me put the parentheses where they're appropriate, so that, that, that, that, and all I did here, the limit, the limit of this sum, that's going to be the sum of the limits, that's going to be the limit of this, plus the limit of that, and then the limit of the products is going to be the same thing as the product of the limits, so I just use those limit properties here, but now let's evaluate them. What's the limit, and I'll do it in different colors, what's this thing right over here? The limit as h approaches zero of f of x plus h, well that's just going to be f of x. Now, and this is the exciting part, what is this? The limit as h approaches zero of g of x plus h minus g of x over h. Well that's just our, that's the definition of our derivative, that's the derivative of g, so this is going to be, this is going to be the derivative of g of x, which is going to be g prime of x, g prime of x, so you're multiplying these two, and then you're going to have plus, what's the limit as h approaches zero of g of x?"}, {"video_title": "Differentiating related functions intro Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Y is equal to square root of x. It's interesting, they're telling us that they're both differentiable functions. Even x is a function, must be a function of something else. Well, they tell us that the derivative of x with respect to t is 12. And they want us to find the derivative of y with respect to t when x is equal to nine. So let's just make sure we can understand this. So they're telling us that both x and y are functions."}, {"video_title": "Differentiating related functions intro Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, they tell us that the derivative of x with respect to t is 12. And they want us to find the derivative of y with respect to t when x is equal to nine. So let's just make sure we can understand this. So they're telling us that both x and y are functions. Arguably, they're both functions of t. Y is a function of x, but then x is a function of t, so y could also be a function of t. One way to think about it is if x is equal to f of t, then y is equal to the square root of x, which would just be f of t. Another way to think about it, if you took t as your input into your function f, you're going to produce x. And then if you took that as your input into the square root function, you are going to produce y. So you could just view this as just one big box here, that y is a function of t. But now let's actually answer their question."}, {"video_title": "Differentiating related functions intro Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So they're telling us that both x and y are functions. Arguably, they're both functions of t. Y is a function of x, but then x is a function of t, so y could also be a function of t. One way to think about it is if x is equal to f of t, then y is equal to the square root of x, which would just be f of t. Another way to think about it, if you took t as your input into your function f, you're going to produce x. And then if you took that as your input into the square root function, you are going to produce y. So you could just view this as just one big box here, that y is a function of t. But now let's actually answer their question. To tackle it, we just have to apply the chain rule. The chain rule tells us that the derivative of y with respect to t is going to be equal to the derivative of y with respect to x times the derivative of x with respect to t. So let's apply it to this particular situation. We're gonna have the derivative of y with respect to t is equal to the derivative of y with respect to x."}, {"video_title": "Differentiating related functions intro Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So you could just view this as just one big box here, that y is a function of t. But now let's actually answer their question. To tackle it, we just have to apply the chain rule. The chain rule tells us that the derivative of y with respect to t is going to be equal to the derivative of y with respect to x times the derivative of x with respect to t. So let's apply it to this particular situation. We're gonna have the derivative of y with respect to t is equal to the derivative of y with respect to x. Well, what's that? Well, y is equal to the principal root of x. You could also write this as y is equal to x to the 1 1 2 power."}, {"video_title": "Differentiating related functions intro Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We're gonna have the derivative of y with respect to t is equal to the derivative of y with respect to x. Well, what's that? Well, y is equal to the principal root of x. You could also write this as y is equal to x to the 1 1 2 power. We could just use the power rule. The derivative of y with respect to x is 1 1 2 x to the negative 1 1 2. So let me write that down."}, {"video_title": "Differentiating related functions intro Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "You could also write this as y is equal to x to the 1 1 2 power. We could just use the power rule. The derivative of y with respect to x is 1 1 2 x to the negative 1 1 2. So let me write that down. 1 1 2 x to the negative 1 1 2. And then times the derivative of x with respect to t. Times the derivative of x with respect to t. So let's see, we wanna find what we have here in orange. That's what the question asks us."}, {"video_title": "Differentiating related functions intro Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let me write that down. 1 1 2 x to the negative 1 1 2. And then times the derivative of x with respect to t. Times the derivative of x with respect to t. So let's see, we wanna find what we have here in orange. That's what the question asks us. They tell us when x is equal to nine and the derivative of x with respect to t is equal to 12. So we have all of the information necessary to solve for this. So this is going to be equal to 1 1 2 times nine to the negative 1 1 2."}, {"video_title": "Differentiating related functions intro Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "That's what the question asks us. They tell us when x is equal to nine and the derivative of x with respect to t is equal to 12. So we have all of the information necessary to solve for this. So this is going to be equal to 1 1 2 times nine to the negative 1 1 2. Nine to the negative 1 1 2. Times dx dt. The derivative of x with respect to t is equal to 12."}, {"video_title": "Differentiating related functions intro Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be equal to 1 1 2 times nine to the negative 1 1 2. Nine to the negative 1 1 2. Times dx dt. The derivative of x with respect to t is equal to 12. Times 12. So let's see, nine to the 1 1 2 would be three. Nine to the negative 1 1 2 would be 1 1 3."}, {"video_title": "Differentiating related functions intro Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "The derivative of x with respect to t is equal to 12. Times 12. So let's see, nine to the 1 1 2 would be three. Nine to the negative 1 1 2 would be 1 1 3. So this is 1 1 3. So this will all simplify to 1 1 2 times 1 1 3 is 1 1 6. So we can have a six in the denominator."}, {"video_title": "Differentiating related functions intro Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Nine to the negative 1 1 2 would be 1 1 3. So this is 1 1 3. So this will all simplify to 1 1 2 times 1 1 3 is 1 1 6. So we can have a six in the denominator. And then we are going to have a 12 in the numerator. So 12 6. So the derivative of y with respect to t when x is equal to nine and derivative of x with respect to t is 12 is two."}, {"video_title": "Limits of combined functions piecewise functions AP Calculus AB Khan Academy.mp3", "Sentence": "So when you do this first one, you might just try to find the limit as x approaches negative two of f of x, and then the limit as x approaches negative two of g of x, and then add those two limits together. But you will quickly find a problem, because when you find the limit as x approaches negative two of f of x, it looks as we are approaching negative two from the left, it looks like we're approaching one. As we approach x equals negative two from the right, it looks like we're approaching three. So it looks like the limit as x approaches negative two of f of x doesn't exist, and the same thing's true of g of x. If we approach from the left, it looks like we're approaching three. If we approach from the right, it looks like we're approaching one. But it turns out that this limit can still exist, as long as the limit as x approaches negative two from the left of the sum, f of x plus g of x, exists and is equal to the limit as x approaches negative two from the right of the sum, f of x plus g of x."}, {"video_title": "Limits of combined functions piecewise functions AP Calculus AB Khan Academy.mp3", "Sentence": "So it looks like the limit as x approaches negative two of f of x doesn't exist, and the same thing's true of g of x. If we approach from the left, it looks like we're approaching three. If we approach from the right, it looks like we're approaching one. But it turns out that this limit can still exist, as long as the limit as x approaches negative two from the left of the sum, f of x plus g of x, exists and is equal to the limit as x approaches negative two from the right of the sum, f of x plus g of x. So what are these things? Well, as we approach negative two from the left, f of x is approaching, looks like one, and g of x is approaching three. So it looks like we're approaching one and three, so it looks like this is approaching, the sum is going to approach four."}, {"video_title": "Limits of combined functions piecewise functions AP Calculus AB Khan Academy.mp3", "Sentence": "But it turns out that this limit can still exist, as long as the limit as x approaches negative two from the left of the sum, f of x plus g of x, exists and is equal to the limit as x approaches negative two from the right of the sum, f of x plus g of x. So what are these things? Well, as we approach negative two from the left, f of x is approaching, looks like one, and g of x is approaching three. So it looks like we're approaching one and three, so it looks like this is approaching, the sum is going to approach four. And if we're coming from the right, f of x looks like it's approaching three, and g of x looks like it is approaching one. And so once again, this is equal to four. And since the left and right-handed limits are approaching the same thing, we would say that this limit exists, and it is equal to four."}, {"video_title": "Limits of combined functions piecewise functions AP Calculus AB Khan Academy.mp3", "Sentence": "So it looks like we're approaching one and three, so it looks like this is approaching, the sum is going to approach four. And if we're coming from the right, f of x looks like it's approaching three, and g of x looks like it is approaching one. And so once again, this is equal to four. And since the left and right-handed limits are approaching the same thing, we would say that this limit exists, and it is equal to four. Now let's do this next example, as x approaches one. Well, we'll do the exact same exercise. And once again, if you look at the individual limits for f of x from the left and the right, as we approach one, this limit doesn't exist, but the limit as x approaches one of the sum might exist."}, {"video_title": "Limits of combined functions piecewise functions AP Calculus AB Khan Academy.mp3", "Sentence": "And since the left and right-handed limits are approaching the same thing, we would say that this limit exists, and it is equal to four. Now let's do this next example, as x approaches one. Well, we'll do the exact same exercise. And once again, if you look at the individual limits for f of x from the left and the right, as we approach one, this limit doesn't exist, but the limit as x approaches one of the sum might exist. So let's try that out. So the limit as x approaches one from the left-hand side of f of x plus g of x, what is that going to be equal to? As we approach, so f of x, as we approach one from the left, looks like this is approaching two."}, {"video_title": "Limits of combined functions piecewise functions AP Calculus AB Khan Academy.mp3", "Sentence": "And once again, if you look at the individual limits for f of x from the left and the right, as we approach one, this limit doesn't exist, but the limit as x approaches one of the sum might exist. So let's try that out. So the limit as x approaches one from the left-hand side of f of x plus g of x, what is that going to be equal to? As we approach, so f of x, as we approach one from the left, looks like this is approaching two. I'm just doing this for shorthand. And g of x, as we approach one from the left, looks like it is approaching zero. So this will be approaching two plus zero, which is two."}, {"video_title": "Limits of combined functions piecewise functions AP Calculus AB Khan Academy.mp3", "Sentence": "As we approach, so f of x, as we approach one from the left, looks like this is approaching two. I'm just doing this for shorthand. And g of x, as we approach one from the left, looks like it is approaching zero. So this will be approaching two plus zero, which is two. And then the limit as x approaches one from the right-hand side of f of x plus g of x is going to be equal to, well, for f of x, as we're approaching one from the right-hand side, looks like it's approaching negative one. And for g of x, as we're approaching one from the right-hand side, looks like we're approaching zero again. And so here, it looks like we're approaching negative one."}, {"video_title": "Limits of combined functions piecewise functions AP Calculus AB Khan Academy.mp3", "Sentence": "So this will be approaching two plus zero, which is two. And then the limit as x approaches one from the right-hand side of f of x plus g of x is going to be equal to, well, for f of x, as we're approaching one from the right-hand side, looks like it's approaching negative one. And for g of x, as we're approaching one from the right-hand side, looks like we're approaching zero again. And so here, it looks like we're approaching negative one. So the left and right-hand limits aren't approaching the same value, so this one does not exist. And then last but not least, x approaches one of f of x times g of x. So we'll do the same drill."}, {"video_title": "Limits of combined functions piecewise functions AP Calculus AB Khan Academy.mp3", "Sentence": "And so here, it looks like we're approaching negative one. So the left and right-hand limits aren't approaching the same value, so this one does not exist. And then last but not least, x approaches one of f of x times g of x. So we'll do the same drill. Limit as x approaches one from the left-hand side of f of x times g of x. Well, here, and we could even use the values here. We see we're approaching one from the left."}, {"video_title": "Limits of combined functions piecewise functions AP Calculus AB Khan Academy.mp3", "Sentence": "So we'll do the same drill. Limit as x approaches one from the left-hand side of f of x times g of x. Well, here, and we could even use the values here. We see we're approaching one from the left. We are approaching two, so this is two. And when we're approaching one from the left here, we're approaching zero. And so this is gonna be two times, we're gonna be approaching two times zero, which is zero."}, {"video_title": "Limits of combined functions piecewise functions AP Calculus AB Khan Academy.mp3", "Sentence": "We see we're approaching one from the left. We are approaching two, so this is two. And when we're approaching one from the left here, we're approaching zero. And so this is gonna be two times, we're gonna be approaching two times zero, which is zero. And then we approach from the right, x approaches one from the right of f of x times g of x. Well, we already saw, when we're approaching one from the right of f of x, we are approaching negative one. But g of x approaching one from the right is still approaching zero, so this is going to be zero again."}, {"video_title": "Limits of combined functions piecewise functions AP Calculus AB Khan Academy.mp3", "Sentence": "And so this is gonna be two times, we're gonna be approaching two times zero, which is zero. And then we approach from the right, x approaches one from the right of f of x times g of x. Well, we already saw, when we're approaching one from the right of f of x, we are approaching negative one. But g of x approaching one from the right is still approaching zero, so this is going to be zero again. So this limit exists. We get the same limit when we approach from the left and the right. It is equal to zero."}, {"video_title": "Vertical distance of bouncing ball Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "And every time it bounces, it goes half as high as the previous bounce. So for example, you drop it from 10 meters. The next time, its peak height is going to be at 5 meters. So the next time around, on the next bounce, let me draw it in that same orange color, and the next bounce, the ball is going to go 5 meters. This distance right over here is going to be 5 meters. And then the bounce after that is going to be half as high. So it's going to go 2.5 meters."}, {"video_title": "Vertical distance of bouncing ball Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "So the next time around, on the next bounce, let me draw it in that same orange color, and the next bounce, the ball is going to go 5 meters. This distance right over here is going to be 5 meters. And then the bounce after that is going to be half as high. So it's going to go 2.5 meters. And it's just going to keep doing that. So it's going to go 2.5 meters right over here. And what I want to think about in this video is what is the total vertical distance that the ball travels."}, {"video_title": "Vertical distance of bouncing ball Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "So it's going to go 2.5 meters. And it's just going to keep doing that. So it's going to go 2.5 meters right over here. And what I want to think about in this video is what is the total vertical distance that the ball travels. So let's think about that a little bit. So it's first going to travel 10 meters straight down. So it's going to travel 10 meters, just like that."}, {"video_title": "Vertical distance of bouncing ball Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "And what I want to think about in this video is what is the total vertical distance that the ball travels. So let's think about that a little bit. So it's first going to travel 10 meters straight down. So it's going to travel 10 meters, just like that. And then it's going to travel half of 10 meters twice. It's going to go up 5 meters, up half of 10 meters, and then down half of 10 meters. So then it's going to go 2 times, let me put it this way, 2, so it's going to go, each of these is going to be 10 meters."}, {"video_title": "Vertical distance of bouncing ball Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "So it's going to travel 10 meters, just like that. And then it's going to travel half of 10 meters twice. It's going to go up 5 meters, up half of 10 meters, and then down half of 10 meters. So then it's going to go 2 times, let me put it this way, 2, so it's going to go, each of these is going to be 10 meters. Actually, I don't have to write the units here. Let me take the units out of the way. So let me write it clear."}, {"video_title": "Vertical distance of bouncing ball Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "So then it's going to go 2 times, let me put it this way, 2, so it's going to go, each of these is going to be 10 meters. Actually, I don't have to write the units here. Let me take the units out of the way. So let me write it clear. So the first bounce, once again, it goes straight down 10 meters. Then on the next bounce, it's going to go up 10 times 1 half. And then it's going to go down 10 times 1 half."}, {"video_title": "Vertical distance of bouncing ball Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "So let me write it clear. So the first bounce, once again, it goes straight down 10 meters. Then on the next bounce, it's going to go up 10 times 1 half. And then it's going to go down 10 times 1 half. So we just care about the total vertical distance. We don't care about the direction. So it's going to go up 10 times 1 half, up 5 meters, and then it's going to go down 5 meters."}, {"video_title": "Vertical distance of bouncing ball Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "And then it's going to go down 10 times 1 half. So we just care about the total vertical distance. We don't care about the direction. So it's going to go up 10 times 1 half, up 5 meters, and then it's going to go down 5 meters. So it's going to travel a total vertical distance of 10 meters, 5 up and 5 down. Now what about on this bounce? Well, here it's going to go half as far as it went there."}, {"video_title": "Vertical distance of bouncing ball Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "So it's going to go up 10 times 1 half, up 5 meters, and then it's going to go down 5 meters. So it's going to travel a total vertical distance of 10 meters, 5 up and 5 down. Now what about on this bounce? Well, here it's going to go half as far as it went there. So it's going to go 10 times 1 half squared up, and then 10 times 1 half squared down. And I think you see a pattern here. This looks an awful lot like a geometric series, an infinite geometric series."}, {"video_title": "Vertical distance of bouncing ball Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "Well, here it's going to go half as far as it went there. So it's going to go 10 times 1 half squared up, and then 10 times 1 half squared down. And I think you see a pattern here. This looks an awful lot like a geometric series, an infinite geometric series. It's just going to keep on going like that forever and ever. So let's try to clean this up a little bit so it looks a little bit more like a traditional geometric series. So if we were to simplify this a little bit, we could rewrite this as 10 plus 20 times 1 half to the first power plus 10 times 1 half squared is going to be 20 times 1 half squared."}, {"video_title": "Vertical distance of bouncing ball Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "This looks an awful lot like a geometric series, an infinite geometric series. It's just going to keep on going like that forever and ever. So let's try to clean this up a little bit so it looks a little bit more like a traditional geometric series. So if we were to simplify this a little bit, we could rewrite this as 10 plus 20 times 1 half to the first power plus 10 times 1 half squared is going to be 20 times 1 half squared. And it will just keep on going on and on. So this would be a little bit clearer if this were a 20 right over here. But we could do that."}, {"video_title": "Vertical distance of bouncing ball Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "So if we were to simplify this a little bit, we could rewrite this as 10 plus 20 times 1 half to the first power plus 10 times 1 half squared is going to be 20 times 1 half squared. And it will just keep on going on and on. So this would be a little bit clearer if this were a 20 right over here. But we could do that. We could rewrite negative 10 as negative 10 plus 20. And then we have plus all of this stuff right over here. Let me just copy and paste that."}, {"video_title": "Vertical distance of bouncing ball Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "But we could do that. We could rewrite negative 10 as negative 10 plus 20. And then we have plus all of this stuff right over here. Let me just copy and paste that. Copy and paste. So plus all of this right over here. And we can even write this first."}, {"video_title": "Vertical distance of bouncing ball Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "Let me just copy and paste that. Copy and paste. So plus all of this right over here. And we can even write this first. We can even write this 20 right over here as 20 times 1 half to the 0 power plus all of this. So now it very clearly looks like an infinite geometric series. We can write our entire sum."}, {"video_title": "Vertical distance of bouncing ball Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "And we can even write this first. We can even write this 20 right over here as 20 times 1 half to the 0 power plus all of this. So now it very clearly looks like an infinite geometric series. We can write our entire sum. And maybe I'll write it up here since I don't want to lose the diagram. We could write it as negative 10. That's that negative 10 right over here."}, {"video_title": "Vertical distance of bouncing ball Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "We can write our entire sum. And maybe I'll write it up here since I don't want to lose the diagram. We could write it as negative 10. That's that negative 10 right over here. Plus the sum from k is equal to 0 to infinity of 20 times our common ratio to the k-th power. So what's this going to be? What's this going to turn out to be?"}, {"video_title": "Vertical distance of bouncing ball Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "That's that negative 10 right over here. Plus the sum from k is equal to 0 to infinity of 20 times our common ratio to the k-th power. So what's this going to be? What's this going to turn out to be? Well, we've already derived in multiple videos already here that the sum of an infinite geometric series. So the sum from k equals 0 to infinity of a times r to the k is equal to a over 1 minus r. So we just apply that right over here. This business right over here is going to be equal to 20 over 1 minus 1 half, which is the same thing as 20 over 1 half, which is the same thing as 20 times 2, or 40."}, {"video_title": "Vertical distance of bouncing ball Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "What's this going to turn out to be? Well, we've already derived in multiple videos already here that the sum of an infinite geometric series. So the sum from k equals 0 to infinity of a times r to the k is equal to a over 1 minus r. So we just apply that right over here. This business right over here is going to be equal to 20 over 1 minus 1 half, which is the same thing as 20 over 1 half, which is the same thing as 20 times 2, or 40. So what's the total vertical distance that our ball travels? It's going to be negative 10 plus 40, which is equal to 30 meters. Our total vertical distance that the ball travels is 30 meters."}, {"video_title": "Partial sums formula for nth term from partial sum Series AP Calculus BC Khan Academy.mp3", "Sentence": "So if I have a sub one plus a sub two plus a sub three, and I keep adding all the way to a sub n minus one plus a sub n, this whole thing, this whole thing that I just wrote out, that is s sub n. This whole thing is s, let me, this whole thing is s sub n, which is equal to n plus one over n plus 10. Now if I want to figure out a sub n, which is the goal of this exercise, well I could subtract out the sum of the first n minus one terms. So I could subtract out this. So that is s, that is s sub n minus one, and what would that be equal to? Well, wherever we see an n, we'd replace it with an n minus one, so it'd be n minus one plus one over n minus one plus 10, which is equal to which is equal to n over n plus nine. So if you subtract the red stuff from the blue stuff, all you're going to be left with is a thing that we want to solve for. You're gonna be left with a sub n. So we could write down a sub n is equal to s sub n, is equal to s sub n minus s sub n minus one, s sub n minus one, or we could write that is equal to this stuff, so this is the n plus one over n plus 10 minus n over n plus nine."}, {"video_title": "Partial sums formula for nth term from partial sum Series AP Calculus BC Khan Academy.mp3", "Sentence": "So that is s, that is s sub n minus one, and what would that be equal to? Well, wherever we see an n, we'd replace it with an n minus one, so it'd be n minus one plus one over n minus one plus 10, which is equal to which is equal to n over n plus nine. So if you subtract the red stuff from the blue stuff, all you're going to be left with is a thing that we want to solve for. You're gonna be left with a sub n. So we could write down a sub n is equal to s sub n, is equal to s sub n minus s sub n minus one, s sub n minus one, or we could write that is equal to this stuff, so this is the n plus one over n plus 10 minus n over n plus nine. And this by itself, this is a rule for a sub n, but we could combine these terms, add these two fractions together, and this is actually going to be the case for n greater than one. For n equals one, s sub one is going to be, well you can just, a sub one is going to be equal to s sub one, but then for any other n, we could use this right over here. And if we want to simplify this, well we can add these two fractions."}, {"video_title": "Partial sums formula for nth term from partial sum Series AP Calculus BC Khan Academy.mp3", "Sentence": "You're gonna be left with a sub n. So we could write down a sub n is equal to s sub n, is equal to s sub n minus s sub n minus one, s sub n minus one, or we could write that is equal to this stuff, so this is the n plus one over n plus 10 minus n over n plus nine. And this by itself, this is a rule for a sub n, but we could combine these terms, add these two fractions together, and this is actually going to be the case for n greater than one. For n equals one, s sub one is going to be, well you can just, a sub one is going to be equal to s sub one, but then for any other n, we could use this right over here. And if we want to simplify this, well we can add these two fractions. We can add these two fractions by having a common denominator. So let's see, if we multiply the numerator and denominator here times n plus nine, we are going to get, so this is equal to n plus one times n plus nine over n plus 10 times n plus nine. And from that, we are going to subtract, let's multiply the numerator and denominator here by n plus 10, so we have n times n plus 10 over n plus nine times n plus 10."}, {"video_title": "Partial sums formula for nth term from partial sum Series AP Calculus BC Khan Academy.mp3", "Sentence": "And if we want to simplify this, well we can add these two fractions. We can add these two fractions by having a common denominator. So let's see, if we multiply the numerator and denominator here times n plus nine, we are going to get, so this is equal to n plus one times n plus nine over n plus 10 times n plus nine. And from that, we are going to subtract, let's multiply the numerator and denominator here by n plus 10, so we have n times n plus 10 over n plus nine times n plus 10. N plus nine times n plus 10. And what does that give us? So let's see, if we simplify up here, we're gonna have, this is n squared plus 10n plus nine."}, {"video_title": "Partial sums formula for nth term from partial sum Series AP Calculus BC Khan Academy.mp3", "Sentence": "And from that, we are going to subtract, let's multiply the numerator and denominator here by n plus 10, so we have n times n plus 10 over n plus nine times n plus 10. N plus nine times n plus 10. And what does that give us? So let's see, if we simplify up here, we're gonna have, this is n squared plus 10n plus nine. That's that. And then this right over here is n squared plus, this is n squared plus 10n. Let me do it in that red color."}, {"video_title": "Partial sums formula for nth term from partial sum Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's see, if we simplify up here, we're gonna have, this is n squared plus 10n plus nine. That's that. And then this right over here is n squared plus, this is n squared plus 10n. Let me do it in that red color. So this is n squared plus 10n, and remember we're gonna subtract this. And so, and we are close to deserving a drum roll, a sub n is going to be equal to our denominator right over here is n plus nine times n plus 10. And we're gonna subtract the red stuff from the blue stuff."}, {"video_title": "Partial sums formula for nth term from partial sum Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let me do it in that red color. So this is n squared plus 10n, and remember we're gonna subtract this. And so, and we are close to deserving a drum roll, a sub n is going to be equal to our denominator right over here is n plus nine times n plus 10. And we're gonna subtract the red stuff from the blue stuff. So you subtract an n squared from an n squared, those cancel out. Subtract a 10n from a 10n, those cancel out. And you're just left with that blue nine."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say that we have a cardboard, a sheet of cardboard that is 20 inches by 30 inches. Let me draw the cardboard as neatly as I can. So it might look something like that. So that is my sheet of cardboard. And just to make sure we know the dimensions, it is 20 inches by 30 inches. What we're going to do is cut out the corners of this cardboard. And all the cut out corners are going to be squares."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So that is my sheet of cardboard. And just to make sure we know the dimensions, it is 20 inches by 30 inches. What we're going to do is cut out the corners of this cardboard. And all the cut out corners are going to be squares. And we're going to cut out an x by x corner from each of the corners of this piece of cardboard. x by x. Over here, x by x."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And all the cut out corners are going to be squares. And we're going to cut out an x by x corner from each of the corners of this piece of cardboard. x by x. Over here, x by x. And then over here, x by x. And what we'll do is after we cut out those corners, we can essentially fold down the flaps. Let me draw the flaps so you can imagine."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Over here, x by x. And then over here, x by x. And what we'll do is after we cut out those corners, we can essentially fold down the flaps. Let me draw the flaps so you can imagine. We can fold right there. We could fold right there. We could fold right there."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Let me draw the flaps so you can imagine. We can fold right there. We could fold right there. We could fold right there. And we would form a box. I guess you could imagine a box without a bottom to it. Or you could view a box without a top to it."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We could fold right there. And we would form a box. I guess you could imagine a box without a bottom to it. Or you could view a box without a top to it. So if we were to fold everything up, we would get a container that looks something like this. Let me make my best attempt to draw it. So it would look like this."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Or you could view a box without a top to it. So if we were to fold everything up, we would get a container that looks something like this. Let me make my best attempt to draw it. So it would look like this. This is one flap folded up. You can imagine this flap right over here. If I were to fold it up like that, it now would look like this."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So it would look like this. This is one flap folded up. You can imagine this flap right over here. If I were to fold it up like that, it now would look like this. The height of the flap is x. So this distance right over here is x. And then if I were to fold this flap, let me do that a little bit neater."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "If I were to fold it up like that, it now would look like this. The height of the flap is x. So this distance right over here is x. And then if I were to fold this flap, let me do that a little bit neater. If I were to fold this flap right over here, if I were to fold that up, then it would look like this. My best attempt to draw it. It would look like that."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And then if I were to fold this flap, let me do that a little bit neater. If I were to fold this flap right over here, if I were to fold that up, then it would look like this. My best attempt to draw it. It would look like that. And then I would fold that back flap up. So the back flap would look something like that. That would be the back flap."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "It would look like that. And then I would fold that back flap up. So the back flap would look something like that. That would be the back flap. It would look something like that. And then this flap over here, if I fold it up, would look something like that. And then, of course, my base of my whole thing."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "That would be the back flap. It would look something like that. And then this flap over here, if I fold it up, would look something like that. And then, of course, my base of my whole thing. So this whole region right over here of my piece of cardboard would be the floor of this box that I'm constructing. And what I want to do is I want to maximize the volume of this box. I want to maximize how much it can hold."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And then, of course, my base of my whole thing. So this whole region right over here of my piece of cardboard would be the floor of this box that I'm constructing. And what I want to do is I want to maximize the volume of this box. I want to maximize how much it can hold. And I want to maximize it by picking my x appropriately. So let's think about what the volume of this box is as a function of x. In order to do that, we have to figure out all the dimensions of this box as a function of x."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "I want to maximize how much it can hold. And I want to maximize it by picking my x appropriately. So let's think about what the volume of this box is as a function of x. In order to do that, we have to figure out all the dimensions of this box as a function of x. We already know that this corner right over here, which is made up of when this side and this side connect, when you fold these two flaps up, that's going to be the same height over there. That's going to be the same height over there. The height of this box is going to be x."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "In order to do that, we have to figure out all the dimensions of this box as a function of x. We already know that this corner right over here, which is made up of when this side and this side connect, when you fold these two flaps up, that's going to be the same height over there. That's going to be the same height over there. The height of this box is going to be x. But what's the width? What is the width of this box going to be? Well, the width of the box is going to be this distance right over here."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "The height of this box is going to be x. But what's the width? What is the width of this box going to be? Well, the width of the box is going to be this distance right over here. And this distance is going to be 20 inches minus not one x, but minus two x's. So this is going to be 20 minus two x. You see it right over here."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the width of the box is going to be this distance right over here. And this distance is going to be 20 inches minus not one x, but minus two x's. So this is going to be 20 minus two x. You see it right over here. This whole distance is 20. You subtract this x, you subtract this x, and you get this distance right over here. So it's 20 minus two x."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "You see it right over here. This whole distance is 20. You subtract this x, you subtract this x, and you get this distance right over here. So it's 20 minus two x. Now the same logic, what is the depth of the box? What is that distance right over there? Well, that distance is this distance right over here."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So it's 20 minus two x. Now the same logic, what is the depth of the box? What is that distance right over there? Well, that distance is this distance right over here. We know that this entire distance is 30 inches. If we subtract out this x and we subtract out this x, we get the distance that we care about. So this is going to be 30 minus two x."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, that distance is this distance right over here. We know that this entire distance is 30 inches. If we subtract out this x and we subtract out this x, we get the distance that we care about. So this is going to be 30 minus two x. So now we have all of the dimensions. So what would the volume be as a function of x? Well, the volume as a function of x is going to be equal to the height, which is x, times the width, which is 20 minus two x, times the depth, which is 30 minus two x."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be 30 minus two x. So now we have all of the dimensions. So what would the volume be as a function of x? Well, the volume as a function of x is going to be equal to the height, which is x, times the width, which is 20 minus two x, times the depth, which is 30 minus two x. Now what are possible values of x that give us a valid volume? Well, x can't be less than zero. You can't make a negative cut here."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the volume as a function of x is going to be equal to the height, which is x, times the width, which is 20 minus two x, times the depth, which is 30 minus two x. Now what are possible values of x that give us a valid volume? Well, x can't be less than zero. You can't make a negative cut here. Somehow we would have to add cardboard or something there. So we know that x is going to be greater than or equal to zero. So let me write this down."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "You can't make a negative cut here. Somehow we would have to add cardboard or something there. So we know that x is going to be greater than or equal to zero. So let me write this down. x is going to be greater than or equal to zero. And what does it have to be less than? Well, I can cut at most, we can see here the length right over here, this pink color, this mauve color, is 20 minus two x."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let me write this down. x is going to be greater than or equal to zero. And what does it have to be less than? Well, I can cut at most, we can see here the length right over here, this pink color, this mauve color, is 20 minus two x. So this has got to be greater than zero. This is always going to be shorter than the 30 minus two x. But the 20 minus two x has to be greater than or equal to zero."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, I can cut at most, we can see here the length right over here, this pink color, this mauve color, is 20 minus two x. So this has got to be greater than zero. This is always going to be shorter than the 30 minus two x. But the 20 minus two x has to be greater than or equal to zero. You can't cut more cardboard than there is. Or you could say that 20 has to be greater than or equal to two x. Or you could say that 10 is going to be greater than or equal to x, which is another way of saying that x is going to be less than or equal to 10."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "But the 20 minus two x has to be greater than or equal to zero. You can't cut more cardboard than there is. Or you could say that 20 has to be greater than or equal to two x. Or you could say that 10 is going to be greater than or equal to x, which is another way of saying that x is going to be less than or equal to 10. It's a different shade of yellow. x is going to be less than or equal to 10. So x has got to be between zero and 10."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Or you could say that 10 is going to be greater than or equal to x, which is another way of saying that x is going to be less than or equal to 10. It's a different shade of yellow. x is going to be less than or equal to 10. So x has got to be between zero and 10. Otherwise, we've cut too much or we're somehow adding cardboard or something. So first let's think about the volume at the endpoints of our domain, of what x can be for our volume. Well, our volume when x is equal to zero is equal to what?"}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So x has got to be between zero and 10. Otherwise, we've cut too much or we're somehow adding cardboard or something. So first let's think about the volume at the endpoints of our domain, of what x can be for our volume. Well, our volume when x is equal to zero is equal to what? We're going to have zero times all of this stuff, and it's pretty obvious. You're not going to have any height here, so you're not going to have any volume, so our volume would be zero. What is our volume when x is equal to 10?"}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, our volume when x is equal to zero is equal to what? We're going to have zero times all of this stuff, and it's pretty obvious. You're not going to have any height here, so you're not going to have any volume, so our volume would be zero. What is our volume when x is equal to 10? Well, if x equaled 10, then the width here that I've drawn in pink would be zero. So once again, we would have no volume. And this term right over here, if we just look at it algebraically, would also be equal to zero, so this whole thing would be equal to zero."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "What is our volume when x is equal to 10? Well, if x equaled 10, then the width here that I've drawn in pink would be zero. So once again, we would have no volume. And this term right over here, if we just look at it algebraically, would also be equal to zero, so this whole thing would be equal to zero. So someplace in between x equals zero and x equals 10, we should achieve our maximum volume. And before we do it analytically with a bit of calculus, let's do it graphically. So I'll get my handy TI-85 out."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And this term right over here, if we just look at it algebraically, would also be equal to zero, so this whole thing would be equal to zero. So someplace in between x equals zero and x equals 10, we should achieve our maximum volume. And before we do it analytically with a bit of calculus, let's do it graphically. So I'll get my handy TI-85 out. First let me set my range appropriately before I attempt to graph it. So I'll put my graph function. Let me set my range."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So I'll get my handy TI-85 out. First let me set my range appropriately before I attempt to graph it. So I'll put my graph function. Let me set my range. My minimum x value, let me make that zero. We know that x cannot be less than zero. My maximum x value, well, 10 seems pretty good."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Let me set my range. My minimum x value, let me make that zero. We know that x cannot be less than zero. My maximum x value, well, 10 seems pretty good. My minimum y value, this is essentially going to be my volume. I'm not going to have negative volume, so let me set that equal zero. And my maximum y value, let's see what would be reasonable here."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "My maximum x value, well, 10 seems pretty good. My minimum y value, this is essentially going to be my volume. I'm not going to have negative volume, so let me set that equal zero. And my maximum y value, let's see what would be reasonable here. I'm just going to pick a random x and see what type of a volume I get. If x were to be 5, it would be 5 times 20 minus 10, which is 10. So that would be, did I do that right?"}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And my maximum y value, let's see what would be reasonable here. I'm just going to pick a random x and see what type of a volume I get. If x were to be 5, it would be 5 times 20 minus 10, which is 10. So that would be, did I do that right? Yeah, 20 minus 2 times 5, so that would be 10. And then times 30 minus 2 times 5, which would be 20. So it would be 5 times 10 times 20."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So that would be, did I do that right? Yeah, 20 minus 2 times 5, so that would be 10. And then times 30 minus 2 times 5, which would be 20. So it would be 5 times 10 times 20. So you'd get a volume of 1,000 cubic inches, and I just randomly picked the number 5. Let me give my maximum y value a little higher than that, just in case that that isn't the maximum value. I just randomly picked."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So it would be 5 times 10 times 20. So you'd get a volume of 1,000 cubic inches, and I just randomly picked the number 5. Let me give my maximum y value a little higher than that, just in case that that isn't the maximum value. I just randomly picked. So let's say y max is 1,500, and if for whatever reason our graph doesn't fit in there, then maybe we can make our y max even larger. So I think this is going to be a decent range. Now let's actually input the function itself."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "I just randomly picked. So let's say y max is 1,500, and if for whatever reason our graph doesn't fit in there, then maybe we can make our y max even larger. So I think this is going to be a decent range. Now let's actually input the function itself. So our volume is equal to x times 20 minus 2x times 30 minus 2x. And that looks about right, so now I think we can graph it. So second, and I want to select that up there, so I'll do graph."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's actually input the function itself. So our volume is equal to x times 20 minus 2x times 30 minus 2x. And that looks about right, so now I think we can graph it. So second, and I want to select that up there, so I'll do graph. And it looks like we did get the right range. So this tells us volume is a function of x between x is 0 and x is 10. And it does look like we hit a maximum point right around there."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So second, and I want to select that up there, so I'll do graph. And it looks like we did get the right range. So this tells us volume is a function of x between x is 0 and x is 10. And it does look like we hit a maximum point right around there. So what I'm going to do is I'm going to use the trace function to figure out roughly what that maximum point is. So let me trace this function so I can still go higher, higher. So over there my volume is 1055.5."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And it does look like we hit a maximum point right around there. So what I'm going to do is I'm going to use the trace function to figure out roughly what that maximum point is. So let me trace this function so I can still go higher, higher. So over there my volume is 1055.5. Then I can get to 1056. So see, this was 1056.20. This is 1056.24."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So over there my volume is 1055.5. Then I can get to 1056. So see, this was 1056.20. This is 1056.24. Then I go back to 1055. So at least based on the level of zoom that I have in my calculator right now, this is a pretty good approximation for the maximum value that my function actually takes on. So it looks like my maximum value is around 1056."}, {"video_title": "Optimization box volume (Part 1) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "This is 1056.24. Then I go back to 1055. So at least based on the level of zoom that I have in my calculator right now, this is a pretty good approximation for the maximum value that my function actually takes on. So it looks like my maximum value is around 1056. And it happens at around x equals 3.89. So it looks like my volume at 3.89 is approximately equal to 1056 cubic inches. Or you could say that we hit a maximum when x is approximately equal to 3.89."}, {"video_title": "Analyzing motion problems position AP Calculus AB Khan Academy.mp3", "Sentence": "Divya received the following problem. A particle moves in a straight line with velocity v of t is equal to the square root of three t minus one meters per second, where t is time in seconds. At t equals two, the particle's distance from the starting point was eight meters in the positive direction. What is the particle's position at t equals seven seconds? Which expression should Divya use to solve the problem? So pause this video and have a go at it. All right, now let's do this together."}, {"video_title": "Analyzing motion problems position AP Calculus AB Khan Academy.mp3", "Sentence": "What is the particle's position at t equals seven seconds? Which expression should Divya use to solve the problem? So pause this video and have a go at it. All right, now let's do this together. So we wanna know the particle's position at t is equal to seven. So what we could do, they tell us what our position is at t equals two. So what the position at t equals seven would be your position at t equals two plus your change in position from t equals two to t is equal to seven."}, {"video_title": "Analyzing motion problems position AP Calculus AB Khan Academy.mp3", "Sentence": "All right, now let's do this together. So we wanna know the particle's position at t is equal to seven. So what we could do, they tell us what our position is at t equals two. So what the position at t equals seven would be your position at t equals two plus your change in position from t equals two to t is equal to seven. And there's another word for this. You could also call this your displacement from t equals two to t equals seven. And we know how to think about displacement."}, {"video_title": "Analyzing motion problems position AP Calculus AB Khan Academy.mp3", "Sentence": "So what the position at t equals seven would be your position at t equals two plus your change in position from t equals two to t is equal to seven. And there's another word for this. You could also call this your displacement from t equals two to t equals seven. And we know how to think about displacement. Velocity is your rate of change of displacement. And so if you wanna figure out your displacement, you're between two times, you would integrate the velocity function, so this is going to be the integral from t equals two to t equals seven of our velocity function, v of t dt. This would be our displacement from time two to time seven."}, {"video_title": "Analyzing motion problems position AP Calculus AB Khan Academy.mp3", "Sentence": "And we know how to think about displacement. Velocity is your rate of change of displacement. And so if you wanna figure out your displacement, you're between two times, you would integrate the velocity function, so this is going to be the integral from t equals two to t equals seven of our velocity function, v of t dt. This would be our displacement from time two to time seven. So if they said what is our change in position from time two to time seven, it would be just this expression. But they want us to figure out, or they want Divya to figure out, what is the particle's position at t equals seven seconds? So what you'd wanna do is your position at t equals two, and we know what our position at t equals two is."}, {"video_title": "Analyzing motion problems position AP Calculus AB Khan Academy.mp3", "Sentence": "This would be our displacement from time two to time seven. So if they said what is our change in position from time two to time seven, it would be just this expression. But they want us to figure out, or they want Divya to figure out, what is the particle's position at t equals seven seconds? So what you'd wanna do is your position at t equals two, and we know what our position at t equals two is. It was eight meters in the positive direction, so we could just call that positive eight meters. So it's going to be eight plus your change in position, which is going to be your displacement. And let's see, we can see this choice right over there, and that's what we would pick."}, {"video_title": "Analyzing motion problems position AP Calculus AB Khan Academy.mp3", "Sentence": "So what you'd wanna do is your position at t equals two, and we know what our position at t equals two is. It was eight meters in the positive direction, so we could just call that positive eight meters. So it's going to be eight plus your change in position, which is going to be your displacement. And let's see, we can see this choice right over there, and that's what we would pick. This first option, v of seven, that just gives us our velocity at time seven, or exactly at seven seconds. Or another way is our rate of change of displacement at seven seconds, so that's not what we want. This one right over here, you have your position at t equals two, but then you have your change in position from t equals zero to t equals seven."}, {"video_title": "Analyzing motion problems position AP Calculus AB Khan Academy.mp3", "Sentence": "And let's see, we can see this choice right over there, and that's what we would pick. This first option, v of seven, that just gives us our velocity at time seven, or exactly at seven seconds. Or another way is our rate of change of displacement at seven seconds, so that's not what we want. This one right over here, you have your position at t equals two, but then you have your change in position from t equals zero to t equals seven. So this doesn't seem, this isn't right. And this is your position at time two plus your v prime, the derivative of velocity, is the acceleration, plus your acceleration at time seven. So that's definitely not gonna give you the particle's position, so we like that second choice."}, {"video_title": "Finding increasing interval given the derivative AP Calculus AB Khan Academy.mp3", "Sentence": "On which intervals is g increasing? Well at first you might say, well they don't even give us g. How do we figure out when g is increasing? Well, the answer is, all we need is g prime, which they do give us. And saying on which intervals is g increasing, that's equivalent to saying on which intervals is the first derivative with respect to x, on which intervals is that going to be greater than zero? If your rate of change with respect to x is greater than zero, if it's positive, then your function itself is going to be increasing. And so there's a couple of ways that we could approach this. You might just want to inspect kind of the structure of this expression and think about, well when is that going to be greater than zero?"}, {"video_title": "Finding increasing interval given the derivative AP Calculus AB Khan Academy.mp3", "Sentence": "And saying on which intervals is g increasing, that's equivalent to saying on which intervals is the first derivative with respect to x, on which intervals is that going to be greater than zero? If your rate of change with respect to x is greater than zero, if it's positive, then your function itself is going to be increasing. And so there's a couple of ways that we could approach this. You might just want to inspect kind of the structure of this expression and think about, well when is that going to be greater than zero? Or we could do it a little bit more methodically. We could say, well let's look at the critical points or the critical values for g. So critical, critical points for g. And just to remind ourselves what critical points are, that is when g prime of x is equal to zero or g prime of x is undefined, is undefined. And we have videos on critical points or critical values."}, {"video_title": "Finding increasing interval given the derivative AP Calculus AB Khan Academy.mp3", "Sentence": "You might just want to inspect kind of the structure of this expression and think about, well when is that going to be greater than zero? Or we could do it a little bit more methodically. We could say, well let's look at the critical points or the critical values for g. So critical, critical points for g. And just to remind ourselves what critical points are, that is when g prime of x is equal to zero or g prime of x is undefined, is undefined. And we have videos on critical points or critical values. And why those are relevant is those are the places, those are the possible places where the sign could change, the sign of g prime could change. So when is g prime of x equal to zero? Well the way to get g prime of x equal to zero is getting the numerator equal to zero and that's only going to happen if x squared is equal to zero or if x is equal to zero."}, {"video_title": "Finding increasing interval given the derivative AP Calculus AB Khan Academy.mp3", "Sentence": "And we have videos on critical points or critical values. And why those are relevant is those are the places, those are the possible places where the sign could change, the sign of g prime could change. So when is g prime of x equal to zero? Well the way to get g prime of x equal to zero is getting the numerator equal to zero and that's only going to happen if x squared is equal to zero or if x is equal to zero. So that's the only place where g prime of x is equal to zero. And where is g prime of x undefined? Well it's going to be undefined if the denominator becomes undefined."}, {"video_title": "Finding increasing interval given the derivative AP Calculus AB Khan Academy.mp3", "Sentence": "Well the way to get g prime of x equal to zero is getting the numerator equal to zero and that's only going to happen if x squared is equal to zero or if x is equal to zero. So that's the only place where g prime of x is equal to zero. And where is g prime of x undefined? Well it's going to be undefined if the denominator becomes undefined. The denominator becomes undefined if the denominator is zero. And so that's going to happen if x minus two is equal to zero x minus two is equal to zero or x is equal to two. So we have two critical points or critical values here."}, {"video_title": "Finding increasing interval given the derivative AP Calculus AB Khan Academy.mp3", "Sentence": "Well it's going to be undefined if the denominator becomes undefined. The denominator becomes undefined if the denominator is zero. And so that's going to happen if x minus two is equal to zero x minus two is equal to zero or x is equal to two. So we have two critical points or critical values here. And what I'm gonna do is I'm gonna graph them, let's put them on a number line. And let's just think about what g prime is doing in the intervals between the critical points. So let's start at zero, one, two, three, and then let's go to negative one."}, {"video_title": "Finding increasing interval given the derivative AP Calculus AB Khan Academy.mp3", "Sentence": "So we have two critical points or critical values here. And what I'm gonna do is I'm gonna graph them, let's put them on a number line. And let's just think about what g prime is doing in the intervals between the critical points. So let's start at zero, one, two, three, and then let's go to negative one. And we have a critical point at, let me do that in magenta. We have a critical point at x equals zero, right over there. And we have a critical point at x equals two, right over there."}, {"video_title": "Finding increasing interval given the derivative AP Calculus AB Khan Academy.mp3", "Sentence": "So let's start at zero, one, two, three, and then let's go to negative one. And we have a critical point at, let me do that in magenta. We have a critical point at x equals zero, right over there. And we have a critical point at x equals two, right over there. And so let's think about what g prime is doing in the intervals between the critical values or on either side of the critical values. So let's think about, let's first think about this interval. Let me do it in this purple color."}, {"video_title": "Finding increasing interval given the derivative AP Calculus AB Khan Academy.mp3", "Sentence": "And we have a critical point at x equals two, right over there. And so let's think about what g prime is doing in the intervals between the critical values or on either side of the critical values. So let's think about, let's first think about this interval. Let me do it in this purple color. Let's think about the interval between negative infinity and zero. So if we think about this interval, so negative infinity and zero, that open interval. Well, if we look at g prime, the numerator is still going to be positive."}, {"video_title": "Finding increasing interval given the derivative AP Calculus AB Khan Academy.mp3", "Sentence": "Let me do it in this purple color. Let's think about the interval between negative infinity and zero. So if we think about this interval, so negative infinity and zero, that open interval. Well, if we look at g prime, the numerator is still going to be positive. You take any negative value, you square it, you're gonna get a positive value. So this is going to be positive. Now what about the denominator?"}, {"video_title": "Finding increasing interval given the derivative AP Calculus AB Khan Academy.mp3", "Sentence": "Well, if we look at g prime, the numerator is still going to be positive. You take any negative value, you square it, you're gonna get a positive value. So this is going to be positive. Now what about the denominator? You take a negative number, you subtract two from it, you're still gonna get a negative number, and then you take it to the third power. Well, a negative number to the third power is gonna be a negative number. So that right over there is going to be negative."}, {"video_title": "Finding increasing interval given the derivative AP Calculus AB Khan Academy.mp3", "Sentence": "Now what about the denominator? You take a negative number, you subtract two from it, you're still gonna get a negative number, and then you take it to the third power. Well, a negative number to the third power is gonna be a negative number. So that right over there is going to be negative. So you're gonna have a positive divided by a negative. So g prime is going to be negative. So let me write that down."}, {"video_title": "Finding increasing interval given the derivative AP Calculus AB Khan Academy.mp3", "Sentence": "So that right over there is going to be negative. So you're gonna have a positive divided by a negative. So g prime is going to be negative. So let me write that down. So on this interval, on this interval, I'll write it like this, g prime of x is less than zero. Or if we cared, if we wanted to know when it's decreasing, we would know it's definitely decreasing over that interval. Now let's take the interval between zero and two, right over here."}, {"video_title": "Finding increasing interval given the derivative AP Calculus AB Khan Academy.mp3", "Sentence": "So let me write that down. So on this interval, on this interval, I'll write it like this, g prime of x is less than zero. Or if we cared, if we wanted to know when it's decreasing, we would know it's definitely decreasing over that interval. Now let's take the interval between zero and two, right over here. So this is the interval from zero to two, the open interval. So what's gonna go on with g prime of x here? Well, once again, x squared, anything greater than zero, and this is, we're not including zero in this interval, well, this is for sure going to be positive."}, {"video_title": "Finding increasing interval given the derivative AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's take the interval between zero and two, right over here. So this is the interval from zero to two, the open interval. So what's gonna go on with g prime of x here? Well, once again, x squared, anything greater than zero, and this is, we're not including zero in this interval, well, this is for sure going to be positive. And so let's see, if we have x minus two, where x is greater than zero but less than two. So if x, we could just say, for example, if x was one, one minus two is negative one. We're still gonna get negative values in this denominator right over here."}, {"video_title": "Finding increasing interval given the derivative AP Calculus AB Khan Academy.mp3", "Sentence": "Well, once again, x squared, anything greater than zero, and this is, we're not including zero in this interval, well, this is for sure going to be positive. And so let's see, if we have x minus two, where x is greater than zero but less than two. So if x, we could just say, for example, if x was one, one minus two is negative one. We're still gonna get negative values in this denominator right over here. So since we're still gonna get negative values in this denominator, the denominator is still going to be, take a negative value to the third power, well, you're gonna still get a negative value. So this is going to be negative. Negative."}, {"video_title": "Finding increasing interval given the derivative AP Calculus AB Khan Academy.mp3", "Sentence": "We're still gonna get negative values in this denominator right over here. So since we're still gonna get negative values in this denominator, the denominator is still going to be, take a negative value to the third power, well, you're gonna still get a negative value. So this is going to be negative. Negative. So you're still going to have g prime as less than zero. So let me write that down. So you still have g prime of x is less than zero."}, {"video_title": "Finding increasing interval given the derivative AP Calculus AB Khan Academy.mp3", "Sentence": "Negative. So you're still going to have g prime as less than zero. So let me write that down. So you still have g prime of x is less than zero. And then let's take the interval above. Let's take the interval from two to infinity. Two to infinity."}, {"video_title": "Finding increasing interval given the derivative AP Calculus AB Khan Academy.mp3", "Sentence": "So you still have g prime of x is less than zero. And then let's take the interval above. Let's take the interval from two to infinity. Two to infinity. Well, the numerator is positive, it's always gonna be positive for any x not being equal to zero. And this denominator, you're taking values greater than two, subtracting two from it, which is still gonna give you a positive value. You take the third power, it's all gonna be positive."}, {"video_title": "Finding increasing interval given the derivative AP Calculus AB Khan Academy.mp3", "Sentence": "Two to infinity. Well, the numerator is positive, it's always gonna be positive for any x not being equal to zero. And this denominator, you're taking values greater than two, subtracting two from it, which is still gonna give you a positive value. You take the third power, it's all gonna be positive. It is all going to be positive. So this is the interval where g prime of x is greater than zero. So on which intervals is g increasing?"}, {"video_title": "Finding increasing interval given the derivative AP Calculus AB Khan Academy.mp3", "Sentence": "You take the third power, it's all gonna be positive. It is all going to be positive. So this is the interval where g prime of x is greater than zero. So on which intervals is g increasing? Well, that's where g prime of x is greater than zero. So it's going to be from two, from two to infinity. Or we could just write it like this."}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say that something is traveling at a constant rate of five meters per second. That's its velocity in one dimension. If it was negative, it would be moving to the left. If it's positive, it's moving to the right. Let's say that we care about what is our change in distance over, the delta symbol represents change, over a change in time of four seconds. I could say from t equals zero to t is equal to four. That's our change in time."}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "If it's positive, it's moving to the right. Let's say that we care about what is our change in distance over, the delta symbol represents change, over a change in time of four seconds. I could say from t equals zero to t is equal to four. That's our change in time. That's our four second interval that we care about. One way to think about it is a rate, by definition, is nothing but a change in some quantity, in this case it's distance, over a change in some other quantity. In this case, we're thinking about time."}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "That's our change in time. That's our four second interval that we care about. One way to think about it is a rate, by definition, is nothing but a change in some quantity, in this case it's distance, over a change in some other quantity. In this case, we're thinking about time. Another way to think about it, if we multiply both sides times change in time, you get your change in distance is equal to your rate times change in time. This is very close to, you might remember from pre-algebra, distance is equal to rate times time. That just comes from the definition of what a rate is."}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "In this case, we're thinking about time. Another way to think about it, if we multiply both sides times change in time, you get your change in distance is equal to your rate times change in time. This is very close to, you might remember from pre-algebra, distance is equal to rate times time. That just comes from the definition of what a rate is. It's a change in one quantity with respect to another quantity. If you just apply this, if you say, my rate is a constant five meters per second, and my delta t is four seconds, so times four seconds, that's just going to give you 20. Let me do it in that same color that I had for the change in distance."}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "That just comes from the definition of what a rate is. It's a change in one quantity with respect to another quantity. If you just apply this, if you say, my rate is a constant five meters per second, and my delta t is four seconds, so times four seconds, that's just going to give you 20. Let me do it in that same color that I had for the change in distance. That's going to be 20, and then the seconds cancel with seconds, 20 meters. My total change in distance over those four seconds is going to be 20 meters. Nothing new here, nothing too fancy."}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "Let me do it in that same color that I had for the change in distance. That's going to be 20, and then the seconds cancel with seconds, 20 meters. My total change in distance over those four seconds is going to be 20 meters. Nothing new here, nothing too fancy. What I want to do now is connect this to the area under the rate function over this time period. Let's graph that. Let's graph it."}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "Nothing new here, nothing too fancy. What I want to do now is connect this to the area under the rate function over this time period. Let's graph that. Let's graph it. That's my rate axis. This is my time axis. This is going to be in seconds."}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "Let's graph it. That's my rate axis. This is my time axis. This is going to be in seconds. This is going to be in meters per second. Let's see, one, two, three, four, five. That seems about enough."}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "This is going to be in seconds. This is going to be in meters per second. Let's see, one, two, three, four, five. That seems about enough. Then I go one, two, three, four, five. Our rate, at least in this example, is a constant. It is a constant five meters per second."}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "That seems about enough. Then I go one, two, three, four, five. Our rate, at least in this example, is a constant. It is a constant five meters per second. That is my r of t in this example. What did we just do here? We just multiplied our change in time times our constant rate."}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "It is a constant five meters per second. That is my r of t in this example. What did we just do here? We just multiplied our change in time times our constant rate. We just multiplied our change in time. It was from time equals zero seconds to four seconds. It's this length here if we think of it on that axis."}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "We just multiplied our change in time times our constant rate. We just multiplied our change in time. It was from time equals zero seconds to four seconds. It's this length here if we think of it on that axis. We multiplied it times our constant rate. We multiplied it times this right over here. If I multiply this base times this height, what am I going to get?"}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "It's this length here if we think of it on that axis. We multiplied it times our constant rate. We multiplied it times this right over here. If I multiply this base times this height, what am I going to get? I'm going to get the area under, I am going to get this area under the rate function. That area is going to be 20. If we went with the units of them, obviously you're used to things of area being something unit squared because it's usually meters times meters or miles times miles or inches times inches."}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "If I multiply this base times this height, what am I going to get? I'm going to get the area under, I am going to get this area under the rate function. That area is going to be 20. If we went with the units of them, obviously you're used to things of area being something unit squared because it's usually meters times meters or miles times miles or inches times inches. It would be inches squared, meters squared or miles squared. Here if we go with the units of the axes, it would be meters per second times seconds, which is going to get you meters. The important thing here is that the units here or the area here is 20."}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "If we went with the units of them, obviously you're used to things of area being something unit squared because it's usually meters times meters or miles times miles or inches times inches. It would be inches squared, meters squared or miles squared. Here if we go with the units of the axes, it would be meters per second times seconds, which is going to get you meters. The important thing here is that the units here or the area here is 20. At least for that very simple example, it looks like the area under the rate curve is equal to the net change over that time period where the rate is something with respect to time. Let's test that a little more. Let's get a little more intuition here."}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "The important thing here is that the units here or the area here is 20. At least for that very simple example, it looks like the area under the rate curve is equal to the net change over that time period where the rate is something with respect to time. Let's test that a little more. Let's get a little more intuition here. Let's say that we had a different rate function. Let's say that, let me make it with a different, let's say instead we had a rate function. I'll still use yellow."}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "Let's get a little more intuition here. Let's say that we had a different rate function. Let's say that, let me make it with a different, let's say instead we had a rate function. I'll still use yellow. Let's say we have a rate function that is, let me make it a little bit interesting. Let's say it's one meter per second for time is greater than or equal, or zero is less than or equal to time, which is less than or equal to two seconds. This is obviously, this is all in seconds where we're talking about time."}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "I'll still use yellow. Let's say we have a rate function that is, let me make it a little bit interesting. Let's say it's one meter per second for time is greater than or equal, or zero is less than or equal to time, which is less than or equal to two seconds. This is obviously, this is all in seconds where we're talking about time. There's two meters per second for t is greater than two seconds. What's that going to look like? Actually, try to graph it yourself and just say, what is the total change in distance over the first, let's say, five seconds?"}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "This is obviously, this is all in seconds where we're talking about time. There's two meters per second for t is greater than two seconds. What's that going to look like? Actually, try to graph it yourself and just say, what is the total change in distance over the first, let's say, five seconds? We want to do delta t over not the first four seconds, but the first five seconds. Let's graph it. This is one meter per second."}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "Actually, try to graph it yourself and just say, what is the total change in distance over the first, let's say, five seconds? We want to do delta t over not the first four seconds, but the first five seconds. Let's graph it. This is one meter per second. That is two meters per second. It's in meters per second. This is my rate axis."}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "This is one meter per second. That is two meters per second. It's in meters per second. This is my rate axis. This right over there is going to be my time axis. One, they're obviously not at the same scale, three, four, five. What does this rate function look like?"}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "This is my rate axis. This right over there is going to be my time axis. One, they're obviously not at the same scale, three, four, five. What does this rate function look like? My rate is one meter per second between time is zero and two, including two seconds. Then the rate jumps. This isn't that realistic."}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "What does this rate function look like? My rate is one meter per second between time is zero and two, including two seconds. Then the rate jumps. This isn't that realistic. Nothing can accelerate instantly like this. You'd need an infinite force or an infinitely small mass, maybe there's some things. Anyway, I don't want to get too complex there."}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "This isn't that realistic. Nothing can accelerate instantly like this. You'd need an infinite force or an infinitely small mass, maybe there's some things. Anyway, I don't want to get too complex there. This is unrealistic. It's not typical for something to just have an instantaneous velocity increase like that. Let's just go with it."}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "Anyway, I don't want to get too complex there. This is unrealistic. It's not typical for something to just have an instantaneous velocity increase like that. Let's just go with it. Then after the two seconds, we are at a constant rate of two meters per second. What is our total change in distance over the first five seconds here? Here we care about the first five seconds."}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "Let's just go with it. Then after the two seconds, we are at a constant rate of two meters per second. What is our total change in distance over the first five seconds here? Here we care about the first five seconds. We can break up the problem. We could say over the first two seconds, change in time is two seconds times our, we have a constant rate over those two seconds. It's going to be two seconds times one meter per second."}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "Here we care about the first five seconds. We can break up the problem. We could say over the first two seconds, change in time is two seconds times our, we have a constant rate over those two seconds. It's going to be two seconds times one meter per second. That's going to give us two meters. This here is going to be, actually let me do that orange color. That's going to give us two meters there."}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "It's going to be two seconds times one meter per second. That's going to give us two meters. This here is going to be, actually let me do that orange color. That's going to give us two meters there. Then we look at the next section. Our change in time here is three seconds. Then we multiply that times our constant two meters per second."}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "That's going to give us two meters there. Then we look at the next section. Our change in time here is three seconds. Then we multiply that times our constant two meters per second. That's going to give us an area of six. If we look at the units, in both cases we're multiplying seconds times meters per second, which is going to give us meters. It's going to be two plus six meters or eight meters."}, {"video_title": "Area under rate function gives the net change AP Calculus AB Khan Academy.mp3", "Sentence": "Then we multiply that times our constant two meters per second. That's going to give us an area of six. If we look at the units, in both cases we're multiplying seconds times meters per second, which is going to give us meters. It's going to be two plus six meters or eight meters. Hopefully this is giving you the intuition that the area under the rate curve or the rate function is going to give you our total net change in whatever that rate thing was finding the rate of. In this case, it is distance per unit time. If you take the area under the rate function, that kind of distance per this speed or this velocity function over some period of time, that area is going to be our total net change in distance."}, {"video_title": "Worked example Quotient rule with table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Let capital F be a function defined as, so capital F is defined as lowercase f of x divided by lowercase g of x. And they want us to evaluate the derivative of capital F at x equals negative one. So the way that we can do that is let's just take the derivative of capital F and then evaluate it as at x equals one. And the way they've set up capital F, this function definition, we can see that it is the quotient of two functions. So if we want to take its derivative, you might say, well maybe the quotient rule is important here. And I'll always give you my aside. The quotient rule, I'm gonna state it right now, and it could be useful to know it, but in case you ever forget it, you can derive it pretty quickly from the product rule, and if you know it, the chain rule combined, you can get the quotient rule pretty quick."}, {"video_title": "Worked example Quotient rule with table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And the way they've set up capital F, this function definition, we can see that it is the quotient of two functions. So if we want to take its derivative, you might say, well maybe the quotient rule is important here. And I'll always give you my aside. The quotient rule, I'm gonna state it right now, and it could be useful to know it, but in case you ever forget it, you can derive it pretty quickly from the product rule, and if you know it, the chain rule combined, you can get the quotient rule pretty quick. But let me just state the quotient rule right now. So if you have some function defined as some function in a numerator divided by some function in the denominator, we can say its derivative, and this is really just a restatement of the quotient rule, its derivative is going to be the derivative of the function in the numerator, so d dx f of x times the function in the denominator, so times g of x minus, minus the function in the numerator, minus f of x, not taking its derivative, times the derivative of the function in the denominator, d dx g of x, all of that over, so all of this is going to be over the function in the denominator squared. So this g of x squared, g of x, g of x squared."}, {"video_title": "Worked example Quotient rule with table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "The quotient rule, I'm gonna state it right now, and it could be useful to know it, but in case you ever forget it, you can derive it pretty quickly from the product rule, and if you know it, the chain rule combined, you can get the quotient rule pretty quick. But let me just state the quotient rule right now. So if you have some function defined as some function in a numerator divided by some function in the denominator, we can say its derivative, and this is really just a restatement of the quotient rule, its derivative is going to be the derivative of the function in the numerator, so d dx f of x times the function in the denominator, so times g of x minus, minus the function in the numerator, minus f of x, not taking its derivative, times the derivative of the function in the denominator, d dx g of x, all of that over, so all of this is going to be over the function in the denominator squared. So this g of x squared, g of x, g of x squared. And you could use different types of notation here, you could say, instead of writing this with the derivative operator, you could say this is the same thing as g prime of x, and likewise, you could say, well, that is the same thing as f prime of x. And so now we just want to evaluate this thing, and you might say, well, how do I evaluate this thing? Well, let's just try it."}, {"video_title": "Worked example Quotient rule with table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So this g of x squared, g of x, g of x squared. And you could use different types of notation here, you could say, instead of writing this with the derivative operator, you could say this is the same thing as g prime of x, and likewise, you could say, well, that is the same thing as f prime of x. And so now we just want to evaluate this thing, and you might say, well, how do I evaluate this thing? Well, let's just try it. Let's just say, well, we want to evaluate f prime when x is equal to negative one. So we can write f prime of negative one is equal to, well, everywhere we see an x, let's put a negative one here. It's going to be f prime of negative one, f prime, lowercase f prime, it's a little confusing, lowercase f prime of negative one times g of negative one, g of negative one, minus f of negative one, f of negative one, times g prime of negative one, g prime of negative one, all of that over, let me do it in that same color, so I take my colors seriously."}, {"video_title": "Worked example Quotient rule with table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, let's just try it. Let's just say, well, we want to evaluate f prime when x is equal to negative one. So we can write f prime of negative one is equal to, well, everywhere we see an x, let's put a negative one here. It's going to be f prime of negative one, f prime, lowercase f prime, it's a little confusing, lowercase f prime of negative one times g of negative one, g of negative one, minus f of negative one, f of negative one, times g prime of negative one, g prime of negative one, all of that over, let me do it in that same color, so I take my colors seriously. All right, all of that over, g of negative one squared, g of negative one squared. Now, can we figure out what f prime of negative one, f of negative one, g of negative one, and g prime of negative one, what they are? Well, some of them they tell us outright."}, {"video_title": "Worked example Quotient rule with table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "It's going to be f prime of negative one, f prime, lowercase f prime, it's a little confusing, lowercase f prime of negative one times g of negative one, g of negative one, minus f of negative one, f of negative one, times g prime of negative one, g prime of negative one, all of that over, let me do it in that same color, so I take my colors seriously. All right, all of that over, g of negative one squared, g of negative one squared. Now, can we figure out what f prime of negative one, f of negative one, g of negative one, and g prime of negative one, what they are? Well, some of them they tell us outright. They tell us f and f prime at negative one, and for g, we can actually solve for those. So let's see, if this is, well, let's just first evaluate g of negative one. g of negative one is going to be two times negative one to the third power."}, {"video_title": "Worked example Quotient rule with table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, some of them they tell us outright. They tell us f and f prime at negative one, and for g, we can actually solve for those. So let's see, if this is, well, let's just first evaluate g of negative one. g of negative one is going to be two times negative one to the third power. Well, negative one to the third power is just negative one times two, so this is negative two, and g prime of x, I'll do it here, g prime of x, which use the power rule, bring that three out front, three times two is six, x, decrement that exponent, three minus one is two, and so g prime of negative one is equal to six times negative one squared. Well, negative one squared is just one, so this is going to be equal to six. So we actually know what all of these values are now."}, {"video_title": "Worked example Quotient rule with table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "g of negative one is going to be two times negative one to the third power. Well, negative one to the third power is just negative one times two, so this is negative two, and g prime of x, I'll do it here, g prime of x, which use the power rule, bring that three out front, three times two is six, x, decrement that exponent, three minus one is two, and so g prime of negative one is equal to six times negative one squared. Well, negative one squared is just one, so this is going to be equal to six. So we actually know what all of these values are now. We know, so first we want to figure out f prime of negative one. Well, they tell us that right over here. f prime of negative one is equal to five, so that is five."}, {"video_title": "Worked example Quotient rule with table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So we actually know what all of these values are now. We know, so first we want to figure out f prime of negative one. Well, they tell us that right over here. f prime of negative one is equal to five, so that is five. g of negative one, well, we figure that right here. g of negative one is negative two, so this is negative two. f of negative one, so f of negative one, they tell us that right over there."}, {"video_title": "Worked example Quotient rule with table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "f prime of negative one is equal to five, so that is five. g of negative one, well, we figure that right here. g of negative one is negative two, so this is negative two. f of negative one, so f of negative one, they tell us that right over there. That is equal to three, and then g prime of negative one, let me just circle it in this green color, g prime of negative one, we figured it out. It is equal to six, so this is equal to six, and then finally g of negative one right over here, we already figured that out. That was equal to negative two, so negative two."}, {"video_title": "Worked example Quotient rule with table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "f of negative one, so f of negative one, they tell us that right over there. That is equal to three, and then g prime of negative one, let me just circle it in this green color, g prime of negative one, we figured it out. It is equal to six, so this is equal to six, and then finally g of negative one right over here, we already figured that out. That was equal to negative two, so negative two. So this is all going to simplify to, so you have five times negative two, which is negative 10, minus three times six, which is 18, all of that over negative two squared. Well, negative two squared is just going to be positive four, so this is going to be equal to negative 28 over positive four, which is equal to negative seven. And there you have it."}, {"video_title": "Functions defined by integrals Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Let's call it capital G of x. And it's equal to the definite integral between t is equal to negative three and t is equal to x of our lower case g of t, g of t dt. So given how we have just defined capital G of x, let's see if we can find some values. So let's evaluate the function capital G at x is equal to four. Let's also evaluate the function capital G at x is equal to eight. And I encourage you to pause the video right now and try to think about these on your own, and then we can work through them together. So let's tackle capital G of four first."}, {"video_title": "Functions defined by integrals Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So let's evaluate the function capital G at x is equal to four. Let's also evaluate the function capital G at x is equal to eight. And I encourage you to pause the video right now and try to think about these on your own, and then we can work through them together. So let's tackle capital G of four first. So this is going to be, well, if x is equal to four, this top boundary is going to be four. So this is going to be the definite integral of g of t between t is equal to negative three and t is equal to four, g of t, g of t dt. Now what's that going to be equal to?"}, {"video_title": "Functions defined by integrals Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So let's tackle capital G of four first. So this is going to be, well, if x is equal to four, this top boundary is going to be four. So this is going to be the definite integral of g of t between t is equal to negative three and t is equal to four, g of t, g of t dt. Now what's that going to be equal to? Well, let's see. Let's look at this graph here. So we have negative three."}, {"video_title": "Functions defined by integrals Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Now what's that going to be equal to? Well, let's see. Let's look at this graph here. So we have negative three. So this t is equal to negative three, which is right over here. t is equal to negative three. And we're going to go all the way to t is equal to four."}, {"video_title": "Functions defined by integrals Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So we have negative three. So this t is equal to negative three, which is right over here. t is equal to negative three. And we're going to go all the way to t is equal to four. Let me circle that in orange. All the way to t is equal to four, which is right over here. So this is, one way to think about this is that this is going to be the area above the t axis and below the graph of g. So it's going to be this area right over here that is above the t axis and below the graph of g of t. But we're not going to add to it this area because this area over here, I'll shade it in yellow, this area I'm shading in yellow over here, this is going to be negative."}, {"video_title": "Functions defined by integrals Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And we're going to go all the way to t is equal to four. Let me circle that in orange. All the way to t is equal to four, which is right over here. So this is, one way to think about this is that this is going to be the area above the t axis and below the graph of g. So it's going to be this area right over here that is above the t axis and below the graph of g of t. But we're not going to add to it this area because this area over here, I'll shade it in yellow, this area I'm shading in yellow over here, this is going to be negative. Why is it going to be negative? Because we want the area that is above t and below g. This is the other way around. This is below the t axis and above the graph of g. So one way to think about it is we could split it up."}, {"video_title": "Functions defined by integrals Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So this is, one way to think about this is that this is going to be the area above the t axis and below the graph of g. So it's going to be this area right over here that is above the t axis and below the graph of g of t. But we're not going to add to it this area because this area over here, I'll shade it in yellow, this area I'm shading in yellow over here, this is going to be negative. Why is it going to be negative? Because we want the area that is above t and below g. This is the other way around. This is below the t axis and above the graph of g. So one way to think about it is we could split it up. So we could, actually let me just clear this out so we have more space. So this right over here is going to be equal to the integral, I'll do that in this purple color, the integral from t equals negative three to zero of g of t dt plus, I'll do this in the yellow color, plus the integral from zero, t equals zero, to t is equal to four of g of t dt. Now what are each of these going to be?"}, {"video_title": "Functions defined by integrals Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "This is below the t axis and above the graph of g. So one way to think about it is we could split it up. So we could, actually let me just clear this out so we have more space. So this right over here is going to be equal to the integral, I'll do that in this purple color, the integral from t equals negative three to zero of g of t dt plus, I'll do this in the yellow color, plus the integral from zero, t equals zero, to t is equal to four of g of t dt. Now what are each of these going to be? Well this is just a triangle where the base is three, the base has length three, the height is length three, so it's going to be three times three, which is nine times one half, because three times three would give us the area of this entire square, but this triangle is half of that. So this right over here is going to be 4.5. And then what's this area in yellow?"}, {"video_title": "Functions defined by integrals Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Now what are each of these going to be? Well this is just a triangle where the base is three, the base has length three, the height is length three, so it's going to be three times three, which is nine times one half, because three times three would give us the area of this entire square, but this triangle is half of that. So this right over here is going to be 4.5. And then what's this area in yellow? Well let's see, we have a triangle, its width here is four, its height is four, four times four is 16, which would be the area of this entire square, we take half of that, it's eight. Now we don't just add it to it, because once again, this is going to be negative area, the graph over here is below the t axis. So this we would say, this integral is going to evaluate to negative eight."}, {"video_title": "Functions defined by integrals Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And then what's this area in yellow? Well let's see, we have a triangle, its width here is four, its height is four, four times four is 16, which would be the area of this entire square, we take half of that, it's eight. Now we don't just add it to it, because once again, this is going to be negative area, the graph over here is below the t axis. So this we would say, this integral is going to evaluate to negative eight. Once again, why is it negative eight? Because the graph is below the t axis. And so what do we get?"}, {"video_title": "Functions defined by integrals Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So this we would say, this integral is going to evaluate to negative eight. Once again, why is it negative eight? Because the graph is below the t axis. And so what do we get? We get g of four, which is this area, essentially minus this area, 4.5 minus eight is going to be, let's see, four minus eight is negative four, add a.5, that's negative 3.5. Now let's try to figure out what g of eight is. So g of eight."}, {"video_title": "Functions defined by integrals Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And so what do we get? We get g of four, which is this area, essentially minus this area, 4.5 minus eight is going to be, let's see, four minus eight is negative four, add a.5, that's negative 3.5. Now let's try to figure out what g of eight is. So g of eight. And if you couldn't figure it out the first time around, try to pause the video again, and now that we've figured out g of four, try to figure out what g of eight is. Well g of eight is, one way to think about it, it's going to be that minus that area, and then we're going to add, and then we're going to figure out the area, and then we're going to have two more areas to think about. We're going to go all the way to eight, so actually let me draw the line there."}, {"video_title": "Functions defined by integrals Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So g of eight. And if you couldn't figure it out the first time around, try to pause the video again, and now that we've figured out g of four, try to figure out what g of eight is. Well g of eight is, one way to think about it, it's going to be that minus that area, and then we're going to add, and then we're going to figure out the area, and then we're going to have two more areas to think about. We're going to go all the way to eight, so actually let me draw the line there. So we're going to have to think about, we're going to think about this whole area now, this whole area now, and then we're going to think about this one. So we could write, this is going to be equal to the integral between, let me do that purple color. So it's going to be this integral, which is the integral between negative t equals negative three and zero, g of t dt, plus this entire yellow region right now, the part that we looked at before, plus this part over here, so I'll just write that as plus the definite integral between t is equal to zero and six, g of t dt, and then finally, plus the definite integral between t is equal to six and t is equal to eight, g of t dt."}, {"video_title": "Functions defined by integrals Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "We're going to go all the way to eight, so actually let me draw the line there. So we're going to have to think about, we're going to think about this whole area now, this whole area now, and then we're going to think about this one. So we could write, this is going to be equal to the integral between, let me do that purple color. So it's going to be this integral, which is the integral between negative t equals negative three and zero, g of t dt, plus this entire yellow region right now, the part that we looked at before, plus this part over here, so I'll just write that as plus the definite integral between t is equal to zero and six, g of t dt, and then finally, plus the definite integral between t is equal to six and t is equal to eight, g of t dt. Now, we already know that this first one evaluates to 4.5, so that one is 4.5. Now what's this one going to be? We have a triangle whose base is length six, height is four, six times four is 24, times 1.5 is going to get us 12."}, {"video_title": "Functions defined by integrals Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So it's going to be this integral, which is the integral between negative t equals negative three and zero, g of t dt, plus this entire yellow region right now, the part that we looked at before, plus this part over here, so I'll just write that as plus the definite integral between t is equal to zero and six, g of t dt, and then finally, plus the definite integral between t is equal to six and t is equal to eight, g of t dt. Now, we already know that this first one evaluates to 4.5, so that one is 4.5. Now what's this one going to be? We have a triangle whose base is length six, height is four, six times four is 24, times 1.5 is going to get us 12. So this is going to be, that over there is 12, and then finally, what is this area? Oh, and we have to be careful. This is below the t-axis and above the graph, so this is going to be a negative 12, and then finally we have this area, which is once again going to be a positive area, it's below the graph and above the t-axis, and so let's see, two times four is eight, times 1.5 is four, so we're going to have plus four."}, {"video_title": "Functions defined by integrals Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "We have a triangle whose base is length six, height is four, six times four is 24, times 1.5 is going to get us 12. So this is going to be, that over there is 12, and then finally, what is this area? Oh, and we have to be careful. This is below the t-axis and above the graph, so this is going to be a negative 12, and then finally we have this area, which is once again going to be a positive area, it's below the graph and above the t-axis, and so let's see, two times four is eight, times 1.5 is four, so we're going to have plus four. Plus four, and so what do we have? We end up with 4.5 plus four is 8.5 minus 12, so this is going to be equal to, 8.8 minus 12 would be negative four plus five, it is negative 3.5 again. Now why did these two things end up being the same?"}, {"video_title": "Arc length intro Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "We've used definite integrals to find areas. What I want to do now is to see if we can use a definite integral to find an arc length. What do I mean by that? Well, if I start at this point on the graph of a function, and if I were to go to this point right over here, not in a straight line, we know already how to find the distance in a straight line, but instead we want to find the distance along the curve. If we were to lay a string along the curve, what would be this distance right over here? That's what I'm talking about by arc length. By arc length."}, {"video_title": "Arc length intro Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "Well, if I start at this point on the graph of a function, and if I were to go to this point right over here, not in a straight line, we know already how to find the distance in a straight line, but instead we want to find the distance along the curve. If we were to lay a string along the curve, what would be this distance right over here? That's what I'm talking about by arc length. By arc length. We could think about it as, okay, well that's going to be from x equals a to x equals b along, along, along this curve. How could we do it? Well, the one thing that integration, integral calculus is teaching us is that when we see something that's changing like this, what we could do is we can break it up into infinitely small parts, infinitely small parts that we can approximate it with things like lines and rectangles, and then we could take the infinite sum of those infinitely small parts."}, {"video_title": "Arc length intro Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "By arc length. We could think about it as, okay, well that's going to be from x equals a to x equals b along, along, along this curve. How could we do it? Well, the one thing that integration, integral calculus is teaching us is that when we see something that's changing like this, what we could do is we can break it up into infinitely small parts, infinitely small parts that we can approximate it with things like lines and rectangles, and then we could take the infinite sum of those infinitely small parts. Let me break up my arc length into, let me break it up into infinitely small sections of arc length. Let me call each of those infinitely small sections of my arc length a, I guess I could say a length of differential, an arc length of differential. I'll call it ds."}, {"video_title": "Arc length intro Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "Well, the one thing that integration, integral calculus is teaching us is that when we see something that's changing like this, what we could do is we can break it up into infinitely small parts, infinitely small parts that we can approximate it with things like lines and rectangles, and then we could take the infinite sum of those infinitely small parts. Let me break up my arc length into, let me break it up into infinitely small sections of arc length. Let me call each of those infinitely small sections of my arc length a, I guess I could say a length of differential, an arc length of differential. I'll call it ds. ds. I'm going to draw it much bigger than when at least I conceptualize what a differential is just so that we can see it. What do I mean by breaking it up into these ds's?"}, {"video_title": "Arc length intro Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "I'll call it ds. ds. I'm going to draw it much bigger than when at least I conceptualize what a differential is just so that we can see it. What do I mean by breaking it up into these ds's? Well, if that's a ds, and then let me do these others in another color, that's another change, infinitely small change in my arc length, another infinitely small change in my arc length. If I summed all of these ds's together, I'm going to get the arc length. The arc length, if I take, is going to be the integral of all of these ds's."}, {"video_title": "Arc length intro Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "What do I mean by breaking it up into these ds's? Well, if that's a ds, and then let me do these others in another color, that's another change, infinitely small change in my arc length, another infinitely small change in my arc length. If I summed all of these ds's together, I'm going to get the arc length. The arc length, if I take, is going to be the integral of all of these ds's. All of these ds's summed together over this interval. We can denote it like this. But this doesn't help me right now."}, {"video_title": "Arc length intro Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "The arc length, if I take, is going to be the integral of all of these ds's. All of these ds's summed together over this interval. We can denote it like this. But this doesn't help me right now. This is in terms of this arc length differential. We know how to do things in terms of dx's and dy's. Let's see if we can re-express this in terms of dx's and dy's."}, {"video_title": "Arc length intro Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "But this doesn't help me right now. This is in terms of this arc length differential. We know how to do things in terms of dx's and dy's. Let's see if we can re-express this in terms of dx's and dy's. If we go on a really, really small scale, once again, we can approximate. This is going to be a line, just the way that we approximated area with rectangles at first, but if you have an infinite number of infinitely small rectangles, you're actually approximating a non-rectangular region, the area of a non-rectangular region. Similarly, we're approximating with lines here, but they're infinitely small and there's an infinite number of them."}, {"video_title": "Arc length intro Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "Let's see if we can re-express this in terms of dx's and dy's. If we go on a really, really small scale, once again, we can approximate. This is going to be a line, just the way that we approximated area with rectangles at first, but if you have an infinite number of infinitely small rectangles, you're actually approximating a non-rectangular region, the area of a non-rectangular region. Similarly, we're approximating with lines here, but they're infinitely small and there's an infinite number of them. You are actually finding the length of the curve. Just focusing on this as a line for now. This distance right over here, I'm just going to try to express it in terms of dx's and dy's."}, {"video_title": "Arc length intro Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "Similarly, we're approximating with lines here, but they're infinitely small and there's an infinite number of them. You are actually finding the length of the curve. Just focusing on this as a line for now. This distance right over here, I'm just going to try to express it in terms of dx's and dy's. This distance right over here, that's dx. You could view this as an infinitely small change in x. This distance right over here, this is a dy."}, {"video_title": "Arc length intro Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "This distance right over here, I'm just going to try to express it in terms of dx's and dy's. This distance right over here, that's dx. You could view this as an infinitely small change in x. This distance right over here, this is a dy. Once again, I'm being loosey-goosey with differentials, really to give you a conceptual understanding, not a rigorous proof, but it'll give you a sense of where the formula for arc length is actually coming from. Based on this, you can see that ds could be expressed, based on the Pythagorean theorem, as equal to dx squared plus dy squared. Or you could rewrite it as the square root of dx squared plus dy squared."}, {"video_title": "Arc length intro Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "This distance right over here, this is a dy. Once again, I'm being loosey-goosey with differentials, really to give you a conceptual understanding, not a rigorous proof, but it'll give you a sense of where the formula for arc length is actually coming from. Based on this, you can see that ds could be expressed, based on the Pythagorean theorem, as equal to dx squared plus dy squared. Or you could rewrite it as the square root of dx squared plus dy squared. We can rewrite this, we could say this is the same thing as the integral of, instead of writing ds, I'm going to write it as the square root of dx squared, dx squared plus dy squared, plus dy squared. Once again, this is straight out of the Pythagorean theorem. Now this is starting to get interesting."}, {"video_title": "Arc length intro Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "Or you could rewrite it as the square root of dx squared plus dy squared. We can rewrite this, we could say this is the same thing as the integral of, instead of writing ds, I'm going to write it as the square root of dx squared, dx squared plus dy squared, plus dy squared. Once again, this is straight out of the Pythagorean theorem. Now this is starting to get interesting. I've written in terms of dx's and dy's, but they're getting squared, they're under a radical sign. What can I do to simplify this? Or at least write it in a way that I know how to integrate."}, {"video_title": "Arc length intro Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "Now this is starting to get interesting. I've written in terms of dx's and dy's, but they're getting squared, they're under a radical sign. What can I do to simplify this? Or at least write it in a way that I know how to integrate. Well, I could factor out a dx squared. So let me just rewrite it. This is going to be the same thing as the integral of the square root, I'm going to factor out a dx squared, dx squared times, times one plus dy over dx squared."}, {"video_title": "Arc length intro Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "Or at least write it in a way that I know how to integrate. Well, I could factor out a dx squared. So let me just rewrite it. This is going to be the same thing as the integral of the square root, I'm going to factor out a dx squared, dx squared times, times one plus dy over dx squared. Notice this and this is the exact same quantity. If I distribute this dx squared, I'm going to get this right up here. And now I can take the dx squared out of the radical."}, {"video_title": "Arc length intro Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "This is going to be the same thing as the integral of the square root, I'm going to factor out a dx squared, dx squared times, times one plus dy over dx squared. Notice this and this is the exact same quantity. If I distribute this dx squared, I'm going to get this right up here. And now I can take the dx squared out of the radical. And so this is going to be, this is going to be the integral of, let me write that in that white color, the integral of one plus dy dx squared. And this is interesting because we know what dy dx is. This is the derivative of our function."}, {"video_title": "Arc length intro Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "And now I can take the dx squared out of the radical. And so this is going to be, this is going to be the integral of, let me write that in that white color, the integral of one plus dy dx squared. And this is interesting because we know what dy dx is. This is the derivative of our function. dy dx squared. And if you take the dx squared out of the radical, the square root of dx squared is just going to be dx. It's just going to be dx."}, {"video_title": "Arc length intro Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "This is the derivative of our function. dy dx squared. And if you take the dx squared out of the radical, the square root of dx squared is just going to be dx. It's just going to be dx. It's just going to be dx. Now this is really interesting because we know how to find this between two bounds. We can now take the definite integral from a to b."}, {"video_title": "Arc length intro Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "It's just going to be dx. It's just going to be dx. Now this is really interesting because we know how to find this between two bounds. We can now take the definite integral from a to b. Since now we are integrating a bunch of dx's or we're integrating with respect to x, we can say, okay, x equals a to x equals b. Let's take the sum of the product of this expression and dx. And this is the formula for arc length."}, {"video_title": "Arc length intro Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "We can now take the definite integral from a to b. Since now we are integrating a bunch of dx's or we're integrating with respect to x, we can say, okay, x equals a to x equals b. Let's take the sum of the product of this expression and dx. And this is the formula for arc length. The formula for arc length. And if this looks complicated, in the next video we'll see that it's actually fairly straightforward to apply, although sometimes the math gets hairy. If you wanted to write it in a slightly different notation, you could write this as equal to the integral from a to b, x equals a to x equals b, of the square root of one plus, instead of dy dx, I could write it as f prime of x, f prime of x squared, f prime of x squared dx."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Let's see if we can use everything we know about differentiation and concativity and maximum and minimum points and inflection points to actually graph a function without using a graphing calculator. So let's say our function, let's say that f of x is equal to 3x to the fourth minus 4x to the third plus 2. And of course you could always graph a function just by trying out a bunch of points. But we want to really focus on the points that are interesting to us and then just to get the general shape of the function, especially we want to focus on the things that we can take out from this function using our calculus toolkit or our derivative toolkit. So the first thing we probably want to do is figure out the critical points. We want to figure out, I'll write it here, critical points. And just as a refresher of what critical points means, it's the points where the derivative of f of x is 0."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "But we want to really focus on the points that are interesting to us and then just to get the general shape of the function, especially we want to focus on the things that we can take out from this function using our calculus toolkit or our derivative toolkit. So the first thing we probably want to do is figure out the critical points. We want to figure out, I'll write it here, critical points. And just as a refresher of what critical points means, it's the points where the derivative of f of x is 0. So critical points are f prime of x is either equal to 0 or it's undefined. This function looks differentiable everywhere, so the critical points that we're worried about are probably, well I can tell you, they're definitely just the points where f prime of x is going to be equal to 0. This derivative, f prime of x, is going to actually be defined over the entire domain."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And just as a refresher of what critical points means, it's the points where the derivative of f of x is 0. So critical points are f prime of x is either equal to 0 or it's undefined. This function looks differentiable everywhere, so the critical points that we're worried about are probably, well I can tell you, they're definitely just the points where f prime of x is going to be equal to 0. This derivative, f prime of x, is going to actually be defined over the entire domain. So let's actually write down the derivative right now. So the derivative of this, f prime of x, this is pretty straightforward, the derivative of 3x to the fourth, 4 times 3 is 12. 12x to the, we'll just decrement the 4 by 1, 3."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "This derivative, f prime of x, is going to actually be defined over the entire domain. So let's actually write down the derivative right now. So the derivative of this, f prime of x, this is pretty straightforward, the derivative of 3x to the fourth, 4 times 3 is 12. 12x to the, we'll just decrement the 4 by 1, 3. You just multiply times the exponent and then decrease the new exponent by 1. Minus 3 times 4 is 12, times x to the, 1 less than 3 is 2. And then the derivative of a constant, the slope of a constant you can almost imagine, is 0."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "12x to the, we'll just decrement the 4 by 1, 3. You just multiply times the exponent and then decrease the new exponent by 1. Minus 3 times 4 is 12, times x to the, 1 less than 3 is 2. And then the derivative of a constant, the slope of a constant you can almost imagine, is 0. It's not changing, a constant by definition isn't changing. So that's f prime of x. So let's figure out the critical points."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And then the derivative of a constant, the slope of a constant you can almost imagine, is 0. It's not changing, a constant by definition isn't changing. So that's f prime of x. So let's figure out the critical points. The critical points are where this thing is either going to be equal to 0 or it's undefined. Now, I could look over the entire domain of real numbers and this thing is defined pretty much anywhere. I could put any number here and it's not going to blow up."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So let's figure out the critical points. The critical points are where this thing is either going to be equal to 0 or it's undefined. Now, I could look over the entire domain of real numbers and this thing is defined pretty much anywhere. I could put any number here and it's not going to blow up. It's going to give me an answer to what the function is. So it's defined everywhere, so let's just figure out where it's equal to 0. So f prime of x is equal to 0."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "I could put any number here and it's not going to blow up. It's going to give me an answer to what the function is. So it's defined everywhere, so let's just figure out where it's equal to 0. So f prime of x is equal to 0. So let's solve which x is, so let's solve, well I don't have to rewrite that, I just wrote that. Let's solve for whether this is equal to 0, and I'll do it in the same color. So 12x to the 3rd minus 12x squared is equal to 0."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So f prime of x is equal to 0. So let's solve which x is, so let's solve, well I don't have to rewrite that, I just wrote that. Let's solve for whether this is equal to 0, and I'll do it in the same color. So 12x to the 3rd minus 12x squared is equal to 0. And so let's see what we can do, what can we do to solve this. Well we could factor out a 12x. So if we factor out a 12x, then this term becomes just x, and then actually let's factor out a 12x squared."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So 12x to the 3rd minus 12x squared is equal to 0. And so let's see what we can do, what can we do to solve this. Well we could factor out a 12x. So if we factor out a 12x, then this term becomes just x, and then actually let's factor out a 12x squared. We factor out a 12x squared, if we divide both of these by 12x squared, this term just becomes an x, and then minus 12x squared divided by 12x squared is just 1, is equal to 0. I just rewrote this top thing like this. You could go the other way, if I distributed this 12x squared times this entire quantity, you would get my derivative right there."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So if we factor out a 12x, then this term becomes just x, and then actually let's factor out a 12x squared. We factor out a 12x squared, if we divide both of these by 12x squared, this term just becomes an x, and then minus 12x squared divided by 12x squared is just 1, is equal to 0. I just rewrote this top thing like this. You could go the other way, if I distributed this 12x squared times this entire quantity, you would get my derivative right there. So the reason why I did that is because to solve for 0, or if I want all of the x's that make this equation equal to 0, I now have written it in a form where I'm multiplying one thing by another thing. And in order for this to be 0, one or both of these things must be equal to 0. So 12x squared or equal to 0, which means that x is equal to 0 will make this quantity equal to 0."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "You could go the other way, if I distributed this 12x squared times this entire quantity, you would get my derivative right there. So the reason why I did that is because to solve for 0, or if I want all of the x's that make this equation equal to 0, I now have written it in a form where I'm multiplying one thing by another thing. And in order for this to be 0, one or both of these things must be equal to 0. So 12x squared or equal to 0, which means that x is equal to 0 will make this quantity equal to 0. And then the other thing that would make this quantity 0 is if x minus 1 is equal to 0. So x minus 1 is equal to 0 when x is equal to 1. So these are our two critical points."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So 12x squared or equal to 0, which means that x is equal to 0 will make this quantity equal to 0. And then the other thing that would make this quantity 0 is if x minus 1 is equal to 0. So x minus 1 is equal to 0 when x is equal to 1. So these are our two critical points. Our two critical points are x is equal to 0 and x is equal to 1. And remember, those are just the points where our first derivative is equal to 0. Where the slope is 0."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So these are our two critical points. Our two critical points are x is equal to 0 and x is equal to 1. And remember, those are just the points where our first derivative is equal to 0. Where the slope is 0. They might be maximum points, they might be minimum points, they might be inflection points, we don't know. They might be, you know, if this was a constant function, they could just be anything. So we really can't say a lot about them just yet, but they are points of interest."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Where the slope is 0. They might be maximum points, they might be minimum points, they might be inflection points, we don't know. They might be, you know, if this was a constant function, they could just be anything. So we really can't say a lot about them just yet, but they are points of interest. I guess that's all we can say, that they are definitely points of interest. But let's keep going and let's try to understand the concativity and maybe we can get a better sense of this graph. So let's figure out the second derivative."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So we really can't say a lot about them just yet, but they are points of interest. I guess that's all we can say, that they are definitely points of interest. But let's keep going and let's try to understand the concativity and maybe we can get a better sense of this graph. So let's figure out the second derivative. So the second, I'll do that in, let me do it in this orange color. So the second derivative of my function f, let's see, 3 times 12 is 36x squared minus 24x. So let's see, well there's a couple of things we can do."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So let's figure out the second derivative. So the second, I'll do that in, let me do it in this orange color. So the second derivative of my function f, let's see, 3 times 12 is 36x squared minus 24x. So let's see, well there's a couple of things we can do. Now that we know the second derivative, we can answer the question, is my graph concave upwards or downwards at either of these points? So let's figure out what, at either of these critical points. And it'll all fit together."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So let's see, well there's a couple of things we can do. Now that we know the second derivative, we can answer the question, is my graph concave upwards or downwards at either of these points? So let's figure out what, at either of these critical points. And it'll all fit together. Remember, if it's concave upwards, then we're kind of in a U shape. If it's concave downwards, then we're in a kind of upside down U shape. So f prime prime, our second derivative at x is equal to 0 is equal to what?"}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And it'll all fit together. Remember, if it's concave upwards, then we're kind of in a U shape. If it's concave downwards, then we're in a kind of upside down U shape. So f prime prime, our second derivative at x is equal to 0 is equal to what? It's equal to 36 zero squared minus 24 times 0. So that's just 0. So f prime prime, so it's just equal to 0."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So f prime prime, our second derivative at x is equal to 0 is equal to what? It's equal to 36 zero squared minus 24 times 0. So that's just 0. So f prime prime, so it's just equal to 0. So we're in neither concave upwards nor concave downwards here. It might be a transition point, it may not. If it is a transition point, then we're dealing with an inflection point."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So f prime prime, so it's just equal to 0. So we're in neither concave upwards nor concave downwards here. It might be a transition point, it may not. If it is a transition point, then we're dealing with an inflection point. We're not sure yet. Now let's see what f prime prime, our second derivative, evaluated at 1 is. So that's 36 times 1."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "If it is a transition point, then we're dealing with an inflection point. We're not sure yet. Now let's see what f prime prime, our second derivative, evaluated at 1 is. So that's 36 times 1. Let me write it down. That's equal to 36 times 1 squared, which is just 36, minus 24 times 1. So it's 36 minus 24, so it's equal to 12."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So that's 36 times 1. Let me write it down. That's equal to 36 times 1 squared, which is just 36, minus 24 times 1. So it's 36 minus 24, so it's equal to 12. So this is positive. Our second derivative is positive here. It's equal to 12, which means our first derivative, our slope, is increasing."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So it's 36 minus 24, so it's equal to 12. So this is positive. Our second derivative is positive here. It's equal to 12, which means our first derivative, our slope, is increasing. The rate of change of our slope is positive here. So at this point, right here, we are concave upwards. Which tells me that this is probably a minimum point."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "It's equal to 12, which means our first derivative, our slope, is increasing. The rate of change of our slope is positive here. So at this point, right here, we are concave upwards. Which tells me that this is probably a minimum point. The slope is 0 here, but we are concave upwards at that point. So that's interesting. So let's see if there are any other potential inflection points here."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Which tells me that this is probably a minimum point. The slope is 0 here, but we are concave upwards at that point. So that's interesting. So let's see if there are any other potential inflection points here. We already know that this is a potential inflection point. Let me circle it in red. It's a potential inflection point."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So let's see if there are any other potential inflection points here. We already know that this is a potential inflection point. Let me circle it in red. It's a potential inflection point. We don't know whether our function actually transitions at that point. We'll have to experiment a little bit to see if that's really the case. But let's see if there are any other inflection points, or potential inflection points."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "It's a potential inflection point. We don't know whether our function actually transitions at that point. We'll have to experiment a little bit to see if that's really the case. But let's see if there are any other inflection points, or potential inflection points. So let's see if this equals 0 anywhere else. So 36x squared minus 24x is equal to 0. Let's solve for x."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "But let's see if there are any other inflection points, or potential inflection points. So let's see if this equals 0 anywhere else. So 36x squared minus 24x is equal to 0. Let's solve for x. Let us factor out 12x times 3x. 3x times 12x is 36x squared. Minus 2 is equal to 0."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Let's solve for x. Let us factor out 12x times 3x. 3x times 12x is 36x squared. Minus 2 is equal to 0. These two are equivalent expressions. If you multiply this out, you'll get this thing up here. So this thing is going to be equal to 0 either if 12x is equal to 0."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Minus 2 is equal to 0. These two are equivalent expressions. If you multiply this out, you'll get this thing up here. So this thing is going to be equal to 0 either if 12x is equal to 0. So 12x is equal to 0. That gives us x is equal to 0. So at x equals 0, this thing equals 0."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So this thing is going to be equal to 0 either if 12x is equal to 0. So 12x is equal to 0. That gives us x is equal to 0. So at x equals 0, this thing equals 0. So the second derivative is 0 there. And we already knew that because we tested that number out. Or this thing."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So at x equals 0, this thing equals 0. So the second derivative is 0 there. And we already knew that because we tested that number out. Or this thing. If this expression was 0, then the entire second derivative would also be 0. So let's write that. So 3x minus 2 is equal to 0."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Or this thing. If this expression was 0, then the entire second derivative would also be 0. So let's write that. So 3x minus 2 is equal to 0. 3x is equal to 2. Just adding 2 to both sides. 3x is equal to 2 thirds."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So 3x minus 2 is equal to 0. 3x is equal to 2. Just adding 2 to both sides. 3x is equal to 2 thirds. So this is another interesting point that we haven't really hit upon before. That might be an inflection point. And the reason why I say it might be is because the second derivative is definitely 0 here."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "3x is equal to 2 thirds. So this is another interesting point that we haven't really hit upon before. That might be an inflection point. And the reason why I say it might be is because the second derivative is definitely 0 here. You put 2 thirds here, you're going to get 0. What we have to do is see whether the second derivative is positive or negative on either sides of 2 thirds. And we already have a sense of that."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And the reason why I say it might be is because the second derivative is definitely 0 here. You put 2 thirds here, you're going to get 0. What we have to do is see whether the second derivative is positive or negative on either sides of 2 thirds. And we already have a sense of that. We could try out a couple of numbers. We know that if we take x is greater than 2 thirds. Let me scroll down a little bit just so we have some space."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And we already have a sense of that. We could try out a couple of numbers. We know that if we take x is greater than 2 thirds. Let me scroll down a little bit just so we have some space. So let's see what happens. When x is greater than 2 thirds, what is f prime prime? What is the second derivative?"}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Let me scroll down a little bit just so we have some space. So let's see what happens. When x is greater than 2 thirds, what is f prime prime? What is the second derivative? So let's try out a value that's pretty close just to get a sense of things. So let me rewrite it. f prime prime of x is equal to 12x times 3x minus 2."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "What is the second derivative? So let's try out a value that's pretty close just to get a sense of things. So let me rewrite it. f prime prime of x is equal to 12x times 3x minus 2. So if x is greater than 2 thirds, this term right here is going to be positive. That's definitely, you know, any positive number times 12 is going to be positive. But what about this term right here?"}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "f prime prime of x is equal to 12x times 3x minus 2. So if x is greater than 2 thirds, this term right here is going to be positive. That's definitely, you know, any positive number times 12 is going to be positive. But what about this term right here? 3 times 2 thirds minus 2 is exactly 0. That's 2 minus 2. But anything larger than that, 3 times, you know, if I had 2.1 thirds here, this is going to be a positive quantity."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "But what about this term right here? 3 times 2 thirds minus 2 is exactly 0. That's 2 minus 2. But anything larger than that, 3 times, you know, if I had 2.1 thirds here, this is going to be a positive quantity. Any value of x greater than 2 thirds will make this thing right here positive. This thing is also going to be positive. So that means that when x is greater than 2 thirds, that tells us that the second derivative is positive."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "But anything larger than that, 3 times, you know, if I had 2.1 thirds here, this is going to be a positive quantity. Any value of x greater than 2 thirds will make this thing right here positive. This thing is also going to be positive. So that means that when x is greater than 2 thirds, that tells us that the second derivative is positive. It is greater than 0. So in our domain, as long as x is larger than 2 thirds, we are concave upwards. And we saw that here."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So that means that when x is greater than 2 thirds, that tells us that the second derivative is positive. It is greater than 0. So in our domain, as long as x is larger than 2 thirds, we are concave upwards. And we saw that here. At x is equal to 1, we were concave upwards. But what about x being less than 2 thirds? So when x is less than 2 thirds, let me write it, let me scroll down a little bit."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And we saw that here. At x is equal to 1, we were concave upwards. But what about x being less than 2 thirds? So when x is less than 2 thirds, let me write it, let me scroll down a little bit. When x is less than 2 thirds, what's going on? I'll rewrite it. f prime prime of x, second derivative, 12x times 3x minus 2."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So when x is less than 2 thirds, let me write it, let me scroll down a little bit. When x is less than 2 thirds, what's going on? I'll rewrite it. f prime prime of x, second derivative, 12x times 3x minus 2. Well, if we go, you know, really far less, we're going to get a negative number here. And this might be negative. But let's see if we just go really, like, you know, right below 2 thirds."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "f prime prime of x, second derivative, 12x times 3x minus 2. Well, if we go, you know, really far less, we're going to get a negative number here. And this might be negative. But let's see if we just go really, like, you know, right below 2 thirds. We're still in the positive domain. So, you know, if this was like 1.9 thirds, which is, you know, a mixture of a decimal and a fraction, or even 1 third, this thing is still going to be positive. Right below 2 thirds, this thing is still going to be positive."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "But let's see if we just go really, like, you know, right below 2 thirds. We're still in the positive domain. So, you know, if this was like 1.9 thirds, which is, you know, a mixture of a decimal and a fraction, or even 1 third, this thing is still going to be positive. Right below 2 thirds, this thing is still going to be positive. We're going to be multiplying 12 by a positive number. But what's going to be going on, what's going on right here? At 2 thirds, we're exactly 0."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Right below 2 thirds, this thing is still going to be positive. We're going to be multiplying 12 by a positive number. But what's going to be going on, what's going on right here? At 2 thirds, we're exactly 0. But as you go, anything less than 2 thirds, 3 times 1 third is only 1. 1 minus 2, you're going to get negative numbers. So when x is less than 2 thirds, this thing right here is going to be negative."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "At 2 thirds, we're exactly 0. But as you go, anything less than 2 thirds, 3 times 1 third is only 1. 1 minus 2, you're going to get negative numbers. So when x is less than 2 thirds, this thing right here is going to be negative. So the second derivative, if x is less than 2 thirds, the second derivative right to the left, or right when you go less than 2 thirds, the second derivative of x is less than 0. Now the fact that we have this transition from when we're less than 2 thirds, we have a negative second derivative, and when we're greater than 2 thirds, we have a positive second derivative, that tells us that this indeed is an inflection point. That x is equal to 2 thirds is definitely an inflection point for our original function up here."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So when x is less than 2 thirds, this thing right here is going to be negative. So the second derivative, if x is less than 2 thirds, the second derivative right to the left, or right when you go less than 2 thirds, the second derivative of x is less than 0. Now the fact that we have this transition from when we're less than 2 thirds, we have a negative second derivative, and when we're greater than 2 thirds, we have a positive second derivative, that tells us that this indeed is an inflection point. That x is equal to 2 thirds is definitely an inflection point for our original function up here. Now we have one more candidate inflection point, and then we're ready to graph. Then once you know all the inflection points and the maximum and minimum, you are ready to graph the function. So let's see if x is equal to 0 is an inflection point."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "That x is equal to 2 thirds is definitely an inflection point for our original function up here. Now we have one more candidate inflection point, and then we're ready to graph. Then once you know all the inflection points and the maximum and minimum, you are ready to graph the function. So let's see if x is equal to 0 is an inflection point. We know that the second derivative is 0 at 0, but what happens above and below the second derivative? So let me do our little test here. Let me draw a line so we don't get confused with all of the stuff that I wrote here."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So let's see if x is equal to 0 is an inflection point. We know that the second derivative is 0 at 0, but what happens above and below the second derivative? So let me do our little test here. Let me draw a line so we don't get confused with all of the stuff that I wrote here. So when x is greater than 0, what's happening to the second derivative? Remember, the second derivative was equal to 12x times 3x minus 2. I like writing it this way because you've kind of decomposed it into two linear expressions, and you can see whether each of them are positive or negative."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Let me draw a line so we don't get confused with all of the stuff that I wrote here. So when x is greater than 0, what's happening to the second derivative? Remember, the second derivative was equal to 12x times 3x minus 2. I like writing it this way because you've kind of decomposed it into two linear expressions, and you can see whether each of them are positive or negative. So if x is greater than 0, this thing right here is definitely going to be positive. And then this thing right here, right when you go right above x is greater than 0. So you have to make sure to be very close to this number."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "I like writing it this way because you've kind of decomposed it into two linear expressions, and you can see whether each of them are positive or negative. So if x is greater than 0, this thing right here is definitely going to be positive. And then this thing right here, right when you go right above x is greater than 0. So you have to make sure to be very close to this number. So if this number is like point 1, so you're right above 0. So this isn't going to be true for all of x greater than 0. We just want to test exactly what happens right when we go right above 0."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So you have to make sure to be very close to this number. So if this number is like point 1, so you're right above 0. So this isn't going to be true for all of x greater than 0. We just want to test exactly what happens right when we go right above 0. So if this was point 1, you would have point 3 minus 2. That would be a negative number. So right as x goes right above 0, this thing right here is negative."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "We just want to test exactly what happens right when we go right above 0. So if this was point 1, you would have point 3 minus 2. That would be a negative number. So right as x goes right above 0, this thing right here is negative. So as x is greater than 0, you will have your second derivative is going to be less than 0, your concave downwards, which makes sense because at some point we're going to be hitting a transition. Remember, we were concave downwards before we got to 2 thirds. So this is consistent."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So right as x goes right above 0, this thing right here is negative. So as x is greater than 0, you will have your second derivative is going to be less than 0, your concave downwards, which makes sense because at some point we're going to be hitting a transition. Remember, we were concave downwards before we got to 2 thirds. So this is consistent. From 0 to 2 thirds, we are concave downwards. And then at 2 thirds, we become concave upwards. Now let's see what happens when x is right less than, when x is just barely less than 0."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So this is consistent. From 0 to 2 thirds, we are concave downwards. And then at 2 thirds, we become concave upwards. Now let's see what happens when x is right less than, when x is just barely less than 0. So once again, f prime, the second derivative of x is equal to 12x times 3x minus 2. Well, right if x was minus point 1 or minus point 001, no matter what, this thing is going to be negative. This expression right here is going to be negative."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Now let's see what happens when x is right less than, when x is just barely less than 0. So once again, f prime, the second derivative of x is equal to 12x times 3x minus 2. Well, right if x was minus point 1 or minus point 001, no matter what, this thing is going to be negative. This expression right here is going to be negative. The 12x, right, you have some negative value here times 12, it's going to be negative. And then what's this going to be? Well, 3 times minus point 1 is going to be minus point 3."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "This expression right here is going to be negative. The 12x, right, you have some negative value here times 12, it's going to be negative. And then what's this going to be? Well, 3 times minus point 1 is going to be minus point 3. Minus 2 is minus 2 point 3. You're definitely going to have a negative. This value right here is going to be negative."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Well, 3 times minus point 1 is going to be minus point 3. Minus 2 is minus 2 point 3. You're definitely going to have a negative. This value right here is going to be negative. And then when you subtract from a negative, it's definitely going to be negative. So that is also going to be negative. But if you multiply a negative times a negative, you're going to get a positive."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "This value right here is going to be negative. And then when you subtract from a negative, it's definitely going to be negative. So that is also going to be negative. But if you multiply a negative times a negative, you're going to get a positive. So actually, right below x is less than 0, the second derivative is positive. Now, this all might have been a little bit confusing, but we should now have the payoff. We now have the payoff."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "But if you multiply a negative times a negative, you're going to get a positive. So actually, right below x is less than 0, the second derivative is positive. Now, this all might have been a little bit confusing, but we should now have the payoff. We now have the payoff. We have all of the interesting things going on. We know that at x is equal to 1, let me write it over here. We figured out at x is equal to 1, the slope is 0."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "We now have the payoff. We have all of the interesting things going on. We know that at x is equal to 1, let me write it over here. We figured out at x is equal to 1, the slope is 0. So f prime prime is, sorry, let me write it this way. I should have said, we know that the slope is 0. Slope is equal to 0, and we figured that out because the first derivative was 0."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "We figured out at x is equal to 1, the slope is 0. So f prime prime is, sorry, let me write it this way. I should have said, we know that the slope is 0. Slope is equal to 0, and we figured that out because the first derivative was 0. This was a critical point. And we know that we're dealing with, the function is concave upwards at this point. And that tells us that this is going to be a minimum point."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Slope is equal to 0, and we figured that out because the first derivative was 0. This was a critical point. And we know that we're dealing with, the function is concave upwards at this point. And that tells us that this is going to be a minimum point. So what is f of, and we should actually get the coordinates so we can actually graph it. That was the whole point of this video. And f of 1 is equal to what?"}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And that tells us that this is going to be a minimum point. So what is f of, and we should actually get the coordinates so we can actually graph it. That was the whole point of this video. And f of 1 is equal to what? f of 1, let's go back to our original function, is 3 times 1. 1 to the 4th is just 1. 3 times 1 minus 4 plus 2."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And f of 1 is equal to what? f of 1, let's go back to our original function, is 3 times 1. 1 to the 4th is just 1. 3 times 1 minus 4 plus 2. So it's 3 times 1 minus 4 times 1, which is minus 1 plus 2. Well, that's just a positive 1. So f of 1 is 1."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "3 times 1 minus 4 plus 2. So it's 3 times 1 minus 4 times 1, which is minus 1 plus 2. Well, that's just a positive 1. So f of 1 is 1. And then we know at x is equal to 0, we also figured out that the slope is equal to 0. But we figured out that this was an inflection point. The concativity switches before and after."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So f of 1 is 1. And then we know at x is equal to 0, we also figured out that the slope is equal to 0. But we figured out that this was an inflection point. The concativity switches before and after. So this is an inflection point. And we are concave below 0. So when x is less than 0, we are upwards."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "The concativity switches before and after. So this is an inflection point. And we are concave below 0. So when x is less than 0, we are upwards. Our second derivative is positive. And when x is greater than 0, we are downwards. We are concave downwards."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So when x is less than 0, we are upwards. Our second derivative is positive. And when x is greater than 0, we are downwards. We are concave downwards. Right above. Not for all of the domain x and 0. Just right above 0."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "We are concave downwards. Right above. Not for all of the domain x and 0. Just right above 0. Downwards. And then what is f of 0? Just so we know."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Just right above 0. Downwards. And then what is f of 0? Just so we know. We want to graph that point. f of 0. Let's see."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Just so we know. We want to graph that point. f of 0. Let's see. f of 0, this is easy. 3 times 0 minus 4 times 0 plus 2. Well, that's just 2. f of 0 is 2."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Let's see. f of 0, this is easy. 3 times 0 minus 4 times 0 plus 2. Well, that's just 2. f of 0 is 2. And then finally, we got the point x is equal to 2 thirds. x is equal to, let me do that in another color. We had the point x is equal to 2 thirds."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Well, that's just 2. f of 0 is 2. And then finally, we got the point x is equal to 2 thirds. x is equal to, let me do that in another color. We had the point x is equal to 2 thirds. We figured out that this was an inflection point. The slope definitely isn't 0 there. Because it wasn't one of the critical points."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "We had the point x is equal to 2 thirds. We figured out that this was an inflection point. The slope definitely isn't 0 there. Because it wasn't one of the critical points. And we know that we are downwards. We know that when x is less than 2 thirds. Or right less than 2 thirds."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Because it wasn't one of the critical points. And we know that we are downwards. We know that when x is less than 2 thirds. Or right less than 2 thirds. We are concave downwards. And when x is greater than 2 thirds. We saw it up here."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Or right less than 2 thirds. We are concave downwards. And when x is greater than 2 thirds. We saw it up here. When x was greater than 2 thirds. Right up here, we were concave upwards. The second derivative was positive."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "We saw it up here. When x was greater than 2 thirds. Right up here, we were concave upwards. The second derivative was positive. We were upwards. And we can actually figure out what's f of 2 thirds. That's actually a little bit complicated."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "The second derivative was positive. We were upwards. And we can actually figure out what's f of 2 thirds. That's actually a little bit complicated. We don't even have to figure that out to graph. I think we can do a pretty good job of graphing it. Just with what we know right now."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "That's actually a little bit complicated. We don't even have to figure that out to graph. I think we can do a pretty good job of graphing it. Just with what we know right now. So that's our take away. Let me do a rough graph. Let's see."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Just with what we know right now. So that's our take away. Let me do a rough graph. Let's see. So let me do my axes. Like that. So let's see."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Let's see. So let me do my axes. Like that. So let's see. We're going to want to graph the point 0, 2. So let's say that the point 0, 2. So this is x is equal to 0."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So let's see. We're going to want to graph the point 0, 2. So let's say that the point 0, 2. So this is x is equal to 0. And we go up 1, 2. So this is the point 0, 2. Maybe I'll do it in that color."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So this is x is equal to 0. And we go up 1, 2. So this is the point 0, 2. Maybe I'll do it in that color. The color I was using. So that's this color. So that's that point right there."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Maybe I'll do it in that color. The color I was using. So that's this color. So that's that point right there. Then we have the point x. We have f of 1. Which is the point 1, 1."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So that's that point right there. Then we have the point x. We have f of 1. Which is the point 1, 1. Right. So this point right here. Is we'll go up 1."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Which is the point 1, 1. Right. So this point right here. Is we'll go up 1. So that's the point 1, 1. This was the point 0, 2. And then we have the x is equal to 2 thirds."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Is we'll go up 1. So that's the point 1, 1. This was the point 0, 2. And then we have the x is equal to 2 thirds. Which is our inflection point. So when x is 2 thirds. We don't know exactly what number f of 2 thirds is."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And then we have the x is equal to 2 thirds. Which is our inflection point. So when x is 2 thirds. We don't know exactly what number f of 2 thirds is. Might be here someplace. Let's say f of 2 thirds is right there. So that's the point 2 thirds."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "We don't know exactly what number f of 2 thirds is. Might be here someplace. Let's say f of 2 thirds is right there. So that's the point 2 thirds. And then whatever f of 2 thirds is. It looks like it's going to be 1 point something. f of 2 thirds."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So that's the point 2 thirds. And then whatever f of 2 thirds is. It looks like it's going to be 1 point something. f of 2 thirds. You could calculate it if you like. I'm going to substitute back in the function. But we're ready to graph this."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "f of 2 thirds. You could calculate it if you like. I'm going to substitute back in the function. But we're ready to graph this. We're ready to graph this thing. So we know that at x is equal to 1. The slope is 0."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "But we're ready to graph this. We're ready to graph this thing. So we know that at x is equal to 1. The slope is 0. We know that the slope is 0. It's flat here. And we know it's concave upwards."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "The slope is 0. We know that the slope is 0. It's flat here. And we know it's concave upwards. So we're dealing. It looks like this. It looks like that over that interval."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And we know it's concave upwards. So we're dealing. It looks like this. It looks like that over that interval. We're concave upwards. And we know we're concave upwards. From x is equal to 2 thirds and on."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "It looks like that over that interval. We're concave upwards. And we know we're concave upwards. From x is equal to 2 thirds and on. Right. Let me do it in that color. We know x is equal to 2 thirds and on."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "From x is equal to 2 thirds and on. Right. Let me do it in that color. We know x is equal to 2 thirds and on. And we're concave upwards. And so that's why I was able to draw this u shape. Now we know that when x is less than 2 thirds."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "We know x is equal to 2 thirds and on. And we're concave upwards. And so that's why I was able to draw this u shape. Now we know that when x is less than 2 thirds. And greater than 0. We're concave downwards. So the graph would look something like this."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Now we know that when x is less than 2 thirds. And greater than 0. We're concave downwards. So the graph would look something like this. Over this interval. We'll be concave downwards. Let me draw it nicely."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So the graph would look something like this. Over this interval. We'll be concave downwards. Let me draw it nicely. Over this interval. The slope is decreasing. And you could see it."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Let me draw it nicely. Over this interval. The slope is decreasing. And you could see it. If you keep drawing tangent lines. It's flattish there. It gets negative."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And you could see it. If you keep drawing tangent lines. It's flattish there. It gets negative. More negative. More negative. More negative."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "It gets negative. More negative. More negative. More negative. Until the inflection point. And then it starts increasing again. Because we go back to concave upwards."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "More negative. Until the inflection point. And then it starts increasing again. Because we go back to concave upwards. And finally. The last interval is below 0. And we know below 0."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Because we go back to concave upwards. And finally. The last interval is below 0. And we know below 0. When x is less than 0. We're concave upwards. So the graph looks like this."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And we know below 0. When x is less than 0. We're concave upwards. So the graph looks like this. The graph looks like that. And we also know that x is equal to 0. Was a critical point."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So the graph looks like this. The graph looks like that. And we also know that x is equal to 0. Was a critical point. The slope was 0. So this graph is actually flat right there too. So this is an inflection point."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Was a critical point. The slope was 0. So this graph is actually flat right there too. So this is an inflection point. Where the slope was also 0. So this is our final graph. We're done."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So this is an inflection point. Where the slope was also 0. So this is our final graph. We're done. After all that work. We were able to use our calculus skills. And our knowledge of inflection points."}, {"video_title": "Graphing using derivatives Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "We're done. After all that work. We were able to use our calculus skills. And our knowledge of inflection points. And concativity. And transitions in concativity. To actually graph this fairly hairy looking graph."}, {"video_title": "Calculating average value of function over interval AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say that we have the function f of x is equal to x squared plus one. And what we want to do is we want to figure out the average value of our function f on the interval, on the closed interval between zero and, let's say between zero and three. And I encourage you to pause this video, and especially if you've seen the other videos on introducing the idea of the average value of a function, figure out what this is. What is the average value of our function f over this interval? So I'm assuming you've had a go at it. Let's just visualize what's going on, and then we can actually find the average. So that's my y-axis."}, {"video_title": "Calculating average value of function over interval AP Calculus AB Khan Academy.mp3", "Sentence": "What is the average value of our function f over this interval? So I'm assuming you've had a go at it. Let's just visualize what's going on, and then we can actually find the average. So that's my y-axis. This is my x-axis. Now over the interval between zero and three, so let's say that this is zero. This is one, two, three."}, {"video_title": "Calculating average value of function over interval AP Calculus AB Khan Academy.mp3", "Sentence": "So that's my y-axis. This is my x-axis. Now over the interval between zero and three, so let's say that this is zero. This is one, two, three. It's a closed interval. When x is zero, f of zero is going to be one. So we're gonna be right over here."}, {"video_title": "Calculating average value of function over interval AP Calculus AB Khan Academy.mp3", "Sentence": "This is one, two, three. It's a closed interval. When x is zero, f of zero is going to be one. So we're gonna be right over here. F of one is two, so it's gonna be, so it's one, two, three. Actually, let me make my scale a little bit smaller on that. I have to go all the way up to 10."}, {"video_title": "Calculating average value of function over interval AP Calculus AB Khan Academy.mp3", "Sentence": "So we're gonna be right over here. F of one is two, so it's gonna be, so it's one, two, three. Actually, let me make my scale a little bit smaller on that. I have to go all the way up to 10. So this is gonna be 10. This is gonna be five. And then one, two, three."}, {"video_title": "Calculating average value of function over interval AP Calculus AB Khan Academy.mp3", "Sentence": "I have to go all the way up to 10. So this is gonna be 10. This is gonna be five. And then one, two, three. Actually, let me, this is the hardest part, is making this even. So let's see, this is gonna be in the middle. Pretty good."}, {"video_title": "Calculating average value of function over interval AP Calculus AB Khan Academy.mp3", "Sentence": "And then one, two, three. Actually, let me, this is the hardest part, is making this even. So let's see, this is gonna be in the middle. Pretty good. And then let's see, in the middle, and then we have that. No, good enough. All right, so we're gonna be there, and we're gonna be there."}, {"video_title": "Calculating average value of function over interval AP Calculus AB Khan Academy.mp3", "Sentence": "Pretty good. And then let's see, in the middle, and then we have that. No, good enough. All right, so we're gonna be there, and we're gonna be there. I have, obviously, different scales for x and y-axes. Two squared plus one is five. Three squared plus one is 10."}, {"video_title": "Calculating average value of function over interval AP Calculus AB Khan Academy.mp3", "Sentence": "All right, so we're gonna be there, and we're gonna be there. I have, obviously, different scales for x and y-axes. Two squared plus one is five. Three squared plus one is 10. So it's gonna look something like this. This is what our function is going to look like. So that's the graph of y is equal to f of x."}, {"video_title": "Calculating average value of function over interval AP Calculus AB Khan Academy.mp3", "Sentence": "Three squared plus one is 10. So it's gonna look something like this. This is what our function is going to look like. So that's the graph of y is equal to f of x. And we care about the average value on the interval, closed interval between zero and three. Between zero and three. So one way to think about it, you could apply the formula, but it's very important to think about what does that formula actually mean?"}, {"video_title": "Calculating average value of function over interval AP Calculus AB Khan Academy.mp3", "Sentence": "So that's the graph of y is equal to f of x. And we care about the average value on the interval, closed interval between zero and three. Between zero and three. So one way to think about it, you could apply the formula, but it's very important to think about what does that formula actually mean? And once again, you shouldn't memorize this formula because it actually kind of falls out out of what it actually means. So the average of our function is going to be, it's going to be equal to the definite integral over this interval. So essentially, the area under this curve."}, {"video_title": "Calculating average value of function over interval AP Calculus AB Khan Academy.mp3", "Sentence": "So one way to think about it, you could apply the formula, but it's very important to think about what does that formula actually mean? And once again, you shouldn't memorize this formula because it actually kind of falls out out of what it actually means. So the average of our function is going to be, it's going to be equal to the definite integral over this interval. So essentially, the area under this curve. So it's going to be the definite integral from zero to three of f of x, which is x squared plus one, dx. And then we're gonna take this area. We're gonna take this area right over here, and we're gonna divide it by the width of our interval to essentially come up with the average height or the average value of our function."}, {"video_title": "Calculating average value of function over interval AP Calculus AB Khan Academy.mp3", "Sentence": "So essentially, the area under this curve. So it's going to be the definite integral from zero to three of f of x, which is x squared plus one, dx. And then we're gonna take this area. We're gonna take this area right over here, and we're gonna divide it by the width of our interval to essentially come up with the average height or the average value of our function. So we're gonna divide it by b minus a, or three minus zero, which is just going to be three. And so now we just have to evaluate this. So this is going to be equal to 1 3rd times, see, the antiderivative of x squared is x to the third over three."}, {"video_title": "Calculating average value of function over interval AP Calculus AB Khan Academy.mp3", "Sentence": "We're gonna take this area right over here, and we're gonna divide it by the width of our interval to essentially come up with the average height or the average value of our function. So we're gonna divide it by b minus a, or three minus zero, which is just going to be three. And so now we just have to evaluate this. So this is going to be equal to 1 3rd times, see, the antiderivative of x squared is x to the third over three. Antiderivative of one is x, and we're going to evaluate it from zero to three. And so this is going to be equal to 1 3rd times, when we evaluate it at three, let me use another color here, when we evaluate it at three, it's going to be three to the third divided by three. Well, that's just going to be 27 divided by three."}, {"video_title": "Calculating average value of function over interval AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be equal to 1 3rd times, see, the antiderivative of x squared is x to the third over three. Antiderivative of one is x, and we're going to evaluate it from zero to three. And so this is going to be equal to 1 3rd times, when we evaluate it at three, let me use another color here, when we evaluate it at three, it's going to be three to the third divided by three. Well, that's just going to be 27 divided by three. That's nine plus three. And then when we evaluate it at zero, minus zero, minus zero. So it's just minus, when you evaluate it at zero, it's just gonna be zero."}, {"video_title": "Calculating average value of function over interval AP Calculus AB Khan Academy.mp3", "Sentence": "Well, that's just going to be 27 divided by three. That's nine plus three. And then when we evaluate it at zero, minus zero, minus zero. So it's just minus, when you evaluate it at zero, it's just gonna be zero. And so we are left with, I'm gonna make the brackets that same color. This is going to be 1 3rd times 12, 1 3rd times 12, which is equal to four. Which is equal to four."}, {"video_title": "Calculating average value of function over interval AP Calculus AB Khan Academy.mp3", "Sentence": "So it's just minus, when you evaluate it at zero, it's just gonna be zero. And so we are left with, I'm gonna make the brackets that same color. This is going to be 1 3rd times 12, 1 3rd times 12, which is equal to four. Which is equal to four. So this is the average value of our function. The average value of our function over this interval, over this interval, is equal, the average value of our function is equal to four. And notice, our function actually hits that value at some point in the interval."}, {"video_title": "Calculating average value of function over interval AP Calculus AB Khan Academy.mp3", "Sentence": "Which is equal to four. So this is the average value of our function. The average value of our function over this interval, over this interval, is equal, the average value of our function is equal to four. And notice, our function actually hits that value at some point in the interval. At some point in the interval, something lower than two, but greater than one, we could maybe call that C. It looks like our function hits that value. And this is actually, this is actually a generally true thing. This is a mean value theorem for integrals, and we'll go into more depth there."}, {"video_title": "Calculating average value of function over interval AP Calculus AB Khan Academy.mp3", "Sentence": "And notice, our function actually hits that value at some point in the interval. At some point in the interval, something lower than two, but greater than one, we could maybe call that C. It looks like our function hits that value. And this is actually, this is actually a generally true thing. This is a mean value theorem for integrals, and we'll go into more depth there. But you can see that this kind of does look like its average value. That if you imagine the box, if you multiply this height, this average value times this width, you would have this area right over here, and this area right over here is the same, this area that I'm highlighting in yellow right over here, is the same as the area under the curve. Because we have the average height times the width is the same thing as the area under the curve."}, {"video_title": "Motion along a curve finding velocity magnitude AP Calculus BC Khan Academy.mp3", "Sentence": "That means that the rate of change of y with respect to t is equal to two. What is the magnitude in units per minute of the particle's velocity vector when the particle is at the point four comma four? So when x is four and y is four. So let's see what's going on. So let's first just remind ourselves what a velocity vector, what the velocity vector will look like. So our velocity is going to be a function of time and it's going to have two components. It's going to be what is the rate of change in the x direction and the rate of change in the y direction?"}, {"video_title": "Motion along a curve finding velocity magnitude AP Calculus BC Khan Academy.mp3", "Sentence": "So let's see what's going on. So let's first just remind ourselves what a velocity vector, what the velocity vector will look like. So our velocity is going to be a function of time and it's going to have two components. It's going to be what is the rate of change in the x direction and the rate of change in the y direction? So the rate of change in the x direction is going to be dx dt and the rate of change in the y direction is going to be dy dt. And they tell us that this is, that dy dt is a constant two units per minute. But they're not even just asking us for just the velocity vector for its components."}, {"video_title": "Motion along a curve finding velocity magnitude AP Calculus BC Khan Academy.mp3", "Sentence": "It's going to be what is the rate of change in the x direction and the rate of change in the y direction? So the rate of change in the x direction is going to be dx dt and the rate of change in the y direction is going to be dy dt. And they tell us that this is, that dy dt is a constant two units per minute. But they're not even just asking us for just the velocity vector for its components. They're asking for the magnitude. They're asking for the magnitude of the particle's velocity vector. Well, if I have some vector, let me do a little bit of an aside here."}, {"video_title": "Motion along a curve finding velocity magnitude AP Calculus BC Khan Academy.mp3", "Sentence": "But they're not even just asking us for just the velocity vector for its components. They're asking for the magnitude. They're asking for the magnitude of the particle's velocity vector. Well, if I have some vector, let me do a little bit of an aside here. If I have some vector, let's say A, that is, that has components, I don't know, B and C, well then the magnitude of my vector, sometimes you'll see it written like that, sometimes you'll see it written with double bars like that, the magnitude of my vector, and this comes straight out of the Pythagorean theorem, this is going to be the square root of B squared plus C squared. The square root of the x component squared plus the y component squared. So if we want to know the magnitude of our velocity vector, the magnitude of the particle's velocity vector, well, I could write that as the magnitude of v, and I could even write it as a function of t. It's going to be equal to the square root of the x component squared, so that's the rate of change of x with respect to time squared, plus the y component squared, which in this case is the rate of change of y with respect to t squared."}, {"video_title": "Motion along a curve finding velocity magnitude AP Calculus BC Khan Academy.mp3", "Sentence": "Well, if I have some vector, let me do a little bit of an aside here. If I have some vector, let's say A, that is, that has components, I don't know, B and C, well then the magnitude of my vector, sometimes you'll see it written like that, sometimes you'll see it written with double bars like that, the magnitude of my vector, and this comes straight out of the Pythagorean theorem, this is going to be the square root of B squared plus C squared. The square root of the x component squared plus the y component squared. So if we want to know the magnitude of our velocity vector, the magnitude of the particle's velocity vector, well, I could write that as the magnitude of v, and I could even write it as a function of t. It's going to be equal to the square root of the x component squared, so that's the rate of change of x with respect to time squared, plus the y component squared, which in this case is the rate of change of y with respect to t squared. So how do we figure out these two things? Well, we already know the rate of change of y with respect to t. They say that's a constant rate of two units per minute, so we already know that this is going to be two or that this whole thing right over here is going to be four, but how do we figure out the rate of change of x with respect to t? Well, we could take our original equation that describes the curve, we could take the derivative of both sides with respect to t, and then that's going to give us an equation that involves x, y, and dx dt and dy dt, so let's do that."}, {"video_title": "Motion along a curve finding velocity magnitude AP Calculus BC Khan Academy.mp3", "Sentence": "So if we want to know the magnitude of our velocity vector, the magnitude of the particle's velocity vector, well, I could write that as the magnitude of v, and I could even write it as a function of t. It's going to be equal to the square root of the x component squared, so that's the rate of change of x with respect to time squared, plus the y component squared, which in this case is the rate of change of y with respect to t squared. So how do we figure out these two things? Well, we already know the rate of change of y with respect to t. They say that's a constant rate of two units per minute, so we already know that this is going to be two or that this whole thing right over here is going to be four, but how do we figure out the rate of change of x with respect to t? Well, we could take our original equation that describes the curve, we could take the derivative of both sides with respect to t, and then that's going to give us an equation that involves x, y, and dx dt and dy dt, so let's do that. So we have xy is equal to 16. I'm going to take the derivative with respect to t of both sides. Let me do that in a different color just for a little bit of variety."}, {"video_title": "Motion along a curve finding velocity magnitude AP Calculus BC Khan Academy.mp3", "Sentence": "Well, we could take our original equation that describes the curve, we could take the derivative of both sides with respect to t, and then that's going to give us an equation that involves x, y, and dx dt and dy dt, so let's do that. So we have xy is equal to 16. I'm going to take the derivative with respect to t of both sides. Let me do that in a different color just for a little bit of variety. So derivative with respect to t of the left-hand side, derivative with respect to t of the right-hand side. Now the left-hand side, if we view this as a product of two functions, if we say, look, x is a function of t and y is also a function of t, this is, we're going to do a little bit of the product rule and a little bit of the chain rule here, and so this is going to be equal to derivative of the first function, which is, so we'll first say the derivative of x with respect to x is one times the derivative of x with respect to t. Remember, we're taking the derivative with respect to t, not with respect to x, times the second function, so times, times the second function, so times y, times y, plus the first function, which is just x, times the derivative of the second function with respect to t. So first, what's the derivative of y with respect to y? Well, that's just one."}, {"video_title": "Motion along a curve finding velocity magnitude AP Calculus BC Khan Academy.mp3", "Sentence": "Let me do that in a different color just for a little bit of variety. So derivative with respect to t of the left-hand side, derivative with respect to t of the right-hand side. Now the left-hand side, if we view this as a product of two functions, if we say, look, x is a function of t and y is also a function of t, this is, we're going to do a little bit of the product rule and a little bit of the chain rule here, and so this is going to be equal to derivative of the first function, which is, so we'll first say the derivative of x with respect to x is one times the derivative of x with respect to t. Remember, we're taking the derivative with respect to t, not with respect to x, times the second function, so times, times the second function, so times y, times y, plus the first function, which is just x, times the derivative of the second function with respect to t. So first, what's the derivative of y with respect to y? Well, that's just one. And then what's the derivative of y with respect to t? Well, that's dy dt, and that is going to be equal to, that is going to be equal to, derivative of a constant is just zero. So let's see, what does this simplify to?"}, {"video_title": "Motion along a curve finding velocity magnitude AP Calculus BC Khan Academy.mp3", "Sentence": "Well, that's just one. And then what's the derivative of y with respect to t? Well, that's dy dt, and that is going to be equal to, that is going to be equal to, derivative of a constant is just zero. So let's see, what does this simplify to? This simplifies to, in fact, we don't even have to simplify it more, we can actually plug in the values to solve for dx dt. We know that dy dt is a constant two, and we want the magnitude of the particle's velocity vector when the particle is at the point four comma four. So when x is equal to four, so when x is equal to four, and y is equal to four, and y is equal to four."}, {"video_title": "Motion along a curve finding velocity magnitude AP Calculus BC Khan Academy.mp3", "Sentence": "So let's see, what does this simplify to? This simplifies to, in fact, we don't even have to simplify it more, we can actually plug in the values to solve for dx dt. We know that dy dt is a constant two, and we want the magnitude of the particle's velocity vector when the particle is at the point four comma four. So when x is equal to four, so when x is equal to four, and y is equal to four, and y is equal to four. So now, it's a little messy right now, but this right here is an equation that we can solve for. There's only one unknown here, the rate of change of x with respect to t, right when we are at the point four comma four. And so if we're able to figure that out, we can substitute that in here and figure out the magnitude of our velocity vector."}, {"video_title": "Motion along a curve finding velocity magnitude AP Calculus BC Khan Academy.mp3", "Sentence": "So when x is equal to four, so when x is equal to four, and y is equal to four, and y is equal to four. So now, it's a little messy right now, but this right here is an equation that we can solve for. There's only one unknown here, the rate of change of x with respect to t, right when we are at the point four comma four. And so if we're able to figure that out, we can substitute that in here and figure out the magnitude of our velocity vector. So let us write it out. So this gives us four, four dx dt plus, what is this, four times two, plus eight is equal to zero. And so we have four dx dt is equal to negative eight, just subtracted eight from both sides."}, {"video_title": "Motion along a curve finding velocity magnitude AP Calculus BC Khan Academy.mp3", "Sentence": "And so if we're able to figure that out, we can substitute that in here and figure out the magnitude of our velocity vector. So let us write it out. So this gives us four, four dx dt plus, what is this, four times two, plus eight is equal to zero. And so we have four dx dt is equal to negative eight, just subtracted eight from both sides. Divide both sides by four, you get dx dt, let me scroll down, is equal to negative two. So when all this stuff is going on, the rate of change of x with respect to t is negative two. And so then you square it, you get a four right over here."}, {"video_title": "Motion along a curve finding velocity magnitude AP Calculus BC Khan Academy.mp3", "Sentence": "And so we have four dx dt is equal to negative eight, just subtracted eight from both sides. Divide both sides by four, you get dx dt, let me scroll down, is equal to negative two. So when all this stuff is going on, the rate of change of x with respect to t is negative two. And so then you square it, you get a four right over here. And so the magnitude of our velocity vector is going to be equal to, equal to the square root of four plus four, which is equal to eight, which is the same thing as four times two. So this is going to be two square roots of two units per minute. So that's the magnitude of the velocity vector."}, {"video_title": "Finding derivative with fundamental theorem of calculus AP\u00ae\ufe0e Calculus AB Khan Academy.mp3", "Sentence": "And what I'm curious about finding or trying to figure out is, what is g prime of 27? What is that equal to? Pause this video and try to think about it. And I'll give you a little bit of a hint. Think about the second fundamental theorem of calculus. All right, now let's work on this together. So we wanna figure out what g prime, we could try to figure out what g prime of x is and then evaluate that at 27."}, {"video_title": "Finding derivative with fundamental theorem of calculus AP\u00ae\ufe0e Calculus AB Khan Academy.mp3", "Sentence": "And I'll give you a little bit of a hint. Think about the second fundamental theorem of calculus. All right, now let's work on this together. So we wanna figure out what g prime, we could try to figure out what g prime of x is and then evaluate that at 27. And the best way that I can think about doing that is by taking the derivative of both sides of this equation. So let's take the derivative of both sides of that equation. So the left-hand side, we'll take the derivative with respect to x of g of x, and the right-hand side, the derivative with respect to x of all of this business."}, {"video_title": "Finding derivative with fundamental theorem of calculus AP\u00ae\ufe0e Calculus AB Khan Academy.mp3", "Sentence": "So we wanna figure out what g prime, we could try to figure out what g prime of x is and then evaluate that at 27. And the best way that I can think about doing that is by taking the derivative of both sides of this equation. So let's take the derivative of both sides of that equation. So the left-hand side, we'll take the derivative with respect to x of g of x, and the right-hand side, the derivative with respect to x of all of this business. Now the left-hand side is pretty straightforward. The derivative with respect to x of g of x, that's just going to be g prime of x. But what is the right-hand side going to be equal to?"}, {"video_title": "Finding derivative with fundamental theorem of calculus AP\u00ae\ufe0e Calculus AB Khan Academy.mp3", "Sentence": "So the left-hand side, we'll take the derivative with respect to x of g of x, and the right-hand side, the derivative with respect to x of all of this business. Now the left-hand side is pretty straightforward. The derivative with respect to x of g of x, that's just going to be g prime of x. But what is the right-hand side going to be equal to? Well, that's where the second fundamental theorem of calculus is useful. I'll write it right over here. Second fundamental, I'll abbreviate a little bit, theorem, theorem of calculus."}, {"video_title": "Finding derivative with fundamental theorem of calculus AP\u00ae\ufe0e Calculus AB Khan Academy.mp3", "Sentence": "But what is the right-hand side going to be equal to? Well, that's where the second fundamental theorem of calculus is useful. I'll write it right over here. Second fundamental, I'll abbreviate a little bit, theorem, theorem of calculus. It tells us, let's say we have some function capital F of x, and it's equal to the definite integral from a, some constant a to x of lowercase f of t dt. The second fundamental theorem of calculus tells us that if our lowercase f, if lowercase f is continuous on the interval from a to x, so I'll write it this way, on the closed interval from a to x, then the derivative of our capital F of x, so capital F prime of x, is just going to be equal to our inner function f evaluated at x instead of t. It's going to become lowercase f of x. Now I know when you first saw this, you thought that, hey, this might be some cryptic thing that you might not use too often."}, {"video_title": "Finding derivative with fundamental theorem of calculus AP\u00ae\ufe0e Calculus AB Khan Academy.mp3", "Sentence": "Second fundamental, I'll abbreviate a little bit, theorem, theorem of calculus. It tells us, let's say we have some function capital F of x, and it's equal to the definite integral from a, some constant a to x of lowercase f of t dt. The second fundamental theorem of calculus tells us that if our lowercase f, if lowercase f is continuous on the interval from a to x, so I'll write it this way, on the closed interval from a to x, then the derivative of our capital F of x, so capital F prime of x, is just going to be equal to our inner function f evaluated at x instead of t. It's going to become lowercase f of x. Now I know when you first saw this, you thought that, hey, this might be some cryptic thing that you might not use too often. But we're going to see that it's actually very, very useful. And even in the future, and some of you might already know, there's multiple ways to try to think about a definite integral like this, and you'll learn it in the future. But this can be extremely simplifying, especially if you have a hairy definite integral like this."}, {"video_title": "Finding derivative with fundamental theorem of calculus AP\u00ae\ufe0e Calculus AB Khan Academy.mp3", "Sentence": "Now I know when you first saw this, you thought that, hey, this might be some cryptic thing that you might not use too often. But we're going to see that it's actually very, very useful. And even in the future, and some of you might already know, there's multiple ways to try to think about a definite integral like this, and you'll learn it in the future. But this can be extremely simplifying, especially if you have a hairy definite integral like this. And so this just tells us, hey, look, the derivative with respect to x of all of this business, first we have to check that our inner function, which would be analogous to our lowercase f here, is this continuous on the interval from 19 to x? Well, no matter what x is, this is going to be continuous over that interval because this is continuous for all x's. And so we meet this first condition, or our major condition."}, {"video_title": "Finding derivative with fundamental theorem of calculus AP\u00ae\ufe0e Calculus AB Khan Academy.mp3", "Sentence": "But this can be extremely simplifying, especially if you have a hairy definite integral like this. And so this just tells us, hey, look, the derivative with respect to x of all of this business, first we have to check that our inner function, which would be analogous to our lowercase f here, is this continuous on the interval from 19 to x? Well, no matter what x is, this is going to be continuous over that interval because this is continuous for all x's. And so we meet this first condition, or our major condition. And so then we could just say, all right, then the derivative of all of this is just going to be this inner function replacing t with x. So we're going to get the cube root. Instead of the cube root of t, you're gonna get the cube root of x."}, {"video_title": "Finding derivative with fundamental theorem of calculus AP\u00ae\ufe0e Calculus AB Khan Academy.mp3", "Sentence": "And so we meet this first condition, or our major condition. And so then we could just say, all right, then the derivative of all of this is just going to be this inner function replacing t with x. So we're going to get the cube root. Instead of the cube root of t, you're gonna get the cube root of x. And so we can go back to our original question. What is g prime of 27 going to be equal to? Well, it's going to be equal to the cube root of 27, which is, of course, equal to three."}, {"video_title": "Integration with partial fractions AP Calculus BC Khan Academy.mp3", "Sentence": "And some of you in attempting this might try to say, all right, is the numerator here the derivative or constant multiple of the derivative of the denominator? In which case, u-substitution might apply, but it's not the case here, so what do we do? And my hint to you would be partial fraction decomposition, which might invoke some memories from a precalculus class or maybe from an algebra two class. But it's a technique to break up this rational expression into the sum of two rational expressions. And a good hint there is the fact that this denominator here is factorable into these two expressions. So what we're gonna try to do with partial fraction decomposition is say, can we express x minus five over two x minus three times x minus one? Can we express it as a sum of two rational expressions where the denominator of the first rational expression is two x minus three, and the denominator of the second rational expression is x minus one?"}, {"video_title": "Integration with partial fractions AP Calculus BC Khan Academy.mp3", "Sentence": "But it's a technique to break up this rational expression into the sum of two rational expressions. And a good hint there is the fact that this denominator here is factorable into these two expressions. So what we're gonna try to do with partial fraction decomposition is say, can we express x minus five over two x minus three times x minus one? Can we express it as a sum of two rational expressions where the denominator of the first rational expression is two x minus three, and the denominator of the second rational expression is x minus one? And I don't have to put parentheses there, is x minus one. Now we don't know the numerators. As you might have learned before, and I encourage you, if this is the first time you're ever seeing partial fraction decomposition, look that up on Khan Academy."}, {"video_title": "Integration with partial fractions AP Calculus BC Khan Academy.mp3", "Sentence": "Can we express it as a sum of two rational expressions where the denominator of the first rational expression is two x minus three, and the denominator of the second rational expression is x minus one? And I don't have to put parentheses there, is x minus one. Now we don't know the numerators. As you might have learned before, and I encourage you, if this is the first time you're ever seeing partial fraction decomposition, look that up on Khan Academy. We have many videos on it. But the general principle here is that your numerator is gonna be one degree less than your denominator. So our denominator's here our first degree, so our numerator's gonna be zero degree, or just constants."}, {"video_title": "Integration with partial fractions AP Calculus BC Khan Academy.mp3", "Sentence": "As you might have learned before, and I encourage you, if this is the first time you're ever seeing partial fraction decomposition, look that up on Khan Academy. We have many videos on it. But the general principle here is that your numerator is gonna be one degree less than your denominator. So our denominator's here our first degree, so our numerator's gonna be zero degree, or just constants. So it's gonna be some unknown constant there, let's call that a, and some unknown constant b. And our goal is to solve for a and b. And this is all a review, this isn't really calculus, this is more pre-calculus now, or algebra."}, {"video_title": "Integration with partial fractions AP Calculus BC Khan Academy.mp3", "Sentence": "So our denominator's here our first degree, so our numerator's gonna be zero degree, or just constants. So it's gonna be some unknown constant there, let's call that a, and some unknown constant b. And our goal is to solve for a and b. And this is all a review, this isn't really calculus, this is more pre-calculus now, or algebra. How do we solve for a and b? Well, let's add them as if we're just adding two fractions with unlike denominators. So we wanna have a common denominator."}, {"video_title": "Integration with partial fractions AP Calculus BC Khan Academy.mp3", "Sentence": "And this is all a review, this isn't really calculus, this is more pre-calculus now, or algebra. How do we solve for a and b? Well, let's add them as if we're just adding two fractions with unlike denominators. So we wanna have a common denominator. So what we wanna do is multiply this first, this first rational expression by x minus one in the numerator and the denominator. So you could call this a times x minus one over two x minus three times x minus one. And then for this second rational expression, we'd multiply the numerator and the denominator by two x minus three."}, {"video_title": "Integration with partial fractions AP Calculus BC Khan Academy.mp3", "Sentence": "So we wanna have a common denominator. So what we wanna do is multiply this first, this first rational expression by x minus one in the numerator and the denominator. So you could call this a times x minus one over two x minus three times x minus one. And then for this second rational expression, we'd multiply the numerator and the denominator by two x minus three. So two x minus three times b over two x minus three times x minus one. And now, since I have the same denominator, I can add them. And the goal is when I add them, and I think about the numerator, I will try to say, well, how do the a's and b's line up to what we have already right over here?"}, {"video_title": "Integration with partial fractions AP Calculus BC Khan Academy.mp3", "Sentence": "And then for this second rational expression, we'd multiply the numerator and the denominator by two x minus three. So two x minus three times b over two x minus three times x minus one. And now, since I have the same denominator, I can add them. And the goal is when I add them, and I think about the numerator, I will try to say, well, how do the a's and b's line up to what we have already right over here? So this is going to be equal to same denominator, so the denominator's two x minus three times x minus one. And in the numerators, actually, let me do it over here. This is the same thing as a x minus a, just distributing the a."}, {"video_title": "Integration with partial fractions AP Calculus BC Khan Academy.mp3", "Sentence": "And the goal is when I add them, and I think about the numerator, I will try to say, well, how do the a's and b's line up to what we have already right over here? So this is going to be equal to same denominator, so the denominator's two x minus three times x minus one. And in the numerators, actually, let me do it over here. This is the same thing as a x minus a, just distributing the a. And if I distribute the b here, this is the same thing as two b x minus three b. And so if we add the numerators, we can add the x terms, a x plus two b x. So we could call that a plus two b times x."}, {"video_title": "Integration with partial fractions AP Calculus BC Khan Academy.mp3", "Sentence": "This is the same thing as a x minus a, just distributing the a. And if I distribute the b here, this is the same thing as two b x minus three b. And so if we add the numerators, we can add the x terms, a x plus two b x. So we could call that a plus two b times x. And then if we add these constant terms, minus a minus three b. And since we have a minus right over here, I'm gonna try to pattern match. So I'm gonna have minus, and I'll write that as minus a plus three b."}, {"video_title": "Integration with partial fractions AP Calculus BC Khan Academy.mp3", "Sentence": "So we could call that a plus two b times x. And then if we add these constant terms, minus a minus three b. And since we have a minus right over here, I'm gonna try to pattern match. So I'm gonna have minus, and I'll write that as minus a plus three b. If you distribute this negative sign, then you're gonna get negative a and negative three b. So these are equivalent. And now we can see the pattern."}, {"video_title": "Integration with partial fractions AP Calculus BC Khan Academy.mp3", "Sentence": "So I'm gonna have minus, and I'll write that as minus a plus three b. If you distribute this negative sign, then you're gonna get negative a and negative three b. So these are equivalent. And now we can see the pattern. We can say, all right, our denominators are not the same, so this thing in our numerator has to be the same thing as x minus five. So whatever our coefficient here is on x, that's gotta be equal to one. That's the coefficient on x."}, {"video_title": "Integration with partial fractions AP Calculus BC Khan Academy.mp3", "Sentence": "And now we can see the pattern. We can say, all right, our denominators are not the same, so this thing in our numerator has to be the same thing as x minus five. So whatever our coefficient here is on x, that's gotta be equal to one. That's the coefficient on x. And this thing that we're subtracting, that's gotta be equal to five. And we're gonna set up a system of two equations with two unknowns to solve for a and b. So we know that a plus, so let me write that down."}, {"video_title": "Integration with partial fractions AP Calculus BC Khan Academy.mp3", "Sentence": "That's the coefficient on x. And this thing that we're subtracting, that's gotta be equal to five. And we're gonna set up a system of two equations with two unknowns to solve for a and b. So we know that a plus, so let me write that down. We know that a plus two b needs to be equal to one. And we know that a plus three b, or a plus three b, is going to be equal to five. And so now to solve for a and b, well, we could do that by elimination."}, {"video_title": "Integration with partial fractions AP Calculus BC Khan Academy.mp3", "Sentence": "So we know that a plus, so let me write that down. We know that a plus two b needs to be equal to one. And we know that a plus three b, or a plus three b, is going to be equal to five. And so now to solve for a and b, well, we could do that by elimination. So let's see, what if we multiply this top equation by negative one? So that'd be negative a, negative two b, negative one. And now we add them together."}, {"video_title": "Integration with partial fractions AP Calculus BC Khan Academy.mp3", "Sentence": "And so now to solve for a and b, well, we could do that by elimination. So let's see, what if we multiply this top equation by negative one? So that'd be negative a, negative two b, negative one. And now we add them together. The whole point was to cancel out the a's, so we're left with negative two b plus three b. That's just going to be b is equal to four. And then we can substitute back in to solve for a."}, {"video_title": "Integration with partial fractions AP Calculus BC Khan Academy.mp3", "Sentence": "And now we add them together. The whole point was to cancel out the a's, so we're left with negative two b plus three b. That's just going to be b is equal to four. And then we can substitute back in to solve for a. So let's say right over here, we know that a plus three times four, so that's gonna be 12. I'm using this equation right over here, is going to be equal to five. Subtract 12 from both sides, you get a is equal to negative seven."}, {"video_title": "Integration with partial fractions AP Calculus BC Khan Academy.mp3", "Sentence": "And then we can substitute back in to solve for a. So let's say right over here, we know that a plus three times four, so that's gonna be 12. I'm using this equation right over here, is going to be equal to five. Subtract 12 from both sides, you get a is equal to negative seven. So just like that, we can rewrite this entire integral. We can say this is going to be equal to the indefinite integral of, open parentheses, a over two x minus three. We now know that a is negative seven."}, {"video_title": "Integration with partial fractions AP Calculus BC Khan Academy.mp3", "Sentence": "Subtract 12 from both sides, you get a is equal to negative seven. So just like that, we can rewrite this entire integral. We can say this is going to be equal to the indefinite integral of, open parentheses, a over two x minus three. We now know that a is negative seven. So it's negative seven over two x minus three. And then we're going to have plus b, b is four, so plus four over x minus one, over x minus one. And close the parentheses, dx."}, {"video_title": "Integration with partial fractions AP Calculus BC Khan Academy.mp3", "Sentence": "We now know that a is negative seven. So it's negative seven over two x minus three. And then we're going to have plus b, b is four, so plus four over x minus one, over x minus one. And close the parentheses, dx. Now, if you are so inspired, I encourage you to pause the video and try to run with it from this point, because we have seen the techniques to solve integrals like this before, but I'll do it step by step. This is going to be the same thing as, well actually let me just, so this is gonna be the same thing as the integral of, so two x minus three. I could write the negative seven here, but I'm gonna take the constant out of the integral."}, {"video_title": "Integration with partial fractions AP Calculus BC Khan Academy.mp3", "Sentence": "And close the parentheses, dx. Now, if you are so inspired, I encourage you to pause the video and try to run with it from this point, because we have seen the techniques to solve integrals like this before, but I'll do it step by step. This is going to be the same thing as, well actually let me just, so this is gonna be the same thing as the integral of, so two x minus three. I could write the negative seven here, but I'm gonna take the constant out of the integral. So I'll put a negative seven here. And to help us solve this, and this could be a one, but to help us solve this, it would be nice if we had a two here. Why is that?"}, {"video_title": "Integration with partial fractions AP Calculus BC Khan Academy.mp3", "Sentence": "I could write the negative seven here, but I'm gonna take the constant out of the integral. So I'll put a negative seven here. And to help us solve this, and this could be a one, but to help us solve this, it would be nice if we had a two here. Why is that? Because two is the derivative of two x minus three. And so then we can do our u substitution, which we have sometimes gotten practice doing a little bit in our head. And so if we want this to be a two, we can't just multiply by two, we've also gotta divide by two."}, {"video_title": "Integration with partial fractions AP Calculus BC Khan Academy.mp3", "Sentence": "Why is that? Because two is the derivative of two x minus three. And so then we can do our u substitution, which we have sometimes gotten practice doing a little bit in our head. And so if we want this to be a two, we can't just multiply by two, we've also gotta divide by two. And I could do that outside, because that's just a constant. D x plus, well the derivative of x minus one is one, so we just want a one up in the numerator. So we could take the four out of the integral."}, {"video_title": "Integration with partial fractions AP Calculus BC Khan Academy.mp3", "Sentence": "And so if we want this to be a two, we can't just multiply by two, we've also gotta divide by two. And I could do that outside, because that's just a constant. D x plus, well the derivative of x minus one is one, so we just want a one up in the numerator. So we could take the four out of the integral. Four times the integral of one over x minus one, d x. So this is going to be equal to, we just have our constant out front, negative seven halves. And since we have this thing in the denominator, and we have its derivative, we can really just think about this as integrating with respect to this thing in the denominator."}, {"video_title": "Integration with partial fractions AP Calculus BC Khan Academy.mp3", "Sentence": "So we could take the four out of the integral. Four times the integral of one over x minus one, d x. So this is going to be equal to, we just have our constant out front, negative seven halves. And since we have this thing in the denominator, and we have its derivative, we can really just think about this as integrating with respect to this thing in the denominator. You can sometimes view u substitution, which I'm not explicitly going to do here, as the reverse chain rule. The antiderivative of one over x is the natural log of the absolute value of x. But here, the antiderivative of this is going to be the natural log of the absolute value of two x minus three."}, {"video_title": "Integration with partial fractions AP Calculus BC Khan Academy.mp3", "Sentence": "And since we have this thing in the denominator, and we have its derivative, we can really just think about this as integrating with respect to this thing in the denominator. You can sometimes view u substitution, which I'm not explicitly going to do here, as the reverse chain rule. The antiderivative of one over x is the natural log of the absolute value of x. But here, the antiderivative of this is going to be the natural log of the absolute value of two x minus three. And then for this part, this is gonna be plus four times the antiderivative here is the natural log of the absolute value of x minus one. And once again, I can do that because the derivative of x minus one is just one. And of course, since we're taking an indefinite integral, we do not want to forget our plus c. And we're done."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "He was a contemporary of Isaac Newton. These two gentlemen together were really the founding fathers of calculus, and they did some of their, most of their major work in the late 1600s. And this right over here is Usain Bolt, Jamaican sprinter, who's continuing to do some of his best work in 2012. And as of early 2012, he's the fastest human alive, and he's probably the fastest human that has ever lived. And you might not, you might have not made the association with these three gentlemen. You might not think that they have a lot in common, but they were all obsessed with the same fundamental question. And this is the same fundamental question that differential calculus addresses."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And as of early 2012, he's the fastest human alive, and he's probably the fastest human that has ever lived. And you might not, you might have not made the association with these three gentlemen. You might not think that they have a lot in common, but they were all obsessed with the same fundamental question. And this is the same fundamental question that differential calculus addresses. And the question is, what is the instantaneous rate of change of something? And in the case of Usain Bolt, how fast is he going right now? Not just what his average speed was for the last second, or his average speed over the next 10 seconds."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And this is the same fundamental question that differential calculus addresses. And the question is, what is the instantaneous rate of change of something? And in the case of Usain Bolt, how fast is he going right now? Not just what his average speed was for the last second, or his average speed over the next 10 seconds. How fast is he going right now? And so this is what differential calculus is all about, instantaneous rates of change. Differential calculus."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Not just what his average speed was for the last second, or his average speed over the next 10 seconds. How fast is he going right now? And so this is what differential calculus is all about, instantaneous rates of change. Differential calculus. Calculus. It's all, and Newton's actual original term for differential calculus was the method of fluxions, which actually sounds a little bit fancier. But it's all about what's happening in this instant."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Differential calculus. Calculus. It's all, and Newton's actual original term for differential calculus was the method of fluxions, which actually sounds a little bit fancier. But it's all about what's happening in this instant. In this instant. And to think about why that is not a super easy problem to address with traditional algebra, let's draw a little graph here. So on this axis, on this axis, I'll have distance."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "But it's all about what's happening in this instant. In this instant. And to think about why that is not a super easy problem to address with traditional algebra, let's draw a little graph here. So on this axis, on this axis, I'll have distance. So and I'll say y is equal to distance. I could have said d is equal to distance, but we'll see, especially later on in calculus, d is reserved for something else. We'll say y is equal to distance."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So on this axis, on this axis, I'll have distance. So and I'll say y is equal to distance. I could have said d is equal to distance, but we'll see, especially later on in calculus, d is reserved for something else. We'll say y is equal to distance. And in this axis, we'll say time. And I could say t is equal to time, but I'll just say x is equal to time. X is equal to time."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "We'll say y is equal to distance. And in this axis, we'll say time. And I could say t is equal to time, but I'll just say x is equal to time. X is equal to time. And so if we were to plot Usain Bolt's distance as a function of time, well at time zero, he hasn't gone anywhere. He is right over there. And we know that this gentleman is capable of traveling 100 meters in 9.58 seconds."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "X is equal to time. And so if we were to plot Usain Bolt's distance as a function of time, well at time zero, he hasn't gone anywhere. He is right over there. And we know that this gentleman is capable of traveling 100 meters in 9.58 seconds. So after 9.58 seconds, we'll assume that this is in seconds right over here, he's capable of going 100 meters. And so using this information, we can actually figure out his average speed. Let me write it this way."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And we know that this gentleman is capable of traveling 100 meters in 9.58 seconds. So after 9.58 seconds, we'll assume that this is in seconds right over here, he's capable of going 100 meters. And so using this information, we can actually figure out his average speed. Let me write it this way. His average speed is just going to be his change in distance over his change in time. And using the variables over here, we're saying y is distance, so this is the same thing as change in y over change in x from this point to that point. And this might look somewhat familiar to you from basic algebra."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Let me write it this way. His average speed is just going to be his change in distance over his change in time. And using the variables over here, we're saying y is distance, so this is the same thing as change in y over change in x from this point to that point. And this might look somewhat familiar to you from basic algebra. This is the slope between these two points. If I have a line that connects these two points, this is the slope of that line. The change in distance is this right over here."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And this might look somewhat familiar to you from basic algebra. This is the slope between these two points. If I have a line that connects these two points, this is the slope of that line. The change in distance is this right over here. Change in y is equal to 100 meters. And our change in time is this right over here. So our change in time is equal to 9.58 seconds."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "The change in distance is this right over here. Change in y is equal to 100 meters. And our change in time is this right over here. So our change in time is equal to 9.58 seconds. We started at 0, we go to 9.58 seconds. Another way to think about it, the rise over the run, you might have heard in your algebra class, is going to be 100 meters over 9.58 seconds. So this is 100 meters over 9.58 seconds."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So our change in time is equal to 9.58 seconds. We started at 0, we go to 9.58 seconds. Another way to think about it, the rise over the run, you might have heard in your algebra class, is going to be 100 meters over 9.58 seconds. So this is 100 meters over 9.58 seconds. And the slope is essentially just the rate of change, or you could view it as the average rate of change, between these two points. And you'll see, if you even just follow the units, it gives you units of speed here. It would be velocity if we also specified the direction."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So this is 100 meters over 9.58 seconds. And the slope is essentially just the rate of change, or you could view it as the average rate of change, between these two points. And you'll see, if you even just follow the units, it gives you units of speed here. It would be velocity if we also specified the direction. And we can figure out what that is. Let me get a calculator out. Let me get the calculator on the screen."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "It would be velocity if we also specified the direction. And we can figure out what that is. Let me get a calculator out. Let me get the calculator on the screen. So we're going 100 meters in 9.58 seconds. So it's 10.4. I'll just write 10.4."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Let me get the calculator on the screen. So we're going 100 meters in 9.58 seconds. So it's 10.4. I'll just write 10.4. I'll round to 10.4. So it's approximately 10.4. And then the units are meters per second."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "I'll just write 10.4. I'll round to 10.4. So it's approximately 10.4. And then the units are meters per second. And that is his average speed. And what we're going to see in a second is how average speed is different than instantaneous speed, how it's different than the speed he might be going at any given moment. And just to have a concept of how fast this is, let me get the calculator back."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And then the units are meters per second. And that is his average speed. And what we're going to see in a second is how average speed is different than instantaneous speed, how it's different than the speed he might be going at any given moment. And just to have a concept of how fast this is, let me get the calculator back. This is in meters per second. If you want to know how many meters he's going in an hour, well, there's 3,600 seconds in an hour. So he'll be able to go this many meters 3,600 times."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And just to have a concept of how fast this is, let me get the calculator back. This is in meters per second. If you want to know how many meters he's going in an hour, well, there's 3,600 seconds in an hour. So he'll be able to go this many meters 3,600 times. So that's how many meters he can, if you were able to somehow keep up that speed in an hour. This is how fast he's going meters per hour. And then if you were to say how many miles per hour, there's roughly 1,600."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So he'll be able to go this many meters 3,600 times. So that's how many meters he can, if you were able to somehow keep up that speed in an hour. This is how fast he's going meters per hour. And then if you were to say how many miles per hour, there's roughly 1,600. And I don't know the exact number, but roughly 1,600 meters per mile. So let's divide it by 1,600. And so you see that this is roughly a little over 23, about 23 and 1 half miles per hour."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And then if you were to say how many miles per hour, there's roughly 1,600. And I don't know the exact number, but roughly 1,600 meters per mile. So let's divide it by 1,600. And so you see that this is roughly a little over 23, about 23 and 1 half miles per hour. So this is approximately, I'll write it this way, this is approximately 23.5 miles per hour. And relative to a car, not so fast. But relative to me, extremely fast."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And so you see that this is roughly a little over 23, about 23 and 1 half miles per hour. So this is approximately, I'll write it this way, this is approximately 23.5 miles per hour. And relative to a car, not so fast. But relative to me, extremely fast. Now, to see how this is different than instantaneous velocity, let's think about a potential plot of his distance relative to time. He's not going to just go this speed immediately. He's not just going to go, as soon as the gun fires, he's not just going to go 23 and 1 half miles per hour all the way."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "But relative to me, extremely fast. Now, to see how this is different than instantaneous velocity, let's think about a potential plot of his distance relative to time. He's not going to just go this speed immediately. He's not just going to go, as soon as the gun fires, he's not just going to go 23 and 1 half miles per hour all the way. He's going to have to accelerate. So at first, he's going to start off going a little bit slower. So his slope is going to be a little bit lower than the average slope."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "He's not just going to go, as soon as the gun fires, he's not just going to go 23 and 1 half miles per hour all the way. He's going to have to accelerate. So at first, he's going to start off going a little bit slower. So his slope is going to be a little bit lower than the average slope. He's going to go a little bit slower. Then he's going to start accelerating. And so his speed, and you'll see the slope here, is getting steeper and steeper and steeper."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So his slope is going to be a little bit lower than the average slope. He's going to go a little bit slower. Then he's going to start accelerating. And so his speed, and you'll see the slope here, is getting steeper and steeper and steeper. And then maybe near the end, he starts tiring off a little bit. And so his distance plotted against time might be a curve that looks something like this. And what we calculated here is just the average slope across this change in time."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And so his speed, and you'll see the slope here, is getting steeper and steeper and steeper. And then maybe near the end, he starts tiring off a little bit. And so his distance plotted against time might be a curve that looks something like this. And what we calculated here is just the average slope across this change in time. But we could see at any given moment, the slope is actually different. In the beginning, he has a slower rate of change of distance. Then over here, then he accelerates."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And what we calculated here is just the average slope across this change in time. But we could see at any given moment, the slope is actually different. In the beginning, he has a slower rate of change of distance. Then over here, then he accelerates. Over here, it seems like his rate of change of distance, which would be roughly, or you could view it as the slope of the tangent line at that point, it looks higher than his average. And then he starts to slow down again. And when you all average it out, it gets to 23 and 1 1 half miles per hour."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Then over here, then he accelerates. Over here, it seems like his rate of change of distance, which would be roughly, or you could view it as the slope of the tangent line at that point, it looks higher than his average. And then he starts to slow down again. And when you all average it out, it gets to 23 and 1 1 half miles per hour. And I looked it up. Usain Bolt's instantaneous velocity, his peak instantaneous velocity, is actually closer to 30 miles per hour. So the slope over here might be 23 whatever miles per hour."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And when you all average it out, it gets to 23 and 1 1 half miles per hour. And I looked it up. Usain Bolt's instantaneous velocity, his peak instantaneous velocity, is actually closer to 30 miles per hour. So the slope over here might be 23 whatever miles per hour. But the instantaneous, his fastest point in this 9.58 seconds, is closer to 30 miles per hour. But you see, it's not a trivial thing to do. You could say, OK, let me try to approximate the slope right over here."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So the slope over here might be 23 whatever miles per hour. But the instantaneous, his fastest point in this 9.58 seconds, is closer to 30 miles per hour. But you see, it's not a trivial thing to do. You could say, OK, let me try to approximate the slope right over here. And you could do that by saying, OK, well, what is the change in y over the change of x right around this? So you could say, well, let me take some change of x and figure out what the change of y is around it, or as we go past that. So you get that."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "You could say, OK, let me try to approximate the slope right over here. And you could do that by saying, OK, well, what is the change in y over the change of x right around this? So you could say, well, let me take some change of x and figure out what the change of y is around it, or as we go past that. So you get that. But that would just be an approximation, because you see that the slope of this curve is constantly changing. So what you want to do is see what happens as your change of x gets smaller and smaller and smaller. As your change of x gets smaller and smaller and smaller, you're going to get a better and better approximation."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So you get that. But that would just be an approximation, because you see that the slope of this curve is constantly changing. So what you want to do is see what happens as your change of x gets smaller and smaller and smaller. As your change of x gets smaller and smaller and smaller, you're going to get a better and better approximation. Your change of y is going to get smaller and smaller and smaller. So what you want to do, and we're going to go into depth into all of this and study it more rigorously, is you want to take the limit as delta x approaches 0 of your change in y over your change in x. And when you do that, you're going to approach that instantaneous rate of change."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "As your change of x gets smaller and smaller and smaller, you're going to get a better and better approximation. Your change of y is going to get smaller and smaller and smaller. So what you want to do, and we're going to go into depth into all of this and study it more rigorously, is you want to take the limit as delta x approaches 0 of your change in y over your change in x. And when you do that, you're going to approach that instantaneous rate of change. You could view it as the instantaneous slope at that point in the curve, or the slope of the tangent line at that point in the curve. Or if we use calculus terminology, we would view that as the derivative. So the instantaneous slope is the derivative."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And when you do that, you're going to approach that instantaneous rate of change. You could view it as the instantaneous slope at that point in the curve, or the slope of the tangent line at that point in the curve. Or if we use calculus terminology, we would view that as the derivative. So the instantaneous slope is the derivative. And the notation we use for the derivative is dy over dx. And that's why I reserved the letter y. And you say, well, how does this relate to the word differential?"}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So the instantaneous slope is the derivative. And the notation we use for the derivative is dy over dx. And that's why I reserved the letter y. And you say, well, how does this relate to the word differential? Well, the word differential is relating this dy is a differential. dx is a differential. And one way to conceptualize it, this is an infinitely small change in y over an infinitely small change in x."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And you say, well, how does this relate to the word differential? Well, the word differential is relating this dy is a differential. dx is a differential. And one way to conceptualize it, this is an infinitely small change in y over an infinitely small change in x. And by getting super, super small changes in y over change in x, you're able to get your instantaneous slope, or in the case of this example, the instantaneous speed of Usain Bolt right at that moment. And notice, you can't just put a 0 here. If you just put change in x is 0, you're going to get something that's undefined."}, {"video_title": "Newton, Leibniz, and Usain Bolt Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And one way to conceptualize it, this is an infinitely small change in y over an infinitely small change in x. And by getting super, super small changes in y over change in x, you're able to get your instantaneous slope, or in the case of this example, the instantaneous speed of Usain Bolt right at that moment. And notice, you can't just put a 0 here. If you just put change in x is 0, you're going to get something that's undefined. You can't divide by 0. So we take the limit as it approaches 0. And we'll define that more rigorously in the next few videos."}, {"video_title": "Breaking up integral interval Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Now what I want to do in this video is introduce a third value, c, that is in between a and b. And it could be equal to a or it could be equal to b. So let me just introduce it. Write it just like that. And I could write that a is less than or equal to c, which is less than or equal to b. And what I want to think about is, how does this definite integral relate to the definite integral from a to c and the definite integral from c to b? So let's think through that."}, {"video_title": "Breaking up integral interval Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Write it just like that. And I could write that a is less than or equal to c, which is less than or equal to b. And what I want to think about is, how does this definite integral relate to the definite integral from a to c and the definite integral from c to b? So let's think through that. So we have the definite integral from a to c of f of x. Actually, I've already used that purple color for the function itself. So I'm going to use green."}, {"video_title": "Breaking up integral interval Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So let's think through that. So we have the definite integral from a to c of f of x. Actually, I've already used that purple color for the function itself. So I'm going to use green. So we have the integral from a to c of f of x dx. And that, of course, is going to represent this area right over here, from a to c under the curve f of x, above the x axis. So that's that."}, {"video_title": "Breaking up integral interval Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So I'm going to use green. So we have the integral from a to c of f of x dx. And that, of course, is going to represent this area right over here, from a to c under the curve f of x, above the x axis. So that's that. And then we could have the integral from c to b of f of x dx. And that, of course, is going to represent this area right over here. Well, the one thing that probably jumps out at you is that the entire area from a to b, this entire area, is just a sum of these two smaller areas."}, {"video_title": "Breaking up integral interval Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So that's that. And then we could have the integral from c to b of f of x dx. And that, of course, is going to represent this area right over here. Well, the one thing that probably jumps out at you is that the entire area from a to b, this entire area, is just a sum of these two smaller areas. So this is just equal to that plus that over there. And once again, you might say, why is this integration property useful? That if I found a c that is in this interval, that it's greater than or equal to a and it's less than or equal to b, why is it useful to be able to break up the integral this way?"}, {"video_title": "Breaking up integral interval Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the one thing that probably jumps out at you is that the entire area from a to b, this entire area, is just a sum of these two smaller areas. So this is just equal to that plus that over there. And once again, you might say, why is this integration property useful? That if I found a c that is in this interval, that it's greater than or equal to a and it's less than or equal to b, why is it useful to be able to break up the integral this way? Well, as you'll see, this is really useful. It can be very useful when you're looking at functions that have discontinuities. If they have step functions, you can break up the larger integral into smaller integrals."}, {"video_title": "Breaking up integral interval Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "That if I found a c that is in this interval, that it's greater than or equal to a and it's less than or equal to b, why is it useful to be able to break up the integral this way? Well, as you'll see, this is really useful. It can be very useful when you're looking at functions that have discontinuities. If they have step functions, you can break up the larger integral into smaller integrals. You'll also see that this is useful when we prove the fundamental theorem of calculus. So in general, this is actually a very, very, very useful technique. Let me actually draw an integral where it might be very useful to utilize that property."}, {"video_title": "Breaking up integral interval Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "If they have step functions, you can break up the larger integral into smaller integrals. You'll also see that this is useful when we prove the fundamental theorem of calculus. So in general, this is actually a very, very, very useful technique. Let me actually draw an integral where it might be very useful to utilize that property. So if this is a, this is b, and let's say the function, I'm just going to make it constant over an interval. So it's constant from there to there, and then it drops down from there to there. Let's say the function looked like this."}, {"video_title": "Breaking up integral interval Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Let me actually draw an integral where it might be very useful to utilize that property. So if this is a, this is b, and let's say the function, I'm just going to make it constant over an interval. So it's constant from there to there, and then it drops down from there to there. Let's say the function looked like this. Well, you could say that the larger integral, which would be the area under the curve, it would be all of this. Let's just say it's a gap right there, or it jumps down there. So this entire area, you can break up into two smaller areas."}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "And like always, pause the video and see if you can work through this. All right, well, our first temptation is to say, well, this is going to be the same thing as the limit of one minus cosine theta as x approaches, or not x, as theta approaches zero, of theta as theta approaches zero over the limit as theta approaches zero of two sine squared theta. Now, both of these expressions, which could be used to define a function, they'd be continuous if you graphed them. They'd be continuous at theta equals zero, so the limit is going to be the same thing as just evaluating them at theta equals zero. So this is going to be equal to one minus cosine of zero over two sine squared of zero. Now, cosine of zero is one, and then one minus one is zero, and sine of zero is zero, and you square it, you still got zero, and you multiply it times two, you still got zero, so you got zero over zero. So once again, we have that indeterminate form."}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "They'd be continuous at theta equals zero, so the limit is going to be the same thing as just evaluating them at theta equals zero. So this is going to be equal to one minus cosine of zero over two sine squared of zero. Now, cosine of zero is one, and then one minus one is zero, and sine of zero is zero, and you square it, you still got zero, and you multiply it times two, you still got zero, so you got zero over zero. So once again, we have that indeterminate form. And once again, this indeterminate form, when you have zero over zero, doesn't mean to give up, it doesn't mean that the limit doesn't exist, it just means, well, maybe there's some other approaches here to work on. If you got some non-zero number divided by zero, then you say, okay, that limit doesn't exist, and you'd say, well, you'd just say it doesn't exist. But let's see what we can do to maybe think about this expression in a different way."}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "So once again, we have that indeterminate form. And once again, this indeterminate form, when you have zero over zero, doesn't mean to give up, it doesn't mean that the limit doesn't exist, it just means, well, maybe there's some other approaches here to work on. If you got some non-zero number divided by zero, then you say, okay, that limit doesn't exist, and you'd say, well, you'd just say it doesn't exist. But let's see what we can do to maybe think about this expression in a different way. So if we said, so let's just say that this, let me use some other colors here. Let's say that this right over here is f of x. So f of x is equal to one minus cosine theta over two sine squared theta."}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "But let's see what we can do to maybe think about this expression in a different way. So if we said, so let's just say that this, let me use some other colors here. Let's say that this right over here is f of x. So f of x is equal to one minus cosine theta over two sine squared theta. And let's see if we can rewrite it in some way, that at least the limit as theta approaches zero isn't going to, we're not gonna get the same zero over zero. Well, we can, we got some trig functions here, so maybe we can use some of our trig identities to simplify this. And the one that jumps out at me is that we have the sine squared of theta, and we know from the Pythagorean identity in trigonometries, comes straight out of the unit circle definition of sine and cosine, we know that sine squared theta plus cosine squared theta is equal to one, or we know that sine squared theta is one minus cosine squared theta."}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "So f of x is equal to one minus cosine theta over two sine squared theta. And let's see if we can rewrite it in some way, that at least the limit as theta approaches zero isn't going to, we're not gonna get the same zero over zero. Well, we can, we got some trig functions here, so maybe we can use some of our trig identities to simplify this. And the one that jumps out at me is that we have the sine squared of theta, and we know from the Pythagorean identity in trigonometries, comes straight out of the unit circle definition of sine and cosine, we know that sine squared theta plus cosine squared theta is equal to one, or we know that sine squared theta is one minus cosine squared theta. One minus cosine squared theta. So we could rewrite this. This is equal to one minus cosine theta over two times one minus cosine squared theta."}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "And the one that jumps out at me is that we have the sine squared of theta, and we know from the Pythagorean identity in trigonometries, comes straight out of the unit circle definition of sine and cosine, we know that sine squared theta plus cosine squared theta is equal to one, or we know that sine squared theta is one minus cosine squared theta. One minus cosine squared theta. So we could rewrite this. This is equal to one minus cosine theta over two times one minus cosine squared theta. Now, this is a one minus cosine theta, this is a one minus cosine squared theta, so it's not completely obvious yet of how you can simplify it, until you realize that this could be viewed as a difference of squares. If you view this as, if you view this as a squared minus b squared, we know that this can be factored as a plus b times a minus b. So I could rewrite this."}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "This is equal to one minus cosine theta over two times one minus cosine squared theta. Now, this is a one minus cosine theta, this is a one minus cosine squared theta, so it's not completely obvious yet of how you can simplify it, until you realize that this could be viewed as a difference of squares. If you view this as, if you view this as a squared minus b squared, we know that this can be factored as a plus b times a minus b. So I could rewrite this. This is equal to one minus cosine theta over two times, I could write this as one plus cosine theta times one minus cosine theta. One plus cosine theta times one minus, one minus cosine theta. And now this is interesting."}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "So I could rewrite this. This is equal to one minus cosine theta over two times, I could write this as one plus cosine theta times one minus cosine theta. One plus cosine theta times one minus, one minus cosine theta. And now this is interesting. I have a one minus cosine theta in the numerator, and I have a one minus cosine theta in the denominator. Now we might be tempted to say, oh, well, let's just cross that out with that, and we would get, we would simplify it and get f of x is equal to one over, and we could distribute this two now, we could say two plus two cosine theta. We could say, well, aren't these the same thing?"}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "And now this is interesting. I have a one minus cosine theta in the numerator, and I have a one minus cosine theta in the denominator. Now we might be tempted to say, oh, well, let's just cross that out with that, and we would get, we would simplify it and get f of x is equal to one over, and we could distribute this two now, we could say two plus two cosine theta. We could say, well, aren't these the same thing? And we would be almost right, because f of x, this one right over here, this is defined, this right over here is defined when theta is equal to zero, while this one is not defined when theta is equal to zero. When theta is equal to zero, you have a zero in the denominator. And so what we need to do in order for this f of x, or in order for this to be the same thing, we have to say theta cannot be equal to zero."}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "We could say, well, aren't these the same thing? And we would be almost right, because f of x, this one right over here, this is defined, this right over here is defined when theta is equal to zero, while this one is not defined when theta is equal to zero. When theta is equal to zero, you have a zero in the denominator. And so what we need to do in order for this f of x, or in order for this to be the same thing, we have to say theta cannot be equal to zero. But now let's think about the limit again. Essentially what we want to do is we want to find the limit as theta approaches zero of f of x. And we can't just do direct substitution into, if we really take this seriously, because we're going to like, oh, well, if I try to put zero here, it says theta cannot be equal to zero."}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "And so what we need to do in order for this f of x, or in order for this to be the same thing, we have to say theta cannot be equal to zero. But now let's think about the limit again. Essentially what we want to do is we want to find the limit as theta approaches zero of f of x. And we can't just do direct substitution into, if we really take this seriously, because we're going to like, oh, well, if I try to put zero here, it says theta cannot be equal to zero. f of x is not defined at zero. This expression is defined at zero, but this tells me, well, I really shouldn't apply zero to this function. But we know that if we can find another function that is defined, that is the exact same thing as f of x, except at zero, and it is continuous at zero."}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "And we can't just do direct substitution into, if we really take this seriously, because we're going to like, oh, well, if I try to put zero here, it says theta cannot be equal to zero. f of x is not defined at zero. This expression is defined at zero, but this tells me, well, I really shouldn't apply zero to this function. But we know that if we can find another function that is defined, that is the exact same thing as f of x, except at zero, and it is continuous at zero. And so we could say g of x is equal to one over two plus two cosine theta. Well, then we know this limit is going to be the exact same thing as the limit of g of x as theta approaches zero. Once again, these two functions are identical, except f of x is not defined at theta equals zero, while g of x is."}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "But we know that if we can find another function that is defined, that is the exact same thing as f of x, except at zero, and it is continuous at zero. And so we could say g of x is equal to one over two plus two cosine theta. Well, then we know this limit is going to be the exact same thing as the limit of g of x as theta approaches zero. Once again, these two functions are identical, except f of x is not defined at theta equals zero, while g of x is. But the limits as theta approaches zero are going to be the same, and we've seen that in previous videos. And I know what a lot of you are thinking, Sal, this seems like a very, you know, why don't I just, you know, do this algebra here, cross these things out, and just substitute zero for theta? Well, you could do that, and you would get the answer, but you need to be clear, or it's important to be mathematically clear of what you are doing."}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "Once again, these two functions are identical, except f of x is not defined at theta equals zero, while g of x is. But the limits as theta approaches zero are going to be the same, and we've seen that in previous videos. And I know what a lot of you are thinking, Sal, this seems like a very, you know, why don't I just, you know, do this algebra here, cross these things out, and just substitute zero for theta? Well, you could do that, and you would get the answer, but you need to be clear, or it's important to be mathematically clear of what you are doing. If you do that, if you just cross these two out, and all of a sudden, your expression becomes defined at zero, you are now dealing with a different expression, or a different function definition. So to be clear, if you want to say this is the function you're finding the limit of, you have to put this constraint in to make sure it has the exact same domain. But lucky for us, we can say if we found another function that's continuous at that point that doesn't have that gap there, that doesn't have that point discontinuity out, the limits are going to be equivalent."}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "Well, you could do that, and you would get the answer, but you need to be clear, or it's important to be mathematically clear of what you are doing. If you do that, if you just cross these two out, and all of a sudden, your expression becomes defined at zero, you are now dealing with a different expression, or a different function definition. So to be clear, if you want to say this is the function you're finding the limit of, you have to put this constraint in to make sure it has the exact same domain. But lucky for us, we can say if we found another function that's continuous at that point that doesn't have that gap there, that doesn't have that point discontinuity out, the limits are going to be equivalent. So the limit as theta approaches zero of g of x, well, that's just going to be, since it's continuous at zero, we could say that's just going to be, we can just substitute, that's going to be equal to g of zero, which is equal to one over two plus cosine two, one over two plus two times cosine of zero. Cosine of zero is one, so it's just one over two plus two, which is equal to, deserve a little bit of a drum roll here, which is equal to 1 4th. And we are done."}, {"video_title": "Derivative of inverse cosine Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So what is, so our goal here is to figure out, I want the derivative with respect to x of the inverse cosine of x. What is this going to be equal to? So assuming you've had a go at it, let's work through it. So just like last time, we could write, let's just set y being equal to this. Y is equal to the inverse cosine of x, which means the same thing as saying that x is equal to the cosine of y. Now let's take the derivative of both sides with respect to x. On the left-hand side, you're just going to have a one."}, {"video_title": "Derivative of inverse cosine Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So just like last time, we could write, let's just set y being equal to this. Y is equal to the inverse cosine of x, which means the same thing as saying that x is equal to the cosine of y. Now let's take the derivative of both sides with respect to x. On the left-hand side, you're just going to have a one. We're just going to have a one. And on the right-hand side, you're going to have the derivative of cosine y with respect to y, which is negative sine of y times the derivative of y with respect to x, which is dy dx. And so we get, let's see, if we divide both sides by negative sine of y, we get dy dx is equal to negative one over sine of y."}, {"video_title": "Derivative of inverse cosine Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "On the left-hand side, you're just going to have a one. We're just going to have a one. And on the right-hand side, you're going to have the derivative of cosine y with respect to y, which is negative sine of y times the derivative of y with respect to x, which is dy dx. And so we get, let's see, if we divide both sides by negative sine of y, we get dy dx is equal to negative one over sine of y. Now, like we've seen before, this is kind of satisfying, but we have our derivative in terms of y. We want it in terms of x. And we know that x is cosine of y."}, {"video_title": "Derivative of inverse cosine Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And so we get, let's see, if we divide both sides by negative sine of y, we get dy dx is equal to negative one over sine of y. Now, like we've seen before, this is kind of satisfying, but we have our derivative in terms of y. We want it in terms of x. And we know that x is cosine of y. So let's see if we can rewrite this bottom expression in terms of cosine of y instead of sine of y. Well, we know, and we saw it in the last video, that from the Pythagorean identity, that cosine squared of y plus sine squared of y is equal to one. We know that sine of y is equal to the square root of one minus cosine squared of y."}, {"video_title": "Derivative of inverse cosine Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And we know that x is cosine of y. So let's see if we can rewrite this bottom expression in terms of cosine of y instead of sine of y. Well, we know, and we saw it in the last video, that from the Pythagorean identity, that cosine squared of y plus sine squared of y is equal to one. We know that sine of y is equal to the square root of one minus cosine squared of y. So this is equal to negative one. This is just a manipulation of the Pythagorean trig identity. This is equal to one minus cosine."}, {"video_title": "Derivative of inverse cosine Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "We know that sine of y is equal to the square root of one minus cosine squared of y. So this is equal to negative one. This is just a manipulation of the Pythagorean trig identity. This is equal to one minus cosine. I could write it like this, cosine squared of y, but I'll write it like this, because it'll make it a little bit clearer. And what is cosine of y? Well, of course, that is x."}, {"video_title": "Derivative of inverse cosine Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "This is equal to one minus cosine. I could write it like this, cosine squared of y, but I'll write it like this, because it'll make it a little bit clearer. And what is cosine of y? Well, of course, that is x. So this is equal to negative one over the square root of one minus, instead of writing cosine y, instead of writing cosine y, I'm trying to switch colors, instead of writing cosine y, we could write one minus x, one minus x squared. So there you have it. The derivative with respect to x of the inverse cosine of x is, I think I lost that color, I'll do it in magenta, is equal to negative one over the square root of one minus x squared."}, {"video_title": "Derivative of inverse cosine Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Well, of course, that is x. So this is equal to negative one over the square root of one minus, instead of writing cosine y, instead of writing cosine y, I'm trying to switch colors, instead of writing cosine y, we could write one minus x, one minus x squared. So there you have it. The derivative with respect to x of the inverse cosine of x is, I think I lost that color, I'll do it in magenta, is equal to negative one over the square root of one minus x squared. So this is a neat thing. This right over here is a neat thing to know. And of course, we should compare it to the inverse, the derivative of the inverse sine."}, {"video_title": "Derivative of inverse cosine Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "The derivative with respect to x of the inverse cosine of x is, I think I lost that color, I'll do it in magenta, is equal to negative one over the square root of one minus x squared. So this is a neat thing. This right over here is a neat thing to know. And of course, we should compare it to the inverse, the derivative of the inverse sine. Actually, let me put them side by side, and we see that the only difference here is the sine. So let me copy and paste that. So copy and paste it."}, {"video_title": "Derivative of inverse cosine Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And of course, we should compare it to the inverse, the derivative of the inverse sine. Actually, let me put them side by side, and we see that the only difference here is the sine. So let me copy and paste that. So copy and paste it. I'm gonna paste it down here. And now let's look at them side by side. So we see, for taking the derivative with respect to x of the inverse cosine function, we have a negative, a negative one over the square root of one minus x squared."}, {"video_title": "Estimating limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "This table gives select values of g. What is a reasonable estimate for the limit as x approaches five of g of x? So pause this video, look at this table. It gives us the x values as we approach five from values less than five, and as we approach five from values greater than five, it even tells us what g of x is at x equals five. And so given that, what is a reasonable estimate for this limit? All right, now let's work through this together. So let's think about what g of x seems to be approaching as x approaches five from values less than five. Let's see, at four, it's at 3.37, 4.9, it's a little higher, it's at 3.5, 4.99, it's at 3.66, 4.999, so very close to five."}, {"video_title": "Estimating limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And so given that, what is a reasonable estimate for this limit? All right, now let's work through this together. So let's think about what g of x seems to be approaching as x approaches five from values less than five. Let's see, at four, it's at 3.37, 4.9, it's a little higher, it's at 3.5, 4.99, it's at 3.66, 4.999, so very close to five. We're only a thousandth away. We're at 3.68. But then at five, all of a sudden, it looks like we're kind of jumping to 6.37."}, {"video_title": "Estimating limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Let's see, at four, it's at 3.37, 4.9, it's a little higher, it's at 3.5, 4.99, it's at 3.66, 4.999, so very close to five. We're only a thousandth away. We're at 3.68. But then at five, all of a sudden, it looks like we're kind of jumping to 6.37. And once again, I'm making an inference here. These are just sample points of this function. We don't know exactly what the function is."}, {"video_title": "Estimating limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "But then at five, all of a sudden, it looks like we're kind of jumping to 6.37. And once again, I'm making an inference here. These are just sample points of this function. We don't know exactly what the function is. But then if we approach five from values greater than five, at six, we're at 3.97, at 5.1, we're at 3.84, 5.01, 3.7, 5.001, we are at 3.68. So a thousandth below five and a thousandth above five, we're at 3.68, but then at five, all of a sudden, we're at 6.37. So my most reasonable estimate would be, well, it looks like we are approaching 3.68 when we are approaching from values less than five, and we're approaching 3.68 from values, as we approach five from values greater than five."}, {"video_title": "Estimating limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "We don't know exactly what the function is. But then if we approach five from values greater than five, at six, we're at 3.97, at 5.1, we're at 3.84, 5.01, 3.7, 5.001, we are at 3.68. So a thousandth below five and a thousandth above five, we're at 3.68, but then at five, all of a sudden, we're at 6.37. So my most reasonable estimate would be, well, it looks like we are approaching 3.68 when we are approaching from values less than five, and we're approaching 3.68 from values, as we approach five from values greater than five. It doesn't matter that the value of five is 6.37. The limit would be 3.68, or a reasonable estimate for the limit would be 3.68. And this is probably the most tempting distractor here, because if you were to just substitute five, what is g of five?"}, {"video_title": "Estimating limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So my most reasonable estimate would be, well, it looks like we are approaching 3.68 when we are approaching from values less than five, and we're approaching 3.68 from values, as we approach five from values greater than five. It doesn't matter that the value of five is 6.37. The limit would be 3.68, or a reasonable estimate for the limit would be 3.68. And this is probably the most tempting distractor here, because if you were to just substitute five, what is g of five? It tells us 6.37. But the limit does not have to be what the actual function equals at that point. Let me draw what this might look like."}, {"video_title": "Estimating limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And this is probably the most tempting distractor here, because if you were to just substitute five, what is g of five? It tells us 6.37. But the limit does not have to be what the actual function equals at that point. Let me draw what this might look like. And so an example of this. So if this is five right over here, at the point five, the value of my function is 6.37. So let's say that this right over here is 6.37."}, {"video_title": "Estimating limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Let me draw what this might look like. And so an example of this. So if this is five right over here, at the point five, the value of my function is 6.37. So let's say that this right over here is 6.37. So that's the value of my function right over there. So 6.37, but as we approach five, so that's four, actually let me spread out a little bit. This obviously is not drawing to scale."}, {"video_title": "Estimating limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say that this right over here is 6.37. So that's the value of my function right over there. So 6.37, but as we approach five, so that's four, actually let me spread out a little bit. This obviously is not drawing to scale. But as we approach five, so if that's 6.37, then at four, 3.37 is about here, and it looks like it's approaching 3.68. So 3.68, actually let me draw that. So 3.68 is gonna be roughly that."}, {"video_title": "Estimating limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "This obviously is not drawing to scale. But as we approach five, so if that's 6.37, then at four, 3.37 is about here, and it looks like it's approaching 3.68. So 3.68, actually let me draw that. So 3.68 is gonna be roughly that. 3.68 is gonna be roughly that. So the graph, the graph might look something like this. We could infer it looks like it's doing something like this, where it's approaching 3.68 from values less than five and values greater than five."}, {"video_title": "Estimating limits from tables Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So 3.68 is gonna be roughly that. 3.68 is gonna be roughly that. So the graph, the graph might look something like this. We could infer it looks like it's doing something like this, where it's approaching 3.68 from values less than five and values greater than five. But right at five, our value is 6.37. I don't know for sure if this is what the graph looks like. Once again, we're just getting some sample points."}, {"video_title": "Worked example Chain rule with table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "The following table lists the values of functions f and g and of their derivatives, f prime and g prime, for the x values negative two and four. And so you can see for x equals negative two, x equals four, they give us the values of f, g, f prime and g prime. Let function capital F be defined as the composition of f and g. It's lowercase f of g of x. And they want us to evaluate f prime of four. So you might immediately recognize that if I have a function that can be viewed as a composition of other functions, that the chain rule will apply here. And so, and I'm just gonna restate the chain rule. The derivative of capital F, it's going to be the derivative of lowercase f, the outside function with respect to the inside function."}, {"video_title": "Worked example Chain rule with table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And they want us to evaluate f prime of four. So you might immediately recognize that if I have a function that can be viewed as a composition of other functions, that the chain rule will apply here. And so, and I'm just gonna restate the chain rule. The derivative of capital F, it's going to be the derivative of lowercase f, the outside function with respect to the inside function. So lowercase f prime of g of x, times the derivative of the inside function with respect to x, times g prime of x. And if we're looking for f prime of four, f prime of four, well everywhere we see an x, we replace it with a four. That's gonna be lowercase f prime of g of four, times g prime of four."}, {"video_title": "Worked example Chain rule with table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "The derivative of capital F, it's going to be the derivative of lowercase f, the outside function with respect to the inside function. So lowercase f prime of g of x, times the derivative of the inside function with respect to x, times g prime of x. And if we're looking for f prime of four, f prime of four, well everywhere we see an x, we replace it with a four. That's gonna be lowercase f prime of g of four, times g prime of four. Now how do we figure this out? They haven't given us explicitly the values of the functions for all x's, but they've given it to us at some interesting points. So the first thing you might wanna figure out is, well what is g of four going to be?"}, {"video_title": "Worked example Chain rule with table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "That's gonna be lowercase f prime of g of four, times g prime of four. Now how do we figure this out? They haven't given us explicitly the values of the functions for all x's, but they've given it to us at some interesting points. So the first thing you might wanna figure out is, well what is g of four going to be? Well they tell us, when x is equal to four, g of four is negative two. This tells us that the value that g of x takes on when x is equal to four is negative two. So this right over here is negative two."}, {"video_title": "Worked example Chain rule with table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So the first thing you might wanna figure out is, well what is g of four going to be? Well they tell us, when x is equal to four, g of four is negative two. This tells us that the value that g of x takes on when x is equal to four is negative two. So this right over here is negative two. And so this first part is f prime of negative two. So what is f prime, what is f prime of negative two? Well when x is equal to negative two, f prime is equal to one."}, {"video_title": "Worked example Chain rule with table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So this right over here is negative two. And so this first part is f prime of negative two. So what is f prime, what is f prime of negative two? Well when x is equal to negative two, f prime is equal to one. So this right over here is f prime of negative two. That is equal to one. And now we just have to figure out what g prime of four is."}, {"video_title": "Worked example Chain rule with table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well when x is equal to negative two, f prime is equal to one. So this right over here is f prime of negative two. That is equal to one. And now we just have to figure out what g prime of four is. Well when, let me circle this, g prime of four, when x is equal to four, and I'll scroll down a little bit, when x is equal to four, g prime takes on the value eight. So there you have it. f prime of four is equal to one times eight, which is equal to eight."}, {"video_title": "Mistakes when finding inflection points not checking candidates AP Calculus AB Khan Academy.mp3", "Sentence": "This is her solution. So we look at her solution and then they ask us, is Olga's work correct? If not, what's her mistake? So pause this video and see if you can figure this out. All right, let's just follow her work. So here she's trying to take the first derivative. So you would apply the chain rule."}, {"video_title": "Mistakes when finding inflection points not checking candidates AP Calculus AB Khan Academy.mp3", "Sentence": "So pause this video and see if you can figure this out. All right, let's just follow her work. So here she's trying to take the first derivative. So you would apply the chain rule. It would be four times x minus two to the third power times the derivative of x minus two, which is just one. So this checks out. Then you take the derivative of this."}, {"video_title": "Mistakes when finding inflection points not checking candidates AP Calculus AB Khan Academy.mp3", "Sentence": "So you would apply the chain rule. It would be four times x minus two to the third power times the derivative of x minus two, which is just one. So this checks out. Then you take the derivative of this. It would be three times four, which would be 12 times x minus two to the second power times the derivative of x minus two, which is just one, which is exactly what she has here, 12 times x minus two to the second power. That checks out. So step one's looking good for Olga."}, {"video_title": "Mistakes when finding inflection points not checking candidates AP Calculus AB Khan Academy.mp3", "Sentence": "Then you take the derivative of this. It would be three times four, which would be 12 times x minus two to the second power times the derivative of x minus two, which is just one, which is exactly what she has here, 12 times x minus two to the second power. That checks out. So step one's looking good for Olga. Step two, the solution of the second derivative equaling zero is x equals two. That looks right. The second derivative is 12 times x minus two squared and we wanna make that equal to zero."}, {"video_title": "Mistakes when finding inflection points not checking candidates AP Calculus AB Khan Academy.mp3", "Sentence": "So step one's looking good for Olga. Step two, the solution of the second derivative equaling zero is x equals two. That looks right. The second derivative is 12 times x minus two squared and we wanna make that equal to zero. This is only going to be true when x is equal to two. So step two is looking good. So step three, Olga says f has an inflection point at x equals two."}, {"video_title": "Mistakes when finding inflection points not checking candidates AP Calculus AB Khan Academy.mp3", "Sentence": "The second derivative is 12 times x minus two squared and we wanna make that equal to zero. This is only going to be true when x is equal to two. So step two is looking good. So step three, Olga says f has an inflection point at x equals two. So she's basing this just on the fact that the second derivative is zero when x is equal to two. She's basing this just on the fact that f prime prime of two is equal to zero. Now I have a problem with this because the fact that your second derivative is zero at x equals two, that makes two a nice candidate to check out, but you can't immediately say that we have an inflection point there."}, {"video_title": "Mistakes when finding inflection points not checking candidates AP Calculus AB Khan Academy.mp3", "Sentence": "So step three, Olga says f has an inflection point at x equals two. So she's basing this just on the fact that the second derivative is zero when x is equal to two. She's basing this just on the fact that f prime prime of two is equal to zero. Now I have a problem with this because the fact that your second derivative is zero at x equals two, that makes two a nice candidate to check out, but you can't immediately say that we have an inflection point there. Remember, an inflection point is where we go from being concave upwards to concave downwards or concave downwards to concave upwards. And speaking in the language of the second derivative, it means that the second derivative changes signs as we go from below x equals two to above x equals two. But we have to test that because it's not necessarily always the case."}, {"video_title": "Mistakes when finding inflection points not checking candidates AP Calculus AB Khan Academy.mp3", "Sentence": "Now I have a problem with this because the fact that your second derivative is zero at x equals two, that makes two a nice candidate to check out, but you can't immediately say that we have an inflection point there. Remember, an inflection point is where we go from being concave upwards to concave downwards or concave downwards to concave upwards. And speaking in the language of the second derivative, it means that the second derivative changes signs as we go from below x equals two to above x equals two. But we have to test that because it's not necessarily always the case. So let's actually test it. Let's think about some intervals. Intervals."}, {"video_title": "Mistakes when finding inflection points not checking candidates AP Calculus AB Khan Academy.mp3", "Sentence": "But we have to test that because it's not necessarily always the case. So let's actually test it. Let's think about some intervals. Intervals. So let's think about the interval when we go from negative infinity to two. And let's think about the interval where we go from two to positive infinity. If you want, you could have some test values."}, {"video_title": "Mistakes when finding inflection points not checking candidates AP Calculus AB Khan Academy.mp3", "Sentence": "Intervals. So let's think about the interval when we go from negative infinity to two. And let's think about the interval where we go from two to positive infinity. If you want, you could have some test values. You could think about the sign, sign of our second derivative. And then based on that, you could think about concavity. Concavity of f. So let's think about what's happening."}, {"video_title": "Mistakes when finding inflection points not checking candidates AP Calculus AB Khan Academy.mp3", "Sentence": "If you want, you could have some test values. You could think about the sign, sign of our second derivative. And then based on that, you could think about concavity. Concavity of f. So let's think about what's happening. So you could take a test value, let's say one is in this interval, and let's say three is in this interval. And you could say one minus two squared is going to be, let's see, that's negative one squared, which is one. And then you're just going to, this is just going to be 12."}, {"video_title": "Mistakes when finding inflection points not checking candidates AP Calculus AB Khan Academy.mp3", "Sentence": "Concavity of f. So let's think about what's happening. So you could take a test value, let's say one is in this interval, and let's say three is in this interval. And you could say one minus two squared is going to be, let's see, that's negative one squared, which is one. And then you're just going to, this is just going to be 12. So this is going to be positive. And if you tried three, three minus two squared is one times 12. Well, that's also going to be positive."}, {"video_title": "Mistakes when finding inflection points not checking candidates AP Calculus AB Khan Academy.mp3", "Sentence": "And then you're just going to, this is just going to be 12. So this is going to be positive. And if you tried three, three minus two squared is one times 12. Well, that's also going to be positive. And so you're going to be concave upwards, at least at these test values, it looks like on either side of two, that the sign of the second derivative is positive on either side of two. And you might say, well, maybe I just need to find closer values. But if you inspect the second derivative here, you can see that this is never going to be negative."}, {"video_title": "Mistakes when finding inflection points not checking candidates AP Calculus AB Khan Academy.mp3", "Sentence": "Well, that's also going to be positive. And so you're going to be concave upwards, at least at these test values, it looks like on either side of two, that the sign of the second derivative is positive on either side of two. And you might say, well, maybe I just need to find closer values. But if you inspect the second derivative here, you can see that this is never going to be negative. In fact, for any value other than x equals two, this value right over here, since we're, even if x minus two is negative, you're squaring it, which will make this entire thing positive, and then multiplying it times a positive value. So for any value other than x equals two, the sign of our second derivative is positive, which means that we're going to be concave upwards. And so we actually don't have an inflection point at x equals two, because we are not switching signs as we go from values less than x equals two to values greater than x equals two."}, {"video_title": "Mistakes when finding inflection points not checking candidates AP Calculus AB Khan Academy.mp3", "Sentence": "But if you inspect the second derivative here, you can see that this is never going to be negative. In fact, for any value other than x equals two, this value right over here, since we're, even if x minus two is negative, you're squaring it, which will make this entire thing positive, and then multiplying it times a positive value. So for any value other than x equals two, the sign of our second derivative is positive, which means that we're going to be concave upwards. And so we actually don't have an inflection point at x equals two, because we are not switching signs as we go from values less than x equals two to values greater than x equals two. Our second derivative is not switching signs. So once again, this is incorrect. We actually don't have an inflection point at x equals two, because our second derivative does not switch signs as we cross x equals two, which means our concavity does not change."}, {"video_title": "Limit properties Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And we're not going to prove it rigorously here. In order to have the rigorous proof of these properties, we need a rigorous definition of what a limit is. And we're not doing that in this tutorial. We'll do that in the tutorial on the epsilon delta definition of limits. But most of these should be fairly intuitive. And they're very helpful for simplifying limit problems in the future. So let's say we know that the limit of some function f of x as x approaches c is equal to capital L. And let's say that we also know that the limit of some other function, let's say g of x as x approaches c, is equal to capital M. Now, given that, what would be the limit of f of x plus g of x as x approaches c?"}, {"video_title": "Limit properties Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "We'll do that in the tutorial on the epsilon delta definition of limits. But most of these should be fairly intuitive. And they're very helpful for simplifying limit problems in the future. So let's say we know that the limit of some function f of x as x approaches c is equal to capital L. And let's say that we also know that the limit of some other function, let's say g of x as x approaches c, is equal to capital M. Now, given that, what would be the limit of f of x plus g of x as x approaches c? Well, and you could look at this visually. If you look at the graphs of two arbitrary functions, you would essentially just add those two functions. It'll be pretty clear that this is going to be equal to, and once again, I'm not doing a rigorous proof."}, {"video_title": "Limit properties Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say we know that the limit of some function f of x as x approaches c is equal to capital L. And let's say that we also know that the limit of some other function, let's say g of x as x approaches c, is equal to capital M. Now, given that, what would be the limit of f of x plus g of x as x approaches c? Well, and you could look at this visually. If you look at the graphs of two arbitrary functions, you would essentially just add those two functions. It'll be pretty clear that this is going to be equal to, and once again, I'm not doing a rigorous proof. I'm just really giving you the properties here. This is going to be the limit of f of x as x approaches c plus the limit of g of x as x approaches c, which is equal to, well, this right over here is, let me do that in that same color, this right here is just equal to L. It's going to be equal to L plus M. This right over here is equal to M. Not too difficult. This is often called the sum rule or the sum property of limits."}, {"video_title": "Limit properties Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "It'll be pretty clear that this is going to be equal to, and once again, I'm not doing a rigorous proof. I'm just really giving you the properties here. This is going to be the limit of f of x as x approaches c plus the limit of g of x as x approaches c, which is equal to, well, this right over here is, let me do that in that same color, this right here is just equal to L. It's going to be equal to L plus M. This right over here is equal to M. Not too difficult. This is often called the sum rule or the sum property of limits. And we could come up with a very similar one with differences. The limit as x approaches c of f of x minus g of x is just going to be L minus M. It's just the limit of f of x as x approaches c minus the limit of g of x as x approaches c. So it's just going to be L minus M. It's often called the difference rule or the difference property of limits. And these, once again, are very, very, hopefully, reasonably intuitive."}, {"video_title": "Limit properties Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "This is often called the sum rule or the sum property of limits. And we could come up with a very similar one with differences. The limit as x approaches c of f of x minus g of x is just going to be L minus M. It's just the limit of f of x as x approaches c minus the limit of g of x as x approaches c. So it's just going to be L minus M. It's often called the difference rule or the difference property of limits. And these, once again, are very, very, hopefully, reasonably intuitive. Now what happens if you take the product of the functions? The limit of f of x times g of x as x approaches c. Well, lucky for us, this is going to be equal to the limit of f of x as x approaches c times the limit of g of x as x approaches c. Lucky for us, this is kind of a fairly intuitive property of limits. So in this case, this is just going to be equal to L times M. Same thing if instead of having a function here, we had a constant."}, {"video_title": "Limit properties Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And these, once again, are very, very, hopefully, reasonably intuitive. Now what happens if you take the product of the functions? The limit of f of x times g of x as x approaches c. Well, lucky for us, this is going to be equal to the limit of f of x as x approaches c times the limit of g of x as x approaches c. Lucky for us, this is kind of a fairly intuitive property of limits. So in this case, this is just going to be equal to L times M. Same thing if instead of having a function here, we had a constant. So if we just had the limit of k times f of x as x approaches c, where k is just some constant, this is going to be the same thing as k times the limit of f of x as x approaches c. And that is just equal to L. So this whole thing simplifies to k times L. And we can do the same thing with difference. This is often called the constant multiple property. We can do the same thing with differences."}, {"video_title": "Limit properties Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So in this case, this is just going to be equal to L times M. Same thing if instead of having a function here, we had a constant. So if we just had the limit of k times f of x as x approaches c, where k is just some constant, this is going to be the same thing as k times the limit of f of x as x approaches c. And that is just equal to L. So this whole thing simplifies to k times L. And we can do the same thing with difference. This is often called the constant multiple property. We can do the same thing with differences. So if we have the limit as x approaches c of f of x divided by g of x, this is the exact same thing as the limit of f of x as x approaches c divided by the limit of g of x as x approaches c, which is going to be equal to L over M. And finally, this is sometimes called the quotient property. Finally, we'll look at the exponent property. So if I have the limit of f of x to some power, and actually let me even write it as a fractional power, to the r over s power, where both r and s are integers, then the limit of f of x to the r over s power as x approaches c is going to be the exact same thing as the limit of f of x as x approaches c raised to the r over s power."}, {"video_title": "Limit properties Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "We can do the same thing with differences. So if we have the limit as x approaches c of f of x divided by g of x, this is the exact same thing as the limit of f of x as x approaches c divided by the limit of g of x as x approaches c, which is going to be equal to L over M. And finally, this is sometimes called the quotient property. Finally, we'll look at the exponent property. So if I have the limit of f of x to some power, and actually let me even write it as a fractional power, to the r over s power, where both r and s are integers, then the limit of f of x to the r over s power as x approaches c is going to be the exact same thing as the limit of f of x as x approaches c raised to the r over s power. Once again, when r and s are both integers, and s is not equal to 0, otherwise this exponent would not make much sense. And this is the same thing as L to the r over s power. This is equal to L to the r over s power."}, {"video_title": "Limit properties Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So if I have the limit of f of x to some power, and actually let me even write it as a fractional power, to the r over s power, where both r and s are integers, then the limit of f of x to the r over s power as x approaches c is going to be the exact same thing as the limit of f of x as x approaches c raised to the r over s power. Once again, when r and s are both integers, and s is not equal to 0, otherwise this exponent would not make much sense. And this is the same thing as L to the r over s power. This is equal to L to the r over s power. So using these, we can actually find the limit of many, many, many things. And what's neat about it is the property of limits are the things that you would naturally want to do. And if you graph some of these functions, you're about to be quite intuitive."}, {"video_title": "Power series of ln(1+x_) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So I'm assuming you've given a go at it, so let's try to work through this together. So the first thing I wanna do is let me just factor out a common factor. This might simplify it in terms of trying to express it. So let's see, if I factor out, it looks like all of these are divisible by three x squared. So I can rewrite this as three x squared times one minus x to the third power plus x to the sixth power minus x to the ninth power, and a pattern is starting to emerge. And let me actually close the brackets with the same color, with that pink right over there. And let's see, this looks like we are taking powers of x to the third."}, {"video_title": "Power series of ln(1+x_) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's see, if I factor out, it looks like all of these are divisible by three x squared. So I can rewrite this as three x squared times one minus x to the third power plus x to the sixth power minus x to the ninth power, and a pattern is starting to emerge. And let me actually close the brackets with the same color, with that pink right over there. And let's see, this looks like we are taking powers of x to the third. So let me write it that way. This is the same thing as three x squared times, we could write this first term, or I guess I could say the zeroth term. This is x to the third to the zeroth power, then minus, this is x to the third to the first power, and then plus, this is x to the third to the second power."}, {"video_title": "Power series of ln(1+x_) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And let's see, this looks like we are taking powers of x to the third. So let me write it that way. This is the same thing as three x squared times, we could write this first term, or I guess I could say the zeroth term. This is x to the third to the zeroth power, then minus, this is x to the third to the first power, and then plus, this is x to the third to the second power. And then I think you see what's going on. This is x to the third to the third power, and of course we can keep going. But now we have to worry about this, the switching of signs that we keep having."}, {"video_title": "Power series of ln(1+x_) Series AP Calculus BC Khan Academy.mp3", "Sentence": "This is x to the third to the zeroth power, then minus, this is x to the third to the first power, and then plus, this is x to the third to the second power. And then I think you see what's going on. This is x to the third to the third power, and of course we can keep going. But now we have to worry about this, the switching of signs that we keep having. So this would be negative one, this is positive, which is the same thing as negative one to the zero power. This is negative, which is negative one to the first power. So let's actually write it this way."}, {"video_title": "Power series of ln(1+x_) Series AP Calculus BC Khan Academy.mp3", "Sentence": "But now we have to worry about this, the switching of signs that we keep having. So this would be negative one, this is positive, which is the same thing as negative one to the zero power. This is negative, which is negative one to the first power. So let's actually write it this way. We can write is as three x squared times this first term. We could write it as negative one, or we could just write it as negative x to the third to the zeroth power. And then you're gonna say plus, plus we could say negative x to the third to the first power Negative one to the first power is negative one."}, {"video_title": "Power series of ln(1+x_) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's actually write it this way. We can write is as three x squared times this first term. We could write it as negative one, or we could just write it as negative x to the third to the zeroth power. And then you're gonna say plus, plus we could say negative x to the third to the first power Negative one to the first power is negative one. X to the third to the first power is x to the third. Plus negative x to the third to the second power. plus negative x to the third to the third power."}, {"video_title": "Power series of ln(1+x_) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And then you're gonna say plus, plus we could say negative x to the third to the first power Negative one to the first power is negative one. X to the third to the first power is x to the third. Plus negative x to the third to the second power. plus negative x to the third to the third power. That's this term right over here. Negative one to the third power is negative one, and of course, x to the third to the third is x to the ninth, and it keeps going. And so this makes it a lot clearer what our common ratio is."}, {"video_title": "Power series of ln(1+x_) Series AP Calculus BC Khan Academy.mp3", "Sentence": "plus negative x to the third to the third power. That's this term right over here. Negative one to the third power is negative one, and of course, x to the third to the third is x to the ninth, and it keeps going. And so this makes it a lot clearer what our common ratio is. Our common ratio here is negative x to the third. And over what interval would this converge? Well, it's going to converge if the absolute value of our common ratio is less than one."}, {"video_title": "Power series of ln(1+x_) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so this makes it a lot clearer what our common ratio is. Our common ratio here is negative x to the third. And over what interval would this converge? Well, it's going to converge if the absolute value of our common ratio is less than one. So we're going to converge, converge, if the absolute value of our common ratio, the absolute value of our common ratio, which is negative x to the third, is less than one. Or another way of saying this is the same thing, the absolute value of a negative is going to be the same thing as the absolute value of a positive. So that's the same way of saying the absolute value of x to the third is less than one, or saying that x to the third is less than one and is greater than negative one."}, {"video_title": "Power series of ln(1+x_) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, it's going to converge if the absolute value of our common ratio is less than one. So we're going to converge, converge, if the absolute value of our common ratio, the absolute value of our common ratio, which is negative x to the third, is less than one. Or another way of saying this is the same thing, the absolute value of a negative is going to be the same thing as the absolute value of a positive. So that's the same way of saying the absolute value of x to the third is less than one, or saying that x to the third is less than one and is greater than negative one. And the way that's going to happen, if you take the cube roots of both sides of this, or all the sides of this inequality, you're going to get that x is going to be between negative one and one. This right over here is our interval of convergence. Interval of convergence."}, {"video_title": "Power series of ln(1+x_) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So that's the same way of saying the absolute value of x to the third is less than one, or saying that x to the third is less than one and is greater than negative one. And the way that's going to happen, if you take the cube roots of both sides of this, or all the sides of this inequality, you're going to get that x is going to be between negative one and one. This right over here is our interval of convergence. Interval of convergence. And if we restrict our x's to that, what is this going to sum to? Well, this infinite geometric series, our common ratio, its absolute value is less than one. And so this is going to sum to, this is going to be equal to our first term, I guess we could say, or the thing that's multiplying by this whole thing, but if you multiply it out, this would be our first term, is going to be three x squared, all of that over one minus our common ratio."}, {"video_title": "Power series of ln(1+x_) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Interval of convergence. And if we restrict our x's to that, what is this going to sum to? Well, this infinite geometric series, our common ratio, its absolute value is less than one. And so this is going to sum to, this is going to be equal to our first term, I guess we could say, or the thing that's multiplying by this whole thing, but if you multiply it out, this would be our first term, is going to be three x squared, all of that over one minus our common ratio. So one minus negative x to the third, well that's just going to be one plus x to the third. So everything we've done so far is we've shown that this, let me actually write it this way, that this thing is equal to this thing over the interval of convergence. So let me write the copy and paste it, so write like that, over the interval of convergence with x's between negative one and one, these two things are the same."}, {"video_title": "Power series of ln(1+x_) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so this is going to sum to, this is going to be equal to our first term, I guess we could say, or the thing that's multiplying by this whole thing, but if you multiply it out, this would be our first term, is going to be three x squared, all of that over one minus our common ratio. So one minus negative x to the third, well that's just going to be one plus x to the third. So everything we've done so far is we've shown that this, let me actually write it this way, that this thing is equal to this thing over the interval of convergence. So let me write the copy and paste it, so write like that, over the interval of convergence with x's between negative one and one, these two things are the same. Now, we can start to put our calculus hat on here because this looks interesting. This, you might remember, this looks like the derivative of something that's familiar, one plus x to the third, what's the derivative of that? Well, that's three x squared."}, {"video_title": "Power series of ln(1+x_) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let me write the copy and paste it, so write like that, over the interval of convergence with x's between negative one and one, these two things are the same. Now, we can start to put our calculus hat on here because this looks interesting. This, you might remember, this looks like the derivative of something that's familiar, one plus x to the third, what's the derivative of that? Well, that's three x squared. So it looks like this right over here is the derivative of the natural log of one plus x to the third or the absolute value of one plus x to the third. And if you don't believe me, let's take the antiderivative of this thing right over here. In fact, for fun, let's take the antiderivative of both sides of this, and if we do that, then we will have shown, essentially, a geometric series representation of whatever the antiderivative of this thing is."}, {"video_title": "Power series of ln(1+x_) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, that's three x squared. So it looks like this right over here is the derivative of the natural log of one plus x to the third or the absolute value of one plus x to the third. And if you don't believe me, let's take the antiderivative of this thing right over here. In fact, for fun, let's take the antiderivative of both sides of this, and if we do that, then we will have shown, essentially, a geometric series representation of whatever the antiderivative of this thing is. So I encourage you to pause the video again and try to take the antiderivative of both sides of this equation. So let's take, so we're gonna take the antiderivative of the left-hand side, and we're gonna take the antiderivative of the right-hand side. Now, on the left-hand side, I mentioned that, hey, it looks like we have an expression and its derivative that just calls out for u substitution."}, {"video_title": "Power series of ln(1+x_) Series AP Calculus BC Khan Academy.mp3", "Sentence": "In fact, for fun, let's take the antiderivative of both sides of this, and if we do that, then we will have shown, essentially, a geometric series representation of whatever the antiderivative of this thing is. So I encourage you to pause the video again and try to take the antiderivative of both sides of this equation. So let's take, so we're gonna take the antiderivative of the left-hand side, and we're gonna take the antiderivative of the right-hand side. Now, on the left-hand side, I mentioned that, hey, it looks like we have an expression and its derivative that just calls out for u substitution. So if we say that u is equal to one plus x to the third, let me write this down, so u is equal to one plus x to the third, then what's du going to be? So then du is going to be equal to three x squared dx. So notice, we have u, and then du, du is this right over here."}, {"video_title": "Power series of ln(1+x_) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Now, on the left-hand side, I mentioned that, hey, it looks like we have an expression and its derivative that just calls out for u substitution. So if we say that u is equal to one plus x to the third, let me write this down, so u is equal to one plus x to the third, then what's du going to be? So then du is going to be equal to three x squared dx. So notice, we have u, and then du, du is this right over here. So this expression right over here could be rewritten as, so let me go over here, this could be rewritten as the integral, the integral of du, du over u, over u, or I could say, actually, let me write it as one over u du, one over u du, which of course is equal to, which is going to be equal to the natural log of the absolute value of u, the natural log of the absolute value of u plus some constant, plus some constant, and we of course know that u is one plus x to the third, so this is going to be equal to the natural log of the absolute value of one plus x to the third, one plus x to the third, plus c, plus c. Now, we're restricting our domain for x being between negative one and one, so for that domain, this thing is always going to be, this thing is always going to be, this whole thing actually is always going to be positive, so let's, so what we can do, we can, we don't have to write the absolute value sign, so this is going to be equal to the natural log, let me write it, the natural log of one plus x to the third, one plus x to the third, plus c, plus c. So that's this left-hand side, and the right-hand side's actually a lot more straightforward. This is just a straightforward polynomial. Now, as you can imagine, we're going to get a plus some type of constant there, so let me differentiate them a little bit, let me call this one c one, and then on the right-hand side, what do we get?"}, {"video_title": "Power series of ln(1+x_) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So notice, we have u, and then du, du is this right over here. So this expression right over here could be rewritten as, so let me go over here, this could be rewritten as the integral, the integral of du, du over u, over u, or I could say, actually, let me write it as one over u du, one over u du, which of course is equal to, which is going to be equal to the natural log of the absolute value of u, the natural log of the absolute value of u plus some constant, plus some constant, and we of course know that u is one plus x to the third, so this is going to be equal to the natural log of the absolute value of one plus x to the third, one plus x to the third, plus c, plus c. Now, we're restricting our domain for x being between negative one and one, so for that domain, this thing is always going to be, this thing is always going to be, this whole thing actually is always going to be positive, so let's, so what we can do, we can, we don't have to write the absolute value sign, so this is going to be equal to the natural log, let me write it, the natural log of one plus x to the third, one plus x to the third, plus c, plus c. So that's this left-hand side, and the right-hand side's actually a lot more straightforward. This is just a straightforward polynomial. Now, as you can imagine, we're going to get a plus some type of constant there, so let me differentiate them a little bit, let me call this one c one, and then on the right-hand side, what do we get? The antiderivative of this is going to be, let's see, the antiderivative of x squared is x to the third divided by three, so this first term, the antiderivative, is just going to be x to the third power. The derivative of x to the third is three x squared. Now, this term right over here, negative three x to the fifth the antiderivative of x to the fifth is x to the sixth over six, x to the sixth over six, but then we have that three over here, three over six is two, so it's negative x to the sixth over two, actually let me do that in a different color just so we can keep track of it."}, {"video_title": "Power series of ln(1+x_) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Now, as you can imagine, we're going to get a plus some type of constant there, so let me differentiate them a little bit, let me call this one c one, and then on the right-hand side, what do we get? The antiderivative of this is going to be, let's see, the antiderivative of x squared is x to the third divided by three, so this first term, the antiderivative, is just going to be x to the third power. The derivative of x to the third is three x squared. Now, this term right over here, negative three x to the fifth the antiderivative of x to the fifth is x to the sixth over six, x to the sixth over six, but then we have that three over here, three over six is two, so it's negative x to the sixth over two, actually let me do that in a different color just so we can keep track of it. So this one right over here is negative, the antiderivative is negative x to the sixth over two, and then, let's see, I'm running out of colors. The antiderivative of x to the eighth is x to the ninth over nine, so it's going to be plus x to the ninth, and then we have this three, three over nine is three, and I think you see a pattern happening, and then let's just do one more of these for fun, x to the twelfth over 12, but we have this three, so negative x to the twelfth over four, and then we keep going, and then we're of course going to have, we're going to have some constant, and actually let me put the constant in the front. So let me copy and paste that, or cut and paste that, so I have some space, so let me, let me right over here, I'll put some other constant, c two, it doesn't have to be the same one, plus all of this."}, {"video_title": "Power series of ln(1+x_) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Now, this term right over here, negative three x to the fifth the antiderivative of x to the fifth is x to the sixth over six, x to the sixth over six, but then we have that three over here, three over six is two, so it's negative x to the sixth over two, actually let me do that in a different color just so we can keep track of it. So this one right over here is negative, the antiderivative is negative x to the sixth over two, and then, let's see, I'm running out of colors. The antiderivative of x to the eighth is x to the ninth over nine, so it's going to be plus x to the ninth, and then we have this three, three over nine is three, and I think you see a pattern happening, and then let's just do one more of these for fun, x to the twelfth over 12, but we have this three, so negative x to the twelfth over four, and then we keep going, and then we're of course going to have, we're going to have some constant, and actually let me put the constant in the front. So let me copy and paste that, or cut and paste that, so I have some space, so let me, let me right over here, I'll put some other constant, c two, it doesn't have to be the same one, plus all of this. Now to simplify this, I could subtract c one from both sides, or essentially from c two, and then I'm going to have, I'm going to have the natural log of one plus x to the third power, one plus x to the third power, and this is kind of neat, what we've just done with a little bit of integration, is equal to c two minus c one, well this is some constant minus some other constant, so that's just going to be some arbitrary constant, some arbitrary constant, plus all of this business, plus all of this business, and we can even figure out what the constant is going to be, by trying out some values of x, that's in our, that's in our restricted domain. Well x equals zero is between negative one and one, so let's see what happens, what x is equal to zero to solve for c. If x is equal to zero, if x is equal to zero, we get natural log, natural log of one, is equal to c, c plus, well all of these terms are going to be zero, zero to the third power minus zero to the sixth, on and on and on, plus zero, plus zero, or another way, and natural log of one, of course, e to the what power is one, well that's zero, so c must be zero, c is equal to zero, so this thing right over here is equal to zero. So what we've just done, using a little bit of integration, is starting with a, let's just appreciate what went on, starting with an arbitrary infinite series, we showed it could be represented as a geometric series, we defined an interval of convergence over which this would converge, over which the common ratio's absolute value is less than one, and then using that, we expressed its sum, and then we took the anti-derivative of both sides to figure out, to figure out an expansion for the natural log of one plus, one plus x to the third power, which, so at least in my mind, that was pretty neat."}, {"video_title": "Power series of ln(1+x_) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let me copy and paste that, or cut and paste that, so I have some space, so let me, let me right over here, I'll put some other constant, c two, it doesn't have to be the same one, plus all of this. Now to simplify this, I could subtract c one from both sides, or essentially from c two, and then I'm going to have, I'm going to have the natural log of one plus x to the third power, one plus x to the third power, and this is kind of neat, what we've just done with a little bit of integration, is equal to c two minus c one, well this is some constant minus some other constant, so that's just going to be some arbitrary constant, some arbitrary constant, plus all of this business, plus all of this business, and we can even figure out what the constant is going to be, by trying out some values of x, that's in our, that's in our restricted domain. Well x equals zero is between negative one and one, so let's see what happens, what x is equal to zero to solve for c. If x is equal to zero, if x is equal to zero, we get natural log, natural log of one, is equal to c, c plus, well all of these terms are going to be zero, zero to the third power minus zero to the sixth, on and on and on, plus zero, plus zero, or another way, and natural log of one, of course, e to the what power is one, well that's zero, so c must be zero, c is equal to zero, so this thing right over here is equal to zero. So what we've just done, using a little bit of integration, is starting with a, let's just appreciate what went on, starting with an arbitrary infinite series, we showed it could be represented as a geometric series, we defined an interval of convergence over which this would converge, over which the common ratio's absolute value is less than one, and then using that, we expressed its sum, and then we took the anti-derivative of both sides to figure out, to figure out an expansion for the natural log of one plus, one plus x to the third power, which, so at least in my mind, that was pretty neat. This is what we, natural log of one plus x to the third power is x to the third power minus x to the sixth over two, plus x to the ninth over three, so on and on and on, and actually, let's just, just to kind of give ourselves some closure here, let's write it in sigma notation. So we could write the natural log of one plus x to the third, one plus x to the third, over our restricted domain where the absolute value of x is less than one, is equal to, is equal to the sum, the sum from, let's say, let's say n equals one to infinity of x to the third, x to the third to the nth power, so to the first power, the second power, third power, over n, this is x to the third over one, x to the third squared over two, oh, and I, of course, have to throw in, let's see, this first one is, we're gonna have to care about the sign, so let me throw in a negative one, let's see, negative one to the first power should be negative, but here it's positive, so I'll say negative one to the n plus one, negative one to the n plus one power, does that work? I think it does, when n is equal to one, this thing just becomes one, this is x to the third over one when n is equal to two, this becomes negative, which it needs to be, and then this becomes x to the sixth and we're over two, and so there you go, we are done."}, {"video_title": "Power series of arctan(2x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And let's just say we want the first four non-zero terms of the power series approximation of arc tangent of 2x centered at zero. So essentially the Maclaurin series of arc tangent of 2x, the first four non-zero terms of it. And if you feel good about it, I encourage you to pause the video and try to work through it yourself. So you might have tried to work through it and you probably took the first derivative of it. You probably saw that, hey, you know, the derivative with respect to x of arc tangent of 2x is equal to, and this is a refresher if you didn't realize it the first time, it's going to be the derivative of arc tangent of x is one over one plus x squared. So this is going to be the derivative of this, which is two over one plus this whole thing squared. So one plus, I would say, one plus four x squared."}, {"video_title": "Power series of arctan(2x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So you might have tried to work through it and you probably took the first derivative of it. You probably saw that, hey, you know, the derivative with respect to x of arc tangent of 2x is equal to, and this is a refresher if you didn't realize it the first time, it's going to be the derivative of arc tangent of x is one over one plus x squared. So this is going to be the derivative of this, which is two over one plus this whole thing squared. So one plus, I would say, one plus four x squared. And then as you try to find more of the terms of the Maclaurin series, you would have taken more derivatives of this and it would have gotten very hairy very fast, especially if you're looking for the first four non-zero terms. So you probably realize, hey, there must be some type of an insight that I hadn't fully appreciated yet when I just tried to just power through, or no pun intended, power through finding the power series, the first four non-zero terms of the power series centered at zero of arc tangent of 2x. And you are right, there is a key insight here."}, {"video_title": "Power series of arctan(2x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So one plus, I would say, one plus four x squared. And then as you try to find more of the terms of the Maclaurin series, you would have taken more derivatives of this and it would have gotten very hairy very fast, especially if you're looking for the first four non-zero terms. So you probably realize, hey, there must be some type of an insight that I hadn't fully appreciated yet when I just tried to just power through, or no pun intended, power through finding the power series, the first four non-zero terms of the power series centered at zero of arc tangent of 2x. And you are right, there is a key insight here. The key insight here is, well, instead of doing it directly, let's see if we could find the power series representation, the first four terms of this thing right over here, and then we can take the antiderivative of that to find the power series of arc tangent of 2x, making sure that we get the constant right, that it satisfies the fact that we are being centered at zero. So, but I know what you're thinking now. Well, that seems like it's getting us to the exact same issue."}, {"video_title": "Power series of arctan(2x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And you are right, there is a key insight here. The key insight here is, well, instead of doing it directly, let's see if we could find the power series representation, the first four terms of this thing right over here, and then we can take the antiderivative of that to find the power series of arc tangent of 2x, making sure that we get the constant right, that it satisfies the fact that we are being centered at zero. So, but I know what you're thinking now. Well, that seems like it's getting us to the exact same issue. If I want to find the power series representation of this, the first four terms of it, I'm still having to take the derivative of this multiple times, which seems just as hard. But this is the key insight, I guess you could say. The key insight is, let's just say f of x, which of course is the derivative of arc tangent of 2x, is 2 over 1 plus 4x squared."}, {"video_title": "Power series of arctan(2x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, that seems like it's getting us to the exact same issue. If I want to find the power series representation of this, the first four terms of it, I'm still having to take the derivative of this multiple times, which seems just as hard. But this is the key insight, I guess you could say. The key insight is, let's just say f of x, which of course is the derivative of arc tangent of 2x, is 2 over 1 plus 4x squared. Now, if we had another function, if we had another function that kind of cleans it up a little bit, so that we don't get all this hairiness when we take the derivatives. So let's say we had another function, g of x, let me just set a color that I have not used. So let's say that I have g of x is equal to 1 over 1 plus x."}, {"video_title": "Power series of arctan(2x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "The key insight is, let's just say f of x, which of course is the derivative of arc tangent of 2x, is 2 over 1 plus 4x squared. Now, if we had another function, if we had another function that kind of cleans it up a little bit, so that we don't get all this hairiness when we take the derivatives. So let's say we had another function, g of x, let me just set a color that I have not used. So let's say that I have g of x is equal to 1 over 1 plus x. So this is an interesting thing, because it's really easy, and this is the same thing as 1 plus x, 1 plus x to the negative 1 power. g of x is interesting because it's really easy to take its derivatives. For example, g prime of x is going to be equal to, chain rule, derivative of 1 plus x is just 1, so it's going to be equal to negative 1 plus x to the negative 2 power."}, {"video_title": "Power series of arctan(2x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's say that I have g of x is equal to 1 over 1 plus x. So this is an interesting thing, because it's really easy, and this is the same thing as 1 plus x, 1 plus x to the negative 1 power. g of x is interesting because it's really easy to take its derivatives. For example, g prime of x is going to be equal to, chain rule, derivative of 1 plus x is just 1, so it's going to be equal to negative 1 plus x to the negative 2 power. If I want to take the second derivative of that, g prime prime of x, that's going to be negative 2 times negative 1, it's 2 times 1 plus x to the negative 3 power. If I want to take the third derivative of that, if I want to take the third derivative of that, that's going to be, let's see, negative 3 times 2, it's negative 6 times 1 plus x to the negative 4 power. I know you're saying, hey, well, Sal, aren't we worried about this?"}, {"video_title": "Power series of arctan(2x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "For example, g prime of x is going to be equal to, chain rule, derivative of 1 plus x is just 1, so it's going to be equal to negative 1 plus x to the negative 2 power. If I want to take the second derivative of that, g prime prime of x, that's going to be negative 2 times negative 1, it's 2 times 1 plus x to the negative 3 power. If I want to take the third derivative of that, if I want to take the third derivative of that, that's going to be, let's see, negative 3 times 2, it's negative 6 times 1 plus x to the negative 4 power. I know you're saying, hey, well, Sal, aren't we worried about this? Why are you doing this? Well, just bear with me for a second. Just that quickly, I was able to find the first three derivatives of g of x, and it's very easy then to find the first four terms of its power series representation, especially its Maclaurin series, if the power series centered at 0."}, {"video_title": "Power series of arctan(2x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "I know you're saying, hey, well, Sal, aren't we worried about this? Why are you doing this? Well, just bear with me for a second. Just that quickly, I was able to find the first three derivatives of g of x, and it's very easy then to find the first four terms of its power series representation, especially its Maclaurin series, if the power series centered at 0. We just have to evaluate each of these at 0. g of 0 is equal to 1. g prime of 0 is equal to negative 1. g prime prime of 0, so 1 plus 0 to the negative 3, that's just 1, times 2 is equal to 2. Then the third derivative, evaluated at 0, is equal to negative 6. I could write that g of x is approximately equal to, I'm just going to do the first four terms here, it's going to be g of 0, which is 1, minus g prime of 0 times x, so that's minus 1 times x, so that's negative x, plus g prime prime of 0, 2, over 2 factorial times x squared."}, {"video_title": "Power series of arctan(2x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Just that quickly, I was able to find the first three derivatives of g of x, and it's very easy then to find the first four terms of its power series representation, especially its Maclaurin series, if the power series centered at 0. We just have to evaluate each of these at 0. g of 0 is equal to 1. g prime of 0 is equal to negative 1. g prime prime of 0, so 1 plus 0 to the negative 3, that's just 1, times 2 is equal to 2. Then the third derivative, evaluated at 0, is equal to negative 6. I could write that g of x is approximately equal to, I'm just going to do the first four terms here, it's going to be g of 0, which is 1, minus g prime of 0 times x, so that's minus 1 times x, so that's negative x, plus g prime prime of 0, 2, over 2 factorial times x squared. Well, this is just 1 times x squared, so let me just write that. That's just going to be plus x squared. Then we have plus g prime prime prime of 0, which is negative 6 over 3 factorial times x to the third."}, {"video_title": "Power series of arctan(2x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "I could write that g of x is approximately equal to, I'm just going to do the first four terms here, it's going to be g of 0, which is 1, minus g prime of 0 times x, so that's minus 1 times x, so that's negative x, plus g prime prime of 0, 2, over 2 factorial times x squared. Well, this is just 1 times x squared, so let me just write that. That's just going to be plus x squared. Then we have plus g prime prime prime of 0, which is negative 6 over 3 factorial times x to the third. Well, 3 factorial is just 6, so negative 6 divided by 6 is just negative 1, so that's going to be negative x to the third. I know what you're thinking, all right, Sal, you just started with a hard problem and you gave yourself a much easier problem to find the power series representation. How is this useful?"}, {"video_title": "Power series of arctan(2x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Then we have plus g prime prime prime of 0, which is negative 6 over 3 factorial times x to the third. Well, 3 factorial is just 6, so negative 6 divided by 6 is just negative 1, so that's going to be negative x to the third. I know what you're thinking, all right, Sal, you just started with a hard problem and you gave yourself a much easier problem to find the power series representation. How is this useful? Well, this is the key insight that I've been promising throughout this video so far. The key insight, the long promised key insight, is that, and I'm finding a suitable color for a key insight, is that we can write f of x, notice f of x is just 2 times g of 4x squared. Notice, you replace your x's with a 4x squared, you're going to have 1 over 1 plus 4x squared, and then you multiply that whole thing times 2, you get this thing right over here."}, {"video_title": "Power series of arctan(2x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "How is this useful? Well, this is the key insight that I've been promising throughout this video so far. The key insight, the long promised key insight, is that, and I'm finding a suitable color for a key insight, is that we can write f of x, notice f of x is just 2 times g of 4x squared. Notice, you replace your x's with a 4x squared, you're going to have 1 over 1 plus 4x squared, and then you multiply that whole thing times 2, you get this thing right over here. If f of x is equal to that, then f of x's power series representation is just going to be taking this power series, or at least the first four terms of it, and replacing the x's with 4x squared, and then multiplying the whole thing times 2. Let's do that. We can write that f of x is going to be approximately equal to 2 times this thing, evaluated when x is equal to 4x squared."}, {"video_title": "Power series of arctan(2x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Notice, you replace your x's with a 4x squared, you're going to have 1 over 1 plus 4x squared, and then you multiply that whole thing times 2, you get this thing right over here. If f of x is equal to that, then f of x's power series representation is just going to be taking this power series, or at least the first four terms of it, and replacing the x's with 4x squared, and then multiplying the whole thing times 2. Let's do that. We can write that f of x is going to be approximately equal to 2 times this thing, evaluated when x is equal to 4x squared. It is 1 minus, instead of an x, I'm going to write a 4x squared, plus x squared. But instead of an x, I have a 4x squared squared. This is plus 4x squared squared."}, {"video_title": "Power series of arctan(2x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "We can write that f of x is going to be approximately equal to 2 times this thing, evaluated when x is equal to 4x squared. It is 1 minus, instead of an x, I'm going to write a 4x squared, plus x squared. But instead of an x, I have a 4x squared squared. This is plus 4x squared squared. That's going to be 16x to the fourth. Let me just write that. That's going to be plus 16x to the fourth."}, {"video_title": "Power series of arctan(2x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "This is plus 4x squared squared. That's going to be 16x to the fourth. Let me just write that. That's going to be plus 16x to the fourth. Finally, minus x to the third, but now an x is 4x squared, so it's minus 4x squared to the third power. That's going to be 64x to the sixth. Let me write that."}, {"video_title": "Power series of arctan(2x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "That's going to be plus 16x to the fourth. Finally, minus x to the third, but now an x is 4x squared, so it's minus 4x squared to the third power. That's going to be 64x to the sixth. Let me write that. Minus 64x to the sixth power. Then we could say that this is going to be, so f of x is approximately equal to, distribute the 2, 2 minus 8x squared, plus 32x to the fourth, minus 128x to the sixth. Just like that, with a little bit of a substitution, I was able to reasonably, in a non-hairy fashion, find the first four non-zero terms of the power series of 2 over 1 plus 4x squared, which is going to be the derivative of the power series of arc tangent of 2x."}, {"video_title": "Power series of arctan(2x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let me write that. Minus 64x to the sixth power. Then we could say that this is going to be, so f of x is approximately equal to, distribute the 2, 2 minus 8x squared, plus 32x to the fourth, minus 128x to the sixth. Just like that, with a little bit of a substitution, I was able to reasonably, in a non-hairy fashion, find the first four non-zero terms of the power series of 2 over 1 plus 4x squared, which is going to be the derivative of the power series of arc tangent of 2x. Let's write this down. I'm going to write down, arc tangent of 2x, which is equal to an antiderivative of f of x, dx, which is going to be equal to an antiderivative of this whole, all of this business. Equal to the antiderivative of 2 minus 8x squared, plus 32x to the fourth, minus 128x to the sixth."}, {"video_title": "Power series of arctan(2x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Just like that, with a little bit of a substitution, I was able to reasonably, in a non-hairy fashion, find the first four non-zero terms of the power series of 2 over 1 plus 4x squared, which is going to be the derivative of the power series of arc tangent of 2x. Let's write this down. I'm going to write down, arc tangent of 2x, which is equal to an antiderivative of f of x, dx, which is going to be equal to an antiderivative of this whole, all of this business. Equal to the antiderivative of 2 minus 8x squared, plus 32x to the fourth, minus 128x to the sixth. Actually, let me make this approximate, because now, of course, we're doing an approximation with the power series, dx. What is this going to be equal to? We get approximately equal to, well, I'm going to have a constant there."}, {"video_title": "Power series of arctan(2x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Equal to the antiderivative of 2 minus 8x squared, plus 32x to the fourth, minus 128x to the sixth. Actually, let me make this approximate, because now, of course, we're doing an approximation with the power series, dx. What is this going to be equal to? We get approximately equal to, well, I'm going to have a constant there. Let me write the constant first, because when we write our power series, or our Maclaurin series, the first term is the constant term. It's our function evaluated at zero. We're going to have a constant, and if I take the antiderivative of 2, that's going to be plus 2x."}, {"video_title": "Power series of arctan(2x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "We get approximately equal to, well, I'm going to have a constant there. Let me write the constant first, because when we write our power series, or our Maclaurin series, the first term is the constant term. It's our function evaluated at zero. We're going to have a constant, and if I take the antiderivative of 2, that's going to be plus 2x. The antiderivative of this, it's the x to the third, divide 8 by 3, so it's negative 8 thirds, x to the third power. Then plus 32x to the fifth over 5, minus 128x to the seventh over 7. We're in the home stretch."}, {"video_title": "Power series of arctan(2x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "We're going to have a constant, and if I take the antiderivative of 2, that's going to be plus 2x. The antiderivative of this, it's the x to the third, divide 8 by 3, so it's negative 8 thirds, x to the third power. Then plus 32x to the fifth over 5, minus 128x to the seventh over 7. We're in the home stretch. We've got at least four non-zero terms. If this is non-zero, this is going to be five non-zero terms. Now, let's just make sure that our constant is appropriate for arctangent of 2x."}, {"video_title": "Power series of arctan(2x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "We're in the home stretch. We've got at least four non-zero terms. If this is non-zero, this is going to be five non-zero terms. Now, let's just make sure that our constant is appropriate for arctangent of 2x. This should essentially evaluate to what arctangent of, when this function, when x is equal to zero. What is arctangent of zero? Remember, this is centered at zero, so we better get it right there."}, {"video_title": "Power series of arctan(2x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Now, let's just make sure that our constant is appropriate for arctangent of 2x. This should essentially evaluate to what arctangent of, when this function, when x is equal to zero. What is arctangent of zero? Remember, this is centered at zero, so we better get it right there. That's just the most basic thing if we're doing the Maclaurin series representation. We're centering at zero, so our approximation evaluated at zero better be the same thing as the function evaluated at zero. Well, arctangent of 2 times zero is just going to be zero."}, {"video_title": "Power series of arctan(2x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Remember, this is centered at zero, so we better get it right there. That's just the most basic thing if we're doing the Maclaurin series representation. We're centering at zero, so our approximation evaluated at zero better be the same thing as the function evaluated at zero. Well, arctangent of 2 times zero is just going to be zero. This thing, when you evaluate it at zero, this gets to c, so this gets to c, must be equal to zero. If we want this thing to be zero when x is equal to zero. Just like that, we're done."}, {"video_title": "Power series of arctan(2x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, arctangent of 2 times zero is just going to be zero. This thing, when you evaluate it at zero, this gets to c, so this gets to c, must be equal to zero. If we want this thing to be zero when x is equal to zero. Just like that, we're done. We've been able to figure out that arctangent of 2x is approximately equal to 2x minus 8 thirds x to the third power plus 32 over 5 x to the fifth minus 128 over 7 x to the seventh. If we wanted more terms, we could have gotten more terms by just doing what we just did, but doing it in four more terms. Hopefully, you enjoyed that fairly hairy problem, but as you saw, it's not as hairy as we thought it was going to be."}, {"video_title": "Limits at infinity of quotients with square roots (odd power) AP Calculus AB Khan Academy.mp3", "Sentence": "So let's think about what these are going to be. Well, once again, and I'm not doing this in an ultra rigorous way, but more in an intuitive way, is to think about what this function approximately equals as we get larger and larger and larger x's. This is the case if we're getting very positive x's, very positive infinity direction, or very negative. It's still the absolute value of those x's that are very, very, very large as we approach positive infinity or negative infinity. Well, in the numerator we only have one term. We have this x term. But in the denominator we have two terms under the radical here."}, {"video_title": "Limits at infinity of quotients with square roots (odd power) AP Calculus AB Khan Academy.mp3", "Sentence": "It's still the absolute value of those x's that are very, very, very large as we approach positive infinity or negative infinity. Well, in the numerator we only have one term. We have this x term. But in the denominator we have two terms under the radical here. And as x gets larger and larger and larger, either in the positive or the negative direction, this x squared term is going to really dominate this one. You can imagine when x is a million, you're going to have a million squared plus 1. The value of the denominator is going to be dictated by this x squared term."}, {"video_title": "Limits at infinity of quotients with square roots (odd power) AP Calculus AB Khan Academy.mp3", "Sentence": "But in the denominator we have two terms under the radical here. And as x gets larger and larger and larger, either in the positive or the negative direction, this x squared term is going to really dominate this one. You can imagine when x is a million, you're going to have a million squared plus 1. The value of the denominator is going to be dictated by this x squared term. So this is going to be approximately equal to x over the square root of x squared. This term right over here, the 1, isn't going to matter so much when we get very, very, very large x's. And this right over here, x over the square root of x squared, or x over the principal root of x squared, this is going to be equal to x over."}, {"video_title": "Limits at infinity of quotients with square roots (odd power) AP Calculus AB Khan Academy.mp3", "Sentence": "The value of the denominator is going to be dictated by this x squared term. So this is going to be approximately equal to x over the square root of x squared. This term right over here, the 1, isn't going to matter so much when we get very, very, very large x's. And this right over here, x over the square root of x squared, or x over the principal root of x squared, this is going to be equal to x over. If I square something and then take the principal root, remember the principal root is the positive square root of something, then I'm essentially taking the absolute value of x. It's going to be equal to x over the absolute value of x for x approaches infinity or for x approaches negative infinity. So another way to say this, another way to restate these limits, is as we approach infinity, this limit, we can restate it as the limit, this is going to be equal to the limit as x approaches infinity of x over the absolute value of x."}, {"video_title": "Limits at infinity of quotients with square roots (odd power) AP Calculus AB Khan Academy.mp3", "Sentence": "And this right over here, x over the square root of x squared, or x over the principal root of x squared, this is going to be equal to x over. If I square something and then take the principal root, remember the principal root is the positive square root of something, then I'm essentially taking the absolute value of x. It's going to be equal to x over the absolute value of x for x approaches infinity or for x approaches negative infinity. So another way to say this, another way to restate these limits, is as we approach infinity, this limit, we can restate it as the limit, this is going to be equal to the limit as x approaches infinity of x over the absolute value of x. Now, for positive x's, the absolute value of x is just going to be x. This is going to be x divided by x. So this is just going to be 1."}, {"video_title": "Limits at infinity of quotients with square roots (odd power) AP Calculus AB Khan Academy.mp3", "Sentence": "So another way to say this, another way to restate these limits, is as we approach infinity, this limit, we can restate it as the limit, this is going to be equal to the limit as x approaches infinity of x over the absolute value of x. Now, for positive x's, the absolute value of x is just going to be x. This is going to be x divided by x. So this is just going to be 1. Similarly, right over here, we're taking the limit as we go to negative infinity. This is going to be the limit of x over the absolute value of x as x approaches negative infinity. Remember, the only reason why I was able to make this statement is that f of x and this thing right over here become very, very similar, or you can kind of say converge to each other, as x gets very, very, very large or x gets very, very, very, very negative."}, {"video_title": "Limits at infinity of quotients with square roots (odd power) AP Calculus AB Khan Academy.mp3", "Sentence": "So this is just going to be 1. Similarly, right over here, we're taking the limit as we go to negative infinity. This is going to be the limit of x over the absolute value of x as x approaches negative infinity. Remember, the only reason why I was able to make this statement is that f of x and this thing right over here become very, very similar, or you can kind of say converge to each other, as x gets very, very, very large or x gets very, very, very, very negative. Now, for negative values of x, the absolute value of x is going to be positive. x is obviously going to be negative, and we're just going to get negative 1. So using this, we can actually try to graph our function."}, {"video_title": "Limits at infinity of quotients with square roots (odd power) AP Calculus AB Khan Academy.mp3", "Sentence": "Remember, the only reason why I was able to make this statement is that f of x and this thing right over here become very, very similar, or you can kind of say converge to each other, as x gets very, very, very large or x gets very, very, very, very negative. Now, for negative values of x, the absolute value of x is going to be positive. x is obviously going to be negative, and we're just going to get negative 1. So using this, we can actually try to graph our function. So let's try to do that. So let's say that is my y-axis. This is my x-axis."}, {"video_title": "Limits at infinity of quotients with square roots (odd power) AP Calculus AB Khan Academy.mp3", "Sentence": "So using this, we can actually try to graph our function. So let's try to do that. So let's say that is my y-axis. This is my x-axis. And we see that we have two horizontal asymptotes. We have one horizontal asymptote at y is equal to 1. So let's say this right over here is y is equal to 1."}, {"video_title": "Limits at infinity of quotients with square roots (odd power) AP Calculus AB Khan Academy.mp3", "Sentence": "This is my x-axis. And we see that we have two horizontal asymptotes. We have one horizontal asymptote at y is equal to 1. So let's say this right over here is y is equal to 1. Let me draw that line as a dotted line. We're going to approach this thing. And then we have another horizontal asymptote at y is equal to negative 1."}, {"video_title": "Limits at infinity of quotients with square roots (odd power) AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say this right over here is y is equal to 1. Let me draw that line as a dotted line. We're going to approach this thing. And then we have another horizontal asymptote at y is equal to negative 1. So that might be right over there. y is equal to negative 1. And if we want to plot at least one point, we could think about what does f of 0 equal."}, {"video_title": "Limits at infinity of quotients with square roots (odd power) AP Calculus AB Khan Academy.mp3", "Sentence": "And then we have another horizontal asymptote at y is equal to negative 1. So that might be right over there. y is equal to negative 1. And if we want to plot at least one point, we could think about what does f of 0 equal. So f of 0 is going to be equal to 0 over the square root of 0 plus 1, or 0 squared plus 1. Well, that's all just going to be equal to 0. So we have this point right over here."}, {"video_title": "Limits at infinity of quotients with square roots (odd power) AP Calculus AB Khan Academy.mp3", "Sentence": "And if we want to plot at least one point, we could think about what does f of 0 equal. So f of 0 is going to be equal to 0 over the square root of 0 plus 1, or 0 squared plus 1. Well, that's all just going to be equal to 0. So we have this point right over here. And we know that as x approaches infinity, we're approaching this blue horizontal asymptote. So it might look something like this. Let me do it a little bit differently."}, {"video_title": "Limits at infinity of quotients with square roots (odd power) AP Calculus AB Khan Academy.mp3", "Sentence": "So we have this point right over here. And we know that as x approaches infinity, we're approaching this blue horizontal asymptote. So it might look something like this. Let me do it a little bit differently. Let me do it a little bit. There you go. Clean this up."}, {"video_title": "Limits at infinity of quotients with square roots (odd power) AP Calculus AB Khan Academy.mp3", "Sentence": "Let me do it a little bit differently. Let me do it a little bit. There you go. Clean this up. So it might look something like this. That's not the color I wanted to use. So it might look something like that."}, {"video_title": "Limits at infinity of quotients with square roots (odd power) AP Calculus AB Khan Academy.mp3", "Sentence": "Clean this up. So it might look something like this. That's not the color I wanted to use. So it might look something like that. We get closer and closer to that asymptote as x gets larger and larger. And then like this. We get closer and closer to this asymptote as x approaches negative infinity."}, {"video_title": "Limits at infinity of quotients with square roots (odd power) AP Calculus AB Khan Academy.mp3", "Sentence": "So it might look something like that. We get closer and closer to that asymptote as x gets larger and larger. And then like this. We get closer and closer to this asymptote as x approaches negative infinity. I'm not drawing it so well. So that right over there is y is equal to f of x. And you can verify this by taking a calculator, trying to plot more points, or using some type of graphing calculator or something."}, {"video_title": "Limits at infinity of quotients with square roots (odd power) AP Calculus AB Khan Academy.mp3", "Sentence": "We get closer and closer to this asymptote as x approaches negative infinity. I'm not drawing it so well. So that right over there is y is equal to f of x. And you can verify this by taking a calculator, trying to plot more points, or using some type of graphing calculator or something. But anyway, I just wanted to tackle another situation where we're approaching infinity or negative infinity, and we're trying to determine the horizontal asymptotes. And remember, the key is just to say what terms dominate as x approaches positive infinity or negative infinity to say, well, what is that function going to approach? And it's going to approach this horizontal asymptote in the positive direction and this one in the negative."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "And what I'm gonna do is I'm going to first just apply the chain rule, and then maybe dig into it a little bit to make sure we draw the connection between what we're doing here and then what you might see in maybe some of your calculus textbooks that explain the chain rule. So if we have a function that is defined as essentially a composite function, notice this expression right here, we are taking something to the third power. It isn't just an x that we're taking to the third power, we are taking a cosine of x to the third power. So we're taking a function, you could view it this way, we're taking the function cosine of x, and then we're inputting it into another function that takes it to the third power. So let me put it this way. If you viewed, if you say, look, we could take an x, we put it into one function, that is, that first function is cosine of x, so first we evaluate the cosine, and so that's going to produce cosine of x, cosine of x, and then we're going to input it into a function that just takes things to the third power. So it just takes things to the third power, and so what are you going to end up with?"}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "So we're taking a function, you could view it this way, we're taking the function cosine of x, and then we're inputting it into another function that takes it to the third power. So let me put it this way. If you viewed, if you say, look, we could take an x, we put it into one function, that is, that first function is cosine of x, so first we evaluate the cosine, and so that's going to produce cosine of x, cosine of x, and then we're going to input it into a function that just takes things to the third power. So it just takes things to the third power, and so what are you going to end up with? Well, you're going to end up with, what are you taking to the third power? You're taking cosine of x. Cosine of x to the third power. This is a composite function."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "So it just takes things to the third power, and so what are you going to end up with? Well, you're going to end up with, what are you taking to the third power? You're taking cosine of x. Cosine of x to the third power. This is a composite function. You could view this, you could view this as the function, let's call this blue one the function v, and let's call this the function u, and so if we're taking x and into u, this is u of x, and then if we're taking u of x into the input, or as the input into the function v, then this output right over here, this is going to be v of, well, what was inputted? V of u of x. V of u of x. Or, another way of writing it, I'm going to write it multiple ways, it's the same thing as v of cosine of x. V of cosine of x."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "This is a composite function. You could view this, you could view this as the function, let's call this blue one the function v, and let's call this the function u, and so if we're taking x and into u, this is u of x, and then if we're taking u of x into the input, or as the input into the function v, then this output right over here, this is going to be v of, well, what was inputted? V of u of x. V of u of x. Or, another way of writing it, I'm going to write it multiple ways, it's the same thing as v of cosine of x. V of cosine of x. And so v, whatever you input into it, it just takes it to the third power. If you were to write v of x, it would be x to the third power. So the chain rule tells us, or the chain rule is what our brain should say, hey, it becomes applicable if we're going to take the derivative of a function that can be expressed as a composite function like this."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "Or, another way of writing it, I'm going to write it multiple ways, it's the same thing as v of cosine of x. V of cosine of x. And so v, whatever you input into it, it just takes it to the third power. If you were to write v of x, it would be x to the third power. So the chain rule tells us, or the chain rule is what our brain should say, hey, it becomes applicable if we're going to take the derivative of a function that can be expressed as a composite function like this. So just to be clear, we can write f of x, f of x is equal to v of u of x. I know I'm essentially saying the same thing over and over again, but I'm saying it in slightly different ways because the first time you learn this, it can be a little bit hard to grok, or really deeply understand. So I'm going to try to write it in different ways. And the chain rule tells us that if you have a situation like this, then the derivative, f prime of x, and this is something that you will see in your textbooks, so this is going to be the derivative of this whole thing with respect to u of x, so we could write that as v prime of u of x, v prime of u of x, times the derivative of u with respect to x, times u prime of x."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "So the chain rule tells us, or the chain rule is what our brain should say, hey, it becomes applicable if we're going to take the derivative of a function that can be expressed as a composite function like this. So just to be clear, we can write f of x, f of x is equal to v of u of x. I know I'm essentially saying the same thing over and over again, but I'm saying it in slightly different ways because the first time you learn this, it can be a little bit hard to grok, or really deeply understand. So I'm going to try to write it in different ways. And the chain rule tells us that if you have a situation like this, then the derivative, f prime of x, and this is something that you will see in your textbooks, so this is going to be the derivative of this whole thing with respect to u of x, so we could write that as v prime of u of x, v prime of u of x, times the derivative of u with respect to x, times u prime of x. This right over here, this is one expression of the chain rule. And so how do we evaluate it in this case? Let me color code it in a similar way."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "And the chain rule tells us that if you have a situation like this, then the derivative, f prime of x, and this is something that you will see in your textbooks, so this is going to be the derivative of this whole thing with respect to u of x, so we could write that as v prime of u of x, v prime of u of x, times the derivative of u with respect to x, times u prime of x. This right over here, this is one expression of the chain rule. And so how do we evaluate it in this case? Let me color code it in a similar way. So the v function, this outer thing that just takes things to the third power, I'll put in blue. So f prime of x, another way of expressing it, and I'll use it with more of the differential notation. You could view this as the derivative of, well, I'll write it a couple of different ways."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "Let me color code it in a similar way. So the v function, this outer thing that just takes things to the third power, I'll put in blue. So f prime of x, another way of expressing it, and I'll use it with more of the differential notation. You could view this as the derivative of, well, I'll write it a couple of different ways. You could view it as the derivative of v, the derivative of v with respect to u, I want to get the colors right, the derivative of v with respect to u, that's what this thing is right over here, times the derivative of u with respect to x. So times the derivative of u with respect to x. And just to make clear, so you're familiar with the different notations you'll see in different textbooks, this is this right over here, just using different notations, and this is this right over here."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "You could view this as the derivative of, well, I'll write it a couple of different ways. You could view it as the derivative of v, the derivative of v with respect to u, I want to get the colors right, the derivative of v with respect to u, that's what this thing is right over here, times the derivative of u with respect to x. So times the derivative of u with respect to x. And just to make clear, so you're familiar with the different notations you'll see in different textbooks, this is this right over here, just using different notations, and this is this right over here. So let's actually evaluate these things. You're probably tired of just talking in the abstract. So this is going to be equal to, this is going to be equal to, and I'm gonna write it out again, this is the derivative, instead of just writing v and u, I'm gonna write it, let me write it this way, this is going to be, I keep wanting to be using the wrong colors, this is going to be the derivative of, I'm gonna leave some space, times the derivative of something else with respect to something else."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "And just to make clear, so you're familiar with the different notations you'll see in different textbooks, this is this right over here, just using different notations, and this is this right over here. So let's actually evaluate these things. You're probably tired of just talking in the abstract. So this is going to be equal to, this is going to be equal to, and I'm gonna write it out again, this is the derivative, instead of just writing v and u, I'm gonna write it, let me write it this way, this is going to be, I keep wanting to be using the wrong colors, this is going to be the derivative of, I'm gonna leave some space, times the derivative of something else with respect to something else. So we're gonna first take the derivative of v. Well v is cosine of x to the third power, cosine of x. We're gonna take the derivative of that with respect to u, which is just cosine of x, and we're gonna multiply that times the derivative of u, which is cosine of x, with respect to x, so this one we have good, we've seen this before. We know that the derivative with respect to x of cosine of x, cosine, let me use it in that same color, the derivative of cosine of x, well that's equal to negative sine of x."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be equal to, this is going to be equal to, and I'm gonna write it out again, this is the derivative, instead of just writing v and u, I'm gonna write it, let me write it this way, this is going to be, I keep wanting to be using the wrong colors, this is going to be the derivative of, I'm gonna leave some space, times the derivative of something else with respect to something else. So we're gonna first take the derivative of v. Well v is cosine of x to the third power, cosine of x. We're gonna take the derivative of that with respect to u, which is just cosine of x, and we're gonna multiply that times the derivative of u, which is cosine of x, with respect to x, so this one we have good, we've seen this before. We know that the derivative with respect to x of cosine of x, cosine, let me use it in that same color, the derivative of cosine of x, well that's equal to negative sine of x. So this one right over here, that is negative sine of x. You might be more familiar with seeing the derivative operator this way, but in theory, you won't see this as often, but this helps my brain really grok what we're doing. We're taking the derivative of cosine of x with respect to x, well that's gonna be negative sine of x."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "We know that the derivative with respect to x of cosine of x, cosine, let me use it in that same color, the derivative of cosine of x, well that's equal to negative sine of x. So this one right over here, that is negative sine of x. You might be more familiar with seeing the derivative operator this way, but in theory, you won't see this as often, but this helps my brain really grok what we're doing. We're taking the derivative of cosine of x with respect to x, well that's gonna be negative sine of x. Well what about taking the derivative of cosine of x to the third power with respect to cosine of x? What does this thing over here mean? Well, if I was taking the derivative, if I was taking the derivative of, let me write it this way, if I was taking the derivative of x to the third power, x to the third power with respect to x, if it was like that, well this is just going to be, and let me put some brackets here to make it a little bit clearer."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "We're taking the derivative of cosine of x with respect to x, well that's gonna be negative sine of x. Well what about taking the derivative of cosine of x to the third power with respect to cosine of x? What does this thing over here mean? Well, if I was taking the derivative, if I was taking the derivative of, let me write it this way, if I was taking the derivative of x to the third power, x to the third power with respect to x, if it was like that, well this is just going to be, and let me put some brackets here to make it a little bit clearer. If I'm taking the derivative of that, that is going to be, that is going to be, we bring the exponent out front, it's going to be three, three times x, three times x to the second power. Three times x to the second power. So the general notion here is, if I'm taking the derivative of something, whatever this something happens to be, let me just add a new color."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "Well, if I was taking the derivative, if I was taking the derivative of, let me write it this way, if I was taking the derivative of x to the third power, x to the third power with respect to x, if it was like that, well this is just going to be, and let me put some brackets here to make it a little bit clearer. If I'm taking the derivative of that, that is going to be, that is going to be, we bring the exponent out front, it's going to be three, three times x, three times x to the second power. Three times x to the second power. So the general notion here is, if I'm taking the derivative of something, whatever this something happens to be, let me just add a new color. It could be, I'm doing the derivative of orange circle to the third power with respect to orange circle, well that's just going to be three times orange or yellow circle. Let me make it an actual orange circle. So the derivative of orange circle to the third power with respect to orange circle, that's going to be three times the orange circle squared."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "So the general notion here is, if I'm taking the derivative of something, whatever this something happens to be, let me just add a new color. It could be, I'm doing the derivative of orange circle to the third power with respect to orange circle, well that's just going to be three times orange or yellow circle. Let me make it an actual orange circle. So the derivative of orange circle to the third power with respect to orange circle, that's going to be three times the orange circle squared. So if I'm taking the derivative of cosine of x to the third power with respect to cosine of x, well that's just going to be, this is just going to be three times cosine of x, cosine of x to the second power. To the second power. Notice, we're just, one way to think about it, is taking the derivative of this outside function with respect to the inside."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "So the derivative of orange circle to the third power with respect to orange circle, that's going to be three times the orange circle squared. So if I'm taking the derivative of cosine of x to the third power with respect to cosine of x, well that's just going to be, this is just going to be three times cosine of x, cosine of x to the second power. To the second power. Notice, we're just, one way to think about it, is taking the derivative of this outside function with respect to the inside. So I would do the same thing as taking the derivative of x to the third power, but instead of an x, I have a cosine of x. So instead of it being three x squared, it is three cosine of x squared. And then the chain rule says, if we want to finally get the derivative with respect to x, we then take the derivative of cosine of x with respect to x."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "Notice, we're just, one way to think about it, is taking the derivative of this outside function with respect to the inside. So I would do the same thing as taking the derivative of x to the third power, but instead of an x, I have a cosine of x. So instead of it being three x squared, it is three cosine of x squared. And then the chain rule says, if we want to finally get the derivative with respect to x, we then take the derivative of cosine of x with respect to x. Now that's a big mouthful, but we are at the home stretch. We've now figured out the derivative. It's going to be this times this."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "And then the chain rule says, if we want to finally get the derivative with respect to x, we then take the derivative of cosine of x with respect to x. Now that's a big mouthful, but we are at the home stretch. We've now figured out the derivative. It's going to be this times this. So let's see, that's going to be negative three, negative three times sine of x times cosine squared of x. And I know that was kind of a long way of saying it. I'm trying to explain the chain rule at the same time."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "It's going to be this times this. So let's see, that's going to be negative three, negative three times sine of x times cosine squared of x. And I know that was kind of a long way of saying it. I'm trying to explain the chain rule at the same time. But once you get the hang of it, you're just going to say, all right, well let me take the derivative of the outside of something to the third power with respect to the inside. Let me just treat that cosine of x like as if it was an x. Well that's going to be, the second I did that, that's going to be three cosine squared of x."}, {"video_title": "Worked example Derivative of cos_(x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "I'm trying to explain the chain rule at the same time. But once you get the hang of it, you're just going to say, all right, well let me take the derivative of the outside of something to the third power with respect to the inside. Let me just treat that cosine of x like as if it was an x. Well that's going to be, the second I did that, that's going to be three cosine squared of x. So that's that part and that part. And then let me take the derivative of the inside with respect to x. Well that is negative sine of x."}, {"video_title": "Worked example power series from cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Remember, the Maclaurin series is just the Taylor series centered at zero. And let's say our goal here is the first five non-zero terms of the Maclaurin series representation, the Maclaurin series approximation of this. So I'm assuming you had paused the video and you had attempted to do this, and there's a good chance that you might have gotten quite frustrated when you did this, because in order to find a Taylor series, Maclaurin series, we need to find the derivatives of this function, and as soon as you start to do that, it starts to get painful. f prime of x is going to be, let's see, product rule, so it's three x squared times cosine of x squared plus x to the third times the derivative of this thing, which is going to be two x times negative sine of x squared. So just that was pretty painful, but it's only going to get more and more painful as we take the second and third and fourth derivatives. We might have to take more, because some of these terms might end up being zeros, because we want the first five non-zero terms. So the second derivative, this is going to be painful."}, {"video_title": "Worked example power series from cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "f prime of x is going to be, let's see, product rule, so it's three x squared times cosine of x squared plus x to the third times the derivative of this thing, which is going to be two x times negative sine of x squared. So just that was pretty painful, but it's only going to get more and more painful as we take the second and third and fourth derivatives. We might have to take more, because some of these terms might end up being zeros, because we want the first five non-zero terms. So the second derivative, this is going to be painful. And then the third and fourth derivatives are going to be even more painful. So what do we do here? You could just do this, and just evaluate them each at zero, and then use those for our coefficients, but you're probably guessing correctly that there's an easier way to do this, and I will give you a hint."}, {"video_title": "Worked example power series from cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So the second derivative, this is going to be painful. And then the third and fourth derivatives are going to be even more painful. So what do we do here? You could just do this, and just evaluate them each at zero, and then use those for our coefficients, but you're probably guessing correctly that there's an easier way to do this, and I will give you a hint. We know what the Maclaurin series for cosine of x is. We've done that in a previous video. If you want to see that again, there's another video."}, {"video_title": "Worked example power series from cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "You could just do this, and just evaluate them each at zero, and then use those for our coefficients, but you're probably guessing correctly that there's an easier way to do this, and I will give you a hint. We know what the Maclaurin series for cosine of x is. We've done that in a previous video. If you want to see that again, there's another video. Look up cosine Taylor series at zero on Khan Academy, and you'll find that. But we already know from that, and this is one of the most famous Maclaurin series, we know this is, let's just say g of x. Let's say g of x is equal to cosine of x."}, {"video_title": "Worked example power series from cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "If you want to see that again, there's another video. Look up cosine Taylor series at zero on Khan Academy, and you'll find that. But we already know from that, and this is one of the most famous Maclaurin series, we know this is, let's just say g of x. Let's say g of x is equal to cosine of x. Well, we know what this is like. The Maclaurin series approximation of that is going to be one minus x squared over two factorial plus x to the fourth over four factorial minus x to the sixth over six factorial plus, and I could keep going on and on and on, you kind of see where this is going, plus x to the eighth over eight factorial, and it just keeps going minus plus, on and on and on and on. But we wanted first five terms, so this is a start."}, {"video_title": "Worked example power series from cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let's say g of x is equal to cosine of x. Well, we know what this is like. The Maclaurin series approximation of that is going to be one minus x squared over two factorial plus x to the fourth over four factorial minus x to the sixth over six factorial plus, and I could keep going on and on and on, you kind of see where this is going, plus x to the eighth over eight factorial, and it just keeps going minus plus, on and on and on and on. But we wanted first five terms, so this is a start. I know we wanted the first five terms of this thing, but bear with me. We'll see how this thing right over here is going to be useful. So now that I've given you a little bit of a reminder on the Maclaurin series representation of cosine of x, my hint to you is can we use this to find the Maclaurin series representation of this right up here?"}, {"video_title": "Worked example power series from cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "But we wanted first five terms, so this is a start. I know we wanted the first five terms of this thing, but bear with me. We'll see how this thing right over here is going to be useful. So now that I've given you a little bit of a reminder on the Maclaurin series representation of cosine of x, my hint to you is can we use this to find the Maclaurin series representation of this right up here? And remember, all this is, this is x to the third times, and I'm just rereading it for you, this is x to the third times g of x squared. So that's a sizable hint. I encourage you to pause the video again and see if you can work through this."}, {"video_title": "Worked example power series from cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So now that I've given you a little bit of a reminder on the Maclaurin series representation of cosine of x, my hint to you is can we use this to find the Maclaurin series representation of this right up here? And remember, all this is, this is x to the third times, and I'm just rereading it for you, this is x to the third times g of x squared. So that's a sizable hint. I encourage you to pause the video again and see if you can work through this. So let me rewrite, I'm assuming you had a go at it, so let me rewrite what I just wrote. So I just told you that g of x, or f of x, if I want to write it as, I guess you could say as a function, or if I want to construct it using g of x, I could rewrite it as x to the third times, instead of cosine of x squared, that's the same thing as g of x squared. So x to the third times g, g of x squared."}, {"video_title": "Worked example power series from cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "I encourage you to pause the video again and see if you can work through this. So let me rewrite, I'm assuming you had a go at it, so let me rewrite what I just wrote. So I just told you that g of x, or f of x, if I want to write it as, I guess you could say as a function, or if I want to construct it using g of x, I could rewrite it as x to the third times, instead of cosine of x squared, that's the same thing as g of x squared. So x to the third times g, g of x squared. g of x squared. This g of x is just cosine, g of x squared is going to be cosine of x squared, and then you're going to multiply that times x to the third. Well, can't we just apply this insight to the approximation?"}, {"video_title": "Worked example power series from cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So x to the third times g, g of x squared. g of x squared. This g of x is just cosine, g of x squared is going to be cosine of x squared, and then you're going to multiply that times x to the third. Well, can't we just apply this insight to the approximation? And you might be asking that question, and my answer to you is yes, you absolutely can. And notice, when you substitute your x's for x squared, you'll just get another polynomial, and if you multiply that by x to the third, you're just going to get another polynomial, and that actually will be the Maclaurin series representation of what we started off with. We actually will get the Maclaurin series representation of this thing right over here."}, {"video_title": "Worked example power series from cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, can't we just apply this insight to the approximation? And you might be asking that question, and my answer to you is yes, you absolutely can. And notice, when you substitute your x's for x squared, you'll just get another polynomial, and if you multiply that by x to the third, you're just going to get another polynomial, and that actually will be the Maclaurin series representation of what we started off with. We actually will get the Maclaurin series representation of this thing right over here. So we could say that f of x is approximately equal to x to the third times, x to the third times, let me give myself some space right over here, so g of x squared. So over here, this is approximation for g of x, and if we kept going on and on forever, it would be a representation of g of x. So every place where we see an x, let's replace it with an x squared."}, {"video_title": "Worked example power series from cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "We actually will get the Maclaurin series representation of this thing right over here. So we could say that f of x is approximately equal to x to the third times, x to the third times, let me give myself some space right over here, so g of x squared. So over here, this is approximation for g of x, and if we kept going on and on forever, it would be a representation of g of x. So every place where we see an x, let's replace it with an x squared. So this is going to be one minus, so x squared squared is x to the fourth, x to the fourth over two factorial, which is really just two, but I like to put the factorial there because you see the pattern, plus, if our x is now x squared, x squared to the fourth power is x to the eighth, x to the eighth power over four factorial, minus x squared to the sixth power is x to the twelfth over six factorial, and then plus x squared to the eighth is x to the sixteenth power over eight factorial, and of course we can keep going on and on and on, minus plus, but we just care about the first five terms, non-zero terms, and we're saying this is an approximation anyway. And so we can say that this is going to be approximately equal to, we distribute the x to the third, and I'm going to do it in magenta just for fun. So distribute the x to the third, we get x to the third minus x to the seventh over two factorial, plus x to the eleventh over four factorial, minus x to the fifteenth over six factorial, plus x to the nineteenth, x to the nineteenth over eight factorial."}, {"video_title": "Worked example power series from cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So every place where we see an x, let's replace it with an x squared. So this is going to be one minus, so x squared squared is x to the fourth, x to the fourth over two factorial, which is really just two, but I like to put the factorial there because you see the pattern, plus, if our x is now x squared, x squared to the fourth power is x to the eighth, x to the eighth power over four factorial, minus x squared to the sixth power is x to the twelfth over six factorial, and then plus x squared to the eighth is x to the sixteenth power over eight factorial, and of course we can keep going on and on and on, minus plus, but we just care about the first five terms, non-zero terms, and we're saying this is an approximation anyway. And so we can say that this is going to be approximately equal to, we distribute the x to the third, and I'm going to do it in magenta just for fun. So distribute the x to the third, we get x to the third minus x to the seventh over two factorial, plus x to the eleventh over four factorial, minus x to the fifteenth over six factorial, plus x to the nineteenth, x to the nineteenth over eight factorial. So that's the first five non-zero terms, and we are done. And when you actually see what we got, you realize it would have taken you forever to do this by, I guess you could say, by brute force, because you would have had to find the nineteenth derivative of all of this madness. But when you realize that, hey, gee, if I can just re-express this function as essentially an x to a power times something that I know the Maclaurin series of, especially, well, if you view it this way, if I can rewrite, let me say it in a less confusing way, if I can rewrite my function so it is equal to some, actually I can even throw a coefficient here, if it's equal to some ax to the n power times some function, let me just set a new color, times, do purple, times g of b, bx to some other power, where I can fairly easily, without too much computation, maybe I already even know, if I know the Maclaurin series representation of g of x, if I know what g of x is going to be, then I can do exactly what I did just in this video."}, {"video_title": "Midpoint sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "What we want to do in this video is get an understanding of how we can approximate the area under a curve. And for the sake of an example, we'll use the curve y is equal to x squared plus one. And let's think about the area under this curve above the x-axis from x equals negative one to x equals two. So that would be this area right over here. And there's many ways that I could tackle this, but what I'm going to do is I'm going to break up this interval into three equal sections that are really the bases of rectangles. And then we're gonna think about the different ways to define the heights of those rectangles. So once again, I'm going to approximate using three rectangles of equal width."}, {"video_title": "Midpoint sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So that would be this area right over here. And there's many ways that I could tackle this, but what I'm going to do is I'm going to break up this interval into three equal sections that are really the bases of rectangles. And then we're gonna think about the different ways to define the heights of those rectangles. So once again, I'm going to approximate using three rectangles of equal width. And then we'll think about the different ways that we can define the heights of the rectangle. So let's first define the height of each rectangle by the value of the function at the midpoint. So we see that right over here."}, {"video_title": "Midpoint sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So once again, I'm going to approximate using three rectangles of equal width. And then we'll think about the different ways that we can define the heights of the rectangle. So let's first define the height of each rectangle by the value of the function at the midpoint. So we see that right over here. So let's just make sure that it actually makes sense to us. So if we look at our first rectangle right over here, actually let's just first appreciate we have split up the interval from x equals negative one to x equals two into three equal sections. And then each of them have a width of one."}, {"video_title": "Midpoint sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So we see that right over here. So let's just make sure that it actually makes sense to us. So if we look at our first rectangle right over here, actually let's just first appreciate we have split up the interval from x equals negative one to x equals two into three equal sections. And then each of them have a width of one. So if we wanted a better approximation, we could do more sections or more rectangles. But let's just see how we would compute this. Well, the width of each of these is one."}, {"video_title": "Midpoint sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And then each of them have a width of one. So if we wanted a better approximation, we could do more sections or more rectangles. But let's just see how we would compute this. Well, the width of each of these is one. The height is based on the value of the function at the midpoint. The midpoint here is negative 1 1 2. The midpoint here is 1 1 2."}, {"video_title": "Midpoint sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the width of each of these is one. The height is based on the value of the function at the midpoint. The midpoint here is negative 1 1 2. The midpoint here is 1 1 2. The midpoint here is 3 1 2. And so this height is going to be negative 1 1 2 squared plus one. So negative 1 1 2 squared is 1 1 4 plus one."}, {"video_title": "Midpoint sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "The midpoint here is 1 1 2. The midpoint here is 3 1 2. And so this height is going to be negative 1 1 2 squared plus one. So negative 1 1 2 squared is 1 1 4 plus one. So that's 5 4ths. So the height here is 5 4ths. So you take 5 4ths times one."}, {"video_title": "Midpoint sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So negative 1 1 2 squared is 1 1 4 plus one. So that's 5 4ths. So the height here is 5 4ths. So you take 5 4ths times one. This area is 5 4ths. Let me write that down. So if we're doing the midpoint to define the height of each rectangle, this first one has an area of 5 4ths."}, {"video_title": "Midpoint sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So you take 5 4ths times one. This area is 5 4ths. Let me write that down. So if we're doing the midpoint to define the height of each rectangle, this first one has an area of 5 4ths. Let me do that in a color you can see. Five over four. The second one, same idea."}, {"video_title": "Midpoint sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So if we're doing the midpoint to define the height of each rectangle, this first one has an area of 5 4ths. Let me do that in a color you can see. Five over four. The second one, same idea. 1 1 2 squared plus one is 5 4ths times the width of one. So 5 4ths there. So let me add that."}, {"video_title": "Midpoint sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "The second one, same idea. 1 1 2 squared plus one is 5 4ths times the width of one. So 5 4ths there. So let me add that. Plus 5 4ths. And then this third rectangle, what's its height? Well, we're gonna take the height at the midpoint."}, {"video_title": "Midpoint sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So let me add that. Plus 5 4ths. And then this third rectangle, what's its height? Well, we're gonna take the height at the midpoint. So 3 1 2 squared is 9 4ths plus one, which is the same thing as 13 4ths. So it has a height of 13 4ths, and then a width of one, so times one, would just give us 13 4ths. So plus 13 4ths, which would give us 23 over four, which is the same thing as 5 3 4ths."}, {"video_title": "Midpoint sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we're gonna take the height at the midpoint. So 3 1 2 squared is 9 4ths plus one, which is the same thing as 13 4ths. So it has a height of 13 4ths, and then a width of one, so times one, would just give us 13 4ths. So plus 13 4ths, which would give us 23 over four, which is the same thing as 5 3 4ths. And so this is often known as a midpoint approximation, where we're using the midpoint of each interval to define the height of our rectangle. But this isn't the only way to do it. We could look at the left endpoint or the right endpoint, and we do that in other videos."}, {"video_title": "Midpoint sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So plus 13 4ths, which would give us 23 over four, which is the same thing as 5 3 4ths. And so this is often known as a midpoint approximation, where we're using the midpoint of each interval to define the height of our rectangle. But this isn't the only way to do it. We could look at the left endpoint or the right endpoint, and we do that in other videos. And then if we wanna do it just for kicks here, let's just do that really fast. So if we wanna look at the left endpoints of our interval, well, here our left endpoint is negative one. Negative one squared plus one is two."}, {"video_title": "Midpoint sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "We could look at the left endpoint or the right endpoint, and we do that in other videos. And then if we wanna do it just for kicks here, let's just do that really fast. So if we wanna look at the left endpoints of our interval, well, here our left endpoint is negative one. Negative one squared plus one is two. Two times one gives us two. And then here, the left part of this interval is x equals zero. Zero squared plus one is one."}, {"video_title": "Midpoint sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Negative one squared plus one is two. Two times one gives us two. And then here, the left part of this interval is x equals zero. Zero squared plus one is one. One times one is one. And now here, our left endpoint is one. One squared plus one is equal to two times one."}, {"video_title": "Midpoint sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Zero squared plus one is one. One times one is one. And now here, our left endpoint is one. One squared plus one is equal to two times one. Our base is equal to two. So here we have a situation where we take our left endpoints, where it is equal to two plus one plus two or five. But we could also look at the right endpoints of our intervals."}, {"video_title": "Midpoint sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "One squared plus one is equal to two times one. Our base is equal to two. So here we have a situation where we take our left endpoints, where it is equal to two plus one plus two or five. But we could also look at the right endpoints of our intervals. So this first rectangle here, clearly under-approximating the area over this first interval. Its right endpoint is zero. Zero squared plus one is one."}, {"video_title": "Midpoint sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "But we could also look at the right endpoints of our intervals. So this first rectangle here, clearly under-approximating the area over this first interval. Its right endpoint is zero. Zero squared plus one is one. So height of one, width of one has an area of one. Second rectangle here, it has a height of, we look at our right endpoint, one squared plus one is two times our width of one. Well, that's just gonna give us two."}, {"video_title": "Midpoint sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Zero squared plus one is one. So height of one, width of one has an area of one. Second rectangle here, it has a height of, we look at our right endpoint, one squared plus one is two times our width of one. Well, that's just gonna give us two. And then here, our right endpoint is two squared plus one is five times our width of one gives us five. So in this case, we get, when we look at our right endpoints of our intervals, we get one plus two plus five is equal to eight. And eyeballing this, it looks like we're definitely over-counting more than under-counting."}, {"video_title": "Midpoint sums Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Well, that's just gonna give us two. And then here, our right endpoint is two squared plus one is five times our width of one gives us five. So in this case, we get, when we look at our right endpoints of our intervals, we get one plus two plus five is equal to eight. And eyeballing this, it looks like we're definitely over-counting more than under-counting. And so this looks like an over-approximation. So the whole idea here is just to appreciate how we can compute these approximations using rectangles. And as you can imagine, if we added more rectangles that had skinnier and skinnier bases but still covered the interval from x equals negative one to x equals two, we would get better and better approximations of the true area."}, {"video_title": "Limits of composite functions internal limit doesn't exist AP Calculus Khan Academy.mp3", "Sentence": "And the function g, we see it defined graphically here on the left, and the function h, we see it defined graphically here on the right. Pause this video and have a go at this. All right, now your first temptation might be to say, all right, well, what is the limit as x approaches negative one of h of x? And if that limit exists, then input that into g. So if you take the limit as x approaches negative one of h of x, you see that you have a different limit as you approach from the right than when you approach from the left. So your temptation might be to give up at this point, but what we'll do in this video is to realize that this composite limit actually exists even though the limit as x approaches negative one of h of x does not exist. So how do we figure this out? Well, what we could do is take right-handed and left-handed limits."}, {"video_title": "Limits of composite functions internal limit doesn't exist AP Calculus Khan Academy.mp3", "Sentence": "And if that limit exists, then input that into g. So if you take the limit as x approaches negative one of h of x, you see that you have a different limit as you approach from the right than when you approach from the left. So your temptation might be to give up at this point, but what we'll do in this video is to realize that this composite limit actually exists even though the limit as x approaches negative one of h of x does not exist. So how do we figure this out? Well, what we could do is take right-handed and left-handed limits. Let's first figure out what is the limit as x approaches negative one from the right-hand side of g of h of x, h of x. Well, to think about that, what is the limit of h as x approaches negative one from the right-hand side? So as we approach negative one from the right-hand side, it looks like h is approaching negative two."}, {"video_title": "Limits of composite functions internal limit doesn't exist AP Calculus Khan Academy.mp3", "Sentence": "Well, what we could do is take right-handed and left-handed limits. Let's first figure out what is the limit as x approaches negative one from the right-hand side of g of h of x, h of x. Well, to think about that, what is the limit of h as x approaches negative one from the right-hand side? So as we approach negative one from the right-hand side, it looks like h is approaching negative two. So another way to think about it is this is going to be equal to the limit as h of x approaches negative two. And what direction is it approaching negative two from? Well, it's approaching negative two from values larger than negative two."}, {"video_title": "Limits of composite functions internal limit doesn't exist AP Calculus Khan Academy.mp3", "Sentence": "So as we approach negative one from the right-hand side, it looks like h is approaching negative two. So another way to think about it is this is going to be equal to the limit as h of x approaches negative two. And what direction is it approaching negative two from? Well, it's approaching negative two from values larger than negative two. H of x is decreasing down to negative two as x approaches negative one from the right. So it's approaching from values larger than negative two of g of h of x, g of h of x. I'm color-coding it to be able to keep track of things. And so this is analogous to saying what is the limit as if you think about it as x approaches negative two from the positive direction of g. Here, h is just the input into g. So the input into g is approaching negative two from above, from the right, I should say, from values larger than negative two."}, {"video_title": "Limits of composite functions internal limit doesn't exist AP Calculus Khan Academy.mp3", "Sentence": "Well, it's approaching negative two from values larger than negative two. H of x is decreasing down to negative two as x approaches negative one from the right. So it's approaching from values larger than negative two of g of h of x, g of h of x. I'm color-coding it to be able to keep track of things. And so this is analogous to saying what is the limit as if you think about it as x approaches negative two from the positive direction of g. Here, h is just the input into g. So the input into g is approaching negative two from above, from the right, I should say, from values larger than negative two. And we can see that g is approaching three. So this right over here is going to be equal to three. Now let's take the limit as x approaches."}, {"video_title": "Limits of composite functions internal limit doesn't exist AP Calculus Khan Academy.mp3", "Sentence": "And so this is analogous to saying what is the limit as if you think about it as x approaches negative two from the positive direction of g. Here, h is just the input into g. So the input into g is approaching negative two from above, from the right, I should say, from values larger than negative two. And we can see that g is approaching three. So this right over here is going to be equal to three. Now let's take the limit as x approaches. So now let's take the limit as x approaches negative one from the left of g of h of x. So what we could do is first think about, well, what is h approaching as x approaches negative one from the left? So as x approaches negative one from the left, it looks like h is approaching negative three."}, {"video_title": "Limits of composite functions internal limit doesn't exist AP Calculus Khan Academy.mp3", "Sentence": "Now let's take the limit as x approaches. So now let's take the limit as x approaches negative one from the left of g of h of x. So what we could do is first think about, well, what is h approaching as x approaches negative one from the left? So as x approaches negative one from the left, it looks like h is approaching negative three. So we could say this is the limit as h of x is approaching negative three, and it is approaching negative three from values greater than negative three. It's going, h of x is approaching negative three from above, or we could say from values greater than negative three. And then of g of h of x."}, {"video_title": "Limits of composite functions internal limit doesn't exist AP Calculus Khan Academy.mp3", "Sentence": "So as x approaches negative one from the left, it looks like h is approaching negative three. So we could say this is the limit as h of x is approaching negative three, and it is approaching negative three from values greater than negative three. It's going, h of x is approaching negative three from above, or we could say from values greater than negative three. And then of g of h of x. So another way to think about it, what is the limit as the input to g approaches negative three from the right? So as we approach negative three from the right, g is right here at three. And so this is going to be equal to three again."}, {"video_title": "Limits of composite functions internal limit doesn't exist AP Calculus Khan Academy.mp3", "Sentence": "And then of g of h of x. So another way to think about it, what is the limit as the input to g approaches negative three from the right? So as we approach negative three from the right, g is right here at three. And so this is going to be equal to three again. And so notice the right-hand limit and the left-hand limit in this case are both equal to three. And so when the right-hand and the left-hand limit is equal to the same thing, we know that the limit is equal to that thing. And this is a pretty cool example because the limit of, you could say it's the internal function right over here of h of x did not exist, but the limit of the composite function still exists."}, {"video_title": "Riemann approximation introduction Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So let's first think about what those rectangles look like. So four rectangles of equal width. So it would look like that, like that, and like that. And I haven't really defined the top of the rectangles just yet. But let's think about what those widths have to be if they're going to be equal width. And we can call that width delta x. So this distance right over here, we're going to call that delta x."}, {"video_title": "Riemann approximation introduction Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And I haven't really defined the top of the rectangles just yet. But let's think about what those widths have to be if they're going to be equal width. And we can call that width delta x. So this distance right over here, we're going to call that delta x. So delta x is going to have to be the total distance that we're traveling in x. So we finish at 3. We started at 1."}, {"video_title": "Riemann approximation introduction Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So this distance right over here, we're going to call that delta x. So delta x is going to have to be the total distance that we're traveling in x. So we finish at 3. We started at 1. And we want four equal width rectangles. So it's going to be equal to 1 half. So for example, this first interval between the boundary between the first rectangle and the second is going to be 1.5."}, {"video_title": "Riemann approximation introduction Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "We started at 1. And we want four equal width rectangles. So it's going to be equal to 1 half. So for example, this first interval between the boundary between the first rectangle and the second is going to be 1.5. Then we go 1 half to 2. Then we go to 2.5. And then we go 1 half to 3."}, {"video_title": "Riemann approximation introduction Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So for example, this first interval between the boundary between the first rectangle and the second is going to be 1.5. Then we go 1 half to 2. Then we go to 2.5. And then we go 1 half to 3. Now let's think about how we'll define the height of the rectangles. For the sake of this video, we'll see in future videos that there's other ways of doing this, I'm going to use the left boundary of the rectangle to define the height, or the function, I should say. I'm going to use the function evaluated at the left boundary to define the height."}, {"video_title": "Riemann approximation introduction Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And then we go 1 half to 3. Now let's think about how we'll define the height of the rectangles. For the sake of this video, we'll see in future videos that there's other ways of doing this, I'm going to use the left boundary of the rectangle to define the height, or the function, I should say. I'm going to use the function evaluated at the left boundary to define the height. So for example, for the first rectangle, this point right over here is f of 1. And so I will say that that is the height of our first rectangle. Then we go over here, the left boundary of the second rectangle."}, {"video_title": "Riemann approximation introduction Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "I'm going to use the function evaluated at the left boundary to define the height. So for example, for the first rectangle, this point right over here is f of 1. And so I will say that that is the height of our first rectangle. Then we go over here, the left boundary of the second rectangle. We're now looking at the function evaluated at 1.5, so that is f of 1.5. That's the height. And so we get our second rectangle."}, {"video_title": "Riemann approximation introduction Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Then we go over here, the left boundary of the second rectangle. We're now looking at the function evaluated at 1.5, so that is f of 1.5. That's the height. And so we get our second rectangle. Then we get, I could keep going like this, we get for this third rectangle, we have the function evaluated at 2. So we have the function evaluated at 2. So that's right over here."}, {"video_title": "Riemann approximation introduction Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And so we get our second rectangle. Then we get, I could keep going like this, we get for this third rectangle, we have the function evaluated at 2. So we have the function evaluated at 2. So that's right over here. That's f of 2. And so then we get our third rectangle. And then finally, we have our fourth rectangle, the function evaluated at 2.5."}, {"video_title": "Riemann approximation introduction Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So that's right over here. That's f of 2. And so then we get our third rectangle. And then finally, we have our fourth rectangle, the function evaluated at 2.5. So the function evaluated at 2.5 is the height. So this is f of 2.5. Remember, in each of these, I am just looking at the left boundary of the rectangle and evaluating the function there to get the height of the rectangle."}, {"video_title": "Riemann approximation introduction Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And then finally, we have our fourth rectangle, the function evaluated at 2.5. So the function evaluated at 2.5 is the height. So this is f of 2.5. Remember, in each of these, I am just looking at the left boundary of the rectangle and evaluating the function there to get the height of the rectangle. Now that I set it up in this way, what is the total approximate area using the sum of these rectangles? And clearly, this isn't going to be a perfect approximation. I'm giving up on a bunch of area here."}, {"video_title": "Riemann approximation introduction Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Remember, in each of these, I am just looking at the left boundary of the rectangle and evaluating the function there to get the height of the rectangle. Now that I set it up in this way, what is the total approximate area using the sum of these rectangles? And clearly, this isn't going to be a perfect approximation. I'm giving up on a bunch of area here. Let me see if I can color that in with a color that I have not used. So I'm giving up this area. I'm giving up this area."}, {"video_title": "Riemann approximation introduction Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "I'm giving up on a bunch of area here. Let me see if I can color that in with a color that I have not used. So I'm giving up this area. I'm giving up this area. I'm giving up that area. I'm giving up that area there. But this is just an approximation."}, {"video_title": "Riemann approximation introduction Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "I'm giving up this area. I'm giving up that area. I'm giving up that area there. But this is just an approximation. And maybe if I had many more rectangles, it would be a better approximation. So let's figure out what the areas of each of the rectangles are. So the area of this first rectangle is going to be the height, which is f of 1 times the base, which is delta x."}, {"video_title": "Riemann approximation introduction Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "But this is just an approximation. And maybe if I had many more rectangles, it would be a better approximation. So let's figure out what the areas of each of the rectangles are. So the area of this first rectangle is going to be the height, which is f of 1 times the base, which is delta x. The area of the second rectangle is going to be the height, which we already said is f of 1.5 times the base times delta x. The height of the third rectangle is going to be the function evaluated at its left boundary, so f of 2, so plus f of 2 times the base times delta x. And then finally, the area of the third rectangle is the function."}, {"video_title": "Riemann approximation introduction Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So the area of this first rectangle is going to be the height, which is f of 1 times the base, which is delta x. The area of the second rectangle is going to be the height, which we already said is f of 1.5 times the base times delta x. The height of the third rectangle is going to be the function evaluated at its left boundary, so f of 2, so plus f of 2 times the base times delta x. And then finally, the area of the third rectangle is the function. The height is a function evaluated at 2.5. So plus, that's a different color than what I wanted to use. Let me use that orange color."}, {"video_title": "Riemann approximation introduction Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And then finally, the area of the third rectangle is the function. The height is a function evaluated at 2.5. So plus, that's a different color than what I wanted to use. Let me use that orange color. So plus the function evaluated at 2.5 times the base. This is going to be equal to our approximate area. Let's make it clear."}, {"video_title": "Riemann approximation introduction Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Let me use that orange color. So plus the function evaluated at 2.5 times the base. This is going to be equal to our approximate area. Let's make it clear. Approximate area under the curve, just the sum of these rectangles. So let's evaluate this. So this is going to be equal to f of, it's going to be equal to the function evaluated at 1."}, {"video_title": "Riemann approximation introduction Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Let's make it clear. Approximate area under the curve, just the sum of these rectangles. So let's evaluate this. So this is going to be equal to f of, it's going to be equal to the function evaluated at 1. 1 squared plus 1 is just 2. So it's going to be 2 times 1 half plus the function evaluated at 1.25. 1.25 squared is 2.25."}, {"video_title": "Riemann approximation introduction Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be equal to f of, it's going to be equal to the function evaluated at 1. 1 squared plus 1 is just 2. So it's going to be 2 times 1 half plus the function evaluated at 1.25. 1.25 squared is 2.25. And then you add 1 to it. It becomes 3.25. So plus 3.25 times 1 half."}, {"video_title": "Riemann approximation introduction Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "1.25 squared is 2.25. And then you add 1 to it. It becomes 3.25. So plus 3.25 times 1 half. And then we have the function evaluated at 2. Well, 2 squared plus 1 is 5. So it's 5 times 1 half."}, {"video_title": "Riemann approximation introduction Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So plus 3.25 times 1 half. And then we have the function evaluated at 2. Well, 2 squared plus 1 is 5. So it's 5 times 1 half. And then finally, you have the function evaluated at 2.5. 2.5 squared is 6.25 times 1 half. 6.25 plus 1."}, {"video_title": "Riemann approximation introduction Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So it's 5 times 1 half. And then finally, you have the function evaluated at 2.5. 2.5 squared is 6.25 times 1 half. 6.25 plus 1. So that's 7.25 times 1 half. And just to make the math simpler, we can factor out the 1 half. So this is going to be equal to 1 half times 2 plus 3.25 plus 5 plus 7.25, which is equal to 1 half times."}, {"video_title": "Riemann approximation introduction Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "6.25 plus 1. So that's 7.25 times 1 half. And just to make the math simpler, we can factor out the 1 half. So this is going to be equal to 1 half times 2 plus 3.25 plus 5 plus 7.25, which is equal to 1 half times. Let's see if I can do this in my head. 2 plus 5 is easy. That's 7."}, {"video_title": "Riemann approximation introduction Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be equal to 1 half times 2 plus 3.25 plus 5 plus 7.25, which is equal to 1 half times. Let's see if I can do this in my head. 2 plus 5 is easy. That's 7. 3 plus 7 is 10. And then we have 0.25 plus 0.25. So it's going to be 10.5 plus 7 is 17.5."}, {"video_title": "Riemann approximation introduction Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "That's 7. 3 plus 7 is 10. And then we have 0.25 plus 0.25. So it's going to be 10.5 plus 7 is 17.5. So 1 half times 17.5, which is equal to 8.75, which once again gives us an approximation. Clearly, the way I've drawn it right over here for the function we're using, it's going to be an underestimate because we've given up all of that pink area that I had colored in before. It's an underestimate, but it's an approximation of the area under the curve."}, {"video_title": "Riemann approximation introduction Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So it's going to be 10.5 plus 7 is 17.5. So 1 half times 17.5, which is equal to 8.75, which once again gives us an approximation. Clearly, the way I've drawn it right over here for the function we're using, it's going to be an underestimate because we've given up all of that pink area that I had colored in before. It's an underestimate, but it's an approximation of the area under the curve. In the next few videos, we're going to try to generalize this to situations where we have an arbitrary function and we have an arbitrary number of rectangles. And we'll also start, in videos after that, we'll look at rectangles where we define the height not by the left boundary, but by the right boundary, or by the midpoint. Or maybe we don't use rectangles at all."}, {"video_title": "_-substitution defining _ AP Calculus AB Khan Academy.mp3", "Sentence": "What we're going to do in this video is give ourselves some practice in the first step of u-substitution, which is often the most difficult for those who are first learning it, and that's recognizing when u-substitution is appropriate and then defining an appropriate u. So let's just start with an example here. So let's say we want to take the indefinite integral of two x plus one times the square root of x squared plus x dx. Does u-substitution apply here? And if it does, how would you define the u? Pause the video and try to think about that. Well, we just have to remind ourselves that u-substitution is really trying to undo the chain rule."}, {"video_title": "_-substitution defining _ AP Calculus AB Khan Academy.mp3", "Sentence": "Does u-substitution apply here? And if it does, how would you define the u? Pause the video and try to think about that. Well, we just have to remind ourselves that u-substitution is really trying to undo the chain rule. If we remind ourselves what the chain rule tells us, it says, look, if we have a composite function, let's say f of g of x, f of g of x, and we take the derivative of that with respect to x, that that is going to be equal to the derivative of the outside function with respect to the inside function, so f prime of g of x times the derivative of the inside function, times the derivative of the inside function. And so u-substitution is all about, well, do we see a pattern like that inside the integral? Do we see a potential inside function, a g of x, where I see its derivative being multiplied?"}, {"video_title": "_-substitution defining _ AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we just have to remind ourselves that u-substitution is really trying to undo the chain rule. If we remind ourselves what the chain rule tells us, it says, look, if we have a composite function, let's say f of g of x, f of g of x, and we take the derivative of that with respect to x, that that is going to be equal to the derivative of the outside function with respect to the inside function, so f prime of g of x times the derivative of the inside function, times the derivative of the inside function. And so u-substitution is all about, well, do we see a pattern like that inside the integral? Do we see a potential inside function, a g of x, where I see its derivative being multiplied? Well, we see that over here. If I look at x squared plus x, if I make that the u, what's the derivative of that? Well, the derivative of x squared plus x is two x plus one."}, {"video_title": "_-substitution defining _ AP Calculus AB Khan Academy.mp3", "Sentence": "Do we see a potential inside function, a g of x, where I see its derivative being multiplied? Well, we see that over here. If I look at x squared plus x, if I make that the u, what's the derivative of that? Well, the derivative of x squared plus x is two x plus one. So we should make that substitution. If we say u is equal to x squared plus x, then we could say du dx, the derivative of u with respect to x, is equal to two x plus one. If we treat our differentials like variables or numbers, we can multiply both sides by dx, which is a little bit of hand-wavy mathematics, but it's appropriate here."}, {"video_title": "_-substitution defining _ AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the derivative of x squared plus x is two x plus one. So we should make that substitution. If we say u is equal to x squared plus x, then we could say du dx, the derivative of u with respect to x, is equal to two x plus one. If we treat our differentials like variables or numbers, we can multiply both sides by dx, which is a little bit of hand-wavy mathematics, but it's appropriate here. So we could say two x plus one times dx. And now what's really interesting is here is we have our u right over there, and notice we have our two x plus one times dx. In fact, it's not conventional to see an integral rewritten the way I'm about to write it, but I will."}, {"video_title": "_-substitution defining _ AP Calculus AB Khan Academy.mp3", "Sentence": "If we treat our differentials like variables or numbers, we can multiply both sides by dx, which is a little bit of hand-wavy mathematics, but it's appropriate here. So we could say two x plus one times dx. And now what's really interesting is here is we have our u right over there, and notice we have our two x plus one times dx. In fact, it's not conventional to see an integral rewritten the way I'm about to write it, but I will. I could rewrite this integral. You should really view this as the product of three things. Oftentimes, people just view the dx as somehow part of the integral operator, but you could rearrange it."}, {"video_title": "_-substitution defining _ AP Calculus AB Khan Academy.mp3", "Sentence": "In fact, it's not conventional to see an integral rewritten the way I'm about to write it, but I will. I could rewrite this integral. You should really view this as the product of three things. Oftentimes, people just view the dx as somehow part of the integral operator, but you could rearrange it. This would actually be legitimate. You could say the integral of the square root of x squared plus x times two x plus one dx. And if you wanted to be really clear, you could even put all of those things in parentheses or something like that."}, {"video_title": "_-substitution defining _ AP Calculus AB Khan Academy.mp3", "Sentence": "Oftentimes, people just view the dx as somehow part of the integral operator, but you could rearrange it. This would actually be legitimate. You could say the integral of the square root of x squared plus x times two x plus one dx. And if you wanted to be really clear, you could even put all of those things in parentheses or something like that. And so here, this is our u, and this right over here is our du. And so we could rewrite this as being equal to the integral of the square root of u, because x squared plus x is u, times du, which is much easier to evaluate. If you are still confused there, you might recognize it if I rewrite this as u to the 1 1\u20442 power, because now we could just use the reverse power rule to evaluate this, and then we would have to undo the substitution."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "But let's see if we can generalize this and come up with a formula that finds us the slope at any point of the curve y is equal to x squared. So let me redraw my function here. Never hurts to have a nice drawing. So that is my y-axis. That is my x-axis right there. My x-axis, let me draw my curve. It looks something like that."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So that is my y-axis. That is my x-axis right there. My x-axis, let me draw my curve. It looks something like that. You've seen that multiple times. This is y is equal to x squared. So let's be very general right now."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "It looks something like that. You've seen that multiple times. This is y is equal to x squared. So let's be very general right now. Remember, if we want to find, let me just write the definition of our derivative. So if we have some point right here, let's call that x. So we want to be very general."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So let's be very general right now. Remember, if we want to find, let me just write the definition of our derivative. So if we have some point right here, let's call that x. So we want to be very general. We want to find the slope at the point x. We want to find a function where you give me an x, and I'll tell you the slope at that point. We're going to call that f prime of x."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So we want to be very general. We want to find the slope at the point x. We want to find a function where you give me an x, and I'll tell you the slope at that point. We're going to call that f prime of x. That's going to be the derivative of f of x. But all it does is, look, when you f of x, it's a function that you give it an x, and it tells you the value of that. We draw the curve here."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "We're going to call that f prime of x. That's going to be the derivative of f of x. But all it does is, look, when you f of x, it's a function that you give it an x, and it tells you the value of that. We draw the curve here. With f of x, you give that same x, but it's not going to tell you the value of the curve. It's not going to say, oh, this is your f of x. It's going to give you the value of the slope of the curve at that point."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "We draw the curve here. With f of x, you give that same x, but it's not going to tell you the value of the curve. It's not going to say, oh, this is your f of x. It's going to give you the value of the slope of the curve at that point. So f of x, if you put it into that function, it should tell you, oh, the slope at that point is equal to, if you put 3 there, you'll say, oh, the slope there is equal to 6. We saw that in the last example. So that's what we want to do."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "It's going to give you the value of the slope of the curve at that point. So f of x, if you put it into that function, it should tell you, oh, the slope at that point is equal to, if you put 3 there, you'll say, oh, the slope there is equal to 6. We saw that in the last example. So that's what we want to do. And we saw in the last, I think it was two videos ago, that we defined f prime of x to be equal to just the, well, I'll write it this way. It's the slope of the secant line between x and some point that's a little bit further away from x. So the slope of the secant line is change in y."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So that's what we want to do. And we saw in the last, I think it was two videos ago, that we defined f prime of x to be equal to just the, well, I'll write it this way. It's the slope of the secant line between x and some point that's a little bit further away from x. So the slope of the secant line is change in y. So it's the y value of the point that's a little bit further away from x. So f of x plus h minus the y value at x, right? Because this is right here."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So the slope of the secant line is change in y. So it's the y value of the point that's a little bit further away from x. So f of x plus h minus the y value at x, right? Because this is right here. This is f of x. So minus f of x. All of that over the change in x."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Because this is right here. This is f of x. So minus f of x. All of that over the change in x. So if this is x plus h here, the change in x is x plus h minus x, or this distance right here is just h. The change in x is going to be equal to h. So that's your slope of the secant line between any two points like that. And we said, hey, we could find the slope of the tangent line if we just take the limit of this as it approaches, as h approaches 0. Limit as h approaches 0."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "All of that over the change in x. So if this is x plus h here, the change in x is x plus h minus x, or this distance right here is just h. The change in x is going to be equal to h. So that's your slope of the secant line between any two points like that. And we said, hey, we could find the slope of the tangent line if we just take the limit of this as it approaches, as h approaches 0. Limit as h approaches 0. And then we'll be finding the slope of the tangent line. Now let's apply this idea to a particular function, y, or f of x is equal to x squared, or y is equal to x squared. So here we could consider this to be the point x squared."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Limit as h approaches 0. And then we'll be finding the slope of the tangent line. Now let's apply this idea to a particular function, y, or f of x is equal to x squared, or y is equal to x squared. So here we could consider this to be the point x squared. So f of x is just equal to x squared. And then this would be the point. Let me do it in a more vibrant color."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So here we could consider this to be the point x squared. So f of x is just equal to x squared. And then this would be the point. Let me do it in a more vibrant color. This is the point x plus h. That's this point right here. It's a little bit further down. And then x plus h squared."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Let me do it in a more vibrant color. This is the point x plus h. That's this point right here. It's a little bit further down. And then x plus h squared. And in the last video, we did this for a particular x. We did it for 3. But now I want a general formula."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And then x plus h squared. And in the last video, we did this for a particular x. We did it for 3. But now I want a general formula. You give me any x, and I won't have to do what I did in the last video for any particular number. I'll have a general function. You give me 7, I'll tell you what the slope is at 7."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "But now I want a general formula. You give me any x, and I won't have to do what I did in the last video for any particular number. I'll have a general function. You give me 7, I'll tell you what the slope is at 7. You give me negative 3, I'll tell you what the slope is at negative 3. You give me 100,000, I'll tell you what the slope is at 100,000. So let's apply it here."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "You give me 7, I'll tell you what the slope is at 7. You give me negative 3, I'll tell you what the slope is at negative 3. You give me 100,000, I'll tell you what the slope is at 100,000. So let's apply it here. So we want to find the change in y over the change in x. So first of all, the change in y is this guy's y value, which is x plus h squared. That's this guy's y value right here."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So let's apply it here. So we want to find the change in y over the change in x. So first of all, the change in y is this guy's y value, which is x plus h squared. That's this guy's y value right here. That's this right here. That's x plus h squared. I just took x plus h, evaluated, I squared it, and that's its point on the curve."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "That's this guy's y value right here. That's this right here. That's x plus h squared. I just took x plus h, evaluated, I squared it, and that's its point on the curve. So it's x plus h squared. So that's there right there. And then what's this value?"}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "I just took x plus h, evaluated, I squared it, and that's its point on the curve. So it's x plus h squared. So that's there right there. And then what's this value? f of x right here is equal to, I know it's getting messy, is equal to x squared. If you take your x, you evaluate the function at that point. You're going to get x squared."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And then what's this value? f of x right here is equal to, I know it's getting messy, is equal to x squared. If you take your x, you evaluate the function at that point. You're going to get x squared. So it's equal to minus x squared. This is your change in y. That's this distance right there."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "You're going to get x squared. So it's equal to minus x squared. This is your change in y. That's this distance right there. Change in y. And just to relate it to our definition of a derivative, this blue thing right here is equivalent to this thing right here. We just evaluated our function."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "That's this distance right there. Change in y. And just to relate it to our definition of a derivative, this blue thing right here is equivalent to this thing right here. We just evaluated our function. Our function is f of x is equal to x squared. We just evaluated when x is equal to x plus h. So if you have to square it, if I put an a there, it would be a squared. If I put an apple there, it would be apple squared."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "We just evaluated our function. Our function is f of x is equal to x squared. We just evaluated when x is equal to x plus h. So if you have to square it, if I put an a there, it would be a squared. If I put an apple there, it would be apple squared. If I put an x plus h in there, it's going to be x plus h squared. So this is that thing. And then this thing right here is just the function evaluated at the point in question, right there."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "If I put an apple there, it would be apple squared. If I put an x plus h in there, it's going to be x plus h squared. So this is that thing. And then this thing right here is just the function evaluated at the point in question, right there. So this is our change in y. And let's divide that by our change in x. If this is x plus h and this is just x, our change in x is just going to be h. So that's where we get that term from."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And then this thing right here is just the function evaluated at the point in question, right there. So this is our change in y. And let's divide that by our change in x. If this is x plus h and this is just x, our change in x is just going to be h. So that's where we get that term from. So this is just a slope between these two points. But of course, we want to find the limit as this point gets closer and closer to this point, as this point gets closer and closer to that point. So this becomes a tangent line."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "If this is x plus h and this is just x, our change in x is just going to be h. So that's where we get that term from. So this is just a slope between these two points. But of course, we want to find the limit as this point gets closer and closer to this point, as this point gets closer and closer to that point. So this becomes a tangent line. So we're going to take the limit as h approaches 0. And this is our f prime of x. And this is the exact same definition of this."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So this becomes a tangent line. So we're going to take the limit as h approaches 0. And this is our f prime of x. And this is the exact same definition of this. Instead of being general and saying for any function, we know what the function was. It was f of x is equal to x squared. So we actually applied it."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And this is the exact same definition of this. Instead of being general and saying for any function, we know what the function was. It was f of x is equal to x squared. So we actually applied it. Instead of f of x, we wrote x squared. Instead of f of x plus h, we wrote x plus h squared. So let's see if we can evaluate this limit."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So we actually applied it. Instead of f of x, we wrote x squared. Instead of f of x plus h, we wrote x plus h squared. So let's see if we can evaluate this limit. So this is going to be equal to the limit. Let me write a little neater than that. The limit as h approaches 0, to square this out."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So let's see if we can evaluate this limit. So this is going to be equal to the limit. Let me write a little neater than that. The limit as h approaches 0, to square this out. I'll do it in the same color. That's x squared plus 2xh plus h squared. And then we have this minus x squared over here."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "The limit as h approaches 0, to square this out. I'll do it in the same color. That's x squared plus 2xh plus h squared. And then we have this minus x squared over here. I just multiplied this guy out over here. And then all of that is divided by h. Now let's see if we can simplify this a little bit. Well, you immediately see you have an x squared and you have a minus x squared."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And then we have this minus x squared over here. I just multiplied this guy out over here. And then all of that is divided by h. Now let's see if we can simplify this a little bit. Well, you immediately see you have an x squared and you have a minus x squared. So those cancel out. And then we could divide the numerator and the denominator by h. So this simplifies too. So we get f prime of x is equal to, if we divide the numerator and denominator by h, we get 2x plus h. Oh, sorry."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Well, you immediately see you have an x squared and you have a minus x squared. So those cancel out. And then we could divide the numerator and the denominator by h. So this simplifies too. So we get f prime of x is equal to, if we divide the numerator and denominator by h, we get 2x plus h. Oh, sorry. I forgot my limit. It equals the limit. Very important."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So we get f prime of x is equal to, if we divide the numerator and denominator by h, we get 2x plus h. Oh, sorry. I forgot my limit. It equals the limit. Very important. Limit as h approaches 0, divide everything by h, and you get 2x plus h squared divided by h is h. If you remember the last video when we did it with a particular x, when we said x is equal to 3, we got 6 plus delta x here, or 6 plus h here. So it's very similar. So if you take the limit as h approaches 0 here, that's just going to disappear."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Very important. Limit as h approaches 0, divide everything by h, and you get 2x plus h squared divided by h is h. If you remember the last video when we did it with a particular x, when we said x is equal to 3, we got 6 plus delta x here, or 6 plus h here. So it's very similar. So if you take the limit as h approaches 0 here, that's just going to disappear. So this is just going to be equal to 2x. So we just figured out that if f of x, this is a big result, this is exciting, that if f of x is equal to x squared, f prime of x is equal to 2x. That's what we just figured out."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So if you take the limit as h approaches 0 here, that's just going to disappear. So this is just going to be equal to 2x. So we just figured out that if f of x, this is a big result, this is exciting, that if f of x is equal to x squared, f prime of x is equal to 2x. That's what we just figured out. And I want to make sure you understand how to interpret this. f of x, if you give me a value, is going to tell you the value of the function at that point. f prime of x is going to tell you the slope at that point."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "That's what we just figured out. And I want to make sure you understand how to interpret this. f of x, if you give me a value, is going to tell you the value of the function at that point. f prime of x is going to tell you the slope at that point. Let me draw that. Because this is a key realization, and it's kind of maybe initially unintuitive to think of a function that gives us the slope at any point of another function. So it looks like this."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "f prime of x is going to tell you the slope at that point. Let me draw that. Because this is a key realization, and it's kind of maybe initially unintuitive to think of a function that gives us the slope at any point of another function. So it looks like this. Let me draw it a little neater than that. That's still not that neat. That's satisfactory."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So it looks like this. Let me draw it a little neater than that. That's still not that neat. That's satisfactory. Let me just draw it in the positive coordinate. Well, I'll just draw the whole thing. The curve looks like something like that."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "That's satisfactory. Let me just draw it in the positive coordinate. Well, I'll just draw the whole thing. The curve looks like something like that. Now, this is the curve of f of x. This is the curve of f of x is equal to x squared. Just like that."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "The curve looks like something like that. Now, this is the curve of f of x. This is the curve of f of x is equal to x squared. Just like that. So if you give me a point, you give me the point 7. You apply, you put it in here, you square it, and it is mapped to the number 49. So you get the number 49 right there."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Just like that. So if you give me a point, you give me the point 7. You apply, you put it in here, you square it, and it is mapped to the number 49. So you get the number 49 right there. This is number 749. You're used to dealing with functions right there. But what is f prime of 7?"}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So you get the number 49 right there. This is number 749. You're used to dealing with functions right there. But what is f prime of 7? f prime of 7, you say 2 times 7 is equal to 14. What is this 14 number here? What is this thing?"}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "But what is f prime of 7? f prime of 7, you say 2 times 7 is equal to 14. What is this 14 number here? What is this thing? Well, this is the slope of the tangent line at x is equal to 7. So if I were to take that point and draw a tangent line, a point that just grazes our curve, if I were to just draw a tangent line, that wasn't tangent enough for me. So that's my tangent line right there."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "What is this thing? Well, this is the slope of the tangent line at x is equal to 7. So if I were to take that point and draw a tangent line, a point that just grazes our curve, if I were to just draw a tangent line, that wasn't tangent enough for me. So that's my tangent line right there. Did you get the idea? The slope of this guy, you do your change in y over your change in x, is going to be equal to 14. The slope of the curve at y is equal to 7 is a pretty steep curve."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So that's my tangent line right there. Did you get the idea? The slope of this guy, you do your change in y over your change in x, is going to be equal to 14. The slope of the curve at y is equal to 7 is a pretty steep curve. If you wanted to find the slope, let's say that this is y, let's say x is equal to 2. I said at x is equal to 7, the slope is 14. At x is equal to 2, what is the slope?"}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "The slope of the curve at y is equal to 7 is a pretty steep curve. If you wanted to find the slope, let's say that this is y, let's say x is equal to 2. I said at x is equal to 7, the slope is 14. At x is equal to 2, what is the slope? Well, you go, you figure out f prime of 2, which is equal to 2 times 2, which is equal to 4. So the slope here is 4. The slope is 4."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "At x is equal to 2, what is the slope? Well, you go, you figure out f prime of 2, which is equal to 2 times 2, which is equal to 4. So the slope here is 4. The slope is 4. You could say m is equal to 4, m for slope. What is f prime of 0? We know that f of 0 is 0, right?"}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "The slope is 4. You could say m is equal to 4, m for slope. What is f prime of 0? We know that f of 0 is 0, right? 0 squared is 0. What is f prime of 0? Well, 2 times 0 is 0."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "We know that f of 0 is 0, right? 0 squared is 0. What is f prime of 0? Well, 2 times 0 is 0. That's also equal to 0. But what does that mean? What's the interpretation?"}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Well, 2 times 0 is 0. That's also equal to 0. But what does that mean? What's the interpretation? It means the slope of the tangent line is 0. So a 0-sloped line looks like this. Looks just like a horizontal line."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "What's the interpretation? It means the slope of the tangent line is 0. So a 0-sloped line looks like this. Looks just like a horizontal line. And that looks about right. That a horizontal line would be tangent to the curve at y equals 0. Let's try another one."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Looks just like a horizontal line. And that looks about right. That a horizontal line would be tangent to the curve at y equals 0. Let's try another one. Let's try the point minus 1. So let's say we're right there. x is equal to minus 1."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Let's try another one. Let's try the point minus 1. So let's say we're right there. x is equal to minus 1. So f of minus 1, you just square it, because we're dealing with x squared. So it's equal to 1. That's that point right there."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "x is equal to minus 1. So f of minus 1, you just square it, because we're dealing with x squared. So it's equal to 1. That's that point right there. What is f prime of minus 1? f prime of minus 1 is 2 times minus 1. 2 times minus is minus 2."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "That's that point right there. What is f prime of minus 1? f prime of minus 1 is 2 times minus 1. 2 times minus is minus 2. What does that mean? It means that the slope of the tangent line at x is equal to 1 to this curve, to the function, is minus 2. So if I were to draw the tangent line here, the tangent line looks like that."}, {"video_title": "The derivative of f(x)=x^2 for any x Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "2 times minus is minus 2. What does that mean? It means that the slope of the tangent line at x is equal to 1 to this curve, to the function, is minus 2. So if I were to draw the tangent line here, the tangent line looks like that. And look, it is a downward sloping line. It makes sense. The slope here is equal to minus 2."}, {"video_title": "Ratio test Series AP Calculus BC Khan Academy.mp3", "Sentence": "There's a few things we've already thought about here. We know what the common ratio is. So the common ratio, which is the ratio between consecutive terms, is going to be r to the k plus one. Actually, let me just write this as r to the n plus one. I don't want to fixate on this k right over here. So r to the n plus one over r to the n. Well, this is just going to be equal to r. Anything to the n plus one over that same thing to the n, that's just going to be equal to r, or r to the first power. And you see that here."}, {"video_title": "Ratio test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Actually, let me just write this as r to the n plus one. I don't want to fixate on this k right over here. So r to the n plus one over r to the n. Well, this is just going to be equal to r. Anything to the n plus one over that same thing to the n, that's just going to be equal to r, or r to the first power. And you see that here. When you go from one term to another, you are just multiplying, you are just multiplying by r. And this is all review. If it's not review, I encourage you to watch the videos on geometric series. And what's interesting about this is we've proven to ourselves in the videos on geometric series that if the common ratio, if the absolute value of the common ratio is less than one, then the series converges."}, {"video_title": "Ratio test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And you see that here. When you go from one term to another, you are just multiplying, you are just multiplying by r. And this is all review. If it's not review, I encourage you to watch the videos on geometric series. And what's interesting about this is we've proven to ourselves in the videos on geometric series that if the common ratio, if the absolute value of the common ratio is less than one, then the series converges. And if the absolute value of r is greater than or equal to one, then the series diverges. And that made sense, we've proven it as well, but it also makes logical sense that look, if the absolute value of r is less than one, then each term here is going to go down by that common, or it's going to be multiplied by that common ratio, and it's going to decrease on and on and on and on and on. And so it makes sense that even though this is an infinite sum, it will converge to a finite value."}, {"video_title": "Ratio test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And what's interesting about this is we've proven to ourselves in the videos on geometric series that if the common ratio, if the absolute value of the common ratio is less than one, then the series converges. And if the absolute value of r is greater than or equal to one, then the series diverges. And that made sense, we've proven it as well, but it also makes logical sense that look, if the absolute value of r is less than one, then each term here is going to go down by that common, or it's going to be multiplied by that common ratio, and it's going to decrease on and on and on and on and on. And so it makes sense that even though this is an infinite sum, it will converge to a finite value. Now, with that out of the way for review, let's tackle something a little bit more interesting. Let's say that we want to look at a, we want to figure out whether a series like this, so starting at n equals five to infinity, of let's say n to the 10th power, the numerator is growing quickly, n to the 10th power over n factorial. And factorial we know also grows very, very quickly, probably it actually grows much faster than even a high degree polynomial like this, or a high degree term like this."}, {"video_title": "Ratio test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so it makes sense that even though this is an infinite sum, it will converge to a finite value. Now, with that out of the way for review, let's tackle something a little bit more interesting. Let's say that we want to look at a, we want to figure out whether a series like this, so starting at n equals five to infinity, of let's say n to the 10th power, the numerator is growing quickly, n to the 10th power over n factorial. And factorial we know also grows very, very quickly, probably it actually grows much faster than even a high degree polynomial like this, or a high degree term like this. But how do we prove that it converges? We could definitely use the divergence test to show that this does not diverge, but how do we prove that this actually converges? And maybe we can use a little bit of our intuition here."}, {"video_title": "Ratio test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And factorial we know also grows very, very quickly, probably it actually grows much faster than even a high degree polynomial like this, or a high degree term like this. But how do we prove that it converges? We could definitely use the divergence test to show that this does not diverge, but how do we prove that this actually converges? And maybe we can use a little bit of our intuition here. Well, let's see if we can come up with a common ratio. So let's see if there's a common ratio here. So let's take the n plus one-th term, which is going to be n plus one to the 10th power over n plus one factorial, and divide that by the n-th term."}, {"video_title": "Ratio test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And maybe we can use a little bit of our intuition here. Well, let's see if we can come up with a common ratio. So let's see if there's a common ratio here. So let's take the n plus one-th term, which is going to be n plus one to the 10th power over n plus one factorial, and divide that by the n-th term. So let's divide that by n to the 10th over n factorial. Well, dividing by a fraction, or dividing by anything, is equivalent to multiplying by its reciprocal. So let's just multiply by the reciprocal."}, {"video_title": "Ratio test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's take the n plus one-th term, which is going to be n plus one to the 10th power over n plus one factorial, and divide that by the n-th term. So let's divide that by n to the 10th over n factorial. Well, dividing by a fraction, or dividing by anything, is equivalent to multiplying by its reciprocal. So let's just multiply by the reciprocal. So times n factorial over n to the 10th. Remember, all I'm trying to do is exactly what we did right over here. See if there's some type of a common ratio."}, {"video_title": "Ratio test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's just multiply by the reciprocal. So times n factorial over n to the 10th. Remember, all I'm trying to do is exactly what we did right over here. See if there's some type of a common ratio. Well, if we do a little algebra here, n plus one factorial, and factorial algebra is always fun, this is the same thing as n plus one times n factorial. Times n factorial. You're not used to seeing order of operations dealing with factorial, but this factorial only applies to this n right over here."}, {"video_title": "Ratio test Series AP Calculus BC Khan Academy.mp3", "Sentence": "See if there's some type of a common ratio. Well, if we do a little algebra here, n plus one factorial, and factorial algebra is always fun, this is the same thing as n plus one times n factorial. Times n factorial. You're not used to seeing order of operations dealing with factorial, but this factorial only applies to this n right over here. And why is that useful? Because this n factorial cancels with that n factorial, and we're left with n plus one to the 10th power over n plus one times n to the 10th. Times n to the 10th."}, {"video_title": "Ratio test Series AP Calculus BC Khan Academy.mp3", "Sentence": "You're not used to seeing order of operations dealing with factorial, but this factorial only applies to this n right over here. And why is that useful? Because this n factorial cancels with that n factorial, and we're left with n plus one to the 10th power over n plus one times n to the 10th. Times n to the 10th. So I know what you're thinking. Hey, wait, look, this isn't a fixed common ratio here. The ratio between consecutive terms here, when I took the n plus one term divided by the nth term, it's changing depending on what my n is."}, {"video_title": "Ratio test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Times n to the 10th. So I know what you're thinking. Hey, wait, look, this isn't a fixed common ratio here. The ratio between consecutive terms here, when I took the n plus one term divided by the nth term, it's changing depending on what my n is. This ratio, I guess you could say, is a function of n, so this doesn't seem too useful. But what if I were to say, okay, look, in any of these series, we really care about the behavior as our n's get really, really, really, really large, as the limit as our n's go to infinity. So what if we were to look at the behavior here, and if this is approaching some actual value as n approaches infinity, well, it would make conceptual sense that we could kind of think of that as the limit of our common ratio."}, {"video_title": "Ratio test Series AP Calculus BC Khan Academy.mp3", "Sentence": "The ratio between consecutive terms here, when I took the n plus one term divided by the nth term, it's changing depending on what my n is. This ratio, I guess you could say, is a function of n, so this doesn't seem too useful. But what if I were to say, okay, look, in any of these series, we really care about the behavior as our n's get really, really, really, really large, as the limit as our n's go to infinity. So what if we were to look at the behavior here, and if this is approaching some actual value as n approaches infinity, well, it would make conceptual sense that we could kind of think of that as the limit of our common ratio. So let's do that. So let's take the limit as n approaches infinity of this thing. So remember what we're doing here."}, {"video_title": "Ratio test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So what if we were to look at the behavior here, and if this is approaching some actual value as n approaches infinity, well, it would make conceptual sense that we could kind of think of that as the limit of our common ratio. So let's do that. So let's take the limit as n approaches infinity of this thing. So remember what we're doing here. This isn't just kind of some voodoo. Conceptually, so let me copy and paste this. Conceptually, we're just trying to think about, well, what is the common ratio approach?"}, {"video_title": "Ratio test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So remember what we're doing here. This isn't just kind of some voodoo. Conceptually, so let me copy and paste this. Conceptually, we're just trying to think about, well, what is the common ratio approach? What is the ratio between consecutive terms, the n plus one-th term and the n-th term, as n gets really, really, really, really, really large? And what we see here is we have this up here. We don't have to multiply it out."}, {"video_title": "Ratio test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Conceptually, we're just trying to think about, well, what is the common ratio approach? What is the ratio between consecutive terms, the n plus one-th term and the n-th term, as n gets really, really, really, really, really large? And what we see here is we have this up here. We don't have to multiply it out. This is going to be n to the 10th plus a bunch of other stuff. It's gonna be a 10th degree polynomial. And this down here, this is going to be, actually, this one, we can figure it out."}, {"video_title": "Ratio test Series AP Calculus BC Khan Academy.mp3", "Sentence": "We don't have to multiply it out. This is going to be n to the 10th plus a bunch of other stuff. It's gonna be a 10th degree polynomial. And this down here, this is going to be, actually, this one, we can figure it out. This is going to be n to the 11th plus n to the 10th power. So the limit as n approaches infinity, well, this denominator's gonna grow faster than that. You could divide the numerator and denominator by n to the 10th and all of, and everything, well, you could divide, actually, the numerator and denominator by n to the 11th, and everything up here is going to go to zero, and this is going to be one."}, {"video_title": "Ratio test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And this down here, this is going to be, actually, this one, we can figure it out. This is going to be n to the 11th plus n to the 10th power. So the limit as n approaches infinity, well, this denominator's gonna grow faster than that. You could divide the numerator and denominator by n to the 10th and all of, and everything, well, you could divide, actually, the numerator and denominator by n to the 11th, and everything up here is going to go to zero, and this is going to be one. And so the limit right over here is the limit, this limit is going to be equal to, this limit is going to be equal to zero. As n approaches infinity, one way, the ratio between consecutive terms approaches zero. So this seems, if we use the logic from our common ratio up here when we looked at geometric series, this is clearly not a geometric series, but it would say, well, look, as n gets really, really, really large, the ratio between consecutive terms gets smaller and smaller and smaller, hey, maybe we can make the same conclusion."}, {"video_title": "Ratio test Series AP Calculus BC Khan Academy.mp3", "Sentence": "You could divide the numerator and denominator by n to the 10th and all of, and everything, well, you could divide, actually, the numerator and denominator by n to the 11th, and everything up here is going to go to zero, and this is going to be one. And so the limit right over here is the limit, this limit is going to be equal to, this limit is going to be equal to zero. As n approaches infinity, one way, the ratio between consecutive terms approaches zero. So this seems, if we use the logic from our common ratio up here when we looked at geometric series, this is clearly not a geometric series, but it would say, well, look, as n gets really, really, really large, the ratio between consecutive terms gets smaller and smaller and smaller, hey, maybe we can make the same conclusion. Maybe we can make the same conclusion that therefore the series actually converges. The series actually converges. So, or n factorial converges."}, {"video_title": "Ratio test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So this seems, if we use the logic from our common ratio up here when we looked at geometric series, this is clearly not a geometric series, but it would say, well, look, as n gets really, really, really large, the ratio between consecutive terms gets smaller and smaller and smaller, hey, maybe we can make the same conclusion. Maybe we can make the same conclusion that therefore the series actually converges. The series actually converges. So, or n factorial converges. So can we make this statement? And the answer is yes, we can. And what allows us to do it is the ratio test."}, {"video_title": "Ratio test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So, or n factorial converges. So can we make this statement? And the answer is yes, we can. And what allows us to do it is the ratio test. Is the ratio test. So let me write that down. The ratio test."}, {"video_title": "Ratio test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And what allows us to do it is the ratio test. Is the ratio test. So let me write that down. The ratio test. And all the ratio test tells us is if we have an infinite series, if we have an infinite series, and I can say from n equals k to infinity, and if we say, if we take the limit as n approaches infinity, as n approaches infinity, of the absolute value of a to the n plus one over a to the n, and over here I didn't take the absolute value of that limit, but we absolutely could have taken the absolute value of that limit. And these were, all of these terms right over here were positive, so the absolute value would be the same thing as itself. If this approaches some limit, if this approaches some limit, so once again, this limit is what is going to be our ratio between consecutive terms as n gets larger and larger."}, {"video_title": "Ratio test Series AP Calculus BC Khan Academy.mp3", "Sentence": "The ratio test. And all the ratio test tells us is if we have an infinite series, if we have an infinite series, and I can say from n equals k to infinity, and if we say, if we take the limit as n approaches infinity, as n approaches infinity, of the absolute value of a to the n plus one over a to the n, and over here I didn't take the absolute value of that limit, but we absolutely could have taken the absolute value of that limit. And these were, all of these terms right over here were positive, so the absolute value would be the same thing as itself. If this approaches some limit, if this approaches some limit, so once again, this limit is what is going to be our ratio between consecutive terms as n gets larger and larger. The ratio tells us if, if L is less than one, which was the situation here, L is clearly less than one, then the series converges. Series converges. Some definitions would say they'll, it's absolutely convergent, it'll convert absolutely, the absolute value of the series converges, but then that also means that the series itself converges."}, {"video_title": "Ratio test Series AP Calculus BC Khan Academy.mp3", "Sentence": "If this approaches some limit, if this approaches some limit, so once again, this limit is what is going to be our ratio between consecutive terms as n gets larger and larger. The ratio tells us if, if L is less than one, which was the situation here, L is clearly less than one, then the series converges. Series converges. Some definitions would say they'll, it's absolutely convergent, it'll convert absolutely, the absolute value of the series converges, but then that also means that the series itself converges. If L is greater than one, series diverges. Series diverges. And if L is equal to one, it's inconclusive, we don't know."}, {"video_title": "Ratio test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Some definitions would say they'll, it's absolutely convergent, it'll convert absolutely, the absolute value of the series converges, but then that also means that the series itself converges. If L is greater than one, series diverges. Series diverges. And if L is equal to one, it's inconclusive, we don't know. We don't know. We would have to do some other test to prove whether it converges or diverges. So that's the essence of the ratio test."}, {"video_title": "Ratio test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And if L is equal to one, it's inconclusive, we don't know. We don't know. We would have to do some other test to prove whether it converges or diverges. So that's the essence of the ratio test. Find the absolute value of the ratio between consecutive terms, take the limit as that n approaches infinity. If that approaches an actual limit, and that limit is less than one, then the series converges. And it's really based on the same fundamental idea that we saw with the common ratio of geometric series."}, {"video_title": "Mean value theorem for integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So the slope between the two endpoints is gonna be your change in y, which is going to be your change in your function value, so f of b minus f of a over, over b minus a. And once again, we do this, we go into much more depth in this when we covered it the first time in differential calculus, but just to give you a visualization of it, because I think it's always handy, the mean value theorem that we learned in differential calculus just tells us, hey look, you know, if this is a, this is b, I've got my function doing something interesting, so this is f of a, this is f of b, so this quantity right over here, where you're taking the change in the value of our function divided by, so this right over here is f of b, f of b minus f of a is this change in the value of our function divided by the change in our x-axis, so it's our change in y over change in x, that gives us the slope, this right over here gives us the slope of this line, the slope of the line that connects, that connects these two points, that's this quantity, and this, the mean value theorem tells us that there's some c in between a and b where you're gonna have the same slope, so it might be at least one place, so it might be right over there where you have the exact same slope, there exists a c where the slope of the tangent line at that point is going to be the same, so this would be a c right over there, and we actually might have a couple of c's, that's another candidate c, there's at least one c where the slope of the tangent line is the same as the average slope across the interval, and once again, we have to assume that f is continuous and f is differentiable. Now, when you see this, it, something might, it might evoke some similarities with what we saw when we saw the, how we defined, I guess you could say, or the formula for the average value of a function. Remember, what we saw for the average value of a function, we said the average, the average value of a function is going to be equal to one over b minus a, notice, one over b minus a, you have a b minus a in the denominator here, times the definite integral from a to b of f of x dx. Now, this is interesting, because here we have a derivative, here we have an integral, but maybe we could connect these, maybe we could connect these two things. Well, one thing that might jump out at you is maybe we could rewrite, maybe we could rewrite this numerator right over here in this form somehow, and I encourage you to pause the video and see if you can, and I'll give you actually a quite huge hint, instead of it being an f of x here, what happens if there's an f prime of x there? So I encourage you to try to do that."}, {"video_title": "Mean value theorem for integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Remember, what we saw for the average value of a function, we said the average, the average value of a function is going to be equal to one over b minus a, notice, one over b minus a, you have a b minus a in the denominator here, times the definite integral from a to b of f of x dx. Now, this is interesting, because here we have a derivative, here we have an integral, but maybe we could connect these, maybe we could connect these two things. Well, one thing that might jump out at you is maybe we could rewrite, maybe we could rewrite this numerator right over here in this form somehow, and I encourage you to pause the video and see if you can, and I'll give you actually a quite huge hint, instead of it being an f of x here, what happens if there's an f prime of x there? So I encourage you to try to do that. So once again, this is, let me rewrite all of this. This is going to be equal to, this over here is the exact same thing as the definite integral from a to b of f prime of x dx. Think about it."}, {"video_title": "Mean value theorem for integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So I encourage you to try to do that. So once again, this is, let me rewrite all of this. This is going to be equal to, this over here is the exact same thing as the definite integral from a to b of f prime of x dx. Think about it. You're gonna take the antiderivative of f prime of x, which is going to be f of x, and you're going to evaluate it at b, f of b, and then from that, you're going to subtract it, evaluate it at a, minus f of a. These two things are identical, and then you can, of course, divide by b minus a. Now this is starting to get interesting."}, {"video_title": "Mean value theorem for integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Think about it. You're gonna take the antiderivative of f prime of x, which is going to be f of x, and you're going to evaluate it at b, f of b, and then from that, you're going to subtract it, evaluate it at a, minus f of a. These two things are identical, and then you can, of course, divide by b minus a. Now this is starting to get interesting. One way to think about it is there must be a c, if we, there must be a c that takes on the average value of, there must be a c that when you evaluate the derivative at c, it takes on the average value of the derivative. Or another way to think about it, another way to think about it, if we were to just write, if we were to just write, let me just say g of x is equal to f prime of x, then we get very close to what we have over here, because this right over here is going to be g of c. Remember, f prime of c is the same thing as g of c, is equal to one over, is equal to one over b minus a, one over b minus a. So there exists a c, where g of c is equal to one over b minus a, times the definite integral from a to b of g of x, g of x, dx."}, {"video_title": "Mean value theorem for integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Now this is starting to get interesting. One way to think about it is there must be a c, if we, there must be a c that takes on the average value of, there must be a c that when you evaluate the derivative at c, it takes on the average value of the derivative. Or another way to think about it, another way to think about it, if we were to just write, if we were to just write, let me just say g of x is equal to f prime of x, then we get very close to what we have over here, because this right over here is going to be g of c. Remember, f prime of c is the same thing as g of c, is equal to one over, is equal to one over b minus a, one over b minus a. So there exists a c, where g of c is equal to one over b minus a, times the definite integral from a to b of g of x, g of x, dx. F prime of x is the same thing as g of x. So another way of thinking about it, and this is actually another form of the mean value theorem, it's called the mean value theorem for integrals. Mean value theorem for integrals."}, {"video_title": "Mean value theorem for integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So there exists a c, where g of c is equal to one over b minus a, times the definite integral from a to b of g of x, g of x, dx. F prime of x is the same thing as g of x. So another way of thinking about it, and this is actually another form of the mean value theorem, it's called the mean value theorem for integrals. Mean value theorem for integrals. Let me, so this is the mean, I'll just write the acronym, mean value theorem for integrals, or integration, which essentially, and to give it a slightly more formal sense, is if you have some function g, so if g is, let me actually go down a little bit, which tells us that if g of x is continuous, continuous on this closed interval, going from a to b, then there exists, then there exists a c in this interval where g of c is equal to, what is this? This is the average value of our function. There exists a c where g of c is equal to the average value of your function over the interval."}, {"video_title": "Mean value theorem for integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Mean value theorem for integrals. Let me, so this is the mean, I'll just write the acronym, mean value theorem for integrals, or integration, which essentially, and to give it a slightly more formal sense, is if you have some function g, so if g is, let me actually go down a little bit, which tells us that if g of x is continuous, continuous on this closed interval, going from a to b, then there exists, then there exists a c in this interval where g of c is equal to, what is this? This is the average value of our function. There exists a c where g of c is equal to the average value of your function over the interval. This was our definition of the average value of a function. So anyway, this is just another way of saying, you might see some of the mean value theorem of integrals, and just to show you that it's really closely tied, it's using different notation, but it's usually, it's essentially the same exact idea as the mean value theorem you learned in differential calculus, but now you have different notation, and I guess you can have a slightly different interpretation. When we're thinking about it in differential calculus, we're thinking about having a point where the slope of the tangent line of the function at that point is the same as the average rate, so that's when we had our kind of differential mode, and we were kind of thinking in terms of slopes and slopes of tangent lines, and now when we're in integral mode, we're thinking much more in terms of average value, average value of the function."}, {"video_title": "Mean value theorem for integrals AP Calculus AB Khan Academy.mp3", "Sentence": "There exists a c where g of c is equal to the average value of your function over the interval. This was our definition of the average value of a function. So anyway, this is just another way of saying, you might see some of the mean value theorem of integrals, and just to show you that it's really closely tied, it's using different notation, but it's usually, it's essentially the same exact idea as the mean value theorem you learned in differential calculus, but now you have different notation, and I guess you can have a slightly different interpretation. When we're thinking about it in differential calculus, we're thinking about having a point where the slope of the tangent line of the function at that point is the same as the average rate, so that's when we had our kind of differential mode, and we were kind of thinking in terms of slopes and slopes of tangent lines, and now when we're in integral mode, we're thinking much more in terms of average value, average value of the function. So there's some c where g of c, there's some c where the function evaluated at that point is equal to the average value. So another way of thinking about it, if I were to draw, if I were to draw g of, if I were to draw g of x, if I were to draw g of x, so that's x, that is my y-axis, this is the graph of y is equal to g of x, which of course was the same thing as f prime of x, but we've, I guess, we've just rewritten it now to be more consistent with our average value formula, and we're talking about the interval from a to b. We've already seen how to calculate the average value."}, {"video_title": "Unbounded limits Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So right over here we have the graph of y is equal to one over x squared. And my question to you is, what is the limit of one over x squared as x approaches zero? Pause this video and see if you can figure that out. Well, when you try to figure it out, you immediately see something interesting happening at x equals zero. The closer we get to zero from the left, you take one over x squared, it just gets larger and larger and larger. It doesn't approach some finite value. It's unbounded, it has no bound."}, {"video_title": "Unbounded limits Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Well, when you try to figure it out, you immediately see something interesting happening at x equals zero. The closer we get to zero from the left, you take one over x squared, it just gets larger and larger and larger. It doesn't approach some finite value. It's unbounded, it has no bound. And the same thing is happening as we approach from the right. As we get values closer and closer to zero from the right, we get larger and larger values for one over x squared without bound. So terminology that folks will sometimes use where they're both going in the same direction but it's unbounded is they'll say this limit is unbounded."}, {"video_title": "Unbounded limits Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "It's unbounded, it has no bound. And the same thing is happening as we approach from the right. As we get values closer and closer to zero from the right, we get larger and larger values for one over x squared without bound. So terminology that folks will sometimes use where they're both going in the same direction but it's unbounded is they'll say this limit is unbounded. In some context, you might hear teachers say that this limit does not exist or and it definitely does not exist if you're thinking about approaching a finite value. In future videos, we'll start to introduce ideas of infinity and notations around limits and infinity where we can get a little bit more specific about what type of limit this is. But with that out of the way, let's look at another scenario."}, {"video_title": "Unbounded limits Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So terminology that folks will sometimes use where they're both going in the same direction but it's unbounded is they'll say this limit is unbounded. In some context, you might hear teachers say that this limit does not exist or and it definitely does not exist if you're thinking about approaching a finite value. In future videos, we'll start to introduce ideas of infinity and notations around limits and infinity where we can get a little bit more specific about what type of limit this is. But with that out of the way, let's look at another scenario. This right over here, you might recognize, is the graph of y is equal to one over x. So I'm gonna ask you the same question. Pause this video and think about what's the limit of one over x as x approaches zero."}, {"video_title": "Unbounded limits Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "But with that out of the way, let's look at another scenario. This right over here, you might recognize, is the graph of y is equal to one over x. So I'm gonna ask you the same question. Pause this video and think about what's the limit of one over x as x approaches zero. Pause this video and figure it out. All right, so here when we approach from the left, we get more and more and more negative values while when we approach from the right, we're getting more and more positive values. So in this situation where we're not getting unbounded in the same direction, the previous example, we were being unbounded in the positive direction but here from the left, we're getting unbounded in the negative direction while from the right, we're getting unbounded in the positive direction."}, {"video_title": "Unbounded limits Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Pause this video and think about what's the limit of one over x as x approaches zero. Pause this video and figure it out. All right, so here when we approach from the left, we get more and more and more negative values while when we approach from the right, we're getting more and more positive values. So in this situation where we're not getting unbounded in the same direction, the previous example, we were being unbounded in the positive direction but here from the left, we're getting unbounded in the negative direction while from the right, we're getting unbounded in the positive direction. And so when you're thinking about the limit as you approach a point, if it's not even approaching the same value or even the same direction, you would just clearly say that this limit does not exist. Does not exist. So this is a situation where you would not even say that this is an unbounded limit or that the limit is unbounded because you're going in two different directions when you approach from the right and when you approach from the left, you would just clearly say does not exist."}, {"video_title": "Worked examples Definite integral properties 1 AP Calculus AB Khan Academy.mp3", "Sentence": "Well, when you look at this, you actually don't even have to look at this graph over here because in general, if I have the definite integral of any function, f of x dx, from, let's say, a to the same value, from one value to the same value, this is always going to be equal to zero. We're going from three to three. We could be going from negative pi to negative pi. It's always going to be zero. One way to think about it, we're starting and stopping here at three, so we're not capturing any area. Let's do another one. So here, we want to find the definite integral from seven to four of f of x dx."}, {"video_title": "Worked examples Definite integral properties 1 AP Calculus AB Khan Academy.mp3", "Sentence": "It's always going to be zero. One way to think about it, we're starting and stopping here at three, so we're not capturing any area. Let's do another one. So here, we want to find the definite integral from seven to four of f of x dx. So we want to go from seven to four. So we want to go from seven to four. Now, you might be tempted to say, okay, well, look, the area between f of x and x is two, so maybe this thing is two."}, {"video_title": "Worked examples Definite integral properties 1 AP Calculus AB Khan Academy.mp3", "Sentence": "So here, we want to find the definite integral from seven to four of f of x dx. So we want to go from seven to four. So we want to go from seven to four. Now, you might be tempted to say, okay, well, look, the area between f of x and x is two, so maybe this thing is two. But the key realization is this area only applies when you have the lower bound as the lower bound and the higher value as the higher bound. So the integral from four to seven of f of x dx, this thing, this thing is equal to two. This thing is depicting that area right over there."}, {"video_title": "Worked examples Definite integral properties 1 AP Calculus AB Khan Academy.mp3", "Sentence": "Now, you might be tempted to say, okay, well, look, the area between f of x and x is two, so maybe this thing is two. But the key realization is this area only applies when you have the lower bound as the lower bound and the higher value as the higher bound. So the integral from four to seven of f of x dx, this thing, this thing is equal to two. This thing is depicting that area right over there. So what about this, where we've switched it? Instead of going from four to seven, we're going from seven to four. Well, the key realization is is if you switch the bounds, and this is a key definite integral property, that's going to give you the negative value."}, {"video_title": "Worked examples Definite integral properties 1 AP Calculus AB Khan Academy.mp3", "Sentence": "This thing is depicting that area right over there. So what about this, where we've switched it? Instead of going from four to seven, we're going from seven to four. Well, the key realization is is if you switch the bounds, and this is a key definite integral property, that's going to give you the negative value. So this is going to be equal to the negative of the integral from four to seven of f of x dx. And so this is going to be negative, and we just figured out the integral from four to seven of f of x dx. Well, now that is this area."}, {"video_title": "Worked examples Definite integral properties 1 AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the key realization is is if you switch the bounds, and this is a key definite integral property, that's going to give you the negative value. So this is going to be equal to the negative of the integral from four to seven of f of x dx. And so this is going to be negative, and we just figured out the integral from four to seven of f of x dx. Well, now that is this area. F of x is above the x-axis. It's a positive area. So it is going to be, so this thing right over here is going to evaluate to positive two, but we have that negative out front, so our original expression would evaluate to negative two."}, {"video_title": "Justification with the intermediate value theorem equation AP Calculus AB Khan Academy.mp3", "Sentence": "If so, write a justification. So in order to even use the intermediate value theorem, you have to be continuous over the interval that you care about, and this interval that we care about is from x equals negative one to one. And one over x is not continuous over that interval. It is not defined when x is equal to zero. And so we could say no, because, because g of x not defined, not defined, or I could say, let me just say not continuous. It's also not defined on every point of the interval, but let's say not continuous over the closed interval from negative one to one. And we could even put parentheses not defined, not defined at x is equal to zero."}, {"video_title": "Justification with the intermediate value theorem equation AP Calculus AB Khan Academy.mp3", "Sentence": "It is not defined when x is equal to zero. And so we could say no, because, because g of x not defined, not defined, or I could say, let me just say not continuous. It's also not defined on every point of the interval, but let's say not continuous over the closed interval from negative one to one. And we could even put parentheses not defined, not defined at x is equal to zero. All right, now let's start asking the second question. Can we use the intermediate value theorem to say that the equation g of x is equal to 3 4ths has a solution where one is less than or equal to x is less than or equal to two? If so, write a justification."}, {"video_title": "Justification with the intermediate value theorem equation AP Calculus AB Khan Academy.mp3", "Sentence": "And we could even put parentheses not defined, not defined at x is equal to zero. All right, now let's start asking the second question. Can we use the intermediate value theorem to say that the equation g of x is equal to 3 4ths has a solution where one is less than or equal to x is less than or equal to two? If so, write a justification. All right, so first let's look at the interval. If we're thinking about the interval from one to two, well, yeah, our function is going to be continuous over that interval. So we could say g of x is continuous, is continuous on the closed interval from one to two."}, {"video_title": "Justification with the intermediate value theorem equation AP Calculus AB Khan Academy.mp3", "Sentence": "If so, write a justification. All right, so first let's look at the interval. If we're thinking about the interval from one to two, well, yeah, our function is going to be continuous over that interval. So we could say g of x is continuous, is continuous on the closed interval from one to two. And if you wanted to put more justification there, you could say g defined, defined for all real numbers, real numbers, such that x does not equal zero, x does not equal zero. I could write g of x defined for all real numbers such that x does not equal to zero. And you could say rational functions like one over x are continuous, are continuous at all points in their domains, at all points in their domain."}, {"video_title": "Justification with the intermediate value theorem equation AP Calculus AB Khan Academy.mp3", "Sentence": "So we could say g of x is continuous, is continuous on the closed interval from one to two. And if you wanted to put more justification there, you could say g defined, defined for all real numbers, real numbers, such that x does not equal zero, x does not equal zero. I could write g of x defined for all real numbers such that x does not equal to zero. And you could say rational functions like one over x are continuous, are continuous at all points in their domains, at all points in their domain. That's going, really establishing that g of x is continuous on that interval. And then we wanna see what values does g take over, or what values does g take on at the endpoints, or actually these are the endpoints that we're looking at right over here. G of one is going to be equal to one over one is one."}, {"video_title": "Justification with the intermediate value theorem equation AP Calculus AB Khan Academy.mp3", "Sentence": "And you could say rational functions like one over x are continuous, are continuous at all points in their domains, at all points in their domain. That's going, really establishing that g of x is continuous on that interval. And then we wanna see what values does g take over, or what values does g take on at the endpoints, or actually these are the endpoints that we're looking at right over here. G of one is going to be equal to one over one is one. And g of two is going to be one over two is equal to one over two. So 3 4ths is between, is between g of one and g of two. So by the intermediate value theorem, there must be an x, there must be an x that is in the interval from, we're talking about the interval from one to two, one to two, such that, such that g of x is equal to 3 4ths."}, {"video_title": "Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And you have some information at your disposal. You know the spot on the ground that is directly below the hot air balloon. Let's say it took off from that point. It's just been going straight up ever since. And you know, you've measured it out, that you're 500 meters away from there. So you know that you are 500 meters away from that. And you're also able to measure the angle between the horizontal and the hot air balloon."}, {"video_title": "Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "It's just been going straight up ever since. And you know, you've measured it out, that you're 500 meters away from there. So you know that you are 500 meters away from that. And you're also able to measure the angle between the horizontal and the hot air balloon. You could do that with, I don't know, I'm not exactly a surveyor, but I guess a viewfinder or something like that. So you're able to, and I'm not sure if that's the right tool, but there are tools that you can measure the angles between the horizontal and something that's not on the horizontal. So you know that this angle right over here is pi over four radians, or 45 degrees."}, {"video_title": "Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And you're also able to measure the angle between the horizontal and the hot air balloon. You could do that with, I don't know, I'm not exactly a surveyor, but I guess a viewfinder or something like that. So you're able to, and I'm not sure if that's the right tool, but there are tools that you can measure the angles between the horizontal and something that's not on the horizontal. So you know that this angle right over here is pi over four radians, or 45 degrees. We're gonna keep it pi over four, because when you take derivatives of trig functions, you assume that you're dealing with radians. So right over here, this is pi over four radians. And you also are able to measure the rate at which this angle is changing."}, {"video_title": "Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So you know that this angle right over here is pi over four radians, or 45 degrees. We're gonna keep it pi over four, because when you take derivatives of trig functions, you assume that you're dealing with radians. So right over here, this is pi over four radians. And you also are able to measure the rate at which this angle is changing. So this is changing, changing, changing at 0.2 radians per minute. Now, my question to you, or the question that you're trying to figure out as you watch this hot air balloon is how fast is it rising right now? How fast is it rising just as the angle between the horizontal and kind of the line between you and the hot air balloon is pi over four radians, and that angle is changing at 0.2 radians per minute?"}, {"video_title": "Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And you also are able to measure the rate at which this angle is changing. So this is changing, changing, changing at 0.2 radians per minute. Now, my question to you, or the question that you're trying to figure out as you watch this hot air balloon is how fast is it rising right now? How fast is it rising just as the angle between the horizontal and kind of the line between you and the hot air balloon is pi over four radians, and that angle is changing at 0.2 radians per minute? So let's think about what we know and what we're trying to figure out. So we know a couple of things. We know that theta is equal to pi over four."}, {"video_title": "Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "How fast is it rising just as the angle between the horizontal and kind of the line between you and the hot air balloon is pi over four radians, and that angle is changing at 0.2 radians per minute? So let's think about what we know and what we're trying to figure out. So we know a couple of things. We know that theta is equal to pi over four. If we call theta the angle right over here. So this is theta. We also know the rate at which theta is changing."}, {"video_title": "Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We know that theta is equal to pi over four. If we call theta the angle right over here. So this is theta. We also know the rate at which theta is changing. We know d theta, let me do this in yellow. We know d theta, dt, is equal to 0.2 radians per minute. Now, what are we trying to figure out?"}, {"video_title": "Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We also know the rate at which theta is changing. We know d theta, let me do this in yellow. We know d theta, dt, is equal to 0.2 radians per minute. Now, what are we trying to figure out? Well, we're trying to figure out the rate at which the height of the balloon is changing. So if you call this distance right over here, this distance right over here, h, what we wanna figure out is dh dt. That's what we don't know."}, {"video_title": "Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Now, what are we trying to figure out? Well, we're trying to figure out the rate at which the height of the balloon is changing. So if you call this distance right over here, this distance right over here, h, what we wanna figure out is dh dt. That's what we don't know. So what we'd wanna come up with is a relationship between dh dt, d theta dt, and maybe theta if we need it. Or another way to think about it, if we can come up with a relationship between h and theta, then we could take the derivative with respect to t, and we'll probably get a relationship between all of this stuff. So what's the relationship between theta and h?"}, {"video_title": "Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "That's what we don't know. So what we'd wanna come up with is a relationship between dh dt, d theta dt, and maybe theta if we need it. Or another way to think about it, if we can come up with a relationship between h and theta, then we could take the derivative with respect to t, and we'll probably get a relationship between all of this stuff. So what's the relationship between theta and h? Well, it's a little bit of trigonometry. We know, we're trying to figure out h. We already know what this length is right over here. We know opposite over adjacent, that's the definition of tangent."}, {"video_title": "Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So what's the relationship between theta and h? Well, it's a little bit of trigonometry. We know, we're trying to figure out h. We already know what this length is right over here. We know opposite over adjacent, that's the definition of tangent. So let's write that down. So we know that the tangent, the tangent of theta, tangent of theta is equal to the opposite side, the opposite side is equal to h, is equal to h over the adjacent side, which we know is going to be a fixed 500, is going to be a fixed 500. So there you have a relationship between theta and h. And then to figure out a relationship between theta, d theta, and d h, or d theta, d t, the rate at which theta changes with respect to t, and the rate at which h changes with respect to t, we just have to take the derivative of this, of both sides of this, with respect to t implicitly."}, {"video_title": "Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We know opposite over adjacent, that's the definition of tangent. So let's write that down. So we know that the tangent, the tangent of theta, tangent of theta is equal to the opposite side, the opposite side is equal to h, is equal to h over the adjacent side, which we know is going to be a fixed 500, is going to be a fixed 500. So there you have a relationship between theta and h. And then to figure out a relationship between theta, d theta, and d h, or d theta, d t, the rate at which theta changes with respect to t, and the rate at which h changes with respect to t, we just have to take the derivative of this, of both sides of this, with respect to t implicitly. So let's do that. And actually, let me move over this h over 500 a little bit. So let me move it over a little bit so I have space to show the derivative operator."}, {"video_title": "Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So there you have a relationship between theta and h. And then to figure out a relationship between theta, d theta, and d h, or d theta, d t, the rate at which theta changes with respect to t, and the rate at which h changes with respect to t, we just have to take the derivative of this, of both sides of this, with respect to t implicitly. So let's do that. And actually, let me move over this h over 500 a little bit. So let me move it over a little bit so I have space to show the derivative operator. So let's write it like that. And now let's take the derivative with respect to t. So d d t. We're gonna take the derivative with respect to t on the left. We're gonna take the derivative with respect to t on the right."}, {"video_title": "Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let me move it over a little bit so I have space to show the derivative operator. So let's write it like that. And now let's take the derivative with respect to t. So d d t. We're gonna take the derivative with respect to t on the left. We're gonna take the derivative with respect to t on the right. So what's the derivative with respect to t of tangent of theta? Well, we're just going to apply the chain rule here. It's going to be first the derivative of the tangent of theta with respect to theta, which is just secant squared of theta, times the derivative of theta with respect to T. Times d theta dt."}, {"video_title": "Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We're gonna take the derivative with respect to t on the right. So what's the derivative with respect to t of tangent of theta? Well, we're just going to apply the chain rule here. It's going to be first the derivative of the tangent of theta with respect to theta, which is just secant squared of theta, times the derivative of theta with respect to T. Times d theta dt. Once again, this is just d, the derivative of the tangent, the tangent of something with respect to that something times the derivative of the something with respect to T. Derivative of tangent theta with respect to theta times the derivative of theta with respect to T gives us the derivative of tangent of theta with respect to t, which is what we'd want when we use this type of a derivative operator. We're taking the derivative with respect to t, not just the derivative with respect to theta. Fair enough, so this is the left-hand side."}, {"video_title": "Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "It's going to be first the derivative of the tangent of theta with respect to theta, which is just secant squared of theta, times the derivative of theta with respect to T. Times d theta dt. Once again, this is just d, the derivative of the tangent, the tangent of something with respect to that something times the derivative of the something with respect to T. Derivative of tangent theta with respect to theta times the derivative of theta with respect to T gives us the derivative of tangent of theta with respect to t, which is what we'd want when we use this type of a derivative operator. We're taking the derivative with respect to t, not just the derivative with respect to theta. Fair enough, so this is the left-hand side. And then the right-hand side becomes, well, it's just going to be one over 500 dh dt. So one over 500 dh dt. We're literally saying it's just one over 500 times the derivative of h with respect to t. But now we have our relationship."}, {"video_title": "Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Fair enough, so this is the left-hand side. And then the right-hand side becomes, well, it's just going to be one over 500 dh dt. So one over 500 dh dt. We're literally saying it's just one over 500 times the derivative of h with respect to t. But now we have our relationship. We have the relationship that we actually care about. We have a relationship between the rate at which the height is changing with respect to time and the rate at which the angle is changing with respect to time and our angle at any moment. So we can just take these values up here, throw it in here, and then solve for the unknown."}, {"video_title": "Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We're literally saying it's just one over 500 times the derivative of h with respect to t. But now we have our relationship. We have the relationship that we actually care about. We have a relationship between the rate at which the height is changing with respect to time and the rate at which the angle is changing with respect to time and our angle at any moment. So we can just take these values up here, throw it in here, and then solve for the unknown. So let's do that, let's do that right over here. So we get secant squared of theta. So we get secant squared, right now our theta is pi over four, secant squared of pi over four."}, {"video_title": "Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So we can just take these values up here, throw it in here, and then solve for the unknown. So let's do that, let's do that right over here. So we get secant squared of theta. So we get secant squared, right now our theta is pi over four, secant squared of pi over four. Let me write those colors in to show you that I'm putting these values in. Secant squared of pi over four. Times d theta dt, well, that is just 0.2."}, {"video_title": "Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So we get secant squared, right now our theta is pi over four, secant squared of pi over four. Let me write those colors in to show you that I'm putting these values in. Secant squared of pi over four. Times d theta dt, well, that is just 0.2. So times 0.2. And then this is going to be equal to, this is going to be equal to one over 500. And we want to make sure, since this is in radians per minute, we're going to get meters per, and this is meters right over here, we're going to get meters per minute right over here."}, {"video_title": "Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Times d theta dt, well, that is just 0.2. So times 0.2. And then this is going to be equal to, this is going to be equal to one over 500. And we want to make sure, since this is in radians per minute, we're going to get meters per, and this is meters right over here, we're going to get meters per minute right over here. We just want to make sure all of our, we know what our units are doing. I haven't written the units here to save some space. But we get one over 500 times dh dt."}, {"video_title": "Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And we want to make sure, since this is in radians per minute, we're going to get meters per, and this is meters right over here, we're going to get meters per minute right over here. We just want to make sure all of our, we know what our units are doing. I haven't written the units here to save some space. But we get one over 500 times dh dt. So if we want to solve for dh dt, you can multiply both sides by 500 and you get the rate at which our height is changing is equal to 500 times, let's see, secant squared of pi over four, that is one over cosine squared of pi over four. Cosine of pi over four, let me write this over here. Cosine of pi over four is square root of two over two."}, {"video_title": "Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "But we get one over 500 times dh dt. So if we want to solve for dh dt, you can multiply both sides by 500 and you get the rate at which our height is changing is equal to 500 times, let's see, secant squared of pi over four, that is one over cosine squared of pi over four. Cosine of pi over four, let me write this over here. Cosine of pi over four is square root of two over two. Cosine squared of pi over four is going to be equal to two over two over four, which is equal to 1 1\u20442. And so secant squared of pi over four is just one over that, is equal to two. So this is going to be equal to, let me rewrite this instead of, so the secant squared of pi over four, let me erase this right over here."}, {"video_title": "Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Cosine of pi over four is square root of two over two. Cosine squared of pi over four is going to be equal to two over two over four, which is equal to 1 1\u20442. And so secant squared of pi over four is just one over that, is equal to two. So this is going to be equal to, let me rewrite this instead of, so the secant squared of pi over four, let me erase this right over here. Secant squared of pi over four, all of this business right over here, simplifies to two, so times two, times 0.2, times 0.2. So what is this going to be? This is going to be 500 times 0.4."}, {"video_title": "Related rates balloon Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be equal to, let me rewrite this instead of, so the secant squared of pi over four, let me erase this right over here. Secant squared of pi over four, all of this business right over here, simplifies to two, so times two, times 0.2, times 0.2. So what is this going to be? This is going to be 500 times 0.4. So this is equal to 500 times, let me write a dot instead, times 0.4, which is equal to, let me make sure I get this right. This would be with two zeros, and one behind the decimal, yep, there you go. It would be 200."}, {"video_title": "Worked example Differentiating related functions AP Calculus AB Khan Academy.mp3", "Sentence": "The differentiable functions x and y are related by the following equation. The sine of x plus cosine of y is equal to square root of two. They also tell us that the derivative of x with respect to t is equal to five. They also ask us, find the derivative of y with respect to t when y is equal to pi over four and zero is less than x is less than pi over two. So given that they are telling us the derivative of x with respect to t and we wanna find the derivative of y with respect to t, it's a safe assumption that both x and y are functions of t. So you could even rewrite this equation right over here. You could rewrite it as sine of x, which is a function of t, plus cosine of y, which is a function of t, is equal to square root of two. Now it might confuse you a little bit."}, {"video_title": "Worked example Differentiating related functions AP Calculus AB Khan Academy.mp3", "Sentence": "They also ask us, find the derivative of y with respect to t when y is equal to pi over four and zero is less than x is less than pi over two. So given that they are telling us the derivative of x with respect to t and we wanna find the derivative of y with respect to t, it's a safe assumption that both x and y are functions of t. So you could even rewrite this equation right over here. You could rewrite it as sine of x, which is a function of t, plus cosine of y, which is a function of t, is equal to square root of two. Now it might confuse you a little bit. You're not used to seeing x as a function of a third variable or y as a function of something other than x. But remember, x and y are just variables. This could be f of t and this could be g of t instead of x of t and y of t, and that might feel a little bit more natural to you."}, {"video_title": "Worked example Differentiating related functions AP Calculus AB Khan Academy.mp3", "Sentence": "Now it might confuse you a little bit. You're not used to seeing x as a function of a third variable or y as a function of something other than x. But remember, x and y are just variables. This could be f of t and this could be g of t instead of x of t and y of t, and that might feel a little bit more natural to you. But needless to say, if we wanna find dy dt, what we wanna do is take the derivative with respect to t of both sides of this equation. So let's do that. So we're gonna do it on the left-hand side."}, {"video_title": "Worked example Differentiating related functions AP Calculus AB Khan Academy.mp3", "Sentence": "This could be f of t and this could be g of t instead of x of t and y of t, and that might feel a little bit more natural to you. But needless to say, if we wanna find dy dt, what we wanna do is take the derivative with respect to t of both sides of this equation. So let's do that. So we're gonna do it on the left-hand side. So it's gonna be, we're gonna take that with respect to t, derivative of that with respect to t. We're gonna take the derivative of that with respect to t. And then we're gonna take the derivative of the right-hand side, this constant, with respect to t. So let's think about each of these things. So what is this, let me do this in a new color, the stuff that I'm doing in this aqua color right over here. How could I write that?"}, {"video_title": "Worked example Differentiating related functions AP Calculus AB Khan Academy.mp3", "Sentence": "So we're gonna do it on the left-hand side. So it's gonna be, we're gonna take that with respect to t, derivative of that with respect to t. We're gonna take the derivative of that with respect to t. And then we're gonna take the derivative of the right-hand side, this constant, with respect to t. So let's think about each of these things. So what is this, let me do this in a new color, the stuff that I'm doing in this aqua color right over here. How could I write that? So I'm taking the derivative with respect to t. I have sine of something, which is itself a function of t. So I would just apply the chain rule here. I'm first going to take the derivative with respect to x of sine of x. I could write sine of x of t, but I'll just revert back to the sine of x here for simplicity. And then I would then multiply that times the derivative of the inside, you could say, with respect to t, times the derivative of x with respect to t. This might be a little bit counterintuitive to how you've applied the chain rule before when we only dealt with x's and y's, but all that's happening, I'm taking the derivative of the outside of the sine of something with respect to the something, in this case it is x, and then I'm taking the derivative of the something, in this case x, with respect to t. Well, we can do the same thing here for this second term here."}, {"video_title": "Worked example Differentiating related functions AP Calculus AB Khan Academy.mp3", "Sentence": "How could I write that? So I'm taking the derivative with respect to t. I have sine of something, which is itself a function of t. So I would just apply the chain rule here. I'm first going to take the derivative with respect to x of sine of x. I could write sine of x of t, but I'll just revert back to the sine of x here for simplicity. And then I would then multiply that times the derivative of the inside, you could say, with respect to t, times the derivative of x with respect to t. This might be a little bit counterintuitive to how you've applied the chain rule before when we only dealt with x's and y's, but all that's happening, I'm taking the derivative of the outside of the sine of something with respect to the something, in this case it is x, and then I'm taking the derivative of the something, in this case x, with respect to t. Well, we can do the same thing here for this second term here. So I'd wanna take the derivative with respect to y of, I guess you could say the outside, of cosine of y, and then I would multiply that times the derivative of y with respect to t. And then all of that is going to be equal to what? Well, the derivative with respect to t of a constant, square root of two is a constant, it's not gonna change as t changes, so its derivative, its rate of change is zero. All right, so now we just have to figure out all of these things."}, {"video_title": "Worked example Differentiating related functions AP Calculus AB Khan Academy.mp3", "Sentence": "And then I would then multiply that times the derivative of the inside, you could say, with respect to t, times the derivative of x with respect to t. This might be a little bit counterintuitive to how you've applied the chain rule before when we only dealt with x's and y's, but all that's happening, I'm taking the derivative of the outside of the sine of something with respect to the something, in this case it is x, and then I'm taking the derivative of the something, in this case x, with respect to t. Well, we can do the same thing here for this second term here. So I'd wanna take the derivative with respect to y of, I guess you could say the outside, of cosine of y, and then I would multiply that times the derivative of y with respect to t. And then all of that is going to be equal to what? Well, the derivative with respect to t of a constant, square root of two is a constant, it's not gonna change as t changes, so its derivative, its rate of change is zero. All right, so now we just have to figure out all of these things. So first of all, the derivative with respect to x of sine of x is cosine of x times derivative of x with respect to t, I'll just write that out here, derivative of x with respect to t. And then we're going to have, it's gonna be a plus here, the derivative of y with respect to t, so plus the derivative of y with respect to t, I'm just swapping the order here so that this goes out front. Now what's the derivative of cosine of y with respect to y? Well, that is negative sine of y."}, {"video_title": "Worked example Differentiating related functions AP Calculus AB Khan Academy.mp3", "Sentence": "All right, so now we just have to figure out all of these things. So first of all, the derivative with respect to x of sine of x is cosine of x times derivative of x with respect to t, I'll just write that out here, derivative of x with respect to t. And then we're going to have, it's gonna be a plus here, the derivative of y with respect to t, so plus the derivative of y with respect to t, I'm just swapping the order here so that this goes out front. Now what's the derivative of cosine of y with respect to y? Well, that is negative sine of y. And so, actually let me just put a sine of y here, and then I wanna have a negative. So let me erase this and put a negative there. And that is all going to be equal to zero."}, {"video_title": "Worked example Differentiating related functions AP Calculus AB Khan Academy.mp3", "Sentence": "Well, that is negative sine of y. And so, actually let me just put a sine of y here, and then I wanna have a negative. So let me erase this and put a negative there. And that is all going to be equal to zero. And so, what can we figure out now? They've told us that the derivative of x with respect to t is equal to five. They tell us that right over here."}, {"video_title": "Worked example Differentiating related functions AP Calculus AB Khan Academy.mp3", "Sentence": "And that is all going to be equal to zero. And so, what can we figure out now? They've told us that the derivative of x with respect to t is equal to five. They tell us that right over here. So this is equal to five. We wanna find the derivative of y with respect to t. They tell us what y is. Y is pi over four."}, {"video_title": "Worked example Differentiating related functions AP Calculus AB Khan Academy.mp3", "Sentence": "They tell us that right over here. So this is equal to five. We wanna find the derivative of y with respect to t. They tell us what y is. Y is pi over four. This, y is pi over four, so we know this is pi over four. And let's see, we have to figure out what, we still have two unknowns here. We don't know what x is, and we don't know what the derivative of y with respect to t is."}, {"video_title": "Worked example Differentiating related functions AP Calculus AB Khan Academy.mp3", "Sentence": "Y is pi over four. This, y is pi over four, so we know this is pi over four. And let's see, we have to figure out what, we still have two unknowns here. We don't know what x is, and we don't know what the derivative of y with respect to t is. And this is what we need to figure out. So what would x be? What would x be when y is pi over four?"}, {"video_title": "Worked example Differentiating related functions AP Calculus AB Khan Academy.mp3", "Sentence": "We don't know what x is, and we don't know what the derivative of y with respect to t is. And this is what we need to figure out. So what would x be? What would x be when y is pi over four? Well, to figure that out, we can go back to this original equation right over here. So when y is pi over four, you get, let me write it down, sine of x plus cosine of pi over four is equal to square root of two. Cosine of pi over four, we revert to our unit, or we think about our unit circle."}, {"video_title": "Worked example Differentiating related functions AP Calculus AB Khan Academy.mp3", "Sentence": "What would x be when y is pi over four? Well, to figure that out, we can go back to this original equation right over here. So when y is pi over four, you get, let me write it down, sine of x plus cosine of pi over four is equal to square root of two. Cosine of pi over four, we revert to our unit, or we think about our unit circle. We're in the first quadrant. If you think in degrees, it's a 45 degree angle. That's going to be square root of two over two."}, {"video_title": "Worked example Differentiating related functions AP Calculus AB Khan Academy.mp3", "Sentence": "Cosine of pi over four, we revert to our unit, or we think about our unit circle. We're in the first quadrant. If you think in degrees, it's a 45 degree angle. That's going to be square root of two over two. And so we can subtract square root of two over two from both sides, which is going to give us sine of x is equal to, well, if you take square root of two over two from square root of two, you're taking half of it away, so you're gonna have half of it less. So square root of two over two. And so what x value, when I take the sine of it, and remember, where the angle, if we're thinking of the unit circle, it's going to be in that first quadrant."}, {"video_title": "Worked example Differentiating related functions AP Calculus AB Khan Academy.mp3", "Sentence": "That's going to be square root of two over two. And so we can subtract square root of two over two from both sides, which is going to give us sine of x is equal to, well, if you take square root of two over two from square root of two, you're taking half of it away, so you're gonna have half of it less. So square root of two over two. And so what x value, when I take the sine of it, and remember, where the angle, if we're thinking of the unit circle, it's going to be in that first quadrant. X is an angle in this case right over here. Well, that's going to be, once again, pi over four. So this tells us that x is equal to pi over four when y is equal to pi over four."}, {"video_title": "Worked example Differentiating related functions AP Calculus AB Khan Academy.mp3", "Sentence": "And so what x value, when I take the sine of it, and remember, where the angle, if we're thinking of the unit circle, it's going to be in that first quadrant. X is an angle in this case right over here. Well, that's going to be, once again, pi over four. So this tells us that x is equal to pi over four when y is equal to pi over four. And so we know that this is pi over four as well. So let me just rewrite this because it's getting a little bit messy. So we know that five times cosine of pi over four minus dy dt, the derivative of y with respect to t, which is what we want to figure out, times sine of pi over four is equal to zero."}, {"video_title": "Worked example Differentiating related functions AP Calculus AB Khan Academy.mp3", "Sentence": "So this tells us that x is equal to pi over four when y is equal to pi over four. And so we know that this is pi over four as well. So let me just rewrite this because it's getting a little bit messy. So we know that five times cosine of pi over four minus dy dt, the derivative of y with respect to t, which is what we want to figure out, times sine of pi over four is equal to zero. Is equal to zero, and let me put some parentheses here just to clarify things a little bit. All right, so let's see. Now it's just a little bit of algebra."}, {"video_title": "Worked example Differentiating related functions AP Calculus AB Khan Academy.mp3", "Sentence": "So we know that five times cosine of pi over four minus dy dt, the derivative of y with respect to t, which is what we want to figure out, times sine of pi over four is equal to zero. Is equal to zero, and let me put some parentheses here just to clarify things a little bit. All right, so let's see. Now it's just a little bit of algebra. Cosine of pi over four, we already know, is square root of two over two. Sine of pi over four is also square root of two over two. And let's see, what if we divide both sides of this equation by square root of two over two?"}, {"video_title": "Worked example Differentiating related functions AP Calculus AB Khan Academy.mp3", "Sentence": "Now it's just a little bit of algebra. Cosine of pi over four, we already know, is square root of two over two. Sine of pi over four is also square root of two over two. And let's see, what if we divide both sides of this equation by square root of two over two? Well, what's that going to give us? Well, then, square root of two over two divided by square root of two over, square root of two over two divided by square root of two over two is going to be one. Square root of two over two divided by square root of two over two is going to be one."}, {"video_title": "Worked example Differentiating related functions AP Calculus AB Khan Academy.mp3", "Sentence": "And let's see, what if we divide both sides of this equation by square root of two over two? Well, what's that going to give us? Well, then, square root of two over two divided by square root of two over, square root of two over two divided by square root of two over two is going to be one. Square root of two over two divided by square root of two over two is going to be one. And then zero divided by square root of two over two is just still going to be zero. And so this whole thing simplifies to five times one, which is just five, minus the derivative of y with respect to t is equal to zero. And so there you have it."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "And in this video in particular, we will explore modeling population. Modeling population. We're actually gonna go into some depth on this eventually, but here we're gonna start with simpler models. And we'll see, we will stumble on, using the logic of differential equations, things that you might have seen in your algebra or your pre-calculus class. So on some level, what we're going to do here is going to be review, but we're gonna get there using the power of modeling with differential equations. So let's just define some variables. Let's say that P is equal to our population."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "And we'll see, we will stumble on, using the logic of differential equations, things that you might have seen in your algebra or your pre-calculus class. So on some level, what we're going to do here is going to be review, but we're gonna get there using the power of modeling with differential equations. So let's just define some variables. Let's say that P is equal to our population. And let's say that T is, let's say that T is equal to the time that has passed in days. In days. It could have been years or months."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "Let's say that P is equal to our population. And let's say that T is, let's say that T is equal to the time that has passed in days. In days. It could have been years or months. But let's say we're doing the population of insects that reproduce quite quickly. So days seem like a nice time span to care about. Now, what would be a reasonable model?"}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "It could have been years or months. But let's say we're doing the population of insects that reproduce quite quickly. So days seem like a nice time span to care about. Now, what would be a reasonable model? Well, we could say that the rate of change, the rate of change of our population with respect to time, with respect to time, is, well, a reasonable thing to say is that it's going to be proportional to the actual population. The actual population. Why is that reasonable?"}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "Now, what would be a reasonable model? Well, we could say that the rate of change, the rate of change of our population with respect to time, with respect to time, is, well, a reasonable thing to say is that it's going to be proportional to the actual population. The actual population. Why is that reasonable? Well, the larger the population, the larger the rate at any given time. If you have a thousand people, the rate at which they're reproducing is going to be more, or a thousand insects, is going to be more insects per second or per day or per year than if you only have 10 insects. So it makes sense that the rate of growth of your population with respect to time is going to be proportional to your population."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "Why is that reasonable? Well, the larger the population, the larger the rate at any given time. If you have a thousand people, the rate at which they're reproducing is going to be more, or a thousand insects, is going to be more insects per second or per day or per year than if you only have 10 insects. So it makes sense that the rate of growth of your population with respect to time is going to be proportional to your population. And so, you know, sometimes you think of differential equations as these daunting, complex things, but notice, we've just been able to express a reasonably not-so-complicated idea. The rate of change of population is going to be proportional to the population. And now, once we've expressed that, we can actually try to solve this differential equation, find a general solution, and then we can try to put some initial conditions on there or some states of the population that we know to actually solve for the constants to find a particular solution."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "So it makes sense that the rate of growth of your population with respect to time is going to be proportional to your population. And so, you know, sometimes you think of differential equations as these daunting, complex things, but notice, we've just been able to express a reasonably not-so-complicated idea. The rate of change of population is going to be proportional to the population. And now, once we've expressed that, we can actually try to solve this differential equation, find a general solution, and then we can try to put some initial conditions on there or some states of the population that we know to actually solve for the constants to find a particular solution. So how do we do that? And I encourage you to pause the video at any time and see if you can solve this differential equation. So assuming you at least maybe have had an attempt at it."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "And now, once we've expressed that, we can actually try to solve this differential equation, find a general solution, and then we can try to put some initial conditions on there or some states of the population that we know to actually solve for the constants to find a particular solution. So how do we do that? And I encourage you to pause the video at any time and see if you can solve this differential equation. So assuming you at least maybe have had an attempt at it. And you might immediately recognize that this is a separable differential equation. And in separable differential equations, we want one variable and all the differentials involving that variable on one side, and the other variable and all the differentials involving the other variable on the other side, and then we can integrate both sides. And once again, dP, the derivative of P with respect to t, this isn't quite a fraction."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "So assuming you at least maybe have had an attempt at it. And you might immediately recognize that this is a separable differential equation. And in separable differential equations, we want one variable and all the differentials involving that variable on one side, and the other variable and all the differentials involving the other variable on the other side, and then we can integrate both sides. And once again, dP, the derivative of P with respect to t, this isn't quite a fraction. This is the limit as our change in P over change in time. This is our instantaneous change. But for the sake of separable differential equations or differential equations in general, you can treat this derivative in Leibniz notations like fractions, and you can treat these differentials like quantities because we will eventually integrate them."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "And once again, dP, the derivative of P with respect to t, this isn't quite a fraction. This is the limit as our change in P over change in time. This is our instantaneous change. But for the sake of separable differential equations or differential equations in general, you can treat this derivative in Leibniz notations like fractions, and you can treat these differentials like quantities because we will eventually integrate them. So let's do that. So we want to put all the P's and dP's on one side, and all the things that involve t or that I guess don't involve P on the other side. So we could divide both sides by P. We could divide both sides by P. And so we'll have one over P. You have one over P here, and then those will cancel."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "But for the sake of separable differential equations or differential equations in general, you can treat this derivative in Leibniz notations like fractions, and you can treat these differentials like quantities because we will eventually integrate them. So let's do that. So we want to put all the P's and dP's on one side, and all the things that involve t or that I guess don't involve P on the other side. So we could divide both sides by P. We could divide both sides by P. And so we'll have one over P. You have one over P here, and then those will cancel. And then you can multiply both sides times dt. We could multiply both sides times dt. Once again, treating the differential like a quantity, which it really isn't a quantity."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "So we could divide both sides by P. We could divide both sides by P. And so we'll have one over P. You have one over P here, and then those will cancel. And then you can multiply both sides times dt. We could multiply both sides times dt. Once again, treating the differential like a quantity, which it really isn't a quantity. You really have to view this as a limit of that change in P over change in time. The limit as we get smaller and smaller and smaller changes in time. But once again, for the sake of this, we can do this."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "Once again, treating the differential like a quantity, which it really isn't a quantity. You really have to view this as a limit of that change in P over change in time. The limit as we get smaller and smaller and smaller changes in time. But once again, for the sake of this, we can do this. And when we do that, we would be left with one over P dP is equal to K dt. Now we can integrate both sides. Because this was a separable differential equation, we were able to completely separate the P's and dP's from the things involving T's, or I guess the things that aren't involving P's."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "But once again, for the sake of this, we can do this. And when we do that, we would be left with one over P dP is equal to K dt. Now we can integrate both sides. Because this was a separable differential equation, we were able to completely separate the P's and dP's from the things involving T's, or I guess the things that aren't involving P's. And then if we integrate this side, we would get the natural log of the absolute value of our population. And we could say plus some constant if we want, but we're gonna get a constant on this side as well. So we could just say that's going to be equal to K times T plus some constant."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "Because this was a separable differential equation, we were able to completely separate the P's and dP's from the things involving T's, or I guess the things that aren't involving P's. And then if we integrate this side, we would get the natural log of the absolute value of our population. And we could say plus some constant if we want, but we're gonna get a constant on this side as well. So we could just say that's going to be equal to K times T plus some constant. I'll just call that C1. And once again, I could have put a plus C2 here, but I could have then subtracted the constant from both sides, and I would just get the constant on the right-hand side. Now, how can I solve for P?"}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "So we could just say that's going to be equal to K times T plus some constant. I'll just call that C1. And once again, I could have put a plus C2 here, but I could have then subtracted the constant from both sides, and I would just get the constant on the right-hand side. Now, how can I solve for P? Well, the natural log of the absolute value of P is equal to this thing right over here. That means, that's the same thing, that means that the absolute value of P is equal to E to all of this business. E to the, let me do the same colors, KT plus C1."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "Now, how can I solve for P? Well, the natural log of the absolute value of P is equal to this thing right over here. That means, that's the same thing, that means that the absolute value of P is equal to E to all of this business. E to the, let me do the same colors, KT plus C1. Now, this right over here is the same thing, just using our exponent properties, this is the same thing as E to the KT, AE to the K times T times E to the C1. Now, this is just E to some constant, so we could just call this, let's just call that the constant C. So, this is all simplified to CE, CE to the KT, to the KT. And if we assume our population at any given time is positive, then we could get rid of this absolute value sign."}, {"video_title": "Modeling population with simple differential equation Khan Academy.mp3", "Sentence": "E to the, let me do the same colors, KT plus C1. Now, this right over here is the same thing, just using our exponent properties, this is the same thing as E to the KT, AE to the K times T times E to the C1. Now, this is just E to some constant, so we could just call this, let's just call that the constant C. So, this is all simplified to CE, CE to the KT, to the KT. And if we assume our population at any given time is positive, then we could get rid of this absolute value sign. Now, we have a general solution to this, frankly, fairly general differential equation, we just said proportional, we haven't given what the proportionality constant is, but we could say if we assume positive population, that the population is going to be equal to some constant C times E to the KT power. And the reason why I said that you've seen this before is this is just an exponential function, and it's very likely that in algebra or in pre-calculus class, you have modeled things with exponential functions, and my guess is that you've modeled things with populated, modeled things like population. The reason why this is interesting is you now see where this is coming from, the underlying logic that's just driven by the actual differential equation, the rate of change with respect to time of the population, well, maybe it's just proportional to population."}, {"video_title": "Justification with the mean value theorem equation AP Calculus AB Khan Academy.mp3", "Sentence": "Can we use the mean value theorem to say that the equation g prime of x is equal to 1 1 2 has a solution where negative one is less than x is less than two? If so, write a justification. All right, pause this video and see if you can figure that out. So the key to using the mean value theorem to even before you even think about using it, you have to make sure that you are continuous over the closed interval and differentiable over the open interval. So this is the open interval here and then the closed interval would include the endpoints. But you might immediately realize that both of these intervals contain x equals zero and at x equals zero, the function is undefined. And if it's undefined there, well, it's not going to be continuous or differentiable at that point."}, {"video_title": "Justification with the mean value theorem equation AP Calculus AB Khan Academy.mp3", "Sentence": "So the key to using the mean value theorem to even before you even think about using it, you have to make sure that you are continuous over the closed interval and differentiable over the open interval. So this is the open interval here and then the closed interval would include the endpoints. But you might immediately realize that both of these intervals contain x equals zero and at x equals zero, the function is undefined. And if it's undefined there, well, it's not going to be continuous or differentiable at that point. And so no, not continuous or differentiable or differentiable over the interval. Over the interval. All right, let's do the second part."}, {"video_title": "Justification with the mean value theorem equation AP Calculus AB Khan Academy.mp3", "Sentence": "And if it's undefined there, well, it's not going to be continuous or differentiable at that point. And so no, not continuous or differentiable or differentiable over the interval. Over the interval. All right, let's do the second part. Can we use the mean value theorem to say that there is a value c such that g prime of c is equal to negative 1 1 2 and one is less than c is less than two? If so, write a justification. So pause the video again."}, {"video_title": "Justification with the mean value theorem equation AP Calculus AB Khan Academy.mp3", "Sentence": "All right, let's do the second part. Can we use the mean value theorem to say that there is a value c such that g prime of c is equal to negative 1 1 2 and one is less than c is less than two? If so, write a justification. So pause the video again. All right, so in this situation, between one and two on both the open and the closed intervals, well, this is a rational function and a rational function is going to be continuous and differentiable at every point in its domain and its domain completely contains this open and closed interval. Or another way to think about it, every point on this open interval and on the closed interval is in the domain. So we can write g of x is a rational function which lets us know that it is continuous and differentiable at every point in its domain."}, {"video_title": "Justification with the mean value theorem equation AP Calculus AB Khan Academy.mp3", "Sentence": "So pause the video again. All right, so in this situation, between one and two on both the open and the closed intervals, well, this is a rational function and a rational function is going to be continuous and differentiable at every point in its domain and its domain completely contains this open and closed interval. Or another way to think about it, every point on this open interval and on the closed interval is in the domain. So we can write g of x is a rational function which lets us know that it is continuous and differentiable at every point in its domain. At every point in its domain. The closed interval from one to two is in domain. And so now let's see what the average rate of change is from one to two."}, {"video_title": "Justification with the mean value theorem equation AP Calculus AB Khan Academy.mp3", "Sentence": "So we can write g of x is a rational function which lets us know that it is continuous and differentiable at every point in its domain. At every point in its domain. The closed interval from one to two is in domain. And so now let's see what the average rate of change is from one to two. And so we get g of two minus g of one over two minus one is equal to 1 1\u20442 minus one over one, which is equal to negative 1 1\u20442. Therefore, therefore, by the mean value theorem, there must be a c where one is less than c is less than two and g prime of c is equal to the average rate of change between the endpoints, negative 1 1\u20442. And we're done."}, {"video_title": "Worked example Derivatives of sin(x) and cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And at first it could look intimidating. We have a sine of x here, we have a cosine of x, we have this crazy expression here, we have a pi over a cube root of x, we're squaring the whole thing, and at first it might seem intimidating. But as we'll see in this video, we can actually do this with the tools already in our toolkit, using our existing derivative properties, using what we know about the power rule, which tells us that the derivative with respect to x of x to the n is equal to n times x to the n minus one. We've seen that multiple times. We also need to use the fact that the derivative of cosine of x is equal to negative sine of x. And the other way around, the derivative with respect to x of sine of x is equal to positive cosine of x. So using just that, we can actually evaluate this, or evaluate g prime of x."}, {"video_title": "Worked example Derivatives of sin(x) and cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "We've seen that multiple times. We also need to use the fact that the derivative of cosine of x is equal to negative sine of x. And the other way around, the derivative with respect to x of sine of x is equal to positive cosine of x. So using just that, we can actually evaluate this, or evaluate g prime of x. So pause the video and see if you can do it. So probably the most intimidating part of this, because we know the derivative of sine of x and cosine of x is this expression here, and we can just rewrite this, or simplify it a little bit, so it takes a form that you might be a little bit more familiar with. So, actually let me just do this on the side here."}, {"video_title": "Worked example Derivatives of sin(x) and cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So using just that, we can actually evaluate this, or evaluate g prime of x. So pause the video and see if you can do it. So probably the most intimidating part of this, because we know the derivative of sine of x and cosine of x is this expression here, and we can just rewrite this, or simplify it a little bit, so it takes a form that you might be a little bit more familiar with. So, actually let me just do this on the side here. So pi, pi over the cube root of x squared. Well that's the same thing. This is equal to pi squared over the cube root of x squared."}, {"video_title": "Worked example Derivatives of sin(x) and cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So, actually let me just do this on the side here. So pi, pi over the cube root of x squared. Well that's the same thing. This is equal to pi squared over the cube root of x squared. This is just exponent properties that we're dealing with. And so this is the same thing. We're gonna take x to the 1 3rd power, and then raise that to the 2nd power."}, {"video_title": "Worked example Derivatives of sin(x) and cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "This is equal to pi squared over the cube root of x squared. This is just exponent properties that we're dealing with. And so this is the same thing. We're gonna take x to the 1 3rd power, and then raise that to the 2nd power. So this is equal to pi squared over, let me write it this way, I'm not gonna skip any steps, because this is a good review of exponent properties. X to the 1 3rd squared, which is the same thing as pi squared over x to the 2 3rd power, which is the same thing as pi squared times x to the negative 2 3rd power. So when you write it like this, it starts to get into a form, you're like, oh, I can see how the power rule could apply there."}, {"video_title": "Worked example Derivatives of sin(x) and cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "We're gonna take x to the 1 3rd power, and then raise that to the 2nd power. So this is equal to pi squared over, let me write it this way, I'm not gonna skip any steps, because this is a good review of exponent properties. X to the 1 3rd squared, which is the same thing as pi squared over x to the 2 3rd power, which is the same thing as pi squared times x to the negative 2 3rd power. So when you write it like this, it starts to get into a form, you're like, oh, I can see how the power rule could apply there. So this thing is just pi squared times x to the negative 2 3rd power. So actually let me delete this. So this thing can be rewritten, this thing can be rewritten as pi squared times x to the negative 2 3rd power."}, {"video_title": "Worked example Derivatives of sin(x) and cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So when you write it like this, it starts to get into a form, you're like, oh, I can see how the power rule could apply there. So this thing is just pi squared times x to the negative 2 3rd power. So actually let me delete this. So this thing can be rewritten, this thing can be rewritten as pi squared times x to the negative 2 3rd power. So now let's take the derivative of each of these, each of these pieces of this expression. So we're gonna take, we wanna evaluate what g prime of x is. So g prime of x is going to be equal to, you could view it as a derivative with respect to x of seven sine of x."}, {"video_title": "Worked example Derivatives of sin(x) and cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So this thing can be rewritten, this thing can be rewritten as pi squared times x to the negative 2 3rd power. So now let's take the derivative of each of these, each of these pieces of this expression. So we're gonna take, we wanna evaluate what g prime of x is. So g prime of x is going to be equal to, you could view it as a derivative with respect to x of seven sine of x. So we could take, let's do the derivative operator on both sides here, just to make it clear what we're doing. So we're gonna apply it there, we're gonna apply it there, and we're going to apply it there. So this derivative, this is the same thing as this is going to be seven times the derivative of sine of x, so this is just going to be seven times cosine of x."}, {"video_title": "Worked example Derivatives of sin(x) and cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So g prime of x is going to be equal to, you could view it as a derivative with respect to x of seven sine of x. So we could take, let's do the derivative operator on both sides here, just to make it clear what we're doing. So we're gonna apply it there, we're gonna apply it there, and we're going to apply it there. So this derivative, this is the same thing as this is going to be seven times the derivative of sine of x, so this is just going to be seven times cosine of x. This one over here, this is going to be three, or we're subtracting, so it's gonna be this minus, we can bring the constant out that we're multiplying the expression by, and the derivative of cosine of x, so it's minus three times, the derivative of cosine of x is negative sine of x, negative sine of x. And then finally, here in the yellow, we just apply the power rule. So we have the negative 2 3rds, actually let's not forget this minus sign, I'm gonna write it out here."}, {"video_title": "Worked example Derivatives of sin(x) and cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So this derivative, this is the same thing as this is going to be seven times the derivative of sine of x, so this is just going to be seven times cosine of x. This one over here, this is going to be three, or we're subtracting, so it's gonna be this minus, we can bring the constant out that we're multiplying the expression by, and the derivative of cosine of x, so it's minus three times, the derivative of cosine of x is negative sine of x, negative sine of x. And then finally, here in the yellow, we just apply the power rule. So we have the negative 2 3rds, actually let's not forget this minus sign, I'm gonna write it out here. So you have the negative 2 3rds, you multiply the exponent times the coefficient, it might look confusing, pi squared, but that's just a number. So it's gonna be negative, and then you have negative 2 3rds times pi squared, times pi squared, times x to the negative 2 3rds minus one power. Negative 2 3rds minus one power."}, {"video_title": "Worked example Derivatives of sin(x) and cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So we have the negative 2 3rds, actually let's not forget this minus sign, I'm gonna write it out here. So you have the negative 2 3rds, you multiply the exponent times the coefficient, it might look confusing, pi squared, but that's just a number. So it's gonna be negative, and then you have negative 2 3rds times pi squared, times pi squared, times x to the negative 2 3rds minus one power. Negative 2 3rds minus one power. So what is this going to be? So we get g prime of x is equal to, is equal to seven cosine of x, and let's see, we have a negative three times a negative sine of x, so that's a positive three sine of x. And then, we're subtracting, but then this is going to be a negative, so that's going to be a positive."}, {"video_title": "Worked example Derivatives of sin(x) and cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Negative 2 3rds minus one power. So what is this going to be? So we get g prime of x is equal to, is equal to seven cosine of x, and let's see, we have a negative three times a negative sine of x, so that's a positive three sine of x. And then, we're subtracting, but then this is going to be a negative, so that's going to be a positive. So we could say plus two pi squared over three, two pi squared over three, that's that part there, times x to the, so negative 2 3rds minus one, we could say negative one and 2 3rds, or we could say negative 5 3rds power. Negative 5 3rds power. And there you have it, we were able to tackle this thing that looked a little bit hairy, but all we had to use was the power rule and what we knew to be the derivatives of sine and cosine."}, {"video_title": "_-substitution definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So we're gonna go from x equals one to x equals two, and the integral is two x times x squared plus one to the third power dx. So I already told you that we're gonna apply u substitution, but it's interesting to be able to recognize when to use it. And the key giveaway here is, well I have this x squared plus one business to the third power, but then I also have the derivative of x squared plus one, which is two x right over here. So I could do the substitution. I could say u is equal to x squared plus one, in which case the derivative of u with respect to x is just two x plus zero, or just two x. I could write that in differential form. A mathematical hand-wavy way of thinking about it is multiplying both sides by dx. And so you get du is equal to two x dx."}, {"video_title": "_-substitution definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So I could do the substitution. I could say u is equal to x squared plus one, in which case the derivative of u with respect to x is just two x plus zero, or just two x. I could write that in differential form. A mathematical hand-wavy way of thinking about it is multiplying both sides by dx. And so you get du is equal to two x dx. And so at least this part of the integral I can rewrite. So let me at least write, so this is going to be, I'll write the integral. We're gonna think about the bounds in a second."}, {"video_title": "_-substitution definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "And so you get du is equal to two x dx. And so at least this part of the integral I can rewrite. So let me at least write, so this is going to be, I'll write the integral. We're gonna think about the bounds in a second. So we have u to the third power. U, do the same orange color. U to the third power."}, {"video_title": "_-substitution definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "We're gonna think about the bounds in a second. So we have u to the third power. U, do the same orange color. U to the third power. That's this stuff right over here. And then two x times dx. Remember, you could just view this as two x times x squared plus one to the third power times dx."}, {"video_title": "_-substitution definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "U to the third power. That's this stuff right over here. And then two x times dx. Remember, you could just view this as two x times x squared plus one to the third power times dx. So two x times dx, well two x times dx, that is du. So that and that together, that is du. Now an interesting question, because this isn't an indefinite integral."}, {"video_title": "_-substitution definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Remember, you could just view this as two x times x squared plus one to the third power times dx. So two x times dx, well two x times dx, that is du. So that and that together, that is du. Now an interesting question, because this isn't an indefinite integral. We're not just trying to find the antiderivative. This is a definite integral. So what happens to our bounds of integration?"}, {"video_title": "_-substitution definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Now an interesting question, because this isn't an indefinite integral. We're not just trying to find the antiderivative. This is a definite integral. So what happens to our bounds of integration? Well, there's two ways that you can approach this. You can change your bounds of integration, because this one is x equals one to x equals two. But now we're integrating with respect to u."}, {"video_title": "_-substitution definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So what happens to our bounds of integration? Well, there's two ways that you can approach this. You can change your bounds of integration, because this one is x equals one to x equals two. But now we're integrating with respect to u. So one way, if you want to keep your bounds of integration, or if you want to keep this a definite integral, I guess you could say, you would change your bounds from u is equal to something to u is equal to something else. And so your bounds of integration, let's see, when x is equal to one, what is u? Well, when x is equal to one, you have one squared plus one, so you have two."}, {"video_title": "_-substitution definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "But now we're integrating with respect to u. So one way, if you want to keep your bounds of integration, or if you want to keep this a definite integral, I guess you could say, you would change your bounds from u is equal to something to u is equal to something else. And so your bounds of integration, let's see, when x is equal to one, what is u? Well, when x is equal to one, you have one squared plus one, so you have two. U is equal to two in that situation. When x is equal to two, what is u? Well, you have two squared, which is four, plus one, which is five."}, {"video_title": "_-substitution definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Well, when x is equal to one, you have one squared plus one, so you have two. U is equal to two in that situation. When x is equal to two, what is u? Well, you have two squared, which is four, plus one, which is five. So u is equal to five. And you won't typically see someone writing u equal to or u equals five. It's often just from two to five, because we're integrating with respect to u."}, {"video_title": "_-substitution definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Well, you have two squared, which is four, plus one, which is five. So u is equal to five. And you won't typically see someone writing u equal to or u equals five. It's often just from two to five, because we're integrating with respect to u. You assume it's u equals two to u equals five. And so we could just rewrite this as being equal to the integral from two to five of u to the third du. But it's really important to realize why we changed our bounds."}, {"video_title": "_-substitution definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "It's often just from two to five, because we're integrating with respect to u. You assume it's u equals two to u equals five. And so we could just rewrite this as being equal to the integral from two to five of u to the third du. But it's really important to realize why we changed our bounds. We are now integrating with respect to u. And the way we did it is we used our substitution right over here. When x equals one, u is equal to two."}, {"video_title": "_-substitution definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "But it's really important to realize why we changed our bounds. We are now integrating with respect to u. And the way we did it is we used our substitution right over here. When x equals one, u is equal to two. When x is equal to two, two squared plus one, u is equal to five. And then we can just evaluate this right over here. Let's see, this is going to be equal to the antiderivative of u to the third power is u to the fourth over four."}, {"video_title": "_-substitution definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "When x equals one, u is equal to two. When x is equal to two, two squared plus one, u is equal to five. And then we can just evaluate this right over here. Let's see, this is going to be equal to the antiderivative of u to the third power is u to the fourth over four. We're gonna evaluate that at five and two. And so this is going to be five to the fourth over four minus two to the fourth over four. And then we could simplify this if we like."}, {"video_title": "_-substitution definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Let's see, this is going to be equal to the antiderivative of u to the third power is u to the fourth over four. We're gonna evaluate that at five and two. And so this is going to be five to the fourth over four minus two to the fourth over four. And then we could simplify this if we like. But we've just evaluated this definite integral. Now another way to do it is to think about the, is to try to solve the indefinite integral in terms of x and use u substitution as an intermediate. So one way to think about this is to say, well let's just try to evaluate what the indefinite integral of two x times x squared plus one to the third power dx is."}, {"video_title": "_-substitution definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "And then we could simplify this if we like. But we've just evaluated this definite integral. Now another way to do it is to think about the, is to try to solve the indefinite integral in terms of x and use u substitution as an intermediate. So one way to think about this is to say, well let's just try to evaluate what the indefinite integral of two x times x squared plus one to the third power dx is. And then whatever this expression ends up being algebraically I'm going to evaluate it at x equals two and at x equals one. And so then you use the u substitution right over here and you would get, this would simplify using the same substitution as the integral of u, u to the third power du, u to the third power du. And once again you're going to evaluate this whole thing from x equals two and then subtract from that, it evaluated at x equals one."}, {"video_title": "_-substitution definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So one way to think about this is to say, well let's just try to evaluate what the indefinite integral of two x times x squared plus one to the third power dx is. And then whatever this expression ends up being algebraically I'm going to evaluate it at x equals two and at x equals one. And so then you use the u substitution right over here and you would get, this would simplify using the same substitution as the integral of u, u to the third power du, u to the third power du. And once again you're going to evaluate this whole thing from x equals two and then subtract from that, it evaluated at x equals one. And so this is going to be equal to, well let's see, this is u to the fourth power over four. And once again evaluating it at x equals two and then subtracting from that at x equals one. And then now we can just back substitute."}, {"video_title": "_-substitution definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "And once again you're going to evaluate this whole thing from x equals two and then subtract from that, it evaluated at x equals one. And so this is going to be equal to, well let's see, this is u to the fourth power over four. And once again evaluating it at x equals two and then subtracting from that at x equals one. And then now we can just back substitute. We could say hey look, u is equal to x squared plus one. So this is the same thing as x squared plus one to the fourth power over four. And we're now going to evaluate that at x equals two and at x equals one."}, {"video_title": "_-substitution definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "And then now we can just back substitute. We could say hey look, u is equal to x squared plus one. So this is the same thing as x squared plus one to the fourth power over four. And we're now going to evaluate that at x equals two and at x equals one. And you will notice that you will get the exact same thing. When you put x equals two here, you get two squared plus one which is five to the fourth power over four, that right over there. And then minus one squared plus one is two to the fourth power over four, that right over there."}, {"video_title": "_-substitution definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "And we're now going to evaluate that at x equals two and at x equals one. And you will notice that you will get the exact same thing. When you put x equals two here, you get two squared plus one which is five to the fourth power over four, that right over there. And then minus one squared plus one is two to the fourth power over four, that right over there. So either way you'll get the same result. You can either keep it a definite integral and then change your bounds of integration and express them in terms of u, that's one way to do it. The other way is to try to evaluate the indefinite integral."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "What I have listed here is several of the derivative rules that we've used in previous videos. If these things look unfamiliar to you, I encourage you maybe to not watch this video, because in this video we're gonna think about when do we apply these rules, what strategies, and can we algebraically convert expressions so that we can use a simpler rule? But just as a quick review, this is, of course, the power rule right over here. Very handy for taking derivatives of x raised to some power. It's also, we can use that with the derivative properties of sums of derivatives or differences of derivatives to take derivatives of polynomials. This right over here is the product rule. If I have an expression that I wanna take the derivative of and I can think of it as the product of two functions, well then the derivative is gonna be the derivative of the first function times the second function plus the first function times the derivative of the second."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Very handy for taking derivatives of x raised to some power. It's also, we can use that with the derivative properties of sums of derivatives or differences of derivatives to take derivatives of polynomials. This right over here is the product rule. If I have an expression that I wanna take the derivative of and I can think of it as the product of two functions, well then the derivative is gonna be the derivative of the first function times the second function plus the first function times the derivative of the second. Once again, if this looks completely unfamiliar to you or you're a little shaky, go watch the videos, do the practice on the power rule or the product rule, or in this case, the quotient rule. And the quotient rule is a little bit more involved and we have practice and videos on that. And I always have mixed feelings about it, because if you don't remember the quotient rule, you can usually, or you can always convert a quotient into a product by expressing this thing at the bottom as f of x, or by expressing this as f of x times g of x to the negative one, so you could take the derivative with a combination of the products and this fourth rule over here, the chain rule."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "If I have an expression that I wanna take the derivative of and I can think of it as the product of two functions, well then the derivative is gonna be the derivative of the first function times the second function plus the first function times the derivative of the second. Once again, if this looks completely unfamiliar to you or you're a little shaky, go watch the videos, do the practice on the power rule or the product rule, or in this case, the quotient rule. And the quotient rule is a little bit more involved and we have practice and videos on that. And I always have mixed feelings about it, because if you don't remember the quotient rule, you can usually, or you can always convert a quotient into a product by expressing this thing at the bottom as f of x, or by expressing this as f of x times g of x to the negative one, so you could take the derivative with a combination of the products and this fourth rule over here, the chain rule. And if any of this is looking unfamiliar again, don't watch this video. This video is for folks who are familiar with each of these derivative rules or derivative techniques and now wanna think about, well, what are strategies for deciding when to apply which? So let's do that."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And I always have mixed feelings about it, because if you don't remember the quotient rule, you can usually, or you can always convert a quotient into a product by expressing this thing at the bottom as f of x, or by expressing this as f of x times g of x to the negative one, so you could take the derivative with a combination of the products and this fourth rule over here, the chain rule. And if any of this is looking unfamiliar again, don't watch this video. This video is for folks who are familiar with each of these derivative rules or derivative techniques and now wanna think about, well, what are strategies for deciding when to apply which? So let's do that. Let's say that I have the expression, let's say I'm interested in taking the derivative of x squared plus x minus two over x minus one. Which of these rules or techniques would you use? Well, you might immediately say, hey, look, this looks like a rational expression."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So let's do that. Let's say that I have the expression, let's say I'm interested in taking the derivative of x squared plus x minus two over x minus one. Which of these rules or techniques would you use? Well, you might immediately say, hey, look, this looks like a rational expression. I could say, look, I could say this is my f of x right over here. I could say this is my g of x right over here, and I could apply the quotient rule. This looks like a quotient of two expressions."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, you might immediately say, hey, look, this looks like a rational expression. I could say, look, I could say this is my f of x right over here. I could say this is my g of x right over here, and I could apply the quotient rule. This looks like a quotient of two expressions. And you could do that, and if you do all the mathematics correctly, you will get the correct answer. But in this case, it's good to just take a little time to realize, well, can I simplify this algebraically so maybe I can do a little bit less work? And if you look at it that way, you might realize, wait, what if I factored this numerator?"}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "This looks like a quotient of two expressions. And you could do that, and if you do all the mathematics correctly, you will get the correct answer. But in this case, it's good to just take a little time to realize, well, can I simplify this algebraically so maybe I can do a little bit less work? And if you look at it that way, you might realize, wait, what if I factored this numerator? I can factor it as x plus two times x minus one, and then I could cancel these two characters out, and I could say, hey, you know what? This is gonna be the same thing as the derivative with respect to x of x plus two. Derivative with respect to x of x plus two, which is much, much, much, much, much more straightforward than trying to apply the quotient rule."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And if you look at it that way, you might realize, wait, what if I factored this numerator? I can factor it as x plus two times x minus one, and then I could cancel these two characters out, and I could say, hey, you know what? This is gonna be the same thing as the derivative with respect to x of x plus two. Derivative with respect to x of x plus two, which is much, much, much, much, much more straightforward than trying to apply the quotient rule. Here, you would just take the derivative with respect to x of x, which is just gonna be one, and the derivative with respect to x of two is just gonna be zero, and so all of this is just going to simplify to one. For taking the derivative of that, you're essentially just using the power rule. And so once again, just a simple algebraic recognition, things become much, much, much, much more simple."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Derivative with respect to x of x plus two, which is much, much, much, much, much more straightforward than trying to apply the quotient rule. Here, you would just take the derivative with respect to x of x, which is just gonna be one, and the derivative with respect to x of two is just gonna be zero, and so all of this is just going to simplify to one. For taking the derivative of that, you're essentially just using the power rule. And so once again, just a simple algebraic recognition, things become much, much, much, much more simple. Let's do another example. So let's say that you were to see, or someone were to ask you to take the derivative with respect to x of, let me see. So let's say you had x squared plus two x minus five over, over x."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And so once again, just a simple algebraic recognition, things become much, much, much, much more simple. Let's do another example. So let's say that you were to see, or someone were to ask you to take the derivative with respect to x of, let me see. So let's say you had x squared plus two x minus five over, over x. So once again, you might be tempted to use the quotient rule. This looks like the quotient of two expressions, but then you might realize, well, there's some algebraic manipulations I can do to make this simpler. You could express this as a product."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say you had x squared plus two x minus five over, over x. So once again, you might be tempted to use the quotient rule. This looks like the quotient of two expressions, but then you might realize, well, there's some algebraic manipulations I can do to make this simpler. You could express this as a product. You could say that this is the same thing as, and I'm just gonna focus on what's inside the parentheses or inside the brackets. This is the same thing as x to the negative one times x squared plus two x minus five. And then you might want to apply the product rule, but there's even a better simplification here."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "You could express this as a product. You could say that this is the same thing as, and I'm just gonna focus on what's inside the parentheses or inside the brackets. This is the same thing as x to the negative one times x squared plus two x minus five. And then you might want to apply the product rule, but there's even a better simplification here. You could just divide each of these terms by x, or one way to think about it, distribute this one over x across all the terms. x to the negative one is the same thing as one over x. And if you do that, x squared divided by x is going to be x."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And then you might want to apply the product rule, but there's even a better simplification here. You could just divide each of these terms by x, or one way to think about it, distribute this one over x across all the terms. x to the negative one is the same thing as one over x. And if you do that, x squared divided by x is going to be x. Two x divided by x is going to be two. And then negative five divided by x, well, you could write that as negative five over x or negative five x to the negative one. And now, taking the derivative of this with respect to x is much easier than using either the quotient or the power rule."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And if you do that, x squared divided by x is going to be x. Two x divided by x is going to be two. And then negative five divided by x, well, you could write that as negative five over x or negative five x to the negative one. And now, taking the derivative of this with respect to x is much easier than using either the quotient or the power rule. This is going to be, let's see, derivative of that is just gonna be one, derivative of two is just gonna be zero. And here, even though you have a negative exponent, it might look a little intimidating. This is just taking it using the power rule."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And now, taking the derivative of this with respect to x is much easier than using either the quotient or the power rule. This is going to be, let's see, derivative of that is just gonna be one, derivative of two is just gonna be zero. And here, even though you have a negative exponent, it might look a little intimidating. This is just taking it using the power rule. So negative one times negative five is positive five. X to the, if we take one less than negative one, we're gonna go to the negative two power. So once again, making this algebraic recognition simplified things a good bit."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "This is just taking it using the power rule. So negative one times negative five is positive five. X to the, if we take one less than negative one, we're gonna go to the negative two power. So once again, making this algebraic recognition simplified things a good bit. Let's do a few more examples of just starting to recognize when we might be able to simplify things to do things a little bit easier. So let's say that someone said, hey, you, take the derivative with respect to x. And I'm using x as our variable that we're taking the derivative with respect to, but obviously this works for any variables that we are using."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So once again, making this algebraic recognition simplified things a good bit. Let's do a few more examples of just starting to recognize when we might be able to simplify things to do things a little bit easier. So let's say that someone said, hey, you, take the derivative with respect to x. And I'm using x as our variable that we're taking the derivative with respect to, but obviously this works for any variables that we are using. So let's say we're saying square root of x over x squared. Pause this video and think about how would you approach this if you wanna take the derivative with respect to x of the square root of x over x squared. Well, once again, you might say this is a quotient of two expressions, might try to apply the quotient rule, or you might recognize, well, look, this is the same thing."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And I'm using x as our variable that we're taking the derivative with respect to, but obviously this works for any variables that we are using. So let's say we're saying square root of x over x squared. Pause this video and think about how would you approach this if you wanna take the derivative with respect to x of the square root of x over x squared. Well, once again, you might say this is a quotient of two expressions, might try to apply the quotient rule, or you might recognize, well, look, this is the same thing. Let me just focus on what's inside the brackets. You could view this as x to the negative two times the square root of x, times the square root of x. And then you might wanna use a product rule, but you could simplify this even better."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, once again, you might say this is a quotient of two expressions, might try to apply the quotient rule, or you might recognize, well, look, this is the same thing. Let me just focus on what's inside the brackets. You could view this as x to the negative two times the square root of x, times the square root of x. And then you might wanna use a product rule, but you could simplify this even better. You could say this is the same thing as x to the negative two times x to the 1 1\u20442 power. And now you're just using our exponent properties, negative two plus 1 1\u20442 is negative 3 1\u20442. So this is the derivative of x to the negative 3 1\u20442 power."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And then you might wanna use a product rule, but you could simplify this even better. You could say this is the same thing as x to the negative two times x to the 1 1\u20442 power. And now you're just using our exponent properties, negative two plus 1 1\u20442 is negative 3 1\u20442. So this is the derivative of x to the negative 3 1\u20442 power. And so here, once again, we took something that we thought we might have to use a quotient rule or use the product rule, and now this just becomes a straightforward using the power rule. So this is just going to be equal to, so bring the negative 3 1\u20442 out front, negative 3 1\u20442, x to the negative 3 1\u20442 minus one is negative 5 1\u20442 power. So once again, just before you just, especially if you're about to apply the quotient rule and sometimes even the product rule, just see, is there an algebraic simplification, sometimes a trigonometric simplification, that you can make that eases your job, that makes things less hairy?"}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So this is the derivative of x to the negative 3 1\u20442 power. And so here, once again, we took something that we thought we might have to use a quotient rule or use the product rule, and now this just becomes a straightforward using the power rule. So this is just going to be equal to, so bring the negative 3 1\u20442 out front, negative 3 1\u20442, x to the negative 3 1\u20442 minus one is negative 5 1\u20442 power. So once again, just before you just, especially if you're about to apply the quotient rule and sometimes even the product rule, just see, is there an algebraic simplification, sometimes a trigonometric simplification, that you can make that eases your job, that makes things less hairy? And as a general tip, I can't say this is gonna be always true, but if you're taking some type of an exam and you're going down some really hairy route, which the quotient rule will often take you, it's a good sign that, hey, take a pause before trying to run through all of that algebra to apply the quotient rule and see if you can simplify things. So let's give another example. And this one, there's not a must, there's not an obvious way, and it really depends on what folks' preferences are."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So once again, just before you just, especially if you're about to apply the quotient rule and sometimes even the product rule, just see, is there an algebraic simplification, sometimes a trigonometric simplification, that you can make that eases your job, that makes things less hairy? And as a general tip, I can't say this is gonna be always true, but if you're taking some type of an exam and you're going down some really hairy route, which the quotient rule will often take you, it's a good sign that, hey, take a pause before trying to run through all of that algebra to apply the quotient rule and see if you can simplify things. So let's give another example. And this one, there's not a must, there's not an obvious way, and it really depends on what folks' preferences are. But let's say you wanna take the derivative with respect to x of one over two x to the negative five. Sorry, one over two x minus five, I should say. Well, here, you could immediately, you could apply the quotient rule here, or the numerator you view that as f of x."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And this one, there's not a must, there's not an obvious way, and it really depends on what folks' preferences are. But let's say you wanna take the derivative with respect to x of one over two x to the negative five. Sorry, one over two x minus five, I should say. Well, here, you could immediately, you could apply the quotient rule here, or the numerator you view that as f of x. You could view this as the same thing as the derivative with respect to x instead of two x minus five. Let me do that in the blue color. Two x minus five to the negative one power."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, here, you could immediately, you could apply the quotient rule here, or the numerator you view that as f of x. You could view this as the same thing as the derivative with respect to x instead of two x minus five. Let me do that in the blue color. Two x minus five to the negative one power. And now this, in this situation, you would use a combination of the power rule and the chain rule. You'd say, okay, my g of x is two x minus five, and f of g of x is going to be this whole expression. And so if you did, if you applied the chain rule, this is going to be the derivative of the outside function, our f of x with respect to the inside function, so, or the derivative of f of g of x with respect to g of x."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Two x minus five to the negative one power. And now this, in this situation, you would use a combination of the power rule and the chain rule. You'd say, okay, my g of x is two x minus five, and f of g of x is going to be this whole expression. And so if you did, if you applied the chain rule, this is going to be the derivative of the outside function, our f of x with respect to the inside function, so, or the derivative of f of g of x with respect to g of x. So it's going to be negative, we'll bring that negative out front, so we're essentially just gonna use the power rule here. Negative two x minus five to the negative two. And then we multiply that times the derivative of, times the derivative of the inside function."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And so if you did, if you applied the chain rule, this is going to be the derivative of the outside function, our f of x with respect to the inside function, so, or the derivative of f of g of x with respect to g of x. So it's going to be negative, we'll bring that negative out front, so we're essentially just gonna use the power rule here. Negative two x minus five to the negative two. And then we multiply that times the derivative of, times the derivative of the inside function. So the inside function's derivative, the derivative of two x is two, derivative of negative five is zero, so it's gonna be times two. And of course you can simplify it, so it's negative two times all of this business. Let me do one more example here, just to hit the point home."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And then we multiply that times the derivative of, times the derivative of the inside function. So the inside function's derivative, the derivative of two x is two, derivative of negative five is zero, so it's gonna be times two. And of course you can simplify it, so it's negative two times all of this business. Let me do one more example here, just to hit the point home. And once again, there isn't a must way, there isn't a way that you have to do this, but just let you appreciate that there's multiple ways to approach these types of derivatives. So let's say someone said, take the derivative of two x plus one squared. Pause the video and think about how you would do that."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Let me do one more example here, just to hit the point home. And once again, there isn't a must way, there isn't a way that you have to do this, but just let you appreciate that there's multiple ways to approach these types of derivatives. So let's say someone said, take the derivative of two x plus one squared. Pause the video and think about how you would do that. Well one way to do it is just to apply the chain rule just like we just did. So you could say, all right, it's gonna be the derivative of the outside with respect to the inside. So it's gonna be two times two x plus one to the first power, taking one less than that, times the derivative of the inside, which is just going to be two."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Pause the video and think about how you would do that. Well one way to do it is just to apply the chain rule just like we just did. So you could say, all right, it's gonna be the derivative of the outside with respect to the inside. So it's gonna be two times two x plus one to the first power, taking one less than that, times the derivative of the inside, which is just going to be two. And so this is going to be equal to four times two x plus one which is equal to, if we wanna distribute the four, we could say it's eight x plus four. That's a completely legitimate way of doing it. Now there are other ways of doing it."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So it's gonna be two times two x plus one to the first power, taking one less than that, times the derivative of the inside, which is just going to be two. And so this is going to be equal to four times two x plus one which is equal to, if we wanna distribute the four, we could say it's eight x plus four. That's a completely legitimate way of doing it. Now there are other ways of doing it. You could expand out two x plus one squared. You could say, hey, this is the same thing as the derivative with respect to x of two x squared is going to be four x squared, and then two times the product of these terms is gonna be plus four x plus one. And now you would just apply the power rule."}, {"video_title": "Manipulating functions before differentiation Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Now there are other ways of doing it. You could expand out two x plus one squared. You could say, hey, this is the same thing as the derivative with respect to x of two x squared is going to be four x squared, and then two times the product of these terms is gonna be plus four x plus one. And now you would just apply the power rule. So a little bit of extra algebra up front, but you can just go straight forward with the power rule, and you're gonna get this exact same thing. So the whole takeaway here is, pause, look at your expression. See if there's a way to simplify it."}, {"video_title": "Differentiating power series Series AP Calculus BC Khan Academy.mp3", "Sentence": "So we're told here that f of x is equal to this infinite series and we need to figure out What is the third derivative of f evaluated at x equals zero? And like always pause this video and see if you can work it out on your own before we do it together All right, so there's two ways to approach this one is we could just take the derivative of this expression while it's in sigma notation The other way you do it is we could just expand out f of x and take the derivative Three times and see if we get an answer that I guess makes sense. So let me do it the second way first So let me just expand it out f of x is equal to let's see when n is equal to 0 This is negative 1 to the 0 which is just 1 times x Times x to the 0 plus 3 so it's going to be x to the third over 2 times 0 So that's 0 plus 1 factorial So that's just going to be over 1 then the next term when n is equal to 1 well now it's going to be negative 1 to the 1 so now it's going to have a negative out front Negative and it's going to be 2 times 1 plus 3 so that's going to be x to the 5th x to the 5th power over 2 times 1 plus 1 so it's going to be 2 plus 1 is 3 factorial So it's going to be x to the 5th over 6 and then when x is equal to 2 If this is going to be positive again, it's going to be x to the 7th power x to the 7th power over 5 Factorial is that right? Yeah 5 factorial and 5 factorial actually Let me just write that out as 5 factorial 5 factorial would be what would be a hundred and twenty It'd be 5 times 4 times 6 would be 120 But we could just keep going minus plus and it goes on and on and on forever Well now let's just take the derivatives. So f prime of x is going to be equal to or should applying the power rule here It's going to be 3x squared minus 5 6 x to the 4th plus 7 over 5 Factorial x to the 6. I'm just applying the power rule Minus plus we just keep going on and on and on forever the second derivative f prime prime of x is going to be equal to Apply the power rule again. It's going to be 6x to the first Minus 4 times 5 over 6."}, {"video_title": "Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let's explore a bit the infinite series from n equals one to infinity of one over n squared, which of course is equal to one plus 1 4th, that's one over two squared, plus one over three squared, which is 1 9th plus 1 16th, and it goes on and on and on forever. So there's a couple of things that we know about it. The first thing is all of the terms here are positive. So all of the terms here are positive. So they're all positive, and they're decreasing. It looks like they're decreasing quite quickly here from one to 1 4th to 1 9th to 1 16th, and so they're quickly approaching zero, which makes us feel pretty good that this thing has a chance of converging. And because they're all positive, we know that this sum right over here, if it does converge, is going to be greater than zero."}, {"video_title": "Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So all of the terms here are positive. So they're all positive, and they're decreasing. It looks like they're decreasing quite quickly here from one to 1 4th to 1 9th to 1 16th, and so they're quickly approaching zero, which makes us feel pretty good that this thing has a chance of converging. And because they're all positive, we know that this sum right over here, if it does converge, is going to be greater than zero. So the only reason why it wouldn't converge is if somehow it goes unbounded towards infinity, which we know if this was one over n, it would be unbounded towards infinity. So this says that's a possibility here. So if we could show that this is bounded, then that'll be a pretty good argument for why this thing right over here converges, because the only reason why you could diverge is if you went to either positive infinity or negative infinity."}, {"video_title": "Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And because they're all positive, we know that this sum right over here, if it does converge, is going to be greater than zero. So the only reason why it wouldn't converge is if somehow it goes unbounded towards infinity, which we know if this was one over n, it would be unbounded towards infinity. So this says that's a possibility here. So if we could show that this is bounded, then that'll be a pretty good argument for why this thing right over here converges, because the only reason why you could diverge is if you went to either positive infinity or negative infinity. We already know that this thing isn't going to go to negative infinity because it's all positive terms, or you could diverge if this thing oscillates, but it's not going to oscillate because all of these terms are just adding to the sum. None of them are taking away because none of these terms are negative. So let's see if we can make a good argument for why this sum right over here is bounded, especially if we can come up with the bound, then that's a pretty good argument that this infinite series should converge."}, {"video_title": "Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So if we could show that this is bounded, then that'll be a pretty good argument for why this thing right over here converges, because the only reason why you could diverge is if you went to either positive infinity or negative infinity. We already know that this thing isn't going to go to negative infinity because it's all positive terms, or you could diverge if this thing oscillates, but it's not going to oscillate because all of these terms are just adding to the sum. None of them are taking away because none of these terms are negative. So let's see if we can make a good argument for why this sum right over here is bounded, especially if we can come up with the bound, then that's a pretty good argument that this infinite series should converge. And the way that we're gonna do that is we're going to explore a related function. So what I wanna do is I wanna explore f of x is equal to one over x squared. You could really view this right over here, one over n squared, as f of n, if I were to write it this way."}, {"video_title": "Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's see if we can make a good argument for why this sum right over here is bounded, especially if we can come up with the bound, then that's a pretty good argument that this infinite series should converge. And the way that we're gonna do that is we're going to explore a related function. So what I wanna do is I wanna explore f of x is equal to one over x squared. You could really view this right over here, one over n squared, as f of n, if I were to write it this way. So why is this interesting? Well, let's graph it. Let's graph it."}, {"video_title": "Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "You could really view this right over here, one over n squared, as f of n, if I were to write it this way. So why is this interesting? Well, let's graph it. Let's graph it. So that's the graph of y is equal to f of x. y is equal to f of x. f of x. And notice this is a continuous positive decreasing function especially over the interval that I've, or over the interval that I care about right over here. I guess we could say over the, for positive values of x, it is a continuous, it is a continuous positive decreasing function."}, {"video_title": "Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let's graph it. So that's the graph of y is equal to f of x. y is equal to f of x. f of x. And notice this is a continuous positive decreasing function especially over the interval that I've, or over the interval that I care about right over here. I guess we could say over the, for positive values of x, it is a continuous, it is a continuous positive decreasing function. And what's interesting is we can use this as really an underestimate for this area right over here. What do I mean by that? Well, one, this first term right over here, you could view that as the area of this block right over here."}, {"video_title": "Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "I guess we could say over the, for positive values of x, it is a continuous, it is a continuous positive decreasing function. And what's interesting is we can use this as really an underestimate for this area right over here. What do I mean by that? Well, one, this first term right over here, you could view that as the area of this block right over here. That is f of n, or I guess you could say f of one high and one wide. So it's going to be one, it's gonna be one times one over one squared, or one. This term, this, let me make sure I'm using different colors."}, {"video_title": "Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, one, this first term right over here, you could view that as the area of this block right over here. That is f of n, or I guess you could say f of one high and one wide. So it's going to be one, it's gonna be one times one over one squared, or one. This term, this, let me make sure I'm using different colors. This term right over here, that could represent the area of this block, which is 1 4th high and one wide. So it is going to be, it's going to have an area of 1 4th. What could this one represent?"}, {"video_title": "Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "This term, this, let me make sure I'm using different colors. This term right over here, that could represent the area of this block, which is 1 4th high and one wide. So it is going to be, it's going to have an area of 1 4th. What could this one represent? Well, the area of the next block if we're trying to estimate the area under the curve. And this might look familiar from when we first got exposed to the integral, or even before we got exposed to the integral when we were taking Riemann sums. So that right over here, that area is going to be equal to 1 9th."}, {"video_title": "Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "What could this one represent? Well, the area of the next block if we're trying to estimate the area under the curve. And this might look familiar from when we first got exposed to the integral, or even before we got exposed to the integral when we were taking Riemann sums. So that right over here, that area is going to be equal to 1 9th. So what's intriguing about this is we know how to find the exact area, especially, or the exact area from one to infinity, from x equals one to infinity. So maybe we can use that somehow. We know what this area is right over here, which we can denote as the improper integral from one to infinity of f of x dx."}, {"video_title": "Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So that right over here, that area is going to be equal to 1 9th. So what's intriguing about this is we know how to find the exact area, especially, or the exact area from one to infinity, from x equals one to infinity. So maybe we can use that somehow. We know what this area is right over here, which we can denote as the improper integral from one to infinity of f of x dx. We know what that is, and I'll figure out in a little bit. And if we know what this is, if we can figure out the value, that's going to be an upper bound for 1 4th plus 1 9th plus 1 16th, on and on and on and on. And so that will allow us to essentially bound what this series evaluates to."}, {"video_title": "Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "We know what this area is right over here, which we can denote as the improper integral from one to infinity of f of x dx. We know what that is, and I'll figure out in a little bit. And if we know what this is, if we can figure out the value, that's going to be an upper bound for 1 4th plus 1 9th plus 1 16th, on and on and on and on. And so that will allow us to essentially bound what this series evaluates to. And as we said earlier, that would be a very good argument for its convergence. So the whole point here, I'm not doing a rigorous proof, but really getting you the underlying conceptual understanding for a very popular test for convergence or divergence, which is called the integral test. Let me just write that down, just so you know what this is kind of the mental foundations for."}, {"video_title": "Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so that will allow us to essentially bound what this series evaluates to. And as we said earlier, that would be a very good argument for its convergence. So the whole point here, I'm not doing a rigorous proof, but really getting you the underlying conceptual understanding for a very popular test for convergence or divergence, which is called the integral test. Let me just write that down, just so you know what this is kind of the mental foundations for. So what do I mean here? So let me write this sum again. Let me write it a little bit different."}, {"video_title": "Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let me just write that down, just so you know what this is kind of the mental foundations for. So what do I mean here? So let me write this sum again. Let me write it a little bit different. So our original series from n equals one to infinity of one over n squared, it's going to be equal to this first block, the area of this first block, plus the area of all of the rest of the blocks, the 1 4th plus 1 9th plus 1 16th, which we could write, which we could write as the, let me do this in a new color, which we could write as the sum from n equals two to infinity of one over n squared. So I just kind of express this as a sum of this plus all of that stuff. Now what's interesting is that this, this, what I just wrote in this blue notation, that's this block plus this block plus the next block, which is going to be less than this definite integral right over here."}, {"video_title": "Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let me write it a little bit different. So our original series from n equals one to infinity of one over n squared, it's going to be equal to this first block, the area of this first block, plus the area of all of the rest of the blocks, the 1 4th plus 1 9th plus 1 16th, which we could write, which we could write as the, let me do this in a new color, which we could write as the sum from n equals two to infinity of one over n squared. So I just kind of express this as a sum of this plus all of that stuff. Now what's interesting is that this, this, what I just wrote in this blue notation, that's this block plus this block plus the next block, which is going to be less than this definite integral right over here. This definite integral, notice, it's an underestimate. It's always below the curve. So it's going to be less than that definite integral."}, {"video_title": "Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Now what's interesting is that this, this, what I just wrote in this blue notation, that's this block plus this block plus the next block, which is going to be less than this definite integral right over here. This definite integral, notice, it's an underestimate. It's always below the curve. So it's going to be less than that definite integral. So we can write that this thing is going to be less than one, one plus, instead of writing this, I'm going to write the definite integral. One plus the definite integral from one to infinity of one over, one over x squared dx. Now why is that useful?"}, {"video_title": "Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So it's going to be less than that definite integral. So we can write that this thing is going to be less than one, one plus, instead of writing this, I'm going to write the definite integral. One plus the definite integral from one to infinity of one over, one over x squared dx. Now why is that useful? Well, we know how to evaluate this. I encourage you to review the section on Khan Academy on improper integrals if this looks unfamiliar, but I'll evaluate this, I'll evaluate this down here. We know that this is the same thing as the limit as, I'm going to introduce a variable here, t approaches infinity of the definite integral from one to t of, and I'll just write this as x to the negative two dx, which is equal to the limit as t approaches infinity of, let's see, of negative x to the negative one, or actually I could write that as negative one over x, negative one over x, and we're going to evaluate that as t and at one, which is equal to the limit as t approaches infinity of negative one over t, and then minus negative one over one, so that would just be plus one, and as t approaches infinity, this term right over here is going to be zero, so this is just going to simplify to one."}, {"video_title": "Integral test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Now why is that useful? Well, we know how to evaluate this. I encourage you to review the section on Khan Academy on improper integrals if this looks unfamiliar, but I'll evaluate this, I'll evaluate this down here. We know that this is the same thing as the limit as, I'm going to introduce a variable here, t approaches infinity of the definite integral from one to t of, and I'll just write this as x to the negative two dx, which is equal to the limit as t approaches infinity of, let's see, of negative x to the negative one, or actually I could write that as negative one over x, negative one over x, and we're going to evaluate that as t and at one, which is equal to the limit as t approaches infinity of negative one over t, and then minus negative one over one, so that would just be plus one, and as t approaches infinity, this term right over here is going to be zero, so this is just going to simplify to one. So this whole thing evaluates to one. So just like that, we were able to place an upper bound on this series. We're able to say that the series under question or in question, so the infinite sum from n equals one to infinity of one over n squared is going to be less than one plus one, or it's going to be less than, it's going to be, let me do this in a new color, it's going to be less than, it's going to be less than two, or another way to think about it, it's going to be the two is this area, that's one right over there, plus this area right over here."}, {"video_title": "Second derivative test Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "And before I even get into the nitty gritty of it, I really just want to get an intuitive feel for what the second derivative test is telling us. So let me just draw some axes here. So let's say that's my y-axis. Let's say this is my x-axis. And let's say I have a function that has a relative maximum value at x equals c. So let's say we have a situation that looks something like that, and x equals c is right over, so that's the point c, f of c. So, I can draw a straighter dotted line. So that is x being equal to c. And we visually see that we have a local maximum point there, and we can use our calculus tools to think about what's going on there. Well, one thing that we know, we know that the slope of the tangent line, at least the way I've drawn it right over here, is equal to zero."}, {"video_title": "Second derivative test Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say this is my x-axis. And let's say I have a function that has a relative maximum value at x equals c. So let's say we have a situation that looks something like that, and x equals c is right over, so that's the point c, f of c. So, I can draw a straighter dotted line. So that is x being equal to c. And we visually see that we have a local maximum point there, and we can use our calculus tools to think about what's going on there. Well, one thing that we know, we know that the slope of the tangent line, at least the way I've drawn it right over here, is equal to zero. So we can say f prime of c is equal to zero. And the other thing we can see is that we are concave downward in the neighborhood around x equals c. So notice our slope is constantly decreasing. And since our slope is, notice it's positive, it's less positive, even less positive, it goes to zero, then it becomes negative, more negative, and even more negative."}, {"video_title": "Second derivative test Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "Well, one thing that we know, we know that the slope of the tangent line, at least the way I've drawn it right over here, is equal to zero. So we can say f prime of c is equal to zero. And the other thing we can see is that we are concave downward in the neighborhood around x equals c. So notice our slope is constantly decreasing. And since our slope is, notice it's positive, it's less positive, even less positive, it goes to zero, then it becomes negative, more negative, and even more negative. So we know that f prime prime, we know that f prime prime of c is less than zero. And so, I haven't done any deep mathematical proof here, but if I have a critical point where f prime, where our critical point at x equals c, so f prime of c is equal to zero, and we also see that the second derivative there is less than zero, well, intuitively, this makes sense that we are at a maximum value. And we could go the other way."}, {"video_title": "Second derivative test Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "And since our slope is, notice it's positive, it's less positive, even less positive, it goes to zero, then it becomes negative, more negative, and even more negative. So we know that f prime prime, we know that f prime prime of c is less than zero. And so, I haven't done any deep mathematical proof here, but if I have a critical point where f prime, where our critical point at x equals c, so f prime of c is equal to zero, and we also see that the second derivative there is less than zero, well, intuitively, this makes sense that we are at a maximum value. And we could go the other way. If we are at a local minimum point at x equals c, or relative minimum point, so our first derivative should still be equal to zero, because our slope of a tangent line right over there is still zero, so f prime of c is equal to zero. But in this second situation, we are concave upwards. The slope is constantly increasing."}, {"video_title": "Second derivative test Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "And we could go the other way. If we are at a local minimum point at x equals c, or relative minimum point, so our first derivative should still be equal to zero, because our slope of a tangent line right over there is still zero, so f prime of c is equal to zero. But in this second situation, we are concave upwards. The slope is constantly increasing. We have an upward opening bowl. And so here, we have a relative minimum value. Or we could say our second derivative is greater than zero."}, {"video_title": "Second derivative test Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "The slope is constantly increasing. We have an upward opening bowl. And so here, we have a relative minimum value. Or we could say our second derivative is greater than zero. Visually, we see it's a relative minimum value, and we can tell just looking at our derivatives, at least the way I've drawn it, first derivative is equal to zero, and we are concave upwards. Second derivative is greater than zero. And so this intuition that we hopefully just built up is what the second derivative test tells us."}, {"video_title": "Second derivative test Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "Or we could say our second derivative is greater than zero. Visually, we see it's a relative minimum value, and we can tell just looking at our derivatives, at least the way I've drawn it, first derivative is equal to zero, and we are concave upwards. Second derivative is greater than zero. And so this intuition that we hopefully just built up is what the second derivative test tells us. So it says, hey, look, if we're dealing with some function f, let's say it's a twice-differentiable function, so that means that over some interval, so that means that you could find its first and second derivatives are defined. And so let's say there's some point, x equals c, where its first derivative is equal to zero. So the slope of the tangent line is equal to zero."}, {"video_title": "Second derivative test Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "And so this intuition that we hopefully just built up is what the second derivative test tells us. So it says, hey, look, if we're dealing with some function f, let's say it's a twice-differentiable function, so that means that over some interval, so that means that you could find its first and second derivatives are defined. And so let's say there's some point, x equals c, where its first derivative is equal to zero. So the slope of the tangent line is equal to zero. And the derivative exists in a neighborhood around c, and most of the functions we deal with, if it's differentiable at c, it tends to be differentiable in the neighborhood around c. And then we also assume that the second derivative exists, it's twice-differentiable. Well then, we might be dealing with a maximum point, we might be dealing with a minimum point, or we might not know what we're dealing with. And it might be neither a minimum or a maximum point."}, {"video_title": "Second derivative test Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "So the slope of the tangent line is equal to zero. And the derivative exists in a neighborhood around c, and most of the functions we deal with, if it's differentiable at c, it tends to be differentiable in the neighborhood around c. And then we also assume that the second derivative exists, it's twice-differentiable. Well then, we might be dealing with a maximum point, we might be dealing with a minimum point, or we might not know what we're dealing with. And it might be neither a minimum or a maximum point. But using the second derivative test, if we take the second derivative, and if we see that the second derivative is indeed less than zero, then we have a relative maximum point. So this is a situation that we started with right up there. If our second derivative is greater than zero, then we are in this situation right here, we're concave upwards."}, {"video_title": "Second derivative test Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "And it might be neither a minimum or a maximum point. But using the second derivative test, if we take the second derivative, and if we see that the second derivative is indeed less than zero, then we have a relative maximum point. So this is a situation that we started with right up there. If our second derivative is greater than zero, then we are in this situation right here, we're concave upwards. Where the slope is zero, that's the bottom of the bowl, we have a relative minimum point. And if our second derivative is zero, it's inconclusive. We don't know what is actually going on at that point, so we can't make any strong statement."}, {"video_title": "Second derivative test Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "If our second derivative is greater than zero, then we are in this situation right here, we're concave upwards. Where the slope is zero, that's the bottom of the bowl, we have a relative minimum point. And if our second derivative is zero, it's inconclusive. We don't know what is actually going on at that point, so we can't make any strong statement. So with that out of the way, let's just do a quick example, just to see if this has gelled. Let's say that I have some twice-differentiable function h. And let's say that I tell you that h of eight is equal to five. I tell you that h prime of eight is equal to zero."}, {"video_title": "Second derivative test Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "We don't know what is actually going on at that point, so we can't make any strong statement. So with that out of the way, let's just do a quick example, just to see if this has gelled. Let's say that I have some twice-differentiable function h. And let's say that I tell you that h of eight is equal to five. I tell you that h prime of eight is equal to zero. And I tell you that the second derivative at x equals eight is equal to negative four. So given this, can you tell me whether the point eight comma five, so the point eight comma five, is it a relative, is it a relative minimum, relative minimum maximum point, or not enough info? Not enough info or inconclusive."}, {"video_title": "Second derivative test Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "I tell you that h prime of eight is equal to zero. And I tell you that the second derivative at x equals eight is equal to negative four. So given this, can you tell me whether the point eight comma five, so the point eight comma five, is it a relative, is it a relative minimum, relative minimum maximum point, or not enough info? Not enough info or inconclusive. And like always, pause the video and see if you can figure it out. Well, we're assuming it's twice-differentiable. I think it's safe to assume that it's, and we'll, for the sake of this problem, we're gonna assume that the derivative exists in a neighborhood around x equals eight."}, {"video_title": "Second derivative test Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "Not enough info or inconclusive. And like always, pause the video and see if you can figure it out. Well, we're assuming it's twice-differentiable. I think it's safe to assume that it's, and we'll, for the sake of this problem, we're gonna assume that the derivative exists in a neighborhood around x equals eight. So this example, c is eight, so the point eight five is definitely on the curve. The derivative is equal to zero, so we're dealing potentially with one of these scenarios. And our second derivative is less than zero."}, {"video_title": "Second derivative test Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "I think it's safe to assume that it's, and we'll, for the sake of this problem, we're gonna assume that the derivative exists in a neighborhood around x equals eight. So this example, c is eight, so the point eight five is definitely on the curve. The derivative is equal to zero, so we're dealing potentially with one of these scenarios. And our second derivative is less than zero. Second derivative is less than zero. So this threw us, so the fact that the second derivative, so h prime prime of eight is less than zero, tells us that we fall into this situation right over here. So just with the information they've given us, we can say that at the point eight comma five, we have a relative maximum value, or that this is a relative maximum point for this."}, {"video_title": "Indefinite integrals sums & multiples AP Calculus AB Khan Academy.mp3", "Sentence": "So what we have listed here are two significant properties of indefinite integrals, and we will see in the future that they are very, very powerful. All this is saying is the indefinite integral of the sum of two different functions is equal to the sum of the indefinite integral of each of those functions. This one right over here says the indefinite integral of a constant, that's not going to be a function of x, of a constant times f of x is the same thing as the constant times the indefinite integral of f of x. So one way to think about it is we took the constant out of the integral, which we'll see in the future. Both of these are very useful techniques. Now, if you're satisfied with them as they are written, then that's fine, you can move on. If you want a little bit of a proof, what I'm going to do here to give an argument for why this is true is use the derivative properties."}, {"video_title": "Indefinite integrals sums & multiples AP Calculus AB Khan Academy.mp3", "Sentence": "So one way to think about it is we took the constant out of the integral, which we'll see in the future. Both of these are very useful techniques. Now, if you're satisfied with them as they are written, then that's fine, you can move on. If you want a little bit of a proof, what I'm going to do here to give an argument for why this is true is use the derivative properties. Take the derivative of both sides and see that the equality holds once we get rid of the integrals. So let's do that. Let's take the derivative with respect to x of both sides of this, derivative with respect to x."}, {"video_title": "Indefinite integrals sums & multiples AP Calculus AB Khan Academy.mp3", "Sentence": "If you want a little bit of a proof, what I'm going to do here to give an argument for why this is true is use the derivative properties. Take the derivative of both sides and see that the equality holds once we get rid of the integrals. So let's do that. Let's take the derivative with respect to x of both sides of this, derivative with respect to x. The left side here, well, this will just become whatever's inside of the indefinite integral. This will just become f of x plus g of x, plus g of x. Now, what would this become?"}, {"video_title": "Indefinite integrals sums & multiples AP Calculus AB Khan Academy.mp3", "Sentence": "Let's take the derivative with respect to x of both sides of this, derivative with respect to x. The left side here, well, this will just become whatever's inside of the indefinite integral. This will just become f of x plus g of x, plus g of x. Now, what would this become? Well, we could just go to our derivative properties. The derivative of the sum of two things, that's just the same thing as the sum of the derivatives. So this will be a little bit lengthy."}, {"video_title": "Indefinite integrals sums & multiples AP Calculus AB Khan Academy.mp3", "Sentence": "Now, what would this become? Well, we could just go to our derivative properties. The derivative of the sum of two things, that's just the same thing as the sum of the derivatives. So this will be a little bit lengthy. So this is going to be the derivative with respect to x of this first part plus the derivative with respect to x of this second part. And so this first part is the integral of f of x dx. We're gonna add it."}, {"video_title": "Indefinite integrals sums & multiples AP Calculus AB Khan Academy.mp3", "Sentence": "So this will be a little bit lengthy. So this is going to be the derivative with respect to x of this first part plus the derivative with respect to x of this second part. And so this first part is the integral of f of x dx. We're gonna add it. And then this is the integral of g of x dx. So let me write it down. This is f of x."}, {"video_title": "Indefinite integrals sums & multiples AP Calculus AB Khan Academy.mp3", "Sentence": "We're gonna add it. And then this is the integral of g of x dx. So let me write it down. This is f of x. And then this is g of x. Now, what are these things? Well, these things, let me just write this equal sign right over here."}, {"video_title": "Indefinite integrals sums & multiples AP Calculus AB Khan Academy.mp3", "Sentence": "This is f of x. And then this is g of x. Now, what are these things? Well, these things, let me just write this equal sign right over here. So in the end, this is going to be equal to, the derivative of this with respect to x is just going to be f of x. And then the derivative with respect to here is just going to be g of x. And this is obviously true."}, {"video_title": "Indefinite integrals sums & multiples AP Calculus AB Khan Academy.mp3", "Sentence": "Well, these things, let me just write this equal sign right over here. So in the end, this is going to be equal to, the derivative of this with respect to x is just going to be f of x. And then the derivative with respect to here is just going to be g of x. And this is obviously true. So now let's tackle this. Well, let's just do the same thing. Let's take the derivative of both sides."}, {"video_title": "Indefinite integrals sums & multiples AP Calculus AB Khan Academy.mp3", "Sentence": "And this is obviously true. So now let's tackle this. Well, let's just do the same thing. Let's take the derivative of both sides. So the derivative with respect to x of that and the derivative with respect to x of that. So the left-hand side will clearly become c times f of x. The right-hand side is going to become, well, we know from our derivative properties, the derivative of a constant times something is the same thing as the constant times the derivative of that something."}, {"video_title": "Indefinite integrals sums & multiples AP Calculus AB Khan Academy.mp3", "Sentence": "Let's take the derivative of both sides. So the derivative with respect to x of that and the derivative with respect to x of that. So the left-hand side will clearly become c times f of x. The right-hand side is going to become, well, we know from our derivative properties, the derivative of a constant times something is the same thing as the constant times the derivative of that something. So then we have the integral, indefinite integral of f of x dx. And then this thing is just going to be f of x. So this is all going to be equal to c times f of x."}, {"video_title": "Indefinite integrals sums & multiples AP Calculus AB Khan Academy.mp3", "Sentence": "The right-hand side is going to become, well, we know from our derivative properties, the derivative of a constant times something is the same thing as the constant times the derivative of that something. So then we have the integral, indefinite integral of f of x dx. And then this thing is just going to be f of x. So this is all going to be equal to c times f of x. So once again, you can see that the equality clearly holds. So hopefully this makes you feel good that those properties are true. But the more important thing is that you know when to use it."}, {"video_title": "Indefinite integrals sums & multiples AP Calculus AB Khan Academy.mp3", "Sentence": "So this is all going to be equal to c times f of x. So once again, you can see that the equality clearly holds. So hopefully this makes you feel good that those properties are true. But the more important thing is that you know when to use it. So for example, if I were to take the integral of, let's say, x squared plus cosine of x, the indefinite integral of that, we now know, and it's going to be useful in the future, say, well, this is the same thing as the integral of x squared dx plus the integral of cosine of x dx. So this is the same thing as that plus that. And then you can separately evaluate them."}, {"video_title": "Indefinite integrals sums & multiples AP Calculus AB Khan Academy.mp3", "Sentence": "But the more important thing is that you know when to use it. So for example, if I were to take the integral of, let's say, x squared plus cosine of x, the indefinite integral of that, we now know, and it's going to be useful in the future, say, well, this is the same thing as the integral of x squared dx plus the integral of cosine of x dx. So this is the same thing as that plus that. And then you can separately evaluate them. And this is helpful because we know that if we are trying to figure out the integral of, let's say, pi times sine of x dx, that we can take this constant out. Pi is in no way dependent on x. It's just going to stay being equal to pi."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Alright, let's see if we can find the limit of one over square root of two sine of theta over cosine of two theta as theta approaches negative pi over four. And like always, try to give it a shot before we go through it together. Well, one take on it is, well, let's just say that this is going to be the same thing as the limit as theta approaches negative pi over four of one plus square root of two sine theta theta over the limit as theta approaches negative pi over four, make sure we can see that negative there, of cosine of two theta. And both of these expressions are, if these were function definitions, or if we were to graph y equals one plus square root of two times sine theta, or y equals cosine of two theta, we would get continuous functions, especially at theta is equal to negative pi over four, so we could just substitute in. We say, well, this is going to be equal to, this expression evaluated at negative pi over four, so one plus square root of two times sine of negative pi over four over cosine of two times negative pi over four. Now, negative pi over four, sine of negative pi over four is going to be negative square root of two over two. So this is negative square root of two over two."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And both of these expressions are, if these were function definitions, or if we were to graph y equals one plus square root of two times sine theta, or y equals cosine of two theta, we would get continuous functions, especially at theta is equal to negative pi over four, so we could just substitute in. We say, well, this is going to be equal to, this expression evaluated at negative pi over four, so one plus square root of two times sine of negative pi over four over cosine of two times negative pi over four. Now, negative pi over four, sine of negative pi over four is going to be negative square root of two over two. So this is negative square root of two over two. We're assuming this is in radians. If we're thinking in degrees, this would be a negative 45 degree angle, so this is one of the trig values that it's good to know. And so if you have one, so let's see."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So this is negative square root of two over two. We're assuming this is in radians. If we're thinking in degrees, this would be a negative 45 degree angle, so this is one of the trig values that it's good to know. And so if you have one, so let's see. Well, actually, let me just rewrite it. So this is going to be equal to one plus square root of two times that is going to be negative two over two. So this is going to be minus one."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And so if you have one, so let's see. Well, actually, let me just rewrite it. So this is going to be equal to one plus square root of two times that is going to be negative two over two. So this is going to be minus one. That's the numerator over here. All of this stuff simplifies to negative one over, this is going to be cosine of negative pi over two. All right, this is negative pi over two."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be minus one. That's the numerator over here. All of this stuff simplifies to negative one over, this is going to be cosine of negative pi over two. All right, this is negative pi over two. Cosine of negative pi over two, if you thought in degrees, that's gonna be negative 90 degrees. Well, cosine of that is just going to be zero. So what we end up with is equal to zero over zero."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "All right, this is negative pi over two. Cosine of negative pi over two, if you thought in degrees, that's gonna be negative 90 degrees. Well, cosine of that is just going to be zero. So what we end up with is equal to zero over zero. And as we've talked about before, if we had something non-zero divided by zero, we'd say, okay, that's undefined. We might as well give up. But when we have this indeterminate form, it does not mean that the limit does not exist."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So what we end up with is equal to zero over zero. And as we've talked about before, if we had something non-zero divided by zero, we'd say, okay, that's undefined. We might as well give up. But when we have this indeterminate form, it does not mean that the limit does not exist. It's usually a clue that we should use some tools in our toolkit, one of which is to do some manipulation here to get an expression that maybe is defined at theta is equal to, or is not an indeterminate form, at theta is equal to negative pi over four, and I will see other tools in our toolkit in the future. So let me algebraically manipulate this a little bit. So if I have one plus the square root of two sine theta over cosine two theta, as you can imagine, the things that might be useful here are our trig identities."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "But when we have this indeterminate form, it does not mean that the limit does not exist. It's usually a clue that we should use some tools in our toolkit, one of which is to do some manipulation here to get an expression that maybe is defined at theta is equal to, or is not an indeterminate form, at theta is equal to negative pi over four, and I will see other tools in our toolkit in the future. So let me algebraically manipulate this a little bit. So if I have one plus the square root of two sine theta over cosine two theta, as you can imagine, the things that might be useful here are our trig identities. And in particular, cosine of two theta seems interesting. Let me write some trig identities involving cosine of two theta. I'll write it over here."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So if I have one plus the square root of two sine theta over cosine two theta, as you can imagine, the things that might be useful here are our trig identities. And in particular, cosine of two theta seems interesting. Let me write some trig identities involving cosine of two theta. I'll write it over here. So we know that cosine of two theta is equal to cosine squared of theta minus sine squared of theta, which is equal to one minus two sine squared of theta, which is equal to two cosine squared theta minus one. And you can go from this one to this one to this one just using the Pythagorean identity, and we proved that in earlier videos in trigonometry on Khan Academy. Now, do any of these look useful?"}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "I'll write it over here. So we know that cosine of two theta is equal to cosine squared of theta minus sine squared of theta, which is equal to one minus two sine squared of theta, which is equal to two cosine squared theta minus one. And you can go from this one to this one to this one just using the Pythagorean identity, and we proved that in earlier videos in trigonometry on Khan Academy. Now, do any of these look useful? Well, all of these three are gonna be differences of squares so we can factor them in interesting ways. And remember, our goal at the end of the day is maybe cancel things out that are making us get this zero over zero. And if I could factor this into something that involves a one plus square root of two sine theta, then I'm gonna be in business."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Now, do any of these look useful? Well, all of these three are gonna be differences of squares so we can factor them in interesting ways. And remember, our goal at the end of the day is maybe cancel things out that are making us get this zero over zero. And if I could factor this into something that involves a one plus square root of two sine theta, then I'm gonna be in business. And it looks like this right over here, that can be factored as one plus square root of two sine theta times one minus square root of two sine theta. So let me use this. Cosine of two theta is the same thing, cosine of two theta is the same thing as one minus two sine squared theta, which is just a difference of squares."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And if I could factor this into something that involves a one plus square root of two sine theta, then I'm gonna be in business. And it looks like this right over here, that can be factored as one plus square root of two sine theta times one minus square root of two sine theta. So let me use this. Cosine of two theta is the same thing, cosine of two theta is the same thing as one minus two sine squared theta, which is just a difference of squares. We can rewrite that as, if this is a squared minus b squared, this is a plus b times a minus b. So I can just replace this with one plus square root of two sine theta times one minus square root of two sine theta. And now we have some nice canceling, or potential canceling that can occur."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Cosine of two theta is the same thing, cosine of two theta is the same thing as one minus two sine squared theta, which is just a difference of squares. We can rewrite that as, if this is a squared minus b squared, this is a plus b times a minus b. So I can just replace this with one plus square root of two sine theta times one minus square root of two sine theta. And now we have some nice canceling, or potential canceling that can occur. So we could say that cancels with that. And we could say that that is going to be equal, and let me do this in a new color, this is going to be equal to, in the numerator we just have one, in the denominator we just are left with one minus square root of two sine theta. And if we want these expressions to truly be equal, we would have to have them to have the same, if you view them as function definitions, as having the same domain."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And now we have some nice canceling, or potential canceling that can occur. So we could say that cancels with that. And we could say that that is going to be equal, and let me do this in a new color, this is going to be equal to, in the numerator we just have one, in the denominator we just are left with one minus square root of two sine theta. And if we want these expressions to truly be equal, we would have to have them to have the same, if you view them as function definitions, as having the same domain. So this one right over here, this one we already saw is not defined at theta is equal to negative pi over four. And so this one, in order for these to be equivalent, we have to say that this one is also not. And actually other places, but let's just say theta does not equal negative pi over four."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And if we want these expressions to truly be equal, we would have to have them to have the same, if you view them as function definitions, as having the same domain. So this one right over here, this one we already saw is not defined at theta is equal to negative pi over four. And so this one, in order for these to be equivalent, we have to say that this one is also not. And actually other places, but let's just say theta does not equal negative pi over four. And we could think about all of this happening in some type of an open interval around negative pi over four if we wanted to get very precise. But if we, for this particular case, well let's just say everything we're doing is in the open interval. So in open interval, in open interval between theta, or say negative one and one."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And actually other places, but let's just say theta does not equal negative pi over four. And we could think about all of this happening in some type of an open interval around negative pi over four if we wanted to get very precise. But if we, for this particular case, well let's just say everything we're doing is in the open interval. So in open interval, in open interval between theta, or say negative one and one. And I think that covers it, because if we have pi over four, that is not going to get us the zero over zero form. And pi over four would make this denominator equal to zero, and it also makes, let's see, pi over four also will make this denominator equal to zero. Because we would get one minus one."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So in open interval, in open interval between theta, or say negative one and one. And I think that covers it, because if we have pi over four, that is not going to get us the zero over zero form. And pi over four would make this denominator equal to zero, and it also makes, let's see, pi over four also will make this denominator equal to zero. Because we would get one minus one. So I think we're good if we're just assuming, if we're restricting to this open interval, and that's okay because we're taking the limit and say it approaches something within this open interval. And I'm being extra precise, because I'm trying to explain it to you, and it's important to be precise. But obviously if you're working this out on a test or notebook, you wouldn't be taking, or taking as much trouble to be putting all of these caveats in."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Because we would get one minus one. So I think we're good if we're just assuming, if we're restricting to this open interval, and that's okay because we're taking the limit and say it approaches something within this open interval. And I'm being extra precise, because I'm trying to explain it to you, and it's important to be precise. But obviously if you're working this out on a test or notebook, you wouldn't be taking, or taking as much trouble to be putting all of these caveats in. So what we've now realized is that, okay, this expression, actually let's think about this. Let's think about the limit, the limit as theta approaches negative pi over four of this thing, without the restriction, of one over one minus the square root of two sine of theta. If we're dealing with this over, you know, in this open interval, or actually even disregarding that, this theta, or this expression is continuous at, it is defined and it is continuous at theta is equal to negative pi over four."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "But obviously if you're working this out on a test or notebook, you wouldn't be taking, or taking as much trouble to be putting all of these caveats in. So what we've now realized is that, okay, this expression, actually let's think about this. Let's think about the limit, the limit as theta approaches negative pi over four of this thing, without the restriction, of one over one minus the square root of two sine of theta. If we're dealing with this over, you know, in this open interval, or actually even disregarding that, this theta, or this expression is continuous at, it is defined and it is continuous at theta is equal to negative pi over four. So this is just going to be equal to one over one minus the square root of two times sine of negative pi over four. Sine of negative pi over four. Sine of negative pi over four, which we've already seen is negative square root of two over two."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "If we're dealing with this over, you know, in this open interval, or actually even disregarding that, this theta, or this expression is continuous at, it is defined and it is continuous at theta is equal to negative pi over four. So this is just going to be equal to one over one minus the square root of two times sine of negative pi over four. Sine of negative pi over four. Sine of negative pi over four, which we've already seen is negative square root of two over two. And so this is going to be equal to one over one minus square root of two times the negative square root of two over two. So negative negative, you get a positive. Square root of two times square root of two is two over two is gonna be one."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Sine of negative pi over four, which we've already seen is negative square root of two over two. And so this is going to be equal to one over one minus square root of two times the negative square root of two over two. So negative negative, you get a positive. Square root of two times square root of two is two over two is gonna be one. So this is going to be equal to 1 1 2. And so, I wanna be very clear. This expression is not the same thing as this expression."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Square root of two times square root of two is two over two is gonna be one. So this is going to be equal to 1 1 2. And so, I wanna be very clear. This expression is not the same thing as this expression. Where the same thing at all values of theta, especially if we're dealing in this open interval, except at theta equals negative pi over four, this one is not defined, and this one is defined. But as we've seen multiple times before, if we find a function that is equal to our original, or an expression that's equal to our original expression, at all values of theta, except where the original one was not defined at a certain point, but this new one is defined and is continuous there, well then these two limits are going to be equal. So if this limit is 1 1 2, then this limit is going to be 1 1 2."}, {"video_title": "Over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Consider the left and right Riemann sums that would approximate the area under y is equal to g of x between x equals two and x equals eight. So we want to approximate this light blue area right over here. Are the approximations overestimations or underestimations? So let's just think about each of them. Let's consider the left and the right Riemann sums. So first the left, and I'm just gonna write left for short, but I'm talking about the left Riemann sum. So they don't tell us how many subdivisions to make for our approximation, so that's up to us to decide."}, {"video_title": "Over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So let's just think about each of them. Let's consider the left and the right Riemann sums. So first the left, and I'm just gonna write left for short, but I'm talking about the left Riemann sum. So they don't tell us how many subdivisions to make for our approximation, so that's up to us to decide. Let's say we went with three subdivisions. Let's say we wanted to make them equal. They don't have to be, but let's say we do."}, {"video_title": "Over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So they don't tell us how many subdivisions to make for our approximation, so that's up to us to decide. Let's say we went with three subdivisions. Let's say we wanted to make them equal. They don't have to be, but let's say we do. So the first one would go from two to four, the next one would go from four to six, and the next one would go from six to eight. If we do a left Riemann sum, you use the left side of each of these subdivisions in order to find the height. You evaluate the function at the left end of each of those subdivisions for the height of our approximating rectangles."}, {"video_title": "Over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "They don't have to be, but let's say we do. So the first one would go from two to four, the next one would go from four to six, and the next one would go from six to eight. If we do a left Riemann sum, you use the left side of each of these subdivisions in order to find the height. You evaluate the function at the left end of each of those subdivisions for the height of our approximating rectangles. So we would use g of two to approximate for, or to set the height of our first approximating rectangle, just like that, and then we would use g of four for the next rectangle. So we would be right over there, and then you'd use g of six to represent the height of our third and our final rectangle right over there. Now, when it's drawn out like this, it's pretty clear that our left Riemann sum is going to be an overestimation."}, {"video_title": "Over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "You evaluate the function at the left end of each of those subdivisions for the height of our approximating rectangles. So we would use g of two to approximate for, or to set the height of our first approximating rectangle, just like that, and then we would use g of four for the next rectangle. So we would be right over there, and then you'd use g of six to represent the height of our third and our final rectangle right over there. Now, when it's drawn out like this, it's pretty clear that our left Riemann sum is going to be an overestimation. Why do we know that? Because these rectangles, the area that they're trying to approximate are always contained in the rectangles, and these rectangles have this surplus area, so they're always going to be larger than the areas that they're trying to approximate. And in general, if you have a function that's decreasing over the interval that we care about right over here, and strictly decreasing the entire time, if you use the left edge of each subdivision to approximate, you're going to have an overestimate, because the left edge, the value of the function there, is going to be higher than the value of the function at any other point in the subdivision."}, {"video_title": "Over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Now, when it's drawn out like this, it's pretty clear that our left Riemann sum is going to be an overestimation. Why do we know that? Because these rectangles, the area that they're trying to approximate are always contained in the rectangles, and these rectangles have this surplus area, so they're always going to be larger than the areas that they're trying to approximate. And in general, if you have a function that's decreasing over the interval that we care about right over here, and strictly decreasing the entire time, if you use the left edge of each subdivision to approximate, you're going to have an overestimate, because the left edge, the value of the function there, is going to be higher than the value of the function at any other point in the subdivision. And so that's why, for a decreasing function, the left Riemann sum is going to be an overestimation. Now, let's think about the right Riemann sum, and you might already guess it's going to be the opposite, but let's visualize that. So let's just go with the same three subdivisions, but now let's use the right side of each of these subdivisions to define the height."}, {"video_title": "Over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And in general, if you have a function that's decreasing over the interval that we care about right over here, and strictly decreasing the entire time, if you use the left edge of each subdivision to approximate, you're going to have an overestimate, because the left edge, the value of the function there, is going to be higher than the value of the function at any other point in the subdivision. And so that's why, for a decreasing function, the left Riemann sum is going to be an overestimation. Now, let's think about the right Riemann sum, and you might already guess it's going to be the opposite, but let's visualize that. So let's just go with the same three subdivisions, but now let's use the right side of each of these subdivisions to define the height. So for this first rectangle, the height is going to be defined by g of four, so that's right over there. And then for the second one, it's going to be g of six, so that is right over there. And for the third one, it's going to be g of eight."}, {"video_title": "Over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So let's just go with the same three subdivisions, but now let's use the right side of each of these subdivisions to define the height. So for this first rectangle, the height is going to be defined by g of four, so that's right over there. And then for the second one, it's going to be g of six, so that is right over there. And for the third one, it's going to be g of eight. And so let me shade these in to make it clear which rectangles we're talking about. This would be the right Riemann sum to approximate the area. And it's very clear here that this is going to be an underestimate, underestimate, because we see in each of these intervals, the Riemann, the right Riemann sum, or the rectangle that we're using for the right Riemann sum is a subset of the area that it's trying to estimate."}, {"video_title": "Over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And for the third one, it's going to be g of eight. And so let me shade these in to make it clear which rectangles we're talking about. This would be the right Riemann sum to approximate the area. And it's very clear here that this is going to be an underestimate, underestimate, because we see in each of these intervals, the Riemann, the right Riemann sum, or the rectangle that we're using for the right Riemann sum is a subset of the area that it's trying to estimate. We're not able to, it doesn't capture this extra area right over here. And once again, that is because this is a strictly decreasing function. So if you use the right endpoint of any one of these, or the right side of any of these subdivisions in order to define the height, that right value of g is going to be the lowest value of g in that subdivision."}, {"video_title": "Over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And it's very clear here that this is going to be an underestimate, underestimate, because we see in each of these intervals, the Riemann, the right Riemann sum, or the rectangle that we're using for the right Riemann sum is a subset of the area that it's trying to estimate. We're not able to, it doesn't capture this extra area right over here. And once again, that is because this is a strictly decreasing function. So if you use the right endpoint of any one of these, or the right side of any of these subdivisions in order to define the height, that right value of g is going to be the lowest value of g in that subdivision. And so it's going to be a lower height than what you could even say is the average height of the value of the function over that interval. So you're going to have an underestimate in this situation. Now, if your function was strictly increasing, then these two things would be swapped around."}, {"video_title": "Addressing treating differentials algebraically AP Calculus AB Khan Academy.mp3", "Sentence": "So when you first learn calculus, you learn that the derivative of some function f could be written as f prime of x is equal to the limit as, and there's multiple ways of doing this, the change in x approaches zero of f of x plus our change in x minus f of x over our change in x. And you learn multiple notations for this. For example, if you know that y is equal to f of x, you might write this as y prime. You might write this as dy dx, which you'll often hear me say is the derivative of y with respect to x, and that you could use the derivative of f with respect to x, because y is equal to our function. But then later on, especially when you start getting into differential equations, you see people start to treat this notation as an actual algebraic expression. For example, you will learn, or you might have already seen, if you're trying to solve the differential equation, the derivative of y with respect to x is equal to y, so the rate of change of y with respect to x is equal to the value of y itself. This is one of the most basic differential equations you might see."}, {"video_title": "Addressing treating differentials algebraically AP Calculus AB Khan Academy.mp3", "Sentence": "You might write this as dy dx, which you'll often hear me say is the derivative of y with respect to x, and that you could use the derivative of f with respect to x, because y is equal to our function. But then later on, especially when you start getting into differential equations, you see people start to treat this notation as an actual algebraic expression. For example, you will learn, or you might have already seen, if you're trying to solve the differential equation, the derivative of y with respect to x is equal to y, so the rate of change of y with respect to x is equal to the value of y itself. This is one of the most basic differential equations you might see. You'll see this technique where people say, well, let's just multiply both sides by dx, just treating dx like as if it's some algebraic expression. So you multiply both sides by dx, and then you have, so that would cancel out algebraically. And so you see people treat it like that."}, {"video_title": "Addressing treating differentials algebraically AP Calculus AB Khan Academy.mp3", "Sentence": "This is one of the most basic differential equations you might see. You'll see this technique where people say, well, let's just multiply both sides by dx, just treating dx like as if it's some algebraic expression. So you multiply both sides by dx, and then you have, so that would cancel out algebraically. And so you see people treat it like that. So you have dy is equal to y times dx, and then they'll say, okay, let's divide both sides by y, which is a reasonable thing to do. Y is an algebraic expression. So if you divide both sides by y, you get one over y dy is equal to dx."}, {"video_title": "Addressing treating differentials algebraically AP Calculus AB Khan Academy.mp3", "Sentence": "And so you see people treat it like that. So you have dy is equal to y times dx, and then they'll say, okay, let's divide both sides by y, which is a reasonable thing to do. Y is an algebraic expression. So if you divide both sides by y, you get one over y dy is equal to dx. And then folks will integrate both sides to find a general solution to this differential equation. But my point on this video isn't to think about how do you solve a differential equation here, but to think about this notion of using what we call differentials, so a dx or a dy, and treating them algebraically like this, treating them as algebraic expressions where I can just multiply both sides by just dx or dy or divide both sides by dx or dy. And I don't normally say this, but the rigor you need to show that this is okay in this situation is not an easy thing to say."}, {"video_title": "Addressing treating differentials algebraically AP Calculus AB Khan Academy.mp3", "Sentence": "So if you divide both sides by y, you get one over y dy is equal to dx. And then folks will integrate both sides to find a general solution to this differential equation. But my point on this video isn't to think about how do you solve a differential equation here, but to think about this notion of using what we call differentials, so a dx or a dy, and treating them algebraically like this, treating them as algebraic expressions where I can just multiply both sides by just dx or dy or divide both sides by dx or dy. And I don't normally say this, but the rigor you need to show that this is okay in this situation is not an easy thing to say. And so to just feel reasonably okay about doing this, this is a little bit hand-wavy. It's not super mathematically rigorous, but it has proven to be a useful tool for us to find these solutions. And conceptually, the way that I think about a dy or a dx is this is the super small change in y in response to a super small change in x."}, {"video_title": "Addressing treating differentials algebraically AP Calculus AB Khan Academy.mp3", "Sentence": "And I don't normally say this, but the rigor you need to show that this is okay in this situation is not an easy thing to say. And so to just feel reasonably okay about doing this, this is a little bit hand-wavy. It's not super mathematically rigorous, but it has proven to be a useful tool for us to find these solutions. And conceptually, the way that I think about a dy or a dx is this is the super small change in y in response to a super small change in x. And that's essentially what this definition of the limit is telling us, especially as delta x approaches zero, we're going to have a super small change in x as delta x approaches zero, and then we're gonna have a resulting super small change in y. So that's one way that you can feel a little bit better of, and this is actually one of the justifications for this type of notation, is you could view this, what's the resulting super small, or what's the super small change in y for a given super small change in x, which is giving us the sense of what's the limiting value of the slope as we go from the slope of a secant line to a tangent line. And if you view it that way, you might feel a little bit better about using the differentials or treating them algebraically, whereas, okay, let me just multiply both sides by that super small change in x."}, {"video_title": "Addressing treating differentials algebraically AP Calculus AB Khan Academy.mp3", "Sentence": "And conceptually, the way that I think about a dy or a dx is this is the super small change in y in response to a super small change in x. And that's essentially what this definition of the limit is telling us, especially as delta x approaches zero, we're going to have a super small change in x as delta x approaches zero, and then we're gonna have a resulting super small change in y. So that's one way that you can feel a little bit better of, and this is actually one of the justifications for this type of notation, is you could view this, what's the resulting super small, or what's the super small change in y for a given super small change in x, which is giving us the sense of what's the limiting value of the slope as we go from the slope of a secant line to a tangent line. And if you view it that way, you might feel a little bit better about using the differentials or treating them algebraically, whereas, okay, let me just multiply both sides by that super small change in x. So the big picture is this is a technique that you will often see in introductory differential equations classes, introductory multivariable classes, and introductory calculus classes, but it's not very mathematically rigorous to just treat differentials like algebraic expressions, but even though it's not very mathematically rigorous to do it willy-nilly like that, it has proven to be very useful. Now, as you get more sophisticated in your mathematics, there are rigorous definitions of a differential where you can get a better sense of where it is mathematically rigorous to use it and where it isn't. But the whole point here is if you felt a little weird feeling about multiplying both sides by dx or dividing both sides by dx or dy, your feeling is mathematically justified because it's not a very rigorous thing to do, at least until you have more rigor behind it."}, {"video_title": "Limits of composite functions external limit doesn't exist AP Calculus Khan Academy.mp3", "Sentence": "On the left here I have the graph of g of x, and on the right here I have the graph of h of x. And what I want to do is figure out what is the limit of g of h of x as x approaches one. Pause this video and see if you can figure that out. All right, now let's do this together. Now the first thing that you might try to say is, all right, let's just figure out first the limit as x approaches one of h of x. And when you look at that, what is that going to be? Well, as we approach one from the left, it looks like h of x is approaching two."}, {"video_title": "Limits of composite functions external limit doesn't exist AP Calculus Khan Academy.mp3", "Sentence": "All right, now let's do this together. Now the first thing that you might try to say is, all right, let's just figure out first the limit as x approaches one of h of x. And when you look at that, what is that going to be? Well, as we approach one from the left, it looks like h of x is approaching two. And as we approach from the right, it looks like h of x is approaching two. So it looks like this is just going to be two. And then we say, okay, well maybe we can then just input that into g. So what is g of two?"}, {"video_title": "Limits of composite functions external limit doesn't exist AP Calculus Khan Academy.mp3", "Sentence": "Well, as we approach one from the left, it looks like h of x is approaching two. And as we approach from the right, it looks like h of x is approaching two. So it looks like this is just going to be two. And then we say, okay, well maybe we can then just input that into g. So what is g of two? Well, g of two is zero, but the limit doesn't seem defined. It looks like when we approach two from the right, we're approaching zero. And when we approach two from the left, we're approaching negative two."}, {"video_title": "Limits of composite functions external limit doesn't exist AP Calculus Khan Academy.mp3", "Sentence": "And then we say, okay, well maybe we can then just input that into g. So what is g of two? Well, g of two is zero, but the limit doesn't seem defined. It looks like when we approach two from the right, we're approaching zero. And when we approach two from the left, we're approaching negative two. So maybe this limit doesn't exist. But if you're thinking that, we haven't fully thought through it. Because what we could do is think about this limit in terms of both left-handed and right-handed limits."}, {"video_title": "Limits of composite functions external limit doesn't exist AP Calculus Khan Academy.mp3", "Sentence": "And when we approach two from the left, we're approaching negative two. So maybe this limit doesn't exist. But if you're thinking that, we haven't fully thought through it. Because what we could do is think about this limit in terms of both left-handed and right-handed limits. So let's think of it this way. First, let's think about what is the limit as x approaches one from the left-hand side of g of h of x. All right, when you think about it this way, if we're approaching one from the left, right over here, we see that we are approaching two from the left, I guess you could say, or we're approaching two from below."}, {"video_title": "Limits of composite functions external limit doesn't exist AP Calculus Khan Academy.mp3", "Sentence": "Because what we could do is think about this limit in terms of both left-handed and right-handed limits. So let's think of it this way. First, let's think about what is the limit as x approaches one from the left-hand side of g of h of x. All right, when you think about it this way, if we're approaching one from the left, right over here, we see that we are approaching two from the left, I guess you could say, or we're approaching two from below. And so the thing that we are inputting into g of x is approaching two from below. So the thing that we are inputting into g is approaching two from below. So if you approach two from below, right over here, what is g approaching?"}, {"video_title": "Limits of composite functions external limit doesn't exist AP Calculus Khan Academy.mp3", "Sentence": "All right, when you think about it this way, if we're approaching one from the left, right over here, we see that we are approaching two from the left, I guess you could say, or we're approaching two from below. And so the thing that we are inputting into g of x is approaching two from below. So the thing that we are inputting into g is approaching two from below. So if you approach two from below, right over here, what is g approaching? It looks like g is approaching negative two. So this looks like it is going to be equal to negative two, at least this left-handed limit. Now let's do a right-handed limit."}, {"video_title": "Limits of composite functions external limit doesn't exist AP Calculus Khan Academy.mp3", "Sentence": "So if you approach two from below, right over here, what is g approaching? It looks like g is approaching negative two. So this looks like it is going to be equal to negative two, at least this left-handed limit. Now let's do a right-handed limit. What is the limit as x approaches one from the right-hand of g of h of x? Well, we can do the same exercise. As we approach one from the right, it looks like h is approaching two from below, from values less than two."}, {"video_title": "Limits of composite functions external limit doesn't exist AP Calculus Khan Academy.mp3", "Sentence": "Now let's do a right-handed limit. What is the limit as x approaches one from the right-hand of g of h of x? Well, we can do the same exercise. As we approach one from the right, it looks like h is approaching two from below, from values less than two. And so if we are approaching two from below, because remember, whatever h is outputting is the input into g. So if the thing that we're inputting into g is approaching two from below, that means that g once again is going to be approaching negative two. So this is a really, really, really interesting case where the limit of g of x as x approaches two does not exist. But because on h of x, when we approach from both the left and the right-hand side, h is approaching two from below."}, {"video_title": "Limits of composite functions external limit doesn't exist AP Calculus Khan Academy.mp3", "Sentence": "As we approach one from the right, it looks like h is approaching two from below, from values less than two. And so if we are approaching two from below, because remember, whatever h is outputting is the input into g. So if the thing that we're inputting into g is approaching two from below, that means that g once again is going to be approaching negative two. So this is a really, really, really interesting case where the limit of g of x as x approaches two does not exist. But because on h of x, when we approach from both the left and the right-hand side, h is approaching two from below. We just have to think about the left-handed limit as we approach two from below or from the left on g. Because in both situations, we are approaching negative two. And so that is going to be our limit. When the left-handed and the right-hand limit are the same, that is going to be your limit."}, {"video_title": "Introduction to integral calculus Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "How do we find the area under this curve? Maybe under the curve and above the x-axis? And let's say between two boundaries. Let's say between x is equal to a and x is equal to b. So let me draw these boundaries right over here. That's our left boundary. This is our right boundary and we wanna think about this area right over here."}, {"video_title": "Introduction to integral calculus Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say between x is equal to a and x is equal to b. So let me draw these boundaries right over here. That's our left boundary. This is our right boundary and we wanna think about this area right over here. Well, without calculus, you could actually get better and better approximations for it. How would you do it? Well, you could divide this section into a bunch of delta x's that go from a to b."}, {"video_title": "Introduction to integral calculus Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "This is our right boundary and we wanna think about this area right over here. Well, without calculus, you could actually get better and better approximations for it. How would you do it? Well, you could divide this section into a bunch of delta x's that go from a to b. They could be equal sections or not, but let's just say for the sake of visualizations, I'm gonna draw roughly equal sections here. So that's the first, that's the second, this is the third, this is the fourth, this is the fifth, and then we have the sixth right over here. And so each of these, this is delta x, let's just call that delta x one, this is delta x two, this width right over here, this is delta x three, all the way to delta x n. I'll try to be general here."}, {"video_title": "Introduction to integral calculus Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Well, you could divide this section into a bunch of delta x's that go from a to b. They could be equal sections or not, but let's just say for the sake of visualizations, I'm gonna draw roughly equal sections here. So that's the first, that's the second, this is the third, this is the fourth, this is the fifth, and then we have the sixth right over here. And so each of these, this is delta x, let's just call that delta x one, this is delta x two, this width right over here, this is delta x three, all the way to delta x n. I'll try to be general here. And so what we could do is let's try to sum up the area of the rectangles defined here. And we could make the height, maybe we make the height based on the value of the function at the right bound. It doesn't have to be, it could be the value of the function someplace in this delta x."}, {"video_title": "Introduction to integral calculus Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And so each of these, this is delta x, let's just call that delta x one, this is delta x two, this width right over here, this is delta x three, all the way to delta x n. I'll try to be general here. And so what we could do is let's try to sum up the area of the rectangles defined here. And we could make the height, maybe we make the height based on the value of the function at the right bound. It doesn't have to be, it could be the value of the function someplace in this delta x. But that's one solution. We're gonna go into a lot more depth into it in future videos. And so we do that."}, {"video_title": "Introduction to integral calculus Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "It doesn't have to be, it could be the value of the function someplace in this delta x. But that's one solution. We're gonna go into a lot more depth into it in future videos. And so we do that. And so now we have an approximation. Or we could say, look, the area of each of these rectangles are going to be f of x sub i, or maybe x sub i is the right boundary the way I've drawn it, times delta x i. That's each of these rectangles."}, {"video_title": "Introduction to integral calculus Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And so we do that. And so now we have an approximation. Or we could say, look, the area of each of these rectangles are going to be f of x sub i, or maybe x sub i is the right boundary the way I've drawn it, times delta x i. That's each of these rectangles. And then we can sum them up. And that would give us an approximation for the area. But as long as we use a finite number, we might say, well, we can always get better by making our delta x's smaller, and then by having more of these rectangles."}, {"video_title": "Introduction to integral calculus Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "That's each of these rectangles. And then we can sum them up. And that would give us an approximation for the area. But as long as we use a finite number, we might say, well, we can always get better by making our delta x's smaller, and then by having more of these rectangles. Or get to a situation, here we're going from i is equal to one to i is equal to n. But what happens is delta x gets thinner and thinner and thinner, and we have, and n gets larger and larger and larger. As delta x gets infinitesimally small, and then as n approaches infinity. And so you're probably sensing something."}, {"video_title": "Introduction to integral calculus Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "But as long as we use a finite number, we might say, well, we can always get better by making our delta x's smaller, and then by having more of these rectangles. Or get to a situation, here we're going from i is equal to one to i is equal to n. But what happens is delta x gets thinner and thinner and thinner, and we have, and n gets larger and larger and larger. As delta x gets infinitesimally small, and then as n approaches infinity. And so you're probably sensing something. Then maybe we could think about the limit as we could say as n approaches infinity, or the limit as delta x becomes very, very, very, very small. And this notion of getting better and better approximations as we take the limit as n approaches infinity, this is the core idea of integral calculus. And it's called integral calculus because the central operation we use, the summing up of an infinite number of infinitesimally thin things is one way to visualize it, is the integral."}, {"video_title": "Introduction to integral calculus Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And so you're probably sensing something. Then maybe we could think about the limit as we could say as n approaches infinity, or the limit as delta x becomes very, very, very, very small. And this notion of getting better and better approximations as we take the limit as n approaches infinity, this is the core idea of integral calculus. And it's called integral calculus because the central operation we use, the summing up of an infinite number of infinitesimally thin things is one way to visualize it, is the integral. That this is going to be the integral, in this case, from a to b. And we're gonna learn a lot more depth. In this case, it is a definite integral of f of x, f of x dx."}, {"video_title": "Introduction to integral calculus Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And it's called integral calculus because the central operation we use, the summing up of an infinite number of infinitesimally thin things is one way to visualize it, is the integral. That this is going to be the integral, in this case, from a to b. And we're gonna learn a lot more depth. In this case, it is a definite integral of f of x, f of x dx. But you can already see the parallels here. You can view the integral sign as like a sigma notation, as a summation sign, but instead of taking the sum of a discrete number of things, you're taking the sum of an infinite number infinitely thin things. Instead of delta x, you now have dx, infinitesimally small things."}, {"video_title": "Introduction to integral calculus Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "In this case, it is a definite integral of f of x, f of x dx. But you can already see the parallels here. You can view the integral sign as like a sigma notation, as a summation sign, but instead of taking the sum of a discrete number of things, you're taking the sum of an infinite number infinitely thin things. Instead of delta x, you now have dx, infinitesimally small things. And this is a notion of an integral. So this right over here is an integral. Now what makes it interesting to calculus, it is using this notion of a limit."}, {"video_title": "Introduction to integral calculus Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Instead of delta x, you now have dx, infinitesimally small things. And this is a notion of an integral. So this right over here is an integral. Now what makes it interesting to calculus, it is using this notion of a limit. But what makes it even more powerful is it's connected to the notion of a derivative, which is one of these beautiful things in mathematics. As we will see in the fundamental theorem of calculus, that integration, the notion of an integral, is closely, tied closely to the notion of a derivative. In fact, the notion of an antiderivative."}, {"video_title": "Introduction to integral calculus Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Now what makes it interesting to calculus, it is using this notion of a limit. But what makes it even more powerful is it's connected to the notion of a derivative, which is one of these beautiful things in mathematics. As we will see in the fundamental theorem of calculus, that integration, the notion of an integral, is closely, tied closely to the notion of a derivative. In fact, the notion of an antiderivative. In differential calculus, we looked at the problem of, hey, if I have some function, I can take its derivative and I can get the derivative of the function. Integral calculus, we're going to be doing a lot of, well, what if we start with a derivative, can we figure out through integration, can we figure out its antiderivative, or the function whose derivative it is? As we will see, all of these are related."}, {"video_title": "Definite integral over a single point AP Calculus AB Khan Academy.mp3", "Sentence": "But let's do something interesting. Let's think about a definite integral of f of x dx. So it's the area under the curve f of x, but instead of it being between two different x values, say a and b, like we've seen multiple times, let's say it's between the same one. Let's say it's between c and c, or let's say c is right over here. What do you think this thing right over here is going to be equal to? What does this represent? What does this equal to?"}, {"video_title": "Definite integral over a single point AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say it's between c and c, or let's say c is right over here. What do you think this thing right over here is going to be equal to? What does this represent? What does this equal to? And I encourage you to pause the video and try to think about it. Well, if you try to visualize it, you're thinking, okay, the area under the curve f of x above the x-axis from x equals c to x equals c. So this region, I guess we could call it, that we think about it, does have a height. The height here is f of c. But what's the width?"}, {"video_title": "Definite integral over a single point AP Calculus AB Khan Academy.mp3", "Sentence": "What does this equal to? And I encourage you to pause the video and try to think about it. Well, if you try to visualize it, you're thinking, okay, the area under the curve f of x above the x-axis from x equals c to x equals c. So this region, I guess we could call it, that we think about it, does have a height. The height here is f of c. But what's the width? Well, there is no width. We're just at a single point. We're not going from c to c plus some delta x, or c plus some even very small change in x, or some other, or c plus some other very small value."}, {"video_title": "Definite integral over a single point AP Calculus AB Khan Academy.mp3", "Sentence": "The height here is f of c. But what's the width? Well, there is no width. We're just at a single point. We're not going from c to c plus some delta x, or c plus some even very small change in x, or some other, or c plus some other very small value. We're just saying at the point c. So we really, when we're thinking about area, we're thinking about a, we're thinking about things, we're thinking about how much two-dimensional space you're taking up. But this idea, this is just a one-dimensional, this is just a one-dimensional, I guess you could think of it as a line segment. What's the area of a line segment?"}, {"video_title": "Definite integral over a single point AP Calculus AB Khan Academy.mp3", "Sentence": "We're not going from c to c plus some delta x, or c plus some even very small change in x, or some other, or c plus some other very small value. We're just saying at the point c. So we really, when we're thinking about area, we're thinking about a, we're thinking about things, we're thinking about how much two-dimensional space you're taking up. But this idea, this is just a one-dimensional, this is just a one-dimensional, I guess you could think of it as a line segment. What's the area of a line segment? Well, a line segment has no area. So this thing right over here is going to be equal to zero. Now, you might say, okay, I get that."}, {"video_title": "Definite integral over a single point AP Calculus AB Khan Academy.mp3", "Sentence": "What's the area of a line segment? Well, a line segment has no area. So this thing right over here is going to be equal to zero. Now, you might say, okay, I get that. I see why that could make sense, why that makes intuitive sense. I have, I'm trying to find the area of a rectangle where I know its height, but it has, its width is zero, so that area is going to be zero, is one way to think about it. But Sal, why are you even pointing this out to me?"}, {"video_title": "Linear approximation of a rational function Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So there are situations where you have some type of a function. This is clearly a nonlinear function. F of x is equal to one over x minus one. This is its graph, or at least part of its graph right over here. But where you want to approximate it with a linear function, especially around a certain value. And so what we're going to do is we want to find an approximation. Let me write this down."}, {"video_title": "Linear approximation of a rational function Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "This is its graph, or at least part of its graph right over here. But where you want to approximate it with a linear function, especially around a certain value. And so what we're going to do is we want to find an approximation. Let me write this down. I want to find an approximation for, and actually let me be clear. I want to find a linear approximation. So I'm going to approximate it with a line."}, {"video_title": "Linear approximation of a rational function Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Let me write this down. I want to find an approximation for, and actually let me be clear. I want to find a linear approximation. So I'm going to approximate it with a line. I want to find a linear approximation, approximation of f, of f around, and you need to know where you're going to be approximating it, around x equals negative one. So what do we mean by that? Well, let's look at this graph over here."}, {"video_title": "Linear approximation of a rational function Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So I'm going to approximate it with a line. I want to find a linear approximation, approximation of f, of f around, and you need to know where you're going to be approximating it, around x equals negative one. So what do we mean by that? Well, let's look at this graph over here. On this curve, when x is equal to negative one, f of negative one is negative, is negative one half, which sticks us right over there. Let me do this in a better color. So it's right over there."}, {"video_title": "Linear approximation of a rational function Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, let's look at this graph over here. On this curve, when x is equal to negative one, f of negative one is negative, is negative one half, which sticks us right over there. Let me do this in a better color. So it's right over there. And what we want to do is approximate it with a line around that. And what we're essentially going to do is we're going to approximate it with the equation of the tangent line. The tangent line is going to look something, something like that."}, {"video_title": "Linear approximation of a rational function Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So it's right over there. And what we want to do is approximate it with a line around that. And what we're essentially going to do is we're going to approximate it with the equation of the tangent line. The tangent line is going to look something, something like that. And as we can see, as we get further and further from x equals negative one, the approximation gets worse and worse. But if we stay around x equals negative one, well, it's a decent, it is as good as you can get for a linear approximation, or at least in this example, it is a very good linear approximation. So when people say, hey, find a linear approximation of f around x equals negative one, or if they say, what is the following is the best approximation, and all of your choices are lines, well, essentially they're asking you to find the equation of the tangent line at x equals negative one."}, {"video_title": "Linear approximation of a rational function Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "The tangent line is going to look something, something like that. And as we can see, as we get further and further from x equals negative one, the approximation gets worse and worse. But if we stay around x equals negative one, well, it's a decent, it is as good as you can get for a linear approximation, or at least in this example, it is a very good linear approximation. So when people say, hey, find a linear approximation of f around x equals negative one, or if they say, what is the following is the best approximation, and all of your choices are lines, well, essentially they're asking you to find the equation of the tangent line at x equals negative one. So let's do that. So in order to find the equation of the tangent line, the equation of a line is y is equal to mx plus b, where m is the slope and b is the y-intercept. There's other ways that you could think about it."}, {"video_title": "Linear approximation of a rational function Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So when people say, hey, find a linear approximation of f around x equals negative one, or if they say, what is the following is the best approximation, and all of your choices are lines, well, essentially they're asking you to find the equation of the tangent line at x equals negative one. So let's do that. So in order to find the equation of the tangent line, the equation of a line is y is equal to mx plus b, where m is the slope and b is the y-intercept. There's other ways that you could think about it. You could think about it in terms of point slope, where you could say y minus some y that sits on that line is equal to the slope times x minus the corresponding x one. So x one comma y one sits on that line someplace. Actually, I like to write this point slope form like this sometimes."}, {"video_title": "Linear approximation of a rational function Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "There's other ways that you could think about it. You could think about it in terms of point slope, where you could say y minus some y that sits on that line is equal to the slope times x minus the corresponding x one. So x one comma y one sits on that line someplace. Actually, I like to write this point slope form like this sometimes. Y minus y one over x minus x one is equal to b. Because this comes straight out of the idea of, look, if x one and y one are on the line, the slope between any other point on the line and that point is going to be your slope of your line. So we could think about it any of these ways."}, {"video_title": "Linear approximation of a rational function Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Actually, I like to write this point slope form like this sometimes. Y minus y one over x minus x one is equal to b. Because this comes straight out of the idea of, look, if x one and y one are on the line, the slope between any other point on the line and that point is going to be your slope of your line. So we could think about it any of these ways. So let's first find the slope of the tangent line, and that's where the derivative is useful. So f, well, actually, let me just write f of x again. So I'm gonna write it as x minus one to the negative one power, because that makes it a little bit clearer that we can use the power rule and a little bit of the chain rule."}, {"video_title": "Linear approximation of a rational function Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So we could think about it any of these ways. So let's first find the slope of the tangent line, and that's where the derivative is useful. So f, well, actually, let me just write f of x again. So I'm gonna write it as x minus one to the negative one power, because that makes it a little bit clearer that we can use the power rule and a little bit of the chain rule. So the derivative of f with respect to x is equal to, so the derivative of x minus one to the negative one with respect to x minus one, well, that's just going to be, we're just going to use the power rule here. It's gonna be negative one times x minus one to the negative two, and then we're gonna multiply that times the derivative of x minus one with respect to x. Well, that's just going to be one."}, {"video_title": "Linear approximation of a rational function Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So I'm gonna write it as x minus one to the negative one power, because that makes it a little bit clearer that we can use the power rule and a little bit of the chain rule. So the derivative of f with respect to x is equal to, so the derivative of x minus one to the negative one with respect to x minus one, well, that's just going to be, we're just going to use the power rule here. It's gonna be negative one times x minus one to the negative two, and then we're gonna multiply that times the derivative of x minus one with respect to x. Well, that's just going to be one. The derivative of x with respect to x is one. Derivative of negative one with respect to x is zero, so we could say times one here if we like, or we could just not write that because it doesn't change the value. And so let's evaluate that when x is equal to negative one."}, {"video_title": "Linear approximation of a rational function Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, that's just going to be one. The derivative of x with respect to x is one. Derivative of negative one with respect to x is zero, so we could say times one here if we like, or we could just not write that because it doesn't change the value. And so let's evaluate that when x is equal to negative one. So f prime of negative one is equal to, I could just write this as negative, all right, look this way, negative one over negative one minus one squared. Minus one squared. And so this is going to be negative two down here."}, {"video_title": "Linear approximation of a rational function Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And so let's evaluate that when x is equal to negative one. So f prime of negative one is equal to, I could just write this as negative, all right, look this way, negative one over negative one minus one squared. Minus one squared. And so this is going to be negative two down here. So this is equal to negative negative 1 1 4. So the slope of our tangent line is, so I could write it this way, m is equal to negative one negative 1 1 4. And so now we just have to write its equation down."}, {"video_title": "Linear approximation of a rational function Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And so this is going to be negative two down here. So this is equal to negative negative 1 1 4. So the slope of our tangent line is, so I could write it this way, m is equal to negative one negative 1 1 4. And so now we just have to write its equation down. So we already know an x one and a y one that sits on the line. In fact, we want to use the point when x equals negative one so we know that the point negative one comma, we could just input it right over here, f of negative one is negative 1 1 2, one over negative one minus one, negative 1 1 2. So we know that this negative one comma negative 1 1 2, that that is on our curve and it is on our line, that's the point at which the tangent and the curve actually intersect."}, {"video_title": "Linear approximation of a rational function Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And so now we just have to write its equation down. So we already know an x one and a y one that sits on the line. In fact, we want to use the point when x equals negative one so we know that the point negative one comma, we could just input it right over here, f of negative one is negative 1 1 2, one over negative one minus one, negative 1 1 2. So we know that this negative one comma negative 1 1 2, that that is on our curve and it is on our line, that's the point at which the tangent and the curve actually intersect. And so we can use any of these to now write the equation of our line. We could say y, I'll do it right here, y minus y one, so minus negative 1 1 2 is going to be equal to, is going to be equal to our slope negative 1 4. I'm just using the point slope version of our equation."}, {"video_title": "Linear approximation of a rational function Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So we know that this negative one comma negative 1 1 2, that that is on our curve and it is on our line, that's the point at which the tangent and the curve actually intersect. And so we can use any of these to now write the equation of our line. We could say y, I'll do it right here, y minus y one, so minus negative 1 1 2 is going to be equal to, is going to be equal to our slope negative 1 4. I'm just using the point slope version of our equation. Is equal to our slope times x minus x one. So x minus our x coordinate that we know sits on this. So minus negative one."}, {"video_title": "Linear approximation of a rational function Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "I'm just using the point slope version of our equation. Is equal to our slope times x minus x one. So x minus our x coordinate that we know sits on this. So minus negative one. And so let me now write all of this in a neutral color. This will be y plus 1 1 2 is equal to, and I can, so this is going to be plus one right over there. So I can distribute the negative 1 4."}, {"video_title": "Linear approximation of a rational function Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So minus negative one. And so let me now write all of this in a neutral color. This will be y plus 1 1 2 is equal to, and I can, so this is going to be plus one right over there. So I can distribute the negative 1 4. So it's negative 1 4 x minus 1 4, minus 1 4, minus 1 4. And then I can subtract 1 1 2 from both sides. So I'm going to get y is equal to negative 1 4 x."}, {"video_title": "Linear approximation of a rational function Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So I can distribute the negative 1 4. So it's negative 1 4 x minus 1 4, minus 1 4, minus 1 4. And then I can subtract 1 1 2 from both sides. So I'm going to get y is equal to negative 1 4 x. And then if I already am subtracting 1 4 and I subtract another half, it's going to be negative 3 4. So minus 3, minus 3 4. So, and that's actually pretty close to what I drew up here."}, {"video_title": "Linear approximation of a rational function Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So I'm going to get y is equal to negative 1 4 x. And then if I already am subtracting 1 4 and I subtract another half, it's going to be negative 3 4. So minus 3, minus 3 4. So, and that's actually pretty close to what I drew up here. This should be intersecting the y axis at negative 3 4ths. So there you have it. This line, or you could even say this equation, is going to be a very good linear approximation, about as good as you can get for a linear approximation, for that non-linear function around x equals negative one."}, {"video_title": "Worked example over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "The continuous function g is graphed. We're interested in the area under the curve between x equals negative seven and x equals seven. And we're considering using Riemann sums to approximate it. So this is the area that we're thinking about in this light blue color. Order the areas from least on top to greatest on bottom. So this is a screenshot from a Khan Academy exercise where you would be expected to actually click and drag these around. But it's just a screenshot."}, {"video_title": "Worked example over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So this is the area that we're thinking about in this light blue color. Order the areas from least on top to greatest on bottom. So this is a screenshot from a Khan Academy exercise where you would be expected to actually click and drag these around. But it's just a screenshot. So what I'm gonna do instead of dragging them around, I'm just gonna write numbers ordering them from least to greatest, where one would be the least and then three would be the greatest. So pause this video and try to think about these. Which of these is the least, which is in the middle, and which is the greatest?"}, {"video_title": "Worked example over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "But it's just a screenshot. So what I'm gonna do instead of dragging them around, I'm just gonna write numbers ordering them from least to greatest, where one would be the least and then three would be the greatest. So pause this video and try to think about these. Which of these is the least, which is in the middle, and which is the greatest? So let's just draw out what a left Riemann sum, a right Riemann sum would actually look like and compare it to the actual area. And we could do an arbitrary number of subdivisions. I would encourage us to do fewer because we're just trying to get a general sense of things."}, {"video_title": "Worked example over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Which of these is the least, which is in the middle, and which is the greatest? So let's just draw out what a left Riemann sum, a right Riemann sum would actually look like and compare it to the actual area. And we could do an arbitrary number of subdivisions. I would encourage us to do fewer because we're just trying to get a general sense of things. And they don't even have to be equal subdivisions. So let's start with a left Riemann sum. So we wanna start at x equals negative seven."}, {"video_title": "Worked example over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "I would encourage us to do fewer because we're just trying to get a general sense of things. And they don't even have to be equal subdivisions. So let's start with a left Riemann sum. So we wanna start at x equals negative seven. And we wanna go to x equals seven. Well, let's say that this is the first rectangle right over here. So this is our first subdivision."}, {"video_title": "Worked example over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So we wanna start at x equals negative seven. And we wanna go to x equals seven. Well, let's say that this is the first rectangle right over here. So this is our first subdivision. And it's a left Riemann sum. So we would use the value of the function at the left end of that subdivision, which is negative seven, x equals negative seven. The value of the function there is 12."}, {"video_title": "Worked example over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So this is our first subdivision. And it's a left Riemann sum. So we would use the value of the function at the left end of that subdivision, which is negative seven, x equals negative seven. The value of the function there is 12. And so this would be our first rectangle. You already get a sense that this is gonna be an overestimate relative to the actual area. And so the next subdivision would start here."}, {"video_title": "Worked example over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "The value of the function there is 12. And so this would be our first rectangle. You already get a sense that this is gonna be an overestimate relative to the actual area. And so the next subdivision would start here. So this would be our height of our rectangle. And once again, they don't have to be equal subdivisions. They often are, but I'm gonna show you unequal subdivisions just to show you that this is still a valid Riemann sum."}, {"video_title": "Worked example over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And so the next subdivision would start here. So this would be our height of our rectangle. And once again, they don't have to be equal subdivisions. They often are, but I'm gonna show you unequal subdivisions just to show you that this is still a valid Riemann sum. And once again, this is an overestimate where the actual area that we're trying to approximate is smaller than the area of this rectangle. And then let's say this third subdivision right over here starts right over there at x equals three. And we use the left end of the subdivision, the value of the function there to define the height of the rectangle."}, {"video_title": "Worked example over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "They often are, but I'm gonna show you unequal subdivisions just to show you that this is still a valid Riemann sum. And once again, this is an overestimate where the actual area that we're trying to approximate is smaller than the area of this rectangle. And then let's say this third subdivision right over here starts right over there at x equals three. And we use the left end of the subdivision, the value of the function there to define the height of the rectangle. And once again, you see it is an overestimate. So the left Riemann sum is clearly an overestimate. And it's pretty clear why."}, {"video_title": "Worked example over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And we use the left end of the subdivision, the value of the function there to define the height of the rectangle. And once again, you see it is an overestimate. So the left Riemann sum is clearly an overestimate. And it's pretty clear why. This function is, this function never increases. It's either decreasing or it looks like it stays flat at certain points. And so for a function like that, the left edge, the value of the function at the left edge is going to be just as high or higher than any other value, than any other value the function takes on over that interval for the subdivision."}, {"video_title": "Worked example over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And it's pretty clear why. This function is, this function never increases. It's either decreasing or it looks like it stays flat at certain points. And so for a function like that, the left edge, the value of the function at the left edge is going to be just as high or higher than any other value, than any other value the function takes on over that interval for the subdivision. And so you get left with all of this extra area that is part of the overestimate or this area that is larger than the actual area that you're trying to approximate. Now let's think about a right Riemann sum. And I'll do different subdivisions."}, {"video_title": "Worked example over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And so for a function like that, the left edge, the value of the function at the left edge is going to be just as high or higher than any other value, than any other value the function takes on over that interval for the subdivision. And so you get left with all of this extra area that is part of the overestimate or this area that is larger than the actual area that you're trying to approximate. Now let's think about a right Riemann sum. And I'll do different subdivisions. Let's say the first subdivision goes from negative seven to negative five. And here we would use the right edge to define the height. So f of negative five or g of negative five, I should say."}, {"video_title": "Worked example over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And I'll do different subdivisions. Let's say the first subdivision goes from negative seven to negative five. And here we would use the right edge to define the height. So f of negative five or g of negative five, I should say. So that's right over there. That's our first rectangle. Maybe our next rectangle, the right edge is zero."}, {"video_title": "Worked example over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So f of negative five or g of negative five, I should say. So that's right over there. That's our first rectangle. Maybe our next rectangle, the right edge is zero. So this would be it right over there. And then maybe we'll do four rectangles. Maybe our third subdivision, the right edge, is at x is equal to three."}, {"video_title": "Worked example over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Maybe our next rectangle, the right edge is zero. So this would be it right over there. And then maybe we'll do four rectangles. Maybe our third subdivision, the right edge, is at x is equal to three. So it would be right over there. And then our fourth subdivision, let's just do it at x equals seven. And we're using the right edge of the subdivisions."}, {"video_title": "Worked example over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Maybe our third subdivision, the right edge, is at x is equal to three. So it would be right over there. And then our fourth subdivision, let's just do it at x equals seven. And we're using the right edge of the subdivisions. Remember, this is a right Riemann sum. So we use the right edge, the value of the function there is just like that. And now you can see for any one of these subdivisions, our rectangles are underestimates of the area under the curve."}, {"video_title": "Worked example over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And we're using the right edge of the subdivisions. Remember, this is a right Riemann sum. So we use the right edge, the value of the function there is just like that. And now you can see for any one of these subdivisions, our rectangles are underestimates of the area under the curve. Under, underestimate. And that's because, once again, in this particular case, the function never increases. It's either decreasing or staying flat."}, {"video_title": "Worked example over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And now you can see for any one of these subdivisions, our rectangles are underestimates of the area under the curve. Under, underestimate. And that's because, once again, in this particular case, the function never increases. It's either decreasing or staying flat. So if you use the value of the function at the right edge, it's going to be smaller, it's never going to be larger than the value that the function takes on in the rest of that subdivision. And so we are continuously underestimating. We're missing all of this area right over there is not being included, so we have an underestimate."}, {"video_title": "Worked example over- and under-estimation of Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "It's either decreasing or staying flat. So if you use the value of the function at the right edge, it's going to be smaller, it's never going to be larger than the value that the function takes on in the rest of that subdivision. And so we are continuously underestimating. We're missing all of this area right over there is not being included, so we have an underestimate. So if we want to rank these from least to greatest, well, the right Riemann sum is the least. It is underestimating it. Then you have the actual area of the curve, which is just the area of the curve."}, {"video_title": "Analyzing mistakes when finding extrema (example 1) AP Calculus AB Khan Academy.mp3", "Sentence": "This is her solution. So step one, it looks like she tried to take the derivative. Step two, she tries to find the solution, find where the derivative is equal to zero, and she found that it happens at x equals two, so she says that's a critical point. And step three, she says she makes a conclusion that therefore h has a relative extremum there. Is Pamela's work correct? If not, what's her mistake? So pause this video and try to work through it yourself and see if Pamela's work is correct."}, {"video_title": "Analyzing mistakes when finding extrema (example 1) AP Calculus AB Khan Academy.mp3", "Sentence": "And step three, she says she makes a conclusion that therefore h has a relative extremum there. Is Pamela's work correct? If not, what's her mistake? So pause this video and try to work through it yourself and see if Pamela's work is correct. All right, well I'm just gonna try to do it again in parallel, so first let me just take the derivative here. So h prime of x, just use the power rule multiple times, is gonna be three x squared for the x to the third. Two times negative six is negative 12 or minus 12x, and the derivative of 12x is plus 12."}, {"video_title": "Analyzing mistakes when finding extrema (example 1) AP Calculus AB Khan Academy.mp3", "Sentence": "So pause this video and try to work through it yourself and see if Pamela's work is correct. All right, well I'm just gonna try to do it again in parallel, so first let me just take the derivative here. So h prime of x, just use the power rule multiple times, is gonna be three x squared for the x to the third. Two times negative six is negative 12 or minus 12x, and the derivative of 12x is plus 12. And let's see, you can factor out a three here, so it's three times x squared minus four x plus four, and this part is indeed equal to x minus two squared, so this is equal to three times x minus two squared. So her step one looks right on target. Okay, step two, the solution of h prime of x is equal to zero is equal to x equals two."}, {"video_title": "Analyzing mistakes when finding extrema (example 1) AP Calculus AB Khan Academy.mp3", "Sentence": "Two times negative six is negative 12 or minus 12x, and the derivative of 12x is plus 12. And let's see, you can factor out a three here, so it's three times x squared minus four x plus four, and this part is indeed equal to x minus two squared, so this is equal to three times x minus two squared. So her step one looks right on target. Okay, step two, the solution of h prime of x is equal to zero is equal to x equals two. Yeah, that works out. If you were to say three times x minus two squared, which is h prime of x, the first derivative, and set that equal to zero, this is going to be true when x is equal to two. And so any point where your first derivative is equal to zero or it's undefined, it is indeed a critical point."}, {"video_title": "Analyzing mistakes when finding extrema (example 1) AP Calculus AB Khan Academy.mp3", "Sentence": "Okay, step two, the solution of h prime of x is equal to zero is equal to x equals two. Yeah, that works out. If you were to say three times x minus two squared, which is h prime of x, the first derivative, and set that equal to zero, this is going to be true when x is equal to two. And so any point where your first derivative is equal to zero or it's undefined, it is indeed a critical point. So this step looks good so far. And then step three, h has a relative extremum at x equals two. All right, so she made a big conclusion here."}, {"video_title": "Analyzing mistakes when finding extrema (example 1) AP Calculus AB Khan Academy.mp3", "Sentence": "And so any point where your first derivative is equal to zero or it's undefined, it is indeed a critical point. So this step looks good so far. And then step three, h has a relative extremum at x equals two. All right, so she made a big conclusion here. She assumed that because the derivative was zero, that we have a relative extremum. So let's just see if you can even just make that conclusion. In order to have a relative extremum, your curve is gonna look something like this, and then you would have a relative extremum right over here."}, {"video_title": "Analyzing mistakes when finding extrema (example 1) AP Calculus AB Khan Academy.mp3", "Sentence": "All right, so she made a big conclusion here. She assumed that because the derivative was zero, that we have a relative extremum. So let's just see if you can even just make that conclusion. In order to have a relative extremum, your curve is gonna look something like this, and then you would have a relative extremum right over here. And over here, your slope goes from being positive, and then it hits zero, and then it goes to being negative. Or you could have a relative extremum like this. This would be a maximum point."}, {"video_title": "Analyzing mistakes when finding extrema (example 1) AP Calculus AB Khan Academy.mp3", "Sentence": "In order to have a relative extremum, your curve is gonna look something like this, and then you would have a relative extremum right over here. And over here, your slope goes from being positive, and then it hits zero, and then it goes to being negative. Or you could have a relative extremum like this. This would be a maximum point. This would be a minimum point right over here. And then in a minimum point, your slope is zero right over there, but right before it, your slope was negative, and it goes to being positive. But you actually have cases where your derivative, your first derivative is zero, but you don't have an extremum."}, {"video_title": "Analyzing mistakes when finding extrema (example 1) AP Calculus AB Khan Academy.mp3", "Sentence": "This would be a maximum point. This would be a minimum point right over here. And then in a minimum point, your slope is zero right over there, but right before it, your slope was negative, and it goes to being positive. But you actually have cases where your derivative, your first derivative is zero, but you don't have an extremum. So for example, you could have a point like this, where right over here, your slope or your derivative could be equal to zero, and so your first derivative would be equal to zero, but notice, your slope is positive, it hits zero, and then it goes back to being positive again. And so you can't make the conclusion just because your derivative is zero that it's definitely an extremum. You could say it's a critical point, and so step two is correct."}, {"video_title": "Analyzing mistakes when finding extrema (example 1) AP Calculus AB Khan Academy.mp3", "Sentence": "But you actually have cases where your derivative, your first derivative is zero, but you don't have an extremum. So for example, you could have a point like this, where right over here, your slope or your derivative could be equal to zero, and so your first derivative would be equal to zero, but notice, your slope is positive, it hits zero, and then it goes back to being positive again. And so you can't make the conclusion just because your derivative is zero that it's definitely an extremum. You could say it's a critical point, and so step two is correct. In order to make this conclusion, you would have to test what the derivative is doing before that point and after that point, and verify that it is switching sides. And we could try to do that. So let's make a little table here."}, {"video_title": "Analyzing mistakes when finding extrema (example 1) AP Calculus AB Khan Academy.mp3", "Sentence": "You could say it's a critical point, and so step two is correct. In order to make this conclusion, you would have to test what the derivative is doing before that point and after that point, and verify that it is switching sides. And we could try to do that. So let's make a little table here. Make a little table, a little bit neater. So x, x, h prime of x right over here. We know at x equals two, h prime of two is zero, that's our critical point."}, {"video_title": "Analyzing mistakes when finding extrema (example 1) AP Calculus AB Khan Academy.mp3", "Sentence": "So let's make a little table here. Make a little table, a little bit neater. So x, x, h prime of x right over here. We know at x equals two, h prime of two is zero, that's our critical point. But let's try, I don't know, let's see what happens when x is equal to one, and then let's see what happens when x equals three. We're just sampling points on either side of two. And let's see, we are going to have, when x is equal to one, h prime of one is three times one minus two squared."}, {"video_title": "Analyzing mistakes when finding extrema (example 1) AP Calculus AB Khan Academy.mp3", "Sentence": "We know at x equals two, h prime of two is zero, that's our critical point. But let's try, I don't know, let's see what happens when x is equal to one, and then let's see what happens when x equals three. We're just sampling points on either side of two. And let's see, we are going to have, when x is equal to one, h prime of one is three times one minus two squared. One minus two, negative one squared is positive one times three is still positive. And then three, well, three minus two squared times three, that's also gonna be three. So this is actually a situation where, like I just drawn it, where our slope is positive before we hit the critical point."}, {"video_title": "Analyzing mistakes when finding extrema (example 1) AP Calculus AB Khan Academy.mp3", "Sentence": "And let's see, we are going to have, when x is equal to one, h prime of one is three times one minus two squared. One minus two, negative one squared is positive one times three is still positive. And then three, well, three minus two squared times three, that's also gonna be three. So this is actually a situation where, like I just drawn it, where our slope is positive before we hit the critical point. It gets to zero, but then it starts becoming positive again. And so that's why you actually have to do this test in order to identify whether it's an extremum. It turns out that this is not an extremum."}, {"video_title": "Analyzing mistakes when finding extrema (example 1) AP Calculus AB Khan Academy.mp3", "Sentence": "So this is actually a situation where, like I just drawn it, where our slope is positive before we hit the critical point. It gets to zero, but then it starts becoming positive again. And so that's why you actually have to do this test in order to identify whether it's an extremum. It turns out that this is not an extremum. This is not a maximum or a minimum point here. So Pamela's work is not correct, and her mistake is step three. In order to make this conclusion, you would have to test on either side of that critical point, test the first derivative."}, {"video_title": "Generalizing disc method around x-axis AP Calculus AB Khan Academy.mp3", "Sentence": "It's not for you to memorize that formula. I highly recommend against that because then you really won't know what's going on. It's really just to do it, it's better to do it from first principles where you find the volume of each of these disks and think of it that way. But let's just generalize what we saw in the last video. So instead of saying that this is y is equal to x squared, let's say that this is the graph, this function right over here, this function right over here, let's just generalize it and call it y is equal to f of x. And instead of saying we're going between 0 and 2, let's just say we're going between a and b. So these are just two endpoints along the x-axis."}, {"video_title": "Generalizing disc method around x-axis AP Calculus AB Khan Academy.mp3", "Sentence": "But let's just generalize what we saw in the last video. So instead of saying that this is y is equal to x squared, let's say that this is the graph, this function right over here, this function right over here, let's just generalize it and call it y is equal to f of x. And instead of saying we're going between 0 and 2, let's just say we're going between a and b. So these are just two endpoints along the x-axis. So how would we find the volume of this? Well, just like the last video, we would still take a disk just like this, but what is the height of the disk now? Well, the height of the disk is not just x squared, we've generalized it."}, {"video_title": "Generalizing disc method around x-axis AP Calculus AB Khan Academy.mp3", "Sentence": "So these are just two endpoints along the x-axis. So how would we find the volume of this? Well, just like the last video, we would still take a disk just like this, but what is the height of the disk now? Well, the height of the disk is not just x squared, we've generalized it. It is going to be whatever the value of our function is at that point. So the height of the disk is going to be f of x. The area of the space of this disk is going to be pi times our radius squared."}, {"video_title": "Generalizing disc method around x-axis AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the height of the disk is not just x squared, we've generalized it. It is going to be whatever the value of our function is at that point. So the height of the disk is going to be f of x. The area of the space of this disk is going to be pi times our radius squared. So our radius is f of x, and we are just going to square it. That's the area of this face right over here. What is the volume of our disk?"}, {"video_title": "Generalizing disc method around x-axis AP Calculus AB Khan Academy.mp3", "Sentence": "The area of the space of this disk is going to be pi times our radius squared. So our radius is f of x, and we are just going to square it. That's the area of this face right over here. What is the volume of our disk? We're just going to have to multiply that times our depth. So it's going to be that times dx. And we want to take the sum of all of these disks from a to b."}, {"video_title": "Generalizing disc method around x-axis AP Calculus AB Khan Academy.mp3", "Sentence": "What is the volume of our disk? We're just going to have to multiply that times our depth. So it's going to be that times dx. And we want to take the sum of all of these disks from a to b. And we're going to take the sum of them and then take the limit as the dx's get smaller and smaller and smaller, and we have an infinite number of these disks. And that means we're just going to take a definite integral. So we're going to take the definite integral of this from a to b."}, {"video_title": "Generalizing disc method around x-axis AP Calculus AB Khan Academy.mp3", "Sentence": "And we want to take the sum of all of these disks from a to b. And we're going to take the sum of them and then take the limit as the dx's get smaller and smaller and smaller, and we have an infinite number of these disks. And that means we're just going to take a definite integral. So we're going to take the definite integral of this from a to b. And this right here is the formula that you will often see in a calculus book for using the disk method when you're rotating around the x-axis. But I just wanted to show you that it comes out of the common sense of finding the volume of this disk. The f of x right over here is just the radius of the disk."}, {"video_title": "Generalizing disc method around x-axis AP Calculus AB Khan Academy.mp3", "Sentence": "So we're going to take the definite integral of this from a to b. And this right here is the formula that you will often see in a calculus book for using the disk method when you're rotating around the x-axis. But I just wanted to show you that it comes out of the common sense of finding the volume of this disk. The f of x right over here is just the radius of the disk. So this part right over here is just really pi r squared. We multiply it times the depth, and then we take the sum from a to b. And we take the sum from a to b of all of the disks."}, {"video_title": "Worked examples Summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Which expression is equal to the sum above? And they tell us choose all answers that apply. So like always, pause the video and see if you can work through this on your own. So when you look at the sum, it's clear you're starting at two, you're adding three each time, and we also are dealing with you have four total terms. Now we could try to construct an equation here or an expression using sigma notation, but instead what I like to do is look at our options. We really only have to look at these two options here and expand them out. What sum would each of these be?"}, {"video_title": "Worked examples Summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So when you look at the sum, it's clear you're starting at two, you're adding three each time, and we also are dealing with you have four total terms. Now we could try to construct an equation here or an expression using sigma notation, but instead what I like to do is look at our options. We really only have to look at these two options here and expand them out. What sum would each of these be? Well this is going to be the same thing as. We're starting at n equals one. So this is going to be three times when n equals one, three times one minus one, and then plus, then we'll go to n equals two, three times two minus one."}, {"video_title": "Worked examples Summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "What sum would each of these be? Well this is going to be the same thing as. We're starting at n equals one. So this is going to be three times when n equals one, three times one minus one, and then plus, then we'll go to n equals two, three times two minus one. Then we're gonna go to n equals three, so plus three times three minus one. And then finally we're gonna go to n equals four. So plus three times, when n equals four, this is three times four minus one."}, {"video_title": "Worked examples Summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be three times when n equals one, three times one minus one, and then plus, then we'll go to n equals two, three times two minus one. Then we're gonna go to n equals three, so plus three times three minus one. And then finally we're gonna go to n equals four. So plus three times, when n equals four, this is three times four minus one. So just to be clear, this is what we did when n equals one. Let me write it down, n equals one. This is what we got when n equals two."}, {"video_title": "Worked examples Summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So plus three times, when n equals four, this is three times four minus one. So just to be clear, this is what we did when n equals one. Let me write it down, n equals one. This is what we got when n equals two. This is what we got when n equals three. And this is what we got when n is equal to four. And we stopped at n equals four because it tells us right over there."}, {"video_title": "Worked examples Summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "This is what we got when n equals two. This is what we got when n equals three. And this is what we got when n is equal to four. And we stopped at n equals four because it tells us right over there. We start at n equals one and we go all the way to n equals four. So what does this equal? So let's see, three times one minus one is two, so this is looking good so far."}, {"video_title": "Worked examples Summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And we stopped at n equals four because it tells us right over there. We start at n equals one and we go all the way to n equals four. So what does this equal? So let's see, three times one minus one is two, so this is looking good so far. Three times two minus one, that's six minus one, that is five, so still looking good. Three times three minus one, that's nine minus one. Once again, still looking good."}, {"video_title": "Worked examples Summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So let's see, three times one minus one is two, so this is looking good so far. Three times two minus one, that's six minus one, that is five, so still looking good. Three times three minus one, that's nine minus one. Once again, still looking good. Three times four minus one, that is 11. So we like this choice, so I would definitely select this one. Now let's do the same thing over here."}, {"video_title": "Worked examples Summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Once again, still looking good. Three times four minus one, that is 11. So we like this choice, so I would definitely select this one. Now let's do the same thing over here. When n is equal to zero, it's going to be two plus three times zero, so that's just going to be two. And then plus, when n is equal to one, it's going to be two plus three times one, which is five, this is starting to look good. Now when n is equal to, this was n equals zero, this was n equals one, so now we're at n equals two."}, {"video_title": "Worked examples Summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's do the same thing over here. When n is equal to zero, it's going to be two plus three times zero, so that's just going to be two. And then plus, when n is equal to one, it's going to be two plus three times one, which is five, this is starting to look good. Now when n is equal to, this was n equals zero, this was n equals one, so now we're at n equals two. So at n equals two, two plus three times two is two plus six, which is eight. And this makes sense. Every time we increase n by one, we are adding another three, which is consistent with what we saw there."}, {"video_title": "Worked examples Summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Now when n is equal to, this was n equals zero, this was n equals one, so now we're at n equals two. So at n equals two, two plus three times two is two plus six, which is eight. And this makes sense. Every time we increase n by one, we are adding another three, which is consistent with what we saw there. We start at two when n equals zero. This is just, that three n is just zero, so you start at two, and then you keep, every time you increase n by one, you are just adding three again. So finally, when n is equal to three, two plus three times three is two plus, or is 11, I should say."}, {"video_title": "Worked examples Summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Every time we increase n by one, we are adding another three, which is consistent with what we saw there. We start at two when n equals zero. This is just, that three n is just zero, so you start at two, and then you keep, every time you increase n by one, you are just adding three again. So finally, when n is equal to three, two plus three times three is two plus, or is 11, I should say. And so this also is exactly the same sum. So I feel good about both of these. Let's do one more example."}, {"video_title": "Worked examples Summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So finally, when n is equal to three, two plus three times three is two plus, or is 11, I should say. And so this also is exactly the same sum. So I feel good about both of these. Let's do one more example. So we're here, we're given the sum, and we're saying choose one answer. Which of these is equivalent to this sum right over here? Well, like we did before, let's just expand it out."}, {"video_title": "Worked examples Summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Let's do one more example. So we're here, we're given the sum, and we're saying choose one answer. Which of these is equivalent to this sum right over here? Well, like we did before, let's just expand it out. And what's different here is that we just have a variable, but that shouldn't make it too much more difficult. So let's do the situation, and I'll write it out. Let's do the situation when n is equal to one."}, {"video_title": "Worked examples Summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Well, like we did before, let's just expand it out. And what's different here is that we just have a variable, but that shouldn't make it too much more difficult. So let's do the situation, and I'll write it out. Let's do the situation when n is equal to one. When n is equal to one, it's gonna be k over one plus one. K over one plus one, and I'll write it out. This is n is equal to one."}, {"video_title": "Worked examples Summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Let's do the situation when n is equal to one. When n is equal to one, it's gonna be k over one plus one. K over one plus one, and I'll write it out. This is n is equal to one. And then plus, when n is equal to two, it's gonna be k over two plus one. This is n is equal to two. And then we're gonna keep going."}, {"video_title": "Worked examples Summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "This is n is equal to one. And then plus, when n is equal to two, it's gonna be k over two plus one. This is n is equal to two. And then we're gonna keep going. When n is equal to three, it's gonna be k over three plus one as n is equal to three. And then plus, finally, because we stop right over here at n equals four. When n equals four, it's going to be k over four plus one."}, {"video_title": "Worked examples Summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And then we're gonna keep going. When n is equal to three, it's gonna be k over three plus one as n is equal to three. And then plus, finally, because we stop right over here at n equals four. When n equals four, it's going to be k over four plus one. That's when n is equal to four. So this is all going to be equal. I'll just write it over here."}, {"video_title": "Worked examples Summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "When n equals four, it's going to be k over four plus one. That's when n is equal to four. So this is all going to be equal. I'll just write it over here. This is equal to k over two plus k over three plus k over four plus k over five, which is exactly this choice right over here. And actually, if I had looked at the choices ahead of time, I might have even been able to save even more time by just saying, well, look, actually, if you just try to compute the first term, when n is equal to one, it would be k over two. Well, only this one is starting with a k over two."}, {"video_title": "Worked examples Summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "I'll just write it over here. This is equal to k over two plus k over three plus k over four plus k over five, which is exactly this choice right over here. And actually, if I had looked at the choices ahead of time, I might have even been able to save even more time by just saying, well, look, actually, if you just try to compute the first term, when n is equal to one, it would be k over two. Well, only this one is starting with a k over two. This one has no k's here, which is sketchy. They're trying to look at the error where you try to replace the k with the number as well, not just the n. So that's what they're trying to do here. Here, they're trying to, let's see, well, this isn't as obvious what they're even trying to do, right over here where they put the k in the denominator."}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Let's now generalize what we did in the last video. So if this is my y-axis, and this is, that's not that straight, this right over here is my x-axis. And let's say I have two functions. So I'm just gonna say it in general terms. So let's say I have a function right over here. So let's say it looks something like this. So that's one function."}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So I'm just gonna say it in general terms. So let's say I have a function right over here. So let's say it looks something like this. So that's one function. So this is y is equal to f of x. And then I have another function that y is equal to g of x. So let's say it looks something like this."}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So that's one function. So this is y is equal to f of x. And then I have another function that y is equal to g of x. So let's say it looks something like this. So y, the blue one right over here is y is equal to g of x. And like we did in the last video, I wanna think about the volume of the solid of revolution we will get if we essentially rotate the area between these two, and if we were to rotate it around the x-axis. So we're saying in very general terms, this could be anything, but in the way we've drawn it now, it would literally be that same truffle shape."}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say it looks something like this. So y, the blue one right over here is y is equal to g of x. And like we did in the last video, I wanna think about the volume of the solid of revolution we will get if we essentially rotate the area between these two, and if we were to rotate it around the x-axis. So we're saying in very general terms, this could be anything, but in the way we've drawn it now, it would literally be that same truffle shape. It'd be a very similar truffle shape where on the outside it looks like a truffle, on the outside it looks like something like a truffle, and on the inside we have carved out a cone. Obviously this visualization is very specific to the way I've drawn these functions, but what we wanna do is generalize at least the mathematics of it. So how do we find a volume?"}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So we're saying in very general terms, this could be anything, but in the way we've drawn it now, it would literally be that same truffle shape. It'd be a very similar truffle shape where on the outside it looks like a truffle, on the outside it looks like something like a truffle, and on the inside we have carved out a cone. Obviously this visualization is very specific to the way I've drawn these functions, but what we wanna do is generalize at least the mathematics of it. So how do we find a volume? Well, we could think of disks, but instead of thinking of disks, we're gonna think about washers now, which is essentially the exact same thing we did in the last video mathematically, but it's a slightly different way of conceptualizing it. So imagine taking a little chunk between these two functions just like that. What is going to be the width of this chunk?"}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So how do we find a volume? Well, we could think of disks, but instead of thinking of disks, we're gonna think about washers now, which is essentially the exact same thing we did in the last video mathematically, but it's a slightly different way of conceptualizing it. So imagine taking a little chunk between these two functions just like that. What is going to be the width of this chunk? Well, it's going to be equal to dx, and let's rotate that whole thing around the x-axis. So if we rotate this thing around the x-axis, we end up with a washer. That's why we're gonna call this the washer method."}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "What is going to be the width of this chunk? Well, it's going to be equal to dx, and let's rotate that whole thing around the x-axis. So if we rotate this thing around the x-axis, we end up with a washer. That's why we're gonna call this the washer method. It's really just kind of the disk method where you're gutting out the inside of a disk. So that's the inside of our washer, and then this is the outside of our washer. Outside of our washer looks something like that."}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "That's why we're gonna call this the washer method. It's really just kind of the disk method where you're gutting out the inside of a disk. So that's the inside of our washer, and then this is the outside of our washer. Outside of our washer looks something like that. Hopefully that makes sense. And so the surface of our washer looks something like that. I know I could have drawn this a little bit better, but hopefully it serves the purpose so that we understand it."}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Outside of our washer looks something like that. Hopefully that makes sense. And so the surface of our washer looks something like that. I know I could have drawn this a little bit better, but hopefully it serves the purpose so that we understand it. So the surface of our washer looks something like that, and it has depth of dx. It has depth dx. So let me see how well I can draw this."}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "I know I could have drawn this a little bit better, but hopefully it serves the purpose so that we understand it. So the surface of our washer looks something like that, and it has depth of dx. It has depth dx. So let me see how well I can draw this. So depth dx, that's the side of this washer. So a washer, you can imagine, is kind of a gutted out, it's a gutted out coin. It's a gutted out coin."}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So let me see how well I can draw this. So depth dx, that's the side of this washer. So a washer, you can imagine, is kind of a gutted out, it's a gutted out coin. It's a gutted out coin. So how do we find the volume? Well, once again, if we know the surface, if we know the area of the face of this washer, we can just multiply that times the depth. So it's gonna be the area of the face of the washer."}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "It's a gutted out coin. So how do we find the volume? Well, once again, if we know the surface, if we know the area of the face of this washer, we can just multiply that times the depth. So it's gonna be the area of the face of the washer. So the area of the face of the washer, well, it could be the area if it wasn't gutted out, and what would that area be? Well, it would be pi times the overall, the outside radius squared. It would be pi times the outside radius squared."}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So it's gonna be the area of the face of the washer. So the area of the face of the washer, well, it could be the area if it wasn't gutted out, and what would that area be? Well, it would be pi times the overall, the outside radius squared. It would be pi times the outside radius squared. Well, what's the outside radius? The radius that goes to the outside of the washer. Well, that's f of x."}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "It would be pi times the outside radius squared. Well, what's the outside radius? The radius that goes to the outside of the washer. Well, that's f of x. So it's gonna be f of x is the radius, and we're just going to square that. So that would give us, this expression right over here would give us the area of the entire face if it wasn't a washer, if it was a coin. But now we have to subtract out the inside."}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Well, that's f of x. So it's gonna be f of x is the radius, and we're just going to square that. So that would give us, this expression right over here would give us the area of the entire face if it wasn't a washer, if it was a coin. But now we have to subtract out the inside. So what's the area of the inside? What's the area of the inside? This part right over here."}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "But now we have to subtract out the inside. So what's the area of the inside? What's the area of the inside? This part right over here. Well, we're gonna subtract it out. It's going to be pi times the radius of the inside squared. The radius of the inside squared."}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "This part right over here. Well, we're gonna subtract it out. It's going to be pi times the radius of the inside squared. The radius of the inside squared. Well, what's the radius of the inside squared? Well, the inside, in this case, is g of x. It's going to be pi times g of x squared."}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "The radius of the inside squared. Well, what's the radius of the inside squared? Well, the inside, in this case, is g of x. It's going to be pi times g of x squared. That's the inner function, at least over the interval that we care about. So the area of this washer, we could just leave it like this, or we could factor out a pi. We could say it's equal to the area is equal to, if we factor out a pi, pi times f of x squared, f of x squared minus g of x squared."}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "It's going to be pi times g of x squared. That's the inner function, at least over the interval that we care about. So the area of this washer, we could just leave it like this, or we could factor out a pi. We could say it's equal to the area is equal to, if we factor out a pi, pi times f of x squared, f of x squared minus g of x squared. Minus, I don't, I have to write a parentheses there. So we could write f of x squared minus g of x squared. And then if we want the volume, put that in that same yellow."}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "We could say it's equal to the area is equal to, if we factor out a pi, pi times f of x squared, f of x squared minus g of x squared. Minus, I don't, I have to write a parentheses there. So we could write f of x squared minus g of x squared. And then if we want the volume, put that in that same yellow. If we want the volume of this thing, we just multiply it times the depth of each of those washers. So the volume of each of these washers are going to be pi times, pi times f of x, f of x squared minus g of x squared, the outer function squared over our interval minus our inner function squared over the interval, and then times our depth. Times our depth."}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "And then if we want the volume, put that in that same yellow. If we want the volume of this thing, we just multiply it times the depth of each of those washers. So the volume of each of these washers are going to be pi times, pi times f of x, f of x squared minus g of x squared, the outer function squared over our interval minus our inner function squared over the interval, and then times our depth. Times our depth. That'd be the volume of each of these washers. Now, for each, and that's going to be defined at a given x in our interval, but for each x at this interval, we can define a new washer. So there could be a washer out here and a washer out here."}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Times our depth. That'd be the volume of each of these washers. Now, for each, and that's going to be defined at a given x in our interval, but for each x at this interval, we can define a new washer. So there could be a washer out here and a washer out here. So we're going to take the sum of all of those washers and take the limit as we have smaller and smaller depths and we have an infinite number of infinitely thin washers. So we're going to take the integral over our interval from where these two things intersect, the interval that we care about. It doesn't have to be where they intersect, but in this case, that's what we'll do."}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So there could be a washer out here and a washer out here. So we're going to take the sum of all of those washers and take the limit as we have smaller and smaller depths and we have an infinite number of infinitely thin washers. So we're going to take the integral over our interval from where these two things intersect, the interval that we care about. It doesn't have to be where they intersect, but in this case, that's what we'll do. So let's say x equals a to x equals b, although it could have been from, this could have been a, that could have been b, but this is our interval. We're saying in general terms from a to b. And this will give our volume."}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "It doesn't have to be where they intersect, but in this case, that's what we'll do. So let's say x equals a to x equals b, although it could have been from, this could have been a, that could have been b, but this is our interval. We're saying in general terms from a to b. And this will give our volume. This right over here is the volume of each washer, and then we're summing up all of the washers and taking the limit as we have an infinite number of them. So let's see if we apply this to the example in the last video, whether we get the exact same answer. Well, in the last video, y equals g of x was equal to x, and y is equal to f of x was equal to the square root of x."}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "And this will give our volume. This right over here is the volume of each washer, and then we're summing up all of the washers and taking the limit as we have an infinite number of them. So let's see if we apply this to the example in the last video, whether we get the exact same answer. Well, in the last video, y equals g of x was equal to x, and y is equal to f of x was equal to the square root of x. So let's evaluate that given what we just were able to derive. So our volume, I'll do it up here. The volume is going to be the integral."}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Well, in the last video, y equals g of x was equal to x, and y is equal to f of x was equal to the square root of x. So let's evaluate that given what we just were able to derive. So our volume, I'll do it up here. The volume is going to be the integral. What are the two intersection points? Well, over here, once again, we could have defined the interval someplace else, like between there and there, and we would have gotten a different shape, but the points that we care about, the way we visualize it is between x is equal to zero, x is equal to zero, and x is equal to one. That's where these two things intersect."}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "The volume is going to be the integral. What are the two intersection points? Well, over here, once again, we could have defined the interval someplace else, like between there and there, and we would have gotten a different shape, but the points that we care about, the way we visualize it is between x is equal to zero, x is equal to zero, and x is equal to one. That's where these two things intersect. We saw that in the last video. Of pi, pi, times, what's f of x squared? f of x squared, square root of x squared is just x, minus g of x squared."}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "That's where these two things intersect. We saw that in the last video. Of pi, pi, times, what's f of x squared? f of x squared, square root of x squared is just x, minus g of x squared. G of x is x, that squared is x squared, and then we multiply times dx. So this is going to be equal to, we can factor out the pi, zero to one, x minus x squared dx, which is equal to pi times, let's see, the antiderivative of x is x squared over two, antiderivative of x squared is x squared over three, x to the third over three, and we're going to evaluate this from zero to one. So this is going to be equal to, I'm running out of my real estate a little bit, let me scroll over to the right a little bit."}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "f of x squared, square root of x squared is just x, minus g of x squared. G of x is x, that squared is x squared, and then we multiply times dx. So this is going to be equal to, we can factor out the pi, zero to one, x minus x squared dx, which is equal to pi times, let's see, the antiderivative of x is x squared over two, antiderivative of x squared is x squared over three, x to the third over three, and we're going to evaluate this from zero to one. So this is going to be equal to, I'm running out of my real estate a little bit, let me scroll over to the right a little bit. So this is going to be equal to pi times, well, when you evaluate this whole thing at one, you get, let's see, you get one half minus one third, one half minus one third, and then you subtract it, evaluate it at zero, but that's just going to be zero. Zero squared over two minus zero to the third power over three, that's just all going to be zero. So when you subtract out zero, you're just left with this expression right over here."}, {"video_title": "Generalizing the washer method Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be equal to, I'm running out of my real estate a little bit, let me scroll over to the right a little bit. So this is going to be equal to pi times, well, when you evaluate this whole thing at one, you get, let's see, you get one half minus one third, one half minus one third, and then you subtract it, evaluate it at zero, but that's just going to be zero. Zero squared over two minus zero to the third power over three, that's just all going to be zero. So when you subtract out zero, you're just left with this expression right over here. What's one half minus one third? Well, that's one sixth, this is one sixth, and so we are left with, this is equal to pi over six, which is the exact same thing that we got in the last video, and that's because we did the exact same thing that we did in the last video, we just conceptualized it a little bit differently. We generalized it in terms of f of x and g of x, and we essentially conceptualized it as, we conceptualized it as a washer, as opposed to kind of doing the disk method for an outer shape and an inner shape like we did in the last video."}, {"video_title": "Mean value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "And so let's just think about some function f. So let's say I have some function f, and we know a few things about this function. We know that it is continuous over the closed interval between x equals a and x is equal to b. And so when we put these brackets here, that just means closed interval. That means we're including, so when I put a bracket here, that means we're including the point a. And if I put the bracket on the right-hand side instead of a parenthesis, that means that we are including the point b. And continuous just means we don't have any gaps or jumps in the function over this closed interval. Now let's also assume that it's differentiable over the open interval between a and b."}, {"video_title": "Mean value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "That means we're including, so when I put a bracket here, that means we're including the point a. And if I put the bracket on the right-hand side instead of a parenthesis, that means that we are including the point b. And continuous just means we don't have any gaps or jumps in the function over this closed interval. Now let's also assume that it's differentiable over the open interval between a and b. So now we're saying, well, it's okay if it's not differentiable right at a, or if it's not differentiable right at b. And differentiable just means that there's a defined derivative, that you can actually take the derivative at those points. So it's differentiable over the open interval between a and b."}, {"video_title": "Mean value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's also assume that it's differentiable over the open interval between a and b. So now we're saying, well, it's okay if it's not differentiable right at a, or if it's not differentiable right at b. And differentiable just means that there's a defined derivative, that you can actually take the derivative at those points. So it's differentiable over the open interval between a and b. So those are the constraints we're gonna put on ourselves for the mean value theorem. And so let's just try to visualize this thing. So this is my function."}, {"video_title": "Mean value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "So it's differentiable over the open interval between a and b. So those are the constraints we're gonna put on ourselves for the mean value theorem. And so let's just try to visualize this thing. So this is my function. That's the y-axis. And this right over here is the x-axis. And I'm gonna, it's the x-axis, and let me draw my interval."}, {"video_title": "Mean value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "So this is my function. That's the y-axis. And this right over here is the x-axis. And I'm gonna, it's the x-axis, and let me draw my interval. So that's a, and then this is b right over here. And so let's say our function looks something like this. Let's say it looks something, draw an arbitrary function right over here."}, {"video_title": "Mean value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "And I'm gonna, it's the x-axis, and let me draw my interval. So that's a, and then this is b right over here. And so let's say our function looks something like this. Let's say it looks something, draw an arbitrary function right over here. Let's say my function looks something like that. So this point right over here, the x-value is a, and the y-value is f of a. F of a, this point right over here, the x-value is b, and the y-value, of course, is f of b. F of b. F of b. So all the mean value theorem tells us is if we take the average rate of change over the interval, that at some point, the instantaneous rate of change, at least at some point in this open interval, the instantaneous change is going to be the same as the average change."}, {"video_title": "Mean value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say it looks something, draw an arbitrary function right over here. Let's say my function looks something like that. So this point right over here, the x-value is a, and the y-value is f of a. F of a, this point right over here, the x-value is b, and the y-value, of course, is f of b. F of b. F of b. So all the mean value theorem tells us is if we take the average rate of change over the interval, that at some point, the instantaneous rate of change, at least at some point in this open interval, the instantaneous change is going to be the same as the average change. Now what does that mean visually? So let's calculate the average change. The average change between point a and point b, well that's going to be the slope of the secant line."}, {"video_title": "Mean value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "So all the mean value theorem tells us is if we take the average rate of change over the interval, that at some point, the instantaneous rate of change, at least at some point in this open interval, the instantaneous change is going to be the same as the average change. Now what does that mean visually? So let's calculate the average change. The average change between point a and point b, well that's going to be the slope of the secant line. That's going to be the slope of the secant line. So this is the secant line, so think about its slope. All the mean value theorem tells us is that at some point in this interval, the slope of the tangent line is going to be the same as the slope of the secant line."}, {"video_title": "Mean value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "The average change between point a and point b, well that's going to be the slope of the secant line. That's going to be the slope of the secant line. So this is the secant line, so think about its slope. All the mean value theorem tells us is that at some point in this interval, the slope of the tangent line is going to be the same as the slope of the secant line. And we can see just visually, it looks like right over here the slope of the tangent line looks like the same as the slope of the secant line. It also looks like the case right over here, the slope of the tangent line is equal to the slope of the secant line. And it makes intuitive sense."}, {"video_title": "Mean value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "All the mean value theorem tells us is that at some point in this interval, the slope of the tangent line is going to be the same as the slope of the secant line. And we can see just visually, it looks like right over here the slope of the tangent line looks like the same as the slope of the secant line. It also looks like the case right over here, the slope of the tangent line is equal to the slope of the secant line. And it makes intuitive sense. At some point, your instantaneous slope is going to be the same as the average slope. Now how would we write that mathematically? Well let's calculate, let's calculate, well let's calculate the average slope over this interval."}, {"video_title": "Mean value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "And it makes intuitive sense. At some point, your instantaneous slope is going to be the same as the average slope. Now how would we write that mathematically? Well let's calculate, let's calculate, well let's calculate the average slope over this interval. Well the average slope over this interval, or the average change, the slope of the secant line, is going to be our change in y, our change in y right over here, over our change in x, over our change in x. Well what is our change in y? Our change in y is f of b minus f of a, minus f of a, and that's going to be over, that is going to be over our change in x."}, {"video_title": "Mean value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "Well let's calculate, let's calculate, well let's calculate the average slope over this interval. Well the average slope over this interval, or the average change, the slope of the secant line, is going to be our change in y, our change in y right over here, over our change in x, over our change in x. Well what is our change in y? Our change in y is f of b minus f of a, minus f of a, and that's going to be over, that is going to be over our change in x. Over b minus, b minus a. Let me do that in that right color. So let's just remind ourselves what's going on here."}, {"video_title": "Mean value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "Our change in y is f of b minus f of a, minus f of a, and that's going to be over, that is going to be over our change in x. Over b minus, b minus a. Let me do that in that right color. So let's just remind ourselves what's going on here. So this right over here, this is the graph of y is equal to f of x. We're saying that the slope of the secant line, or our average rate of change over the interval from a to b, is our change in, is our change in y, our change in y, that the Greek letter delta is just shorthand for change in y over our change in x. Over our change in x, which of course is equal to this."}, {"video_title": "Mean value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "So let's just remind ourselves what's going on here. So this right over here, this is the graph of y is equal to f of x. We're saying that the slope of the secant line, or our average rate of change over the interval from a to b, is our change in, is our change in y, our change in y, that the Greek letter delta is just shorthand for change in y over our change in x. Over our change in x, which of course is equal to this. And the mean value theorem tells us that there exists, so if we know these two things about the function, then there exists, there exists some, some x value in between a and b. So in the open interval between a and b, there exists some c, there exists some c, and we could say it's a member of the open interval between a and b, between a and b. Or we could say some c such that a is less than, such that a is less than c, which is less than b."}, {"video_title": "Mean value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "Over our change in x, which of course is equal to this. And the mean value theorem tells us that there exists, so if we know these two things about the function, then there exists, there exists some, some x value in between a and b. So in the open interval between a and b, there exists some c, there exists some c, and we could say it's a member of the open interval between a and b, between a and b. Or we could say some c such that a is less than, such that a is less than c, which is less than b. So some c in this interval. So some c, some c in between it, where the instantaneous rate of change at, at that x value is the same as the average rate of change. So there exists some c in this open interval where the average rate of change is equal to the instantaneous rate of change at that point."}, {"video_title": "Mean value theorem Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "Or we could say some c such that a is less than, such that a is less than c, which is less than b. So some c in this interval. So some c, some c in between it, where the instantaneous rate of change at, at that x value is the same as the average rate of change. So there exists some c in this open interval where the average rate of change is equal to the instantaneous rate of change at that point. That's all it's saying. And as we saw in this diagram right over here, this could be our c, or this could be our c as well. So nothing, nothing really, it looks, you know, you would say f is continuous over a, b, differentiable over, over, f is continuous over the closed interval, differentiable over the open interval, and you see all of this notation, you're like, what is that telling us?"}, {"video_title": "Integration by parts intro AP Calculus BC Khan Academy.mp3", "Sentence": "What we're going to do in this video is review the product rule that you probably learned a while ago, and from that we're going to derive the formula for integration by parts, which could really be viewed as the inverse product rule, integration by parts. So let's say that I start with some function that can be expressed as the product f of x. It can be expressed as a product of two other functions, f of x times g of x. Now let's take the derivative of this function. Let's apply the derivative operator right over here. And this, once again, just review of the product rule. It's going to be the derivative of the first function times the second function."}, {"video_title": "Integration by parts intro AP Calculus BC Khan Academy.mp3", "Sentence": "Now let's take the derivative of this function. Let's apply the derivative operator right over here. And this, once again, just review of the product rule. It's going to be the derivative of the first function times the second function. So it's going to be f. Now I'm going to do that blue color. It's going to be f. That's not blue. It's going to be f prime of x times g of x plus the first function times the derivative of the second."}, {"video_title": "Integration by parts intro AP Calculus BC Khan Academy.mp3", "Sentence": "It's going to be the derivative of the first function times the second function. So it's going to be f. Now I'm going to do that blue color. It's going to be f. That's not blue. It's going to be f prime of x times g of x plus the first function times the derivative of the second. This is all a review right over here. So the derivative of the first times the second function plus the first function times the derivative of the second function. Now let's take the antiderivative of both sides of this equation."}, {"video_title": "Integration by parts intro AP Calculus BC Khan Academy.mp3", "Sentence": "It's going to be f prime of x times g of x plus the first function times the derivative of the second. This is all a review right over here. So the derivative of the first times the second function plus the first function times the derivative of the second function. Now let's take the antiderivative of both sides of this equation. Well, if I take the antiderivative of what I have here on the left, I get f of x times g of x. We won't think about the constant for now. We can ignore that for now."}, {"video_title": "Integration by parts intro AP Calculus BC Khan Academy.mp3", "Sentence": "Now let's take the antiderivative of both sides of this equation. Well, if I take the antiderivative of what I have here on the left, I get f of x times g of x. We won't think about the constant for now. We can ignore that for now. And that's going to be equal to, well, what's the antiderivative of this? It's going to be the antiderivative of f prime of x times g of x times g of x dx plus the antiderivative of f of x g prime of x dx. Now what I want to do is solve."}, {"video_title": "Integration by parts intro AP Calculus BC Khan Academy.mp3", "Sentence": "We can ignore that for now. And that's going to be equal to, well, what's the antiderivative of this? It's going to be the antiderivative of f prime of x times g of x times g of x dx plus the antiderivative of f of x g prime of x dx. Now what I want to do is solve. I'm going to solve for this part right over here. And to solve for that, I just have to subtract this business from both sides. And then I am left with f of x times g of x minus this minus the antiderivative of f prime of x g of x."}, {"video_title": "Integration by parts intro AP Calculus BC Khan Academy.mp3", "Sentence": "Now what I want to do is solve. I'm going to solve for this part right over here. And to solve for that, I just have to subtract this business from both sides. And then I am left with f of x times g of x minus this minus the antiderivative of f prime of x g of x. And let me do that in a pink color. g of x dx is equal to what I wanted to solve for, is equal to the antiderivative of f of x g prime of x dx. And to make it a little bit clearer, let me swap sides here."}, {"video_title": "Integration by parts intro AP Calculus BC Khan Academy.mp3", "Sentence": "And then I am left with f of x times g of x minus this minus the antiderivative of f prime of x g of x. And let me do that in a pink color. g of x dx is equal to what I wanted to solve for, is equal to the antiderivative of f of x g prime of x dx. And to make it a little bit clearer, let me swap sides here. So let me copy and paste this. So let me copy and then paste it. There you go."}, {"video_title": "Integration by parts intro AP Calculus BC Khan Academy.mp3", "Sentence": "And to make it a little bit clearer, let me swap sides here. So let me copy and paste this. So let me copy and then paste it. There you go. And then let me copy and paste the other side. So let me copy and paste it. So I'm just switching the sides, just to give it in a form that you might be more used to seeing in a calculus book."}, {"video_title": "Integration by parts intro AP Calculus BC Khan Academy.mp3", "Sentence": "There you go. And then let me copy and paste the other side. So let me copy and paste it. So I'm just switching the sides, just to give it in a form that you might be more used to seeing in a calculus book. So this tells us, this is essentially the formula for integration by parts. I will square it off. You'll often see it squared off in a traditional textbook."}, {"video_title": "Integration by parts intro AP Calculus BC Khan Academy.mp3", "Sentence": "So I'm just switching the sides, just to give it in a form that you might be more used to seeing in a calculus book. So this tells us, this is essentially the formula for integration by parts. I will square it off. You'll often see it squared off in a traditional textbook. So I will do the same. So this right over here tells us that if we have an integral or an antiderivative of the form f of x times the derivative of some other function, we can apply this right over here. And you might say, well, this doesn't seem that useful."}, {"video_title": "Limits of trigonometric functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "What we're going to do in this video is think about limits involving trigonometric functions. So let's just start with a fairly straightforward one. Let's find the limit as x approaches pi of sine of x. Pause the video and see if you can figure this out. Well, with both sine of x and cosine of x, they're defined for all real numbers, so their domain is all real numbers. You can put any real number in here for x and it will give you an output. It is defined."}, {"video_title": "Limits of trigonometric functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Pause the video and see if you can figure this out. Well, with both sine of x and cosine of x, they're defined for all real numbers, so their domain is all real numbers. You can put any real number in here for x and it will give you an output. It is defined. And they are also continuous over their entire domain. In fact, all of the trigonometric functions are continuous over their entire domain. And so for sine of x, because it's continuous and it's defined at sine of pi, we would say that this is the same thing as sine of pi, and sine of pi, you might already know, is equal to zero."}, {"video_title": "Limits of trigonometric functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "It is defined. And they are also continuous over their entire domain. In fact, all of the trigonometric functions are continuous over their entire domain. And so for sine of x, because it's continuous and it's defined at sine of pi, we would say that this is the same thing as sine of pi, and sine of pi, you might already know, is equal to zero. And we could do a similar exercise with cosine of x. So if I were to say, what's the limit as x approaches, I'll just take an arbitrary angle, x approaches pi over four of cosine of x? Well, once again, cosine of x is defined for all real numbers."}, {"video_title": "Limits of trigonometric functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And so for sine of x, because it's continuous and it's defined at sine of pi, we would say that this is the same thing as sine of pi, and sine of pi, you might already know, is equal to zero. And we could do a similar exercise with cosine of x. So if I were to say, what's the limit as x approaches, I'll just take an arbitrary angle, x approaches pi over four of cosine of x? Well, once again, cosine of x is defined for all real numbers. X can be any real number. It's also continuous. So for cosine of x, this limit is just going to be cosine of pi over four."}, {"video_title": "Limits of trigonometric functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Well, once again, cosine of x is defined for all real numbers. X can be any real number. It's also continuous. So for cosine of x, this limit is just going to be cosine of pi over four. And that is going to be equal to square root of two over two. This is one of those useful angles to know the sine and cosine of. If you think in degrees, this is a 45-degree angle."}, {"video_title": "Limits of trigonometric functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So for cosine of x, this limit is just going to be cosine of pi over four. And that is going to be equal to square root of two over two. This is one of those useful angles to know the sine and cosine of. If you think in degrees, this is a 45-degree angle. And in general, if I'm dealing with a sine or a cosine, the limit as x approaches a of sine of x is equal to sine of a. Once again, this is going to be true for any a, any real number a. And I can make a similar statement about cosine of x."}, {"video_title": "Limits of trigonometric functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "If you think in degrees, this is a 45-degree angle. And in general, if I'm dealing with a sine or a cosine, the limit as x approaches a of sine of x is equal to sine of a. Once again, this is going to be true for any a, any real number a. And I can make a similar statement about cosine of x. Limit as x approaches a of cosine of x is equal to cosine of a. Now, I've been saying it over and over. That's because both of their domains are all real numbers."}, {"video_title": "Limits of trigonometric functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And I can make a similar statement about cosine of x. Limit as x approaches a of cosine of x is equal to cosine of a. Now, I've been saying it over and over. That's because both of their domains are all real numbers. They are defined for all real numbers that you put in. And they're continuous on their entire domain. But now let's do slightly more involved trigonometric functions, or ones that aren't defined for all real numbers, that their domains are constrained just a little bit more."}, {"video_title": "Limits of trigonometric functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "That's because both of their domains are all real numbers. They are defined for all real numbers that you put in. And they're continuous on their entire domain. But now let's do slightly more involved trigonometric functions, or ones that aren't defined for all real numbers, that their domains are constrained just a little bit more. So let's say if we were to take the limit as x approaches pi of tangent of x. What is this going to be equal to? Well, this is the same thing as the limit as x approaches pi."}, {"video_title": "Limits of trigonometric functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "But now let's do slightly more involved trigonometric functions, or ones that aren't defined for all real numbers, that their domains are constrained just a little bit more. So let's say if we were to take the limit as x approaches pi of tangent of x. What is this going to be equal to? Well, this is the same thing as the limit as x approaches pi. Tangent of x is sine of x over cosine of x. And so both of these are defined for pi. And so we could just substitute pi in."}, {"video_title": "Limits of trigonometric functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Well, this is the same thing as the limit as x approaches pi. Tangent of x is sine of x over cosine of x. And so both of these are defined for pi. And so we could just substitute pi in. And we just want to ensure that we don't get a zero in the denominator, because that would make it undefined. So we get sine of pi over cosine of pi, which is equal to zero over negative one, which is completely fine. If it was negative one over zero, we'd be in trouble."}, {"video_title": "Limits of trigonometric functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And so we could just substitute pi in. And we just want to ensure that we don't get a zero in the denominator, because that would make it undefined. So we get sine of pi over cosine of pi, which is equal to zero over negative one, which is completely fine. If it was negative one over zero, we'd be in trouble. But this is just going to be equal to zero. So that works out. But if I were to ask you, what is the limit as x approaches pi over two of tangent of x?"}, {"video_title": "Limits of trigonometric functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "If it was negative one over zero, we'd be in trouble. But this is just going to be equal to zero. So that works out. But if I were to ask you, what is the limit as x approaches pi over two of tangent of x? Pause the video and try to work that out. Well, think about it. This is the limit as x approaches pi over two of sine of x over cosine of x."}, {"video_title": "Limits of trigonometric functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "But if I were to ask you, what is the limit as x approaches pi over two of tangent of x? Pause the video and try to work that out. Well, think about it. This is the limit as x approaches pi over two of sine of x over cosine of x. Now sine of pi over two is one, but cosine of pi over two is zero. So if you were to just substitute it in, this would give you one over zero. And one way to think about it is, pi over two is not in the domain of tangent of x."}, {"video_title": "Limits of trigonometric functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "This is the limit as x approaches pi over two of sine of x over cosine of x. Now sine of pi over two is one, but cosine of pi over two is zero. So if you were to just substitute it in, this would give you one over zero. And one way to think about it is, pi over two is not in the domain of tangent of x. And so this limit actually turns out it doesn't exist. In general, if we're dealing with sine cosine tangent or cosecant secant or cotangent, if we're taking a limit to a point that is in their domain, then the value of the limit is going to be the same thing as the value of the function at that point. If you're taking a limit to a point that's not in their domain, there's a good chance that we're not going to have a limit."}, {"video_title": "Limits of trigonometric functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And one way to think about it is, pi over two is not in the domain of tangent of x. And so this limit actually turns out it doesn't exist. In general, if we're dealing with sine cosine tangent or cosecant secant or cotangent, if we're taking a limit to a point that is in their domain, then the value of the limit is going to be the same thing as the value of the function at that point. If you're taking a limit to a point that's not in their domain, there's a good chance that we're not going to have a limit. So here, there is no limit. And a way to deduce that is that pi over two is not in tangent of x's domain. If you were to graph tan of x, you would see a vertical asymptote at pi over two."}, {"video_title": "Limits of trigonometric functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "If you're taking a limit to a point that's not in their domain, there's a good chance that we're not going to have a limit. So here, there is no limit. And a way to deduce that is that pi over two is not in tangent of x's domain. If you were to graph tan of x, you would see a vertical asymptote at pi over two. Let's do one more of these. So let's say the limit as x approaches pi of cotangent of x. Pause the video and see if you can figure out what that's going to be."}, {"video_title": "Limits of trigonometric functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "If you were to graph tan of x, you would see a vertical asymptote at pi over two. Let's do one more of these. So let's say the limit as x approaches pi of cotangent of x. Pause the video and see if you can figure out what that's going to be. Well, one way to think about it, cotangent of x, it's one over tangent of x, it's cosine of x over sine of x. It's sine of x. This is the limit as x approaches pi of this."}, {"video_title": "Limits of trigonometric functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Pause the video and see if you can figure out what that's going to be. Well, one way to think about it, cotangent of x, it's one over tangent of x, it's cosine of x over sine of x. It's sine of x. This is the limit as x approaches pi of this. And is pi in the domain of cotangent of x? Well, no, if you were to just substitute pi in, you're gonna get negative one over zero. And so that is not in the domain of cotangent of x."}, {"video_title": "Limits of trigonometric functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "This is the limit as x approaches pi of this. And is pi in the domain of cotangent of x? Well, no, if you were to just substitute pi in, you're gonna get negative one over zero. And so that is not in the domain of cotangent of x. If you were to plot it, you would see a vertical asymptote right over there. And so we have no limit. We have no limit."}, {"video_title": "Limits of trigonometric functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And so that is not in the domain of cotangent of x. If you were to plot it, you would see a vertical asymptote right over there. And so we have no limit. We have no limit. So once again, this is not in the domain of that. And so good chance that we have no limit. When the thing we're taking the limit to is in the domain of the trigonometric function, we're going to have a defined limit."}, {"video_title": "Limits of trigonometric functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "We have no limit. So once again, this is not in the domain of that. And so good chance that we have no limit. When the thing we're taking the limit to is in the domain of the trigonometric function, we're going to have a defined limit. And sine and cosine in particular are defined for all real numbers. And they're continuous over all real numbers. So you take the limit to anything for them, it's going to be defined, and it's going to be the value of the function at that point."}, {"video_title": "Finding specific antiderivatives rational function AP Calculus AB Khan Academy.mp3", "Sentence": "Alright, so they give us the derivative in terms of x. So maybe we could take the antiderivative of the derivative to find our original function. So let's do that. So we could say that f of x, f of x is going to be equal to the antiderivative, or we could say the indefinite integral of f prime of x, which is equal to 24 over x to the third. I could write it over like this, 24 over x to the third. But to help me process it a little bit more, I'm gonna write it, I'm gonna write this as 24 x to the negative three, because then it'll become a little harder how to take that antiderivative d dx. And so what is the antiderivative of 24 x to the negative three?"}, {"video_title": "Finding specific antiderivatives rational function AP Calculus AB Khan Academy.mp3", "Sentence": "So we could say that f of x, f of x is going to be equal to the antiderivative, or we could say the indefinite integral of f prime of x, which is equal to 24 over x to the third. I could write it over like this, 24 over x to the third. But to help me process it a little bit more, I'm gonna write it, I'm gonna write this as 24 x to the negative three, because then it'll become a little harder how to take that antiderivative d dx. And so what is the antiderivative of 24 x to the negative three? Well, we're just going to do the power rule in reverse. So what we're going to do is we're going to increase the exponent. So let me just rewrite it."}, {"video_title": "Finding specific antiderivatives rational function AP Calculus AB Khan Academy.mp3", "Sentence": "And so what is the antiderivative of 24 x to the negative three? Well, we're just going to do the power rule in reverse. So what we're going to do is we're going to increase the exponent. So let me just rewrite it. It's gonna be 24 x to the, we're going to increase the exponent by one. So it's gonna be x to the negative three plus one. And then we're going to divide by that increased exponent."}, {"video_title": "Finding specific antiderivatives rational function AP Calculus AB Khan Academy.mp3", "Sentence": "So let me just rewrite it. It's gonna be 24 x to the, we're going to increase the exponent by one. So it's gonna be x to the negative three plus one. And then we're going to divide by that increased exponent. So negative three plus one. And so that is going to be negative three plus one is x to the negative two, and then we divide by negative two. And if you're in doubt about what we just did, where we're kind of doing the power rule in reverse, now take the power rule, take the derivative of this using the power rule."}, {"video_title": "Finding specific antiderivatives rational function AP Calculus AB Khan Academy.mp3", "Sentence": "And then we're going to divide by that increased exponent. So negative three plus one. And so that is going to be negative three plus one is x to the negative two, and then we divide by negative two. And if you're in doubt about what we just did, where we're kind of doing the power rule in reverse, now take the power rule, take the derivative of this using the power rule. Negative two times 24 over negative two is just gonna be 24, and then you decrement that exponent, you go to negative three. So are we done here? Is this f of x?"}, {"video_title": "Finding specific antiderivatives rational function AP Calculus AB Khan Academy.mp3", "Sentence": "And if you're in doubt about what we just did, where we're kind of doing the power rule in reverse, now take the power rule, take the derivative of this using the power rule. Negative two times 24 over negative two is just gonna be 24, and then you decrement that exponent, you go to negative three. So are we done here? Is this f of x? Well, f of x might involve a constant. So let's put a constant out here, because notice, if you were to take the derivative of this thing here, the derivative of 24x to the negative two over negative two, we already established is 24x to the negative three. But then if you take the derivative of a constant, well, that just disappears, so you don't see it when you look at the derivative."}, {"video_title": "Finding specific antiderivatives rational function AP Calculus AB Khan Academy.mp3", "Sentence": "Is this f of x? Well, f of x might involve a constant. So let's put a constant out here, because notice, if you were to take the derivative of this thing here, the derivative of 24x to the negative two over negative two, we already established is 24x to the negative three. But then if you take the derivative of a constant, well, that just disappears, so you don't see it when you look at the derivative. So we have to make sure that there might be a constant. And I have a feeling, based on the information that they've given us, we're going to make use of that constant. So let me rewrite f of x."}, {"video_title": "Finding specific antiderivatives rational function AP Calculus AB Khan Academy.mp3", "Sentence": "But then if you take the derivative of a constant, well, that just disappears, so you don't see it when you look at the derivative. So we have to make sure that there might be a constant. And I have a feeling, based on the information that they've given us, we're going to make use of that constant. So let me rewrite f of x. So we know that f of x can be expressed as 24 divided by negative two. It's negative 12x to the negative two plus some constant. So how do we figure out that constant?"}, {"video_title": "Finding specific antiderivatives rational function AP Calculus AB Khan Academy.mp3", "Sentence": "So let me rewrite f of x. So we know that f of x can be expressed as 24 divided by negative two. It's negative 12x to the negative two plus some constant. So how do we figure out that constant? Well, they have told us what f of two is. F of two is equal to 12. So let's write this down."}, {"video_title": "Finding specific antiderivatives rational function AP Calculus AB Khan Academy.mp3", "Sentence": "So how do we figure out that constant? Well, they have told us what f of two is. F of two is equal to 12. So let's write this down. So when, so f, so f of two is equal to 12, which is equal to, well, we just have to put two in everywhere we see an x. That's going to be negative two times two to the negative two power plus c. And so 12 is equal to, what is this? Two to the negative two."}, {"video_title": "Finding specific antiderivatives rational function AP Calculus AB Khan Academy.mp3", "Sentence": "So let's write this down. So when, so f, so f of two is equal to 12, which is equal to, well, we just have to put two in everywhere we see an x. That's going to be negative two times two to the negative two power plus c. And so 12 is equal to, what is this? Two to the negative two. Two to the negative two is equal to one over two squared, which is equal to 1 4th. So this is negative 12 times 1 4th. Negative 12 times 1 4th is negative three."}, {"video_title": "Finding specific antiderivatives rational function AP Calculus AB Khan Academy.mp3", "Sentence": "Two to the negative two. Two to the negative two is equal to one over two squared, which is equal to 1 4th. So this is negative 12 times 1 4th. Negative 12 times 1 4th is negative three. So it's negative three plus c. Now we can add three to both sides to solve for c. We get 15 is equal to, 15 is equal to c. So, or c is equal to 15. That is equal to 15. And so now we can write our f of x as we get f of x is equal to negative 12, and I could even write that as negative 12 over x squared if we like."}, {"video_title": "Finding specific antiderivatives rational function AP Calculus AB Khan Academy.mp3", "Sentence": "Negative 12 times 1 4th is negative three. So it's negative three plus c. Now we can add three to both sides to solve for c. We get 15 is equal to, 15 is equal to c. So, or c is equal to 15. That is equal to 15. And so now we can write our f of x as we get f of x is equal to negative 12, and I could even write that as negative 12 over x squared if we like. Negative 12 over x squared plus 15. And now using that, we can evaluate f of negative one. f of negative one."}, {"video_title": "Finding specific antiderivatives rational function AP Calculus AB Khan Academy.mp3", "Sentence": "And so now we can write our f of x as we get f of x is equal to negative 12, and I could even write that as negative 12 over x squared if we like. Negative 12 over x squared plus 15. And now using that, we can evaluate f of negative one. f of negative one. Wherever we see an x, we put a negative one there. So this is going to be negative one squared. So f of negative one is equal to 12 divided by, or negative 12 divided by negative one squared."}, {"video_title": "Finding specific antiderivatives rational function AP Calculus AB Khan Academy.mp3", "Sentence": "f of negative one. Wherever we see an x, we put a negative one there. So this is going to be negative one squared. So f of negative one is equal to 12 divided by, or negative 12 divided by negative one squared. Well, negative one squared is just one. So it's gonna be negative 12 plus 15, which is equal to three. And we're done."}, {"video_title": "Integration by parts \u00c2\u00bax_cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "Let's see if that really is the case. So let's say I want to take the antiderivative of x times cosine of x dx. Now, if you look at this formula right over here, you want to assign part of this to f of x and some part of it to g prime of x. And the question is, well, do I assign f of x to x and g prime of x to cosine of x, or the other way around? Do I make f of x cosine of x and g prime of x x? And the thing to realize is to look at the other part of the formula and realize that you're essentially going to have to solve this right over here. And here, we have the derivative of f of x times g of x."}, {"video_title": "Integration by parts \u00c2\u00bax_cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "And the question is, well, do I assign f of x to x and g prime of x to cosine of x, or the other way around? Do I make f of x cosine of x and g prime of x x? And the thing to realize is to look at the other part of the formula and realize that you're essentially going to have to solve this right over here. And here, we have the derivative of f of x times g of x. So what you want to do is assign f of x so that the derivative of f of x is actually simpler than f of x. And assign g prime of x that, if you were to take its antiderivative, it doesn't really become any more complicated. So in this case, if we assign f of x to be equal to x, f prime of x is definitely simpler."}, {"video_title": "Integration by parts \u00c2\u00bax_cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "And here, we have the derivative of f of x times g of x. So what you want to do is assign f of x so that the derivative of f of x is actually simpler than f of x. And assign g prime of x that, if you were to take its antiderivative, it doesn't really become any more complicated. So in this case, if we assign f of x to be equal to x, f prime of x is definitely simpler. f prime of x is equal to 1. If we assign g prime of x to be cosine of x, once again, if we take its antiderivative, that's sine of x. It's not any more complicated."}, {"video_title": "Integration by parts \u00c2\u00bax_cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "So in this case, if we assign f of x to be equal to x, f prime of x is definitely simpler. f prime of x is equal to 1. If we assign g prime of x to be cosine of x, once again, if we take its antiderivative, that's sine of x. It's not any more complicated. If we did it the other way around, if we said f of x to be cosine of x, then we're taking its derivative here. That's not that much more complicated. But if we said g prime of x equaling to x, and then we had to take its antiderivative, we get x squared over 2, that is more complicated."}, {"video_title": "Integration by parts \u00c2\u00bax_cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "It's not any more complicated. If we did it the other way around, if we said f of x to be cosine of x, then we're taking its derivative here. That's not that much more complicated. But if we said g prime of x equaling to x, and then we had to take its antiderivative, we get x squared over 2, that is more complicated. So let me make it clear over here. We are assigning f of x to be equal to x. And that means that the derivative of f is going to be equal to 1."}, {"video_title": "Integration by parts \u00c2\u00bax_cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "But if we said g prime of x equaling to x, and then we had to take its antiderivative, we get x squared over 2, that is more complicated. So let me make it clear over here. We are assigning f of x to be equal to x. And that means that the derivative of f is going to be equal to 1. We are assigning g prime of x to be equal to cosine of x, which means g of x is equal to sine of x, the antiderivative of cosine of x. Now, let's see. Given these assumptions, let's see if we can apply this formula."}, {"video_title": "Integration by parts \u00c2\u00bax_cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "And that means that the derivative of f is going to be equal to 1. We are assigning g prime of x to be equal to cosine of x, which means g of x is equal to sine of x, the antiderivative of cosine of x. Now, let's see. Given these assumptions, let's see if we can apply this formula. So this has all of this. Let's see. Right-hand side says f of x times g of x."}, {"video_title": "Integration by parts \u00c2\u00bax_cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "Given these assumptions, let's see if we can apply this formula. So this has all of this. Let's see. Right-hand side says f of x times g of x. So f of x is x. g of x is sine of x. g of x is sine of x. And then from that, we are going to subtract the antiderivative of f prime of x. Well, that's just 1."}, {"video_title": "Integration by parts \u00c2\u00bax_cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "Right-hand side says f of x times g of x. So f of x is x. g of x is sine of x. g of x is sine of x. And then from that, we are going to subtract the antiderivative of f prime of x. Well, that's just 1. That's just 1. Times g of x. Times sine of x."}, {"video_title": "Integration by parts \u00c2\u00bax_cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "Well, that's just 1. That's just 1. Times g of x. Times sine of x. Times sine of x dx. Now, this was a huge simplification. Now I just have to, I went from trying to solve the antiderivative of x cosine of x, to now I just have to find the antiderivative of sine of x."}, {"video_title": "Integration by parts \u00c2\u00bax_cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "Times sine of x. Times sine of x dx. Now, this was a huge simplification. Now I just have to, I went from trying to solve the antiderivative of x cosine of x, to now I just have to find the antiderivative of sine of x. And we know the antiderivative of sine of x dx is just equal to negative cosine of x. And so, and of course we can throw the plus c in now, now that we're pretty done with taking all of our antiderivatives. So all of this is going to be equal to x sine of x. x times sine of x."}, {"video_title": "Integration by parts \u00c2\u00bax_cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "Now I just have to, I went from trying to solve the antiderivative of x cosine of x, to now I just have to find the antiderivative of sine of x. And we know the antiderivative of sine of x dx is just equal to negative cosine of x. And so, and of course we can throw the plus c in now, now that we're pretty done with taking all of our antiderivatives. So all of this is going to be equal to x sine of x. x times sine of x. Sine of x. Minus the antiderivative of this, which is just negative cosine of x. Minus negative cosine of x."}, {"video_title": "Integration by parts \u00c2\u00bax_cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "So all of this is going to be equal to x sine of x. x times sine of x. Sine of x. Minus the antiderivative of this, which is just negative cosine of x. Minus negative cosine of x. And then we could throw in a plus c right at the end of it. And it doesn't matter if we subtract a c or add the c. We're saying this is some arbitrary constant, which could even be negative. And so this is all going to be equal to, just we get our drum roll now."}, {"video_title": "Integration by parts \u00c2\u00bax_cos(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "Minus negative cosine of x. And then we could throw in a plus c right at the end of it. And it doesn't matter if we subtract a c or add the c. We're saying this is some arbitrary constant, which could even be negative. And so this is all going to be equal to, just we get our drum roll now. It's going to be x times sine of x. x times sine of x. Subtract a negative, that becomes a positive. Plus cosine of x. Plus c. And we are done."}, {"video_title": "Worked example coefficient in Maclaurin polynomial Series AP Calculus BC Khan Academy.mp3", "Sentence": "The nth derivative of g at x equals zero is given by, so the nth derivative of g, evaluated at x equals zero, is equal to the square root of n plus seven over n to the third, for n is greater than or equal to one. What is the coefficient for the term containing x squared in the Maclaurin series of g? So let's just think about the Maclaurin series for g. So if I were to have my function g of x, the Maclaurin series, I could say approximately equal to, especially if I'm not gonna list out all of the terms, is going to be equal to, well, it's going to be equal to g of zero plus g prime of zero times x plus g prime prime of zero divided by, I could say two factorial, but that's just two, times x squared. And that's about as far as we go because we just have to think about what is the coefficient for the term containing x squared? I could, if they said what's the coefficient for the term containing x to the third, I would keep going. I'd go g prime, I would take the third derivative, evaluate it at zero, over three factorial. I could view this as a factorial, too, but that just evaluates to two."}, {"video_title": "Worked example coefficient in Maclaurin polynomial Series AP Calculus BC Khan Academy.mp3", "Sentence": "And that's about as far as we go because we just have to think about what is the coefficient for the term containing x squared? I could, if they said what's the coefficient for the term containing x to the third, I would keep going. I'd go g prime, I would take the third derivative, evaluate it at zero, over three factorial. I could view this as a factorial, too, but that just evaluates to two. I could view this as one factorial. I could view this as zero factorial, just so you see it's a consistent idea here. And I could, of course, keep on going."}, {"video_title": "Worked example coefficient in Maclaurin polynomial Series AP Calculus BC Khan Academy.mp3", "Sentence": "I could view this as a factorial, too, but that just evaluates to two. I could view this as one factorial. I could view this as zero factorial, just so you see it's a consistent idea here. And I could, of course, keep on going. But we just care about, they're just asking us, what is the coefficient for the term containing x squared? So they just want us to figure out this. What is this thing right over here?"}, {"video_title": "Worked example coefficient in Maclaurin polynomial Series AP Calculus BC Khan Academy.mp3", "Sentence": "And I could, of course, keep on going. But we just care about, they're just asking us, what is the coefficient for the term containing x squared? So they just want us to figure out this. What is this thing right over here? So to know that, we need to figure out what is the second derivative of g evaluated at x equals zero? Well, they tell us that over here. It's a little bit unconventional, where they give us a formula, a general formula for any derivative evaluated at x equals zero but that's what they're telling us here."}, {"video_title": "Worked example coefficient in Maclaurin polynomial Series AP Calculus BC Khan Academy.mp3", "Sentence": "What is this thing right over here? So to know that, we need to figure out what is the second derivative of g evaluated at x equals zero? Well, they tell us that over here. It's a little bit unconventional, where they give us a formula, a general formula for any derivative evaluated at x equals zero but that's what they're telling us here. So in this case, the n isn't zero. The n is the derivative we're taking and that's going to be our second derivative. So this is, so if I wanted to figure out g, if I am figuring out the second derivative and I could write it like that, evaluated at zero, or I could write it like this, just so the notation is consistent."}, {"video_title": "Worked example coefficient in Maclaurin polynomial Series AP Calculus BC Khan Academy.mp3", "Sentence": "It's a little bit unconventional, where they give us a formula, a general formula for any derivative evaluated at x equals zero but that's what they're telling us here. So in this case, the n isn't zero. The n is the derivative we're taking and that's going to be our second derivative. So this is, so if I wanted to figure out g, if I am figuring out the second derivative and I could write it like that, evaluated at zero, or I could write it like this, just so the notation is consistent. I could write it like that. The second derivative evaluated at x equals zero is going to be equal to, well, our n is two. So this is going to be the square root of two plus seven over two to the third power."}, {"video_title": "Worked example coefficient in Maclaurin polynomial Series AP Calculus BC Khan Academy.mp3", "Sentence": "So this is, so if I wanted to figure out g, if I am figuring out the second derivative and I could write it like that, evaluated at zero, or I could write it like this, just so the notation is consistent. I could write it like that. The second derivative evaluated at x equals zero is going to be equal to, well, our n is two. So this is going to be the square root of two plus seven over two to the third power. So two plus seven is nine, take the principal root of that. It's gonna give us positive three over two to the third, which is eight. So this part right over here is 3 8ths."}, {"video_title": "Worked example coefficient in Maclaurin polynomial Series AP Calculus BC Khan Academy.mp3", "Sentence": "So this is going to be the square root of two plus seven over two to the third power. So two plus seven is nine, take the principal root of that. It's gonna give us positive three over two to the third, which is eight. So this part right over here is 3 8ths. So the whole coefficient is going to be 3 8ths, that's this numerator, divided by two, which of course is equal to three over 16. And we're done. They didn't want us to figure out a couple of terms of this which we could call the Maclaurin polynomial, an nth degree Maclaurin polynomial."}, {"video_title": "_-substitution multiplying by a constant AP Calculus AB Khan Academy.mp3", "Sentence": "Let's take the indefinite integral of the square root of seven x plus nine dx. So my first question to you is, is this going to be a good case for u substitution? Well, when you look here, we, maybe the natural thing to set to be equal to u is seven x plus nine, but do I see its derivative anywhere over here? Well, let's see. If I set u to be equal to seven x plus nine, what is the derivative of u with respect to x going to be? Derivative of u with respect to x is just going to be equal to seven. Derivative of seven x is seven, derivative of nine is zero."}, {"video_title": "_-substitution multiplying by a constant AP Calculus AB Khan Academy.mp3", "Sentence": "Well, let's see. If I set u to be equal to seven x plus nine, what is the derivative of u with respect to x going to be? Derivative of u with respect to x is just going to be equal to seven. Derivative of seven x is seven, derivative of nine is zero. So do we see a seven lying around anywhere over here? Well, we don't. But what could we do in order to have a seven lying around but not change the value of the integral?"}, {"video_title": "_-substitution multiplying by a constant AP Calculus AB Khan Academy.mp3", "Sentence": "Derivative of seven x is seven, derivative of nine is zero. So do we see a seven lying around anywhere over here? Well, we don't. But what could we do in order to have a seven lying around but not change the value of the integral? Well, the neat thing, and we've seen this multiple times, is when you're evaluating integrals, scalars can go in and outside of the integral very easily. Just to remind ourselves, if I have the integral of, let's say, some scalar a times f of x dx, dx, this is the same thing as a times the integral of f of x dx. That the integral of the scalar times the function is equal to the scalar times the integral of the function."}, {"video_title": "_-substitution multiplying by a constant AP Calculus AB Khan Academy.mp3", "Sentence": "But what could we do in order to have a seven lying around but not change the value of the integral? Well, the neat thing, and we've seen this multiple times, is when you're evaluating integrals, scalars can go in and outside of the integral very easily. Just to remind ourselves, if I have the integral of, let's say, some scalar a times f of x dx, dx, this is the same thing as a times the integral of f of x dx. That the integral of the scalar times the function is equal to the scalar times the integral of the function. So let me put this aside right over here. So with that in mind, can we multiply and divide by something that will have a seven showing up? Well, we can multiply and divide by seven."}, {"video_title": "_-substitution multiplying by a constant AP Calculus AB Khan Academy.mp3", "Sentence": "That the integral of the scalar times the function is equal to the scalar times the integral of the function. So let me put this aside right over here. So with that in mind, can we multiply and divide by something that will have a seven showing up? Well, we can multiply and divide by seven. So imagine doing this. Let's rewrite our original integral. The original, so let me draw a little arrow here just to go around that aside."}, {"video_title": "_-substitution multiplying by a constant AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we can multiply and divide by seven. So imagine doing this. Let's rewrite our original integral. The original, so let me draw a little arrow here just to go around that aside. We could write our original integral as being equal to the integral of 1 7th times seven times the square root of seven x plus nine dx. And if we want to, we could take the 1 7th outside of the integral. We don't have to, but we could rewrite this as 1 7th times the integral of seven times the square root of seven x plus nine dx."}, {"video_title": "_-substitution multiplying by a constant AP Calculus AB Khan Academy.mp3", "Sentence": "The original, so let me draw a little arrow here just to go around that aside. We could write our original integral as being equal to the integral of 1 7th times seven times the square root of seven x plus nine dx. And if we want to, we could take the 1 7th outside of the integral. We don't have to, but we could rewrite this as 1 7th times the integral of seven times the square root of seven x plus nine dx. So now if we set u equal to seven x plus nine, do we have its derivative laying around? Well, sure, the seven is right over here. We know that du, if we want to write in differential form, du is equal to seven times dx."}, {"video_title": "_-substitution multiplying by a constant AP Calculus AB Khan Academy.mp3", "Sentence": "We don't have to, but we could rewrite this as 1 7th times the integral of seven times the square root of seven x plus nine dx. So now if we set u equal to seven x plus nine, do we have its derivative laying around? Well, sure, the seven is right over here. We know that du, if we want to write in differential form, du is equal to seven times dx. So du is equal to seven times dx. That part right over there is equal to du. And if we want to care about u, well, that's just going to be the seven x plus nine."}, {"video_title": "_-substitution multiplying by a constant AP Calculus AB Khan Academy.mp3", "Sentence": "We know that du, if we want to write in differential form, du is equal to seven times dx. So du is equal to seven times dx. That part right over there is equal to du. And if we want to care about u, well, that's just going to be the seven x plus nine. That is our u. So let's rewrite this indefinite integral in terms of u. It's going to be equal to 1 7th times the integral of, and I'll just take the seven and put it in the back so we can just write the square root of u, square root of u, du, seven times dx is du."}, {"video_title": "_-substitution multiplying by a constant AP Calculus AB Khan Academy.mp3", "Sentence": "And if we want to care about u, well, that's just going to be the seven x plus nine. That is our u. So let's rewrite this indefinite integral in terms of u. It's going to be equal to 1 7th times the integral of, and I'll just take the seven and put it in the back so we can just write the square root of u, square root of u, du, seven times dx is du. And we can rewrite this if we want as u to the 1 1\u20442 power. It makes it a little bit easier for us to kind of do the reverse power rule here. So we can rewrite this as equal to 1 7th times the integral of u to the 1 1\u20442 power du."}, {"video_title": "_-substitution multiplying by a constant AP Calculus AB Khan Academy.mp3", "Sentence": "It's going to be equal to 1 7th times the integral of, and I'll just take the seven and put it in the back so we can just write the square root of u, square root of u, du, seven times dx is du. And we can rewrite this if we want as u to the 1 1\u20442 power. It makes it a little bit easier for us to kind of do the reverse power rule here. So we can rewrite this as equal to 1 7th times the integral of u to the 1 1\u20442 power du. And let me just make it clear. This u I could have written in white if I wanted the same color, and this du is the same du right over here. So what is the antiderivative of u to the 1 1\u20442 power?"}, {"video_title": "_-substitution multiplying by a constant AP Calculus AB Khan Academy.mp3", "Sentence": "So we can rewrite this as equal to 1 7th times the integral of u to the 1 1\u20442 power du. And let me just make it clear. This u I could have written in white if I wanted the same color, and this du is the same du right over here. So what is the antiderivative of u to the 1 1\u20442 power? Well, we increment u's power by one. So this is going to be equal to, and let me not forget this 1 7th out front. So it's going to be 1 7th times, 1 7th times, if we increment the power here, it's going to be u to the 3 1\u20442."}, {"video_title": "_-substitution multiplying by a constant AP Calculus AB Khan Academy.mp3", "Sentence": "So what is the antiderivative of u to the 1 1\u20442 power? Well, we increment u's power by one. So this is going to be equal to, and let me not forget this 1 7th out front. So it's going to be 1 7th times, 1 7th times, if we increment the power here, it's going to be u to the 3 1\u20442. 1 1\u20442 plus one is 1 1\u20442 or 3 1\u20442. So it's going to be u to the 3 1\u20442, u to the 3 1\u20442. And then we're gonna multiply this new thing times the reciprocal of 3 1\u20442, which is 2 1\u20443."}, {"video_title": "_-substitution multiplying by a constant AP Calculus AB Khan Academy.mp3", "Sentence": "So it's going to be 1 7th times, 1 7th times, if we increment the power here, it's going to be u to the 3 1\u20442. 1 1\u20442 plus one is 1 1\u20442 or 3 1\u20442. So it's going to be u to the 3 1\u20442, u to the 3 1\u20442. And then we're gonna multiply this new thing times the reciprocal of 3 1\u20442, which is 2 1\u20443. And I encourage you to verify the derivative of 2 1\u20443 u to the 3 1\u20442 is indeed u to the 1 1\u20442. And so we have that, and since we're multiplying 1 7th times this entire indefinite integral, we could also throw in a plus c right over here. There might have been a constant."}, {"video_title": "_-substitution multiplying by a constant AP Calculus AB Khan Academy.mp3", "Sentence": "And then we're gonna multiply this new thing times the reciprocal of 3 1\u20442, which is 2 1\u20443. And I encourage you to verify the derivative of 2 1\u20443 u to the 3 1\u20442 is indeed u to the 1 1\u20442. And so we have that, and since we're multiplying 1 7th times this entire indefinite integral, we could also throw in a plus c right over here. There might have been a constant. And if we want, we can distribute the 1 7th. So it would get 1 7th times 2 1\u20443 is 2 over 21, 2 over 21 u to the 3 1\u20442. And 1 7th times some constant, well, that's just going to be some constant."}, {"video_title": "_-substitution multiplying by a constant AP Calculus AB Khan Academy.mp3", "Sentence": "There might have been a constant. And if we want, we can distribute the 1 7th. So it would get 1 7th times 2 1\u20443 is 2 over 21, 2 over 21 u to the 3 1\u20442. And 1 7th times some constant, well, that's just going to be some constant. And so I could write a constant like that. I could call that c one, and then I could call this c two, but it really is just some arbitrary constant. And we're done."}, {"video_title": "_-substitution multiplying by a constant AP Calculus AB Khan Academy.mp3", "Sentence": "And 1 7th times some constant, well, that's just going to be some constant. And so I could write a constant like that. I could call that c one, and then I could call this c two, but it really is just some arbitrary constant. And we're done. Oh, actually, no, we aren't done. We still just have our entire thing in terms of u. So now let's unsubstitute it."}, {"video_title": "_-substitution multiplying by a constant AP Calculus AB Khan Academy.mp3", "Sentence": "And we're done. Oh, actually, no, we aren't done. We still just have our entire thing in terms of u. So now let's unsubstitute it. So this is going to be equal to 2 over 21 times u to the 3 1\u20442. And we already know what u is equal to. u is equal to 7x plus nine."}, {"video_title": "_-substitution multiplying by a constant AP Calculus AB Khan Academy.mp3", "Sentence": "So now let's unsubstitute it. So this is going to be equal to 2 over 21 times u to the 3 1\u20442. And we already know what u is equal to. u is equal to 7x plus nine. Let me put a new color in here just to ease the monotony. So it's going to be 2 over 21 times 7x plus nine to the 3 1\u20442 power, to the 3 1\u20442 power plus c, plus c. And we are done. We were able to take a kind of hairy-looking integral and realize that even though it wasn't completely obvious at first that u-substitution is applicable."}, {"video_title": "Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let's take the derivative of this. So we can view this as the product of two functions. So the product rule tells us that this is going to be the derivative with respect to x of e to the cosine of x, e to the cosine of x times cosine, times cosine of e to the x plus, plus the first function, just e to the cosine of x, e to the cosine of x times the derivative of the second function, times the derivative with respect to x of cosine of e to the x, cosine of e to the x. And so we just need to figure out what these two derivatives are. And so you can imagine the chain rule might be applicable here. So let me make it clear. This we got from the product rule."}, {"video_title": "Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And so we just need to figure out what these two derivatives are. And so you can imagine the chain rule might be applicable here. So let me make it clear. This we got from the product rule. Product, product rule. But then to evaluate each of these derivatives, we need to use the chain rule. So let's think about this a little bit."}, {"video_title": "Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "This we got from the product rule. Product, product rule. But then to evaluate each of these derivatives, we need to use the chain rule. So let's think about this a little bit. So the derivative, let me copy and paste this so I don't have to rewrite it. So copy and paste. So let's think about what the derivative of e to the cosine of x is."}, {"video_title": "Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let's think about this a little bit. So the derivative, let me copy and paste this so I don't have to rewrite it. So copy and paste. So let's think about what the derivative of e to the cosine of x is. E to the cosine of x. So we could view our outer function as e to the something, as e to the something. And the derivative of e to the something with respect to something is just going to be e to that something."}, {"video_title": "Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let's think about what the derivative of e to the cosine of x is. E to the cosine of x. So we could view our outer function as e to the something, as e to the something. And the derivative of e to the something with respect to something is just going to be e to that something. So it's going to be e to the cosine of x. So let me do that in that same blue color. So it's going to be e, actually let me do it in a new color."}, {"video_title": "Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And the derivative of e to the something with respect to something is just going to be e to that something. So it's going to be e to the cosine of x. So let me do that in that same blue color. So it's going to be e, actually let me do it in a new color. Let me do it in magenta. So the derivative of e to the something with respect to something is just e to the something. It's just e to the cosine of x."}, {"video_title": "Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So it's going to be e, actually let me do it in a new color. Let me do it in magenta. So the derivative of e to the something with respect to something is just e to the something. It's just e to the cosine of x. And we have to multiply that times the derivative of the something with respect to x. So what's the derivative of cosine of x with respect to x? Well, that's just negative sine of x."}, {"video_title": "Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "It's just e to the cosine of x. And we have to multiply that times the derivative of the something with respect to x. So what's the derivative of cosine of x with respect to x? Well, that's just negative sine of x. So it's times negative sine of x. And so we figured out this first derivative. Let me make it clear."}, {"video_title": "Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, that's just negative sine of x. So it's times negative sine of x. And so we figured out this first derivative. Let me make it clear. This right over here is the derivative of e to the cosine of x with respect to cosine of x. And this right over here is the derivative of cosine of x with respect to x. And we just took the product of the two."}, {"video_title": "Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Let me make it clear. This right over here is the derivative of e to the cosine of x with respect to cosine of x. And this right over here is the derivative of cosine of x with respect to x. And we just took the product of the two. That's what the chain rule tells us. Fair enough. Now let's figure out this derivative out here."}, {"video_title": "Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And we just took the product of the two. That's what the chain rule tells us. Fair enough. Now let's figure out this derivative out here. So we want to find the derivative with respect to x of cosine e to the x. So once again, let me copy and paste it. So we need to figure out this thing right over here."}, {"video_title": "Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's figure out this derivative out here. So we want to find the derivative with respect to x of cosine e to the x. So once again, let me copy and paste it. So we need to figure out this thing right over here. So first, just like we did, we're just going to apply the chain rule again. We need to figure out the derivative of cosine of something, in this case, e to the x, with respect to that something. So this is going to be equal to derivative of cosine of something with respect to that something is equal to the negative sine of that something."}, {"video_title": "Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So we need to figure out this thing right over here. So first, just like we did, we're just going to apply the chain rule again. We need to figure out the derivative of cosine of something, in this case, e to the x, with respect to that something. So this is going to be equal to derivative of cosine of something with respect to that something is equal to the negative sine of that something. Negative sine of e to the x. Once again, we can view this as the derivative of cosine of e to the x with respect to e to the x. And then we multiply that times the derivative of the something with respect to x."}, {"video_title": "Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be equal to derivative of cosine of something with respect to that something is equal to the negative sine of that something. Negative sine of e to the x. Once again, we can view this as the derivative of cosine of e to the x with respect to e to the x. And then we multiply that times the derivative of the something with respect to x. So let me do this in this. I'm running out of colors. Let me do this in this green color."}, {"video_title": "Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And then we multiply that times the derivative of the something with respect to x. So let me do this in this. I'm running out of colors. Let me do this in this green color. So times the derivative of e to the x with respect to x is just e to the x. So that right over there is the derivative of e to the x with respect to x. And so we're essentially done."}, {"video_title": "Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Let me do this in this green color. So times the derivative of e to the x with respect to x is just e to the x. So that right over there is the derivative of e to the x with respect to x. And so we're essentially done. We just have to substitute what we found using the chain rule back into our original expression. The derivative of this business up here is going to be equal to, let me just copy and paste everything just to make everything nice and clean. So copy and paste."}, {"video_title": "Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And so we're essentially done. We just have to substitute what we found using the chain rule back into our original expression. The derivative of this business up here is going to be equal to, let me just copy and paste everything just to make everything nice and clean. So copy and paste. So that is going to be equal to this times cosine e to the x. So this is going to be, let's see, we could put the e to the x out front. We could put the negative out front."}, {"video_title": "Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So copy and paste. So that is going to be equal to this times cosine e to the x. So this is going to be, let's see, we could put the e to the x out front. We could put the negative out front. So we could write it as negative e to the cosine x times sine of x times cosine of e to the x. So that's this first term here. Plus e to the cosine x times all of this stuff."}, {"video_title": "Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We could put the negative out front. So we could write it as negative e to the cosine x times sine of x times cosine of e to the x. So that's this first term here. Plus e to the cosine x times all of this stuff. And so let's see, we could put the negative out front again. So let's put that negative out front. So we have a negative."}, {"video_title": "Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Plus e to the cosine x times all of this stuff. And so let's see, we could put the negative out front again. So let's put that negative out front. So we have a negative. Now we have e to the cosine of x times e to the x. So I could write it this way. e to the x times e to the cosine x."}, {"video_title": "Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So we have a negative. Now we have e to the cosine of x times e to the x. So I could write it this way. e to the x times e to the cosine x. And you could simplify that or combine it since you're multiplying two things with the same base. But I'll just leave it like this. e to the x times e to the cosine x times the sine."}, {"video_title": "Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "e to the x times e to the cosine x. And you could simplify that or combine it since you're multiplying two things with the same base. But I'll just leave it like this. e to the x times e to the cosine x times the sine. We already have the negative. So then we have sine of e to the x. Sine of e to the x."}, {"video_title": "Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "e to the x times e to the cosine x times the sine. We already have the negative. So then we have sine of e to the x. Sine of e to the x. Let me write it over here. Times sine of e to the x. We had negative sine e to the x times e to the x."}, {"video_title": "Derivative of e_____cos(e_) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Sine of e to the x. Let me write it over here. Times sine of e to the x. We had negative sine e to the x times e to the x. Negative sine of e to the x times e to the x. And then that was multiplied by e to the cosine of x. So we have the exact same thing right over here."}, {"video_title": "Inflection points (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "The graph of g is given right over here, given below. How many inflection points does the graph of g have? And so let's just remind ourselves what are inflection points. So inflection points are where we change concavity. So we go from concave, concave upwards, upwards, actually let me just draw it graphically. We're going from concave upwards to concave downwards, or concave downwards to concave upwards. So, or another way you could think about it, you could say we're going from our slope increasing, increasing, increasing to our slope decreasing, to our slope decreasing, or the other way around."}, {"video_title": "Inflection points (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "So inflection points are where we change concavity. So we go from concave, concave upwards, upwards, actually let me just draw it graphically. We're going from concave upwards to concave downwards, or concave downwards to concave upwards. So, or another way you could think about it, you could say we're going from our slope increasing, increasing, increasing to our slope decreasing, to our slope decreasing, or the other way around. Any points where your slope goes from decreasing, our slope goes from decreasing to increasing, to increasing. So let's think about that. So as we start off right over here, so at the extreme left, it seems like we have a very high slope."}, {"video_title": "Inflection points (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "So, or another way you could think about it, you could say we're going from our slope increasing, increasing, increasing to our slope decreasing, to our slope decreasing, or the other way around. Any points where your slope goes from decreasing, our slope goes from decreasing to increasing, to increasing. So let's think about that. So as we start off right over here, so at the extreme left, it seems like we have a very high slope. It's a very steep curve. And then it stays increasing, but it's getting less positive. So it's getting a little bit, it's getting a little bit flatter."}, {"video_title": "Inflection points (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "So as we start off right over here, so at the extreme left, it seems like we have a very high slope. It's a very steep curve. And then it stays increasing, but it's getting less positive. So it's getting a little bit, it's getting a little bit flatter. So our slope is at a very high level, but it's decreasing. It's decreasing, decreasing, decreasing. Slope is decreasing, decreasing even more."}, {"video_title": "Inflection points (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "So it's getting a little bit, it's getting a little bit flatter. So our slope is at a very high level, but it's decreasing. It's decreasing, decreasing, decreasing. Slope is decreasing, decreasing even more. It's even more. And then it's actually going to zero. Our slope is zero."}, {"video_title": "Inflection points (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "Slope is decreasing, decreasing even more. It's even more. And then it's actually going to zero. Our slope is zero. And then it becomes negative. So our slope is still decreasing. And then it's becoming more and more and more negative."}, {"video_title": "Inflection points (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "Our slope is zero. And then it becomes negative. So our slope is still decreasing. And then it's becoming more and more and more negative. And then right around, and then right around here, it looks like it starts becoming less negative, or it starts increasing. So our slope is increasing, increasing. It's really just becoming less and less negative."}, {"video_title": "Inflection points (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "And then it's becoming more and more and more negative. And then right around, and then right around here, it looks like it starts becoming less negative, or it starts increasing. So our slope is increasing, increasing. It's really just becoming less and less negative. And then it's going close to zero, approaching zero. It looks like our slope is zero right over here. But then it looks like right over there, our slope begins decreasing again."}, {"video_title": "Inflection points (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "It's really just becoming less and less negative. And then it's going close to zero, approaching zero. It looks like our slope is zero right over here. But then it looks like right over there, our slope begins decreasing again. So it looks like our slope is decreasing again. So it looks like our slope is decreasing. It's becoming more and more and more and more negative."}, {"video_title": "Inflection points (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "But then it looks like right over there, our slope begins decreasing again. So it looks like our slope is decreasing again. So it looks like our slope is decreasing. It's becoming more and more and more and more negative. And so it looks like something interesting happened right over there. We had a transition point. And then right around here, it looks like it starts, the slope starts increasing again."}, {"video_title": "Inflection points (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "It's becoming more and more and more and more negative. And so it looks like something interesting happened right over there. We had a transition point. And then right around here, it looks like it starts, the slope starts increasing again. So it looks like the slope starts increasing. It's negative, but it's becoming less and less and less negative. And then it becomes zero."}, {"video_title": "Inflection points (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "And then right around here, it looks like it starts, the slope starts increasing again. So it looks like the slope starts increasing. It's negative, but it's becoming less and less and less negative. And then it becomes zero. And then it becomes positive. And then more and more and more and more positive. So inflection points are where we go from slope increasing to slope decreasing."}, {"video_title": "Inflection points (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "And then it becomes zero. And then it becomes positive. And then more and more and more and more positive. So inflection points are where we go from slope increasing to slope decreasing. So concave upwards to concave downwards. And so slope increasing was here to slope decreasing. So this was an inflection point."}, {"video_title": "Inflection points (graphical) AP Calculus AB Khan Academy.mp3", "Sentence": "So inflection points are where we go from slope increasing to slope decreasing. So concave upwards to concave downwards. And so slope increasing was here to slope decreasing. So this was an inflection point. And also from slope decreasing to slope increasing. So that's slope decreasing to slope increasing. And this is also slope decreasing to slope increasing."}, {"video_title": "Undefined limits by direct substitution Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Let's see if we can figure out the limit of x over natural log of x as x approaches one. And like always, pause this video and see if you can figure it out on your own. Well, we know from our limit properties this is going to be the same thing as the limit as x approaches one of x over, over the limit, the limit as x approaches one of the natural log of x. Now, this top limit, the one I have in magenta, this is pretty straightforward. This, if we had the graph of y equals x, that would be continuous everywhere. It's defined for all real numbers and it's continuous at all real numbers. And so it's continuous, the limit as x approaches one of x is just going to be this evaluated at x equals one."}, {"video_title": "Undefined limits by direct substitution Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Now, this top limit, the one I have in magenta, this is pretty straightforward. This, if we had the graph of y equals x, that would be continuous everywhere. It's defined for all real numbers and it's continuous at all real numbers. And so it's continuous, the limit as x approaches one of x is just going to be this evaluated at x equals one. So this is just going to be one. We just put a one in for this x. So the numerator here, we just evaluate to a one."}, {"video_title": "Undefined limits by direct substitution Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And so it's continuous, the limit as x approaches one of x is just going to be this evaluated at x equals one. So this is just going to be one. We just put a one in for this x. So the numerator here, we just evaluate to a one. And then the denominator. Natural log of x is not defined for all x's and therefore it isn't continuous everywhere, but it is continuous at x equals one. And since it is continuous at x equals one, then the limit here is just going to be the natural log evaluated at x equals one."}, {"video_title": "Undefined limits by direct substitution Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So the numerator here, we just evaluate to a one. And then the denominator. Natural log of x is not defined for all x's and therefore it isn't continuous everywhere, but it is continuous at x equals one. And since it is continuous at x equals one, then the limit here is just going to be the natural log evaluated at x equals one. So this is just going to be the natural log, the natural log of one, which of course is zero. E to the zero power is one. So this is all going to be equal to, this is going to be equal to, we just evaluate it, one over, one over zero."}, {"video_title": "Undefined limits by direct substitution Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And since it is continuous at x equals one, then the limit here is just going to be the natural log evaluated at x equals one. So this is just going to be the natural log, the natural log of one, which of course is zero. E to the zero power is one. So this is all going to be equal to, this is going to be equal to, we just evaluate it, one over, one over zero. And now we face a bit of a conundrum. One over zero is not defined. If it was zero over zero, we wouldn't necessarily be done yet."}, {"video_title": "Undefined limits by direct substitution Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So this is all going to be equal to, this is going to be equal to, we just evaluate it, one over, one over zero. And now we face a bit of a conundrum. One over zero is not defined. If it was zero over zero, we wouldn't necessarily be done yet. That's indeterminate form. As we will learn in the future, there are tools we can apply when we're trying to find limits and we evaluate it like this and we get zero over zero. But one over zero, this is undefined, which tells us that this limit does not exist."}, {"video_title": "Proving a sequence converges using the formal definition Series AP Calculus BC Khan Academy.mp3", "Sentence": "I made a claim that for this sequence, and this was in a previous video, that for this sequence right over here that can be defined explicitly in this way, that the limit of the sequence, and so I could write this as negative one to the n plus one over n. That's one way of defining our sequence explicitly. The limit of this as n approaches infinity is equal to zero. And it seems that way. As n gets larger and larger and larger, this thing gets, even though the numerator oscillates between negative one and one, it seems like it'll get smaller and smaller and smaller. But I didn't prove it, and that's what I wanna do in this video. In order to prove it, this is going to be true if and only if, if and only if, for any epsilon greater than zero, there is an M, a capital M greater than zero, a capital M greater than zero such that if lowercase n, if our index is greater than capital M, then the nth term in our sequence is going to be within epsilon of our limit, within epsilon of zero, is going to be within epsilon of zero. So what does that say?"}, {"video_title": "Proving a sequence converges using the formal definition Series AP Calculus BC Khan Academy.mp3", "Sentence": "As n gets larger and larger and larger, this thing gets, even though the numerator oscillates between negative one and one, it seems like it'll get smaller and smaller and smaller. But I didn't prove it, and that's what I wanna do in this video. In order to prove it, this is going to be true if and only if, if and only if, for any epsilon greater than zero, there is an M, a capital M greater than zero, a capital M greater than zero such that if lowercase n, if our index is greater than capital M, then the nth term in our sequence is going to be within epsilon of our limit, within epsilon of zero, is going to be within epsilon of zero. So what does that say? That says, hey, give me, our limit is zero. Let me do this in a new color. So our limit right over here is zero."}, {"video_title": "Proving a sequence converges using the formal definition Series AP Calculus BC Khan Academy.mp3", "Sentence": "So what does that say? That says, hey, give me, our limit is zero. Let me do this in a new color. So our limit right over here is zero. That's our limit. So our limit right over here is, we're saying the sequence is converging to zero. We're saying, give us an epsilon around zero."}, {"video_title": "Proving a sequence converges using the formal definition Series AP Calculus BC Khan Academy.mp3", "Sentence": "So our limit right over here is zero. That's our limit. So our limit right over here is, we're saying the sequence is converging to zero. We're saying, give us an epsilon around zero. So let's say that this right over here is zero plus epsilon. That is zero plus epsilon. The way I've drawn it here, it looks like epsilon would be.5."}, {"video_title": "Proving a sequence converges using the formal definition Series AP Calculus BC Khan Academy.mp3", "Sentence": "We're saying, give us an epsilon around zero. So let's say that this right over here is zero plus epsilon. That is zero plus epsilon. The way I've drawn it here, it looks like epsilon would be.5. This would be zero minus epsilon. Let me make it a little bit neater. So this would be zero minus epsilon."}, {"video_title": "Proving a sequence converges using the formal definition Series AP Calculus BC Khan Academy.mp3", "Sentence": "The way I've drawn it here, it looks like epsilon would be.5. This would be zero minus epsilon. Let me make it a little bit neater. So this would be zero minus epsilon. So this is negative epsilon, zero minus epsilon, zero plus epsilon. Our limit in this case, our claim of a limit is zero. Now this is saying for any epsilon, we need to find an M such that if N is greater than M, the distance between our sequence and our limit is going to be less than epsilon."}, {"video_title": "Proving a sequence converges using the formal definition Series AP Calculus BC Khan Academy.mp3", "Sentence": "So this would be zero minus epsilon. So this is negative epsilon, zero minus epsilon, zero plus epsilon. Our limit in this case, our claim of a limit is zero. Now this is saying for any epsilon, we need to find an M such that if N is greater than M, the distance between our sequence and our limit is going to be less than epsilon. So if the distance between our sequence and our limit is less than epsilon, that means that the value of our sequence for a given N is going to be within these two bounds. It's gotta be within this range right over here that I'm shading above a certain N. So if I pick an N right over here, it looks like anything larger than that is going to be the case that we're going to be within those bounds. But how do we prove it?"}, {"video_title": "Proving a sequence converges using the formal definition Series AP Calculus BC Khan Academy.mp3", "Sentence": "Now this is saying for any epsilon, we need to find an M such that if N is greater than M, the distance between our sequence and our limit is going to be less than epsilon. So if the distance between our sequence and our limit is less than epsilon, that means that the value of our sequence for a given N is going to be within these two bounds. It's gotta be within this range right over here that I'm shading above a certain N. So if I pick an N right over here, it looks like anything larger than that is going to be the case that we're going to be within those bounds. But how do we prove it? Well, let's just think about what needs to happen for this to be true. So what needs to happen for A sub N minus zero, the absolute value of A sub N minus zero, what needs to be true for this to be less than epsilon? Well, this is another way of saying that the absolute value of A sub N has to be less than epsilon."}, {"video_title": "Proving a sequence converges using the formal definition Series AP Calculus BC Khan Academy.mp3", "Sentence": "But how do we prove it? Well, let's just think about what needs to happen for this to be true. So what needs to happen for A sub N minus zero, the absolute value of A sub N minus zero, what needs to be true for this to be less than epsilon? Well, this is another way of saying that the absolute value of A sub N has to be less than epsilon. And A sub N is just this business right here, so it's another way of saying that the absolute value of negative one to the N plus one over N has to be less than epsilon, which is another way of saying, because this negative one to the N plus one, this numerator just swaps us between a negative and a positive version of one over N. But if you take the absolute value of it, this is always just going to be positive. So this is the same thing as one over N, as the absolute value of one over N has to be less than epsilon. Now, N is always going to be positive."}, {"video_title": "Proving a sequence converges using the formal definition Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, this is another way of saying that the absolute value of A sub N has to be less than epsilon. And A sub N is just this business right here, so it's another way of saying that the absolute value of negative one to the N plus one over N has to be less than epsilon, which is another way of saying, because this negative one to the N plus one, this numerator just swaps us between a negative and a positive version of one over N. But if you take the absolute value of it, this is always just going to be positive. So this is the same thing as one over N, as the absolute value of one over N has to be less than epsilon. Now, N is always going to be positive. N starts at one and goes to infinity. So this value is always going to be positive. So this is saying the same thing, that one over N has to be less than epsilon in order for this stuff to be true."}, {"video_title": "Proving a sequence converges using the formal definition Series AP Calculus BC Khan Academy.mp3", "Sentence": "Now, N is always going to be positive. N starts at one and goes to infinity. So this value is always going to be positive. So this is saying the same thing, that one over N has to be less than epsilon in order for this stuff to be true. And now we can take the reciprocal of both sides. And if you take the reciprocal of both sides of an inequality, you would have that N, if you take the reciprocal of both sides of an inequality, you swap the inequality. So this is, for this to be true, N has to be greater than one over epsilon."}, {"video_title": "Proving a sequence converges using the formal definition Series AP Calculus BC Khan Academy.mp3", "Sentence": "So this is saying the same thing, that one over N has to be less than epsilon in order for this stuff to be true. And now we can take the reciprocal of both sides. And if you take the reciprocal of both sides of an inequality, you would have that N, if you take the reciprocal of both sides of an inequality, you swap the inequality. So this is, for this to be true, N has to be greater than one over epsilon. And we are essentially have proven it now. So now we've said, look, for this particular sequence, you give me any epsilon, you give me any epsilon, and I'm going to set M, I'm going to be set M to be one over epsilon. Because if N is greater than M, which is one over epsilon, if N is greater than one over epsilon, then we know that this right over here is going to be true."}, {"video_title": "Proving a sequence converges using the formal definition Series AP Calculus BC Khan Academy.mp3", "Sentence": "So this is, for this to be true, N has to be greater than one over epsilon. And we are essentially have proven it now. So now we've said, look, for this particular sequence, you give me any epsilon, you give me any epsilon, and I'm going to set M, I'm going to be set M to be one over epsilon. Because if N is greater than M, which is one over epsilon, if N is greater than one over epsilon, then we know that this right over here is going to be true. That is going to be true. So the limit does definitely exist. And so over here, for this particular epsilon, it looks like we picked 0.5 or 1.5 as our epsilon."}, {"video_title": "Proving a sequence converges using the formal definition Series AP Calculus BC Khan Academy.mp3", "Sentence": "Because if N is greater than M, which is one over epsilon, if N is greater than one over epsilon, then we know that this right over here is going to be true. That is going to be true. So the limit does definitely exist. And so over here, for this particular epsilon, it looks like we picked 0.5 or 1.5 as our epsilon. So as long as N is greater than one over 1.5, which is two, so in this case, we could say, look, you gave me 1.5, I just put, my M is going to be a function of epsilon, it's going to be defined for any epsilon you give me greater than zero. So here, one over 1.5 is right over here, I'm going to make my M right over here. And you see, it is indeed the case that my sequence is within the range as we pass for any N greater than two."}, {"video_title": "Proving a sequence converges using the formal definition Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so over here, for this particular epsilon, it looks like we picked 0.5 or 1.5 as our epsilon. So as long as N is greater than one over 1.5, which is two, so in this case, we could say, look, you gave me 1.5, I just put, my M is going to be a function of epsilon, it's going to be defined for any epsilon you give me greater than zero. So here, one over 1.5 is right over here, I'm going to make my M right over here. And you see, it is indeed the case that my sequence is within the range as we pass for any N greater than two. So for N is equal to three, it's in the range, for N is equal to four, it's in the range, for N equals five, and it keeps going and going. And we're not just taking our word for it, we've proven it right over here. So we've made the proof."}, {"video_title": "Proving a sequence converges using the formal definition Series AP Calculus BC Khan Academy.mp3", "Sentence": "And you see, it is indeed the case that my sequence is within the range as we pass for any N greater than two. So for N is equal to three, it's in the range, for N is equal to four, it's in the range, for N equals five, and it keeps going and going. And we're not just taking our word for it, we've proven it right over here. So we've made the proof. You give me any other, any epsilon, I said M is equal to one over that thing. And so for N greater than that, this is going to be true. So this is definitely the case."}, {"video_title": "Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And like always, pause this video and see if you can figure it out on your own before we do it together. Alright, now first let's think about what's the limit of f of x as x approaches six. So as x, let me do this in a color you can see, as x approaches six from both sides, well as we approach six from the left-hand side, from values less than six, it looks like our f of x is approaching one, and as we approach x equals six from the right-hand side, it looks like our f of x is once again approaching one. And in order for this limit to exist, we need to be approaching the same value from both the left and the right-hand side. And so here, at least graphically, so you never are sure with the graph, but this is a pretty good estimate, it looks like we are approaching one. Right over there, let me do it in a darker color. Now let's do this next one."}, {"video_title": "Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And in order for this limit to exist, we need to be approaching the same value from both the left and the right-hand side. And so here, at least graphically, so you never are sure with the graph, but this is a pretty good estimate, it looks like we are approaching one. Right over there, let me do it in a darker color. Now let's do this next one. The limit of f of x as x approaches four. So as we approach four from the left-hand side, what is going on? Well as we approach four from the left-hand side, it looks like our function, the value of our function, it looks like it is approaching three."}, {"video_title": "Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's do this next one. The limit of f of x as x approaches four. So as we approach four from the left-hand side, what is going on? Well as we approach four from the left-hand side, it looks like our function, the value of our function, it looks like it is approaching three. Remember, you can have a limit exist at an x value where the function itself is not defined. The function, if you said what is f of four, it's not defined. But it looks like when we approach it from the left, when we approach x equals four from the left, it looks like f is approaching three."}, {"video_title": "Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Well as we approach four from the left-hand side, it looks like our function, the value of our function, it looks like it is approaching three. Remember, you can have a limit exist at an x value where the function itself is not defined. The function, if you said what is f of four, it's not defined. But it looks like when we approach it from the left, when we approach x equals four from the left, it looks like f is approaching three. And when we approach four from the right, once again, it looks like our function is approaching three. So here, I would say, at least from what we can tell from the graph, it looks like the limit of f of x as x approaches four is three, even though the function itself is not defined there. Now let's think about the limit as x approaches two."}, {"video_title": "Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "But it looks like when we approach it from the left, when we approach x equals four from the left, it looks like f is approaching three. And when we approach four from the right, once again, it looks like our function is approaching three. So here, I would say, at least from what we can tell from the graph, it looks like the limit of f of x as x approaches four is three, even though the function itself is not defined there. Now let's think about the limit as x approaches two. So this is interesting. The function is defined there. F of two is two."}, {"video_title": "Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's think about the limit as x approaches two. So this is interesting. The function is defined there. F of two is two. But see, when we approach from the left-hand side, it looks like our function is approaching the value of two. But when we approach from the right-hand side, when we approach x equals two from the right-hand side, our function is getting closer and closer to five. It's not quite getting to five, but as we go from 2.1, 2.01, 2.001, it looks like our function, the value of our function is getting closer and closer to five."}, {"video_title": "Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "F of two is two. But see, when we approach from the left-hand side, it looks like our function is approaching the value of two. But when we approach from the right-hand side, when we approach x equals two from the right-hand side, our function is getting closer and closer to five. It's not quite getting to five, but as we go from 2.1, 2.01, 2.001, it looks like our function, the value of our function is getting closer and closer to five. And since we are approaching two different values from the left-hand side and the right-hand side as x approaches two from the left-hand side and the right-hand side, we would say that this limit does not exist. So does not exist, which is interesting. In this first case, the function is defined at six, and the limit is equal to the value of the function at x equals six."}, {"video_title": "Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "It's not quite getting to five, but as we go from 2.1, 2.01, 2.001, it looks like our function, the value of our function is getting closer and closer to five. And since we are approaching two different values from the left-hand side and the right-hand side as x approaches two from the left-hand side and the right-hand side, we would say that this limit does not exist. So does not exist, which is interesting. In this first case, the function is defined at six, and the limit is equal to the value of the function at x equals six. Here, the function was not defined at x equals four, but the limit does exist. Here, the function is defined at x equals two, but the limit does not exist as we approach x equals two. Let's do another function, just to get more cases of looking at graphical limits."}, {"video_title": "Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "In this first case, the function is defined at six, and the limit is equal to the value of the function at x equals six. Here, the function was not defined at x equals four, but the limit does exist. Here, the function is defined at x equals two, but the limit does not exist as we approach x equals two. Let's do another function, just to get more cases of looking at graphical limits. So here, we have the graph of y is equal to g of x. And once again, pause this video and have a go at it. See if you can figure out these limits graphically."}, {"video_title": "Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Let's do another function, just to get more cases of looking at graphical limits. So here, we have the graph of y is equal to g of x. And once again, pause this video and have a go at it. See if you can figure out these limits graphically. So first, we have the limit as x approaches five of g of x. So as we approach five from the left-hand side, it looks like we are approaching this value. So let me see if I can draw a straight line that takes us, so it looks like we're approaching this value."}, {"video_title": "Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "See if you can figure out these limits graphically. So first, we have the limit as x approaches five of g of x. So as we approach five from the left-hand side, it looks like we are approaching this value. So let me see if I can draw a straight line that takes us, so it looks like we're approaching this value. And as we approach five from the right-hand side, it also looks like we are approaching that same value. And so this value, just eyeballing it off of here, it looks like it's about.4. So I'll say this limit definitely exists, just when we're looking at a graph, it's not that precise."}, {"video_title": "Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So let me see if I can draw a straight line that takes us, so it looks like we're approaching this value. And as we approach five from the right-hand side, it also looks like we are approaching that same value. And so this value, just eyeballing it off of here, it looks like it's about.4. So I'll say this limit definitely exists, just when we're looking at a graph, it's not that precise. So I would say it's approximately 0.4. It might be 0.41. It might be 0.41456789."}, {"video_title": "Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So I'll say this limit definitely exists, just when we're looking at a graph, it's not that precise. So I would say it's approximately 0.4. It might be 0.41. It might be 0.41456789. We don't know exactly just looking at this graph. But it looks like a value roughly around there. Now let's think about the limit of g of x as x approaches seven."}, {"video_title": "Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "It might be 0.41456789. We don't know exactly just looking at this graph. But it looks like a value roughly around there. Now let's think about the limit of g of x as x approaches seven. So let's do the same exercise. What happens as we approach from the left from values less than seven? 6.9, 6.99, 6.999."}, {"video_title": "Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's think about the limit of g of x as x approaches seven. So let's do the same exercise. What happens as we approach from the left from values less than seven? 6.9, 6.99, 6.999. Well, it looks like the value of our function is approaching two. It doesn't matter that the actual function is defined. The g of seven is five."}, {"video_title": "Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "6.9, 6.99, 6.999. Well, it looks like the value of our function is approaching two. It doesn't matter that the actual function is defined. The g of seven is five. But as we approach from the left, as x goes 6.9, 6.99, and so on, it looks like our value of our function is approaching two. And as we approach x equals seven from the right-hand side, it seems like the same thing is happening. It seems like we are approaching two."}, {"video_title": "Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "The g of seven is five. But as we approach from the left, as x goes 6.9, 6.99, and so on, it looks like our value of our function is approaching two. And as we approach x equals seven from the right-hand side, it seems like the same thing is happening. It seems like we are approaching two. And so I would say that this is going to be equal to two. And so once again, the function is defined there, and the limit exists there, but the g of seven is different than the value of the limit of g of x as x approaches seven. Now let's do one more."}, {"video_title": "Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "It seems like we are approaching two. And so I would say that this is going to be equal to two. And so once again, the function is defined there, and the limit exists there, but the g of seven is different than the value of the limit of g of x as x approaches seven. Now let's do one more. What's the limit as x approaches one? Well, we'll do the same thing. From the left-hand side, it looks like we're going unbounded."}, {"video_title": "Limits from graphs Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's do one more. What's the limit as x approaches one? Well, we'll do the same thing. From the left-hand side, it looks like we're going unbounded. As x goes.9, 0.99, 0.999, 0.99999, it looks like we're just going unbounded towards infinity. And as we approach from the right-hand side, it looks like the same thing, same thing is happening. We're going unbounded to infinity."}, {"video_title": "Motion along a curve finding rate of change Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "We're told that a particle moves along the curve x squared, y squared is equal to 16, so that the x coordinate is changing at a constant rate of negative two units per minute. What is the rate of change in units per minute of the particle's y coordinate when the particle is at the point one comma four? So let's just repeat or rewrite what they told us. So the curve is described by x squared, y squared is equal to 16, they tell us that up there. They tell us that the x coordinate is changing at a constant rate. Let me underline that. The x coordinate is changing at a constant rate of negative two units per minute."}, {"video_title": "Motion along a curve finding rate of change Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "So the curve is described by x squared, y squared is equal to 16, they tell us that up there. They tell us that the x coordinate is changing at a constant rate. Let me underline that. The x coordinate is changing at a constant rate of negative two units per minute. So we could say that dx, I'll write it over here on the right hand side, dx dt, the rate of change of the x coordinate with respect to time, is equal to negative two, and they're saying units, be some units, some unit of distance, units divided by minute, units per minute, and what they want us to figure out is what is the rate of change of the particle's y coordinate? So let me underline that. What is the rate of change of the particle's y coordinate?"}, {"video_title": "Motion along a curve finding rate of change Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "The x coordinate is changing at a constant rate of negative two units per minute. So we could say that dx, I'll write it over here on the right hand side, dx dt, the rate of change of the x coordinate with respect to time, is equal to negative two, and they're saying units, be some units, some unit of distance, units divided by minute, units per minute, and what they want us to figure out is what is the rate of change of the particle's y coordinate? So let me underline that. What is the rate of change of the particle's y coordinate? So what they want us to find is what is dy dt? What is that equal to? And they say when the particle is at the point one comma four so when x is equal to one, so x is equal to one, and y is equal to four, y is equal to four."}, {"video_title": "Motion along a curve finding rate of change Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "What is the rate of change of the particle's y coordinate? So what they want us to find is what is dy dt? What is that equal to? And they say when the particle is at the point one comma four so when x is equal to one, so x is equal to one, and y is equal to four, y is equal to four. So can we set up some equation that involves the rate of change of x with respect to t, y with respect to t, x and y? Well what if we were to take the derivative of this relation that describes the curve, what if we were to take the derivative with respect to t on both sides? So let me write that down."}, {"video_title": "Motion along a curve finding rate of change Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "And they say when the particle is at the point one comma four so when x is equal to one, so x is equal to one, and y is equal to four, y is equal to four. So can we set up some equation that involves the rate of change of x with respect to t, y with respect to t, x and y? Well what if we were to take the derivative of this relation that describes the curve, what if we were to take the derivative with respect to t on both sides? So let me write that down. So we're gonna take the derivative, actually let me just, let me just erase this so I have a little bit more space. Alright, and so that way I can just add it. So let's take the derivative with respect to t of both sides of that."}, {"video_title": "Motion along a curve finding rate of change Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "So let me write that down. So we're gonna take the derivative, actually let me just, let me just erase this so I have a little bit more space. Alright, and so that way I can just add it. So let's take the derivative with respect to t of both sides of that. And if at any point you get inspired, I encourage you to pause the video and try to work through it. Well on the left-hand side, if we view this as a product of two functions right over here, we could take the derivative of the first function, which is going to be the derivative of x squared with respect to x, so that is two x, and remember we're not just taking the derivative with respect to x. We're taking the derivative with respect to t so we have to apply the chain rule."}, {"video_title": "Motion along a curve finding rate of change Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "So let's take the derivative with respect to t of both sides of that. And if at any point you get inspired, I encourage you to pause the video and try to work through it. Well on the left-hand side, if we view this as a product of two functions right over here, we could take the derivative of the first function, which is going to be the derivative of x squared with respect to x, so that is two x, and remember we're not just taking the derivative with respect to x. We're taking the derivative with respect to t so we have to apply the chain rule. So it's gonna be the derivative of x squared with respect to x, which is two x, times the derivative of x with respect to t, So times dx dt. And then we're gonna multiply that times the second function. So times y squared."}, {"video_title": "Motion along a curve finding rate of change Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "We're taking the derivative with respect to t so we have to apply the chain rule. So it's gonna be the derivative of x squared with respect to x, which is two x, times the derivative of x with respect to t, So times dx dt. And then we're gonna multiply that times the second function. So times y squared. Times y squared. And then that's gonna be plus the first function, which is just x squared times the derivative of the second function with respect to t. And so once again, we're going to apply the chain rule. The derivative of y squared with respect to y is two y."}, {"video_title": "Motion along a curve finding rate of change Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "So times y squared. Times y squared. And then that's gonna be plus the first function, which is just x squared times the derivative of the second function with respect to t. And so once again, we're going to apply the chain rule. The derivative of y squared with respect to y is two y. Let me do that in that orange color. It is equal to two y. And then times the derivative of y with respect to t. Times dy dt."}, {"video_title": "Motion along a curve finding rate of change Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "The derivative of y squared with respect to y is two y. Let me do that in that orange color. It is equal to two y. And then times the derivative of y with respect to t. Times dy dt. And then that is going to be equal to, that is going to be equal to the derivative with respect to t of 16. Well, that doesn't change over time. So that's just going to be equal to zero."}, {"video_title": "Motion along a curve finding rate of change Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "And then times the derivative of y with respect to t. Times dy dt. And then that is going to be equal to, that is going to be equal to the derivative with respect to t of 16. Well, that doesn't change over time. So that's just going to be equal to zero. And so here we have it. We need to simplify this a little bit, but we have an equation that gives a relationship between x, derivative of x with respect to t, y, and derivative of y with respect to t. So actually, let me just rewrite it one more time so it's a little bit simplified. So this is two x y squared dx dt plus, actually, I don't even have to rewrite it again."}, {"video_title": "Motion along a curve finding rate of change Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "So that's just going to be equal to zero. And so here we have it. We need to simplify this a little bit, but we have an equation that gives a relationship between x, derivative of x with respect to t, y, and derivative of y with respect to t. So actually, let me just rewrite it one more time so it's a little bit simplified. So this is two x y squared dx dt plus, actually, I don't even have to rewrite it again. We can, all we're doing, trying to do is solve for dy dt. So let's actually just substitute the values in. So we know, we want to figure out what's going on when x is equal to one."}, {"video_title": "Motion along a curve finding rate of change Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "So this is two x y squared dx dt plus, actually, I don't even have to rewrite it again. We can, all we're doing, trying to do is solve for dy dt. So let's actually just substitute the values in. So we know, we want to figure out what's going on when x is equal to one. So we know that the x's here are equal to one. This x, x squared, well, that's just going to be one squared so that's going to be equal to one. We know that y is equal to four."}, {"video_title": "Motion along a curve finding rate of change Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "So we know, we want to figure out what's going on when x is equal to one. So we know that the x's here are equal to one. This x, x squared, well, that's just going to be one squared so that's going to be equal to one. We know that y is equal to four. So this is going to be 16 and this is going to be eight. We know the derivative of x with respect to t is negative two they tell us that in the problem statement, negative two. And so now this is a good time to simplify this thing."}, {"video_title": "Motion along a curve finding rate of change Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "We know that y is equal to four. So this is going to be 16 and this is going to be eight. We know the derivative of x with respect to t is negative two they tell us that in the problem statement, negative two. And so now this is a good time to simplify this thing. So this will simplify to, let's see, all of this is going to be two times one times negative two so that is negative four times 16, so that is negative 64. And then we have, let me do this in a color you can see, and then we have all of this. Well, this is just going to be one times eight times dy dt."}, {"video_title": "Motion along a curve finding rate of change Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "And so now this is a good time to simplify this thing. So this will simplify to, let's see, all of this is going to be two times one times negative two so that is negative four times 16, so that is negative 64. And then we have, let me do this in a color you can see, and then we have all of this. Well, this is just going to be one times eight times dy dt. So this is going to be eight dy dt. So plus eight times the derivative of y with respect to t is equal to zero. Add 64 to both sides and we get, I'll switch to a neutral color, eight times the derivative of y with respect to t is equal to 64."}, {"video_title": "Motion along a curve finding rate of change Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "Well, this is just going to be one times eight times dy dt. So this is going to be eight dy dt. So plus eight times the derivative of y with respect to t is equal to zero. Add 64 to both sides and we get, I'll switch to a neutral color, eight times the derivative of y with respect to t is equal to 64. Divide both sides by eight and you get the derivative of y with respect to t is equal to, 64 divided by eight is just eight. And if you want to look at the units, it will also be in units per minute, some units of distance per minute. And we are done."}, {"video_title": "Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So what do we mean by that? And this is a very, what I just said is not that rigorous, or not rigorous at all, is that, well, think about the point right over here, let's say that's our C. If I can draw the graph at that point, the value of the function at that point without picking up my pencil, or my pen, then it's continuous there. So I could just start here, and I don't have to pick up my pencil, and there you go, I can draw, I can go through that point, so we could say that our function is continuous there. But if I had a function that looked somewhat different than that, if I had a function that looked like this, let's say that it is defined up until then, and then there's a bit of a jump, and then it goes like this, well, this would be very hard to draw, this function would be very hard to draw going through X equals C without picking up my pen. Let's see, my pen is touching the screen, touching the screen, touching the screen, how do I keep drawing this function without picking up my pen? I would have to pick it up, and then move back down here, and so that is an intuitive sense that we are not continuous in this case right over here. But let's actually come up with a formal definition for continuity, and then see if it feels intuitive for us."}, {"video_title": "Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "But if I had a function that looked somewhat different than that, if I had a function that looked like this, let's say that it is defined up until then, and then there's a bit of a jump, and then it goes like this, well, this would be very hard to draw, this function would be very hard to draw going through X equals C without picking up my pen. Let's see, my pen is touching the screen, touching the screen, touching the screen, how do I keep drawing this function without picking up my pen? I would have to pick it up, and then move back down here, and so that is an intuitive sense that we are not continuous in this case right over here. But let's actually come up with a formal definition for continuity, and then see if it feels intuitive for us. So the formal definition of continuity, let's start here, we'll start with continuity at a point. So we could say the function F is continuous, continuous at X equals C, continuous at X equals C, if and only if, I'll draw this two-way arrow to show if and only if, the two-sided limit of F of X as X approaches C is equal to F of C. So this seems very technical, but let's just think about what it's saying. It's saying, look, if the limit as we approach C from the left and the right of F of X, if that's actually the value of our function there, then we are continuous at that point."}, {"video_title": "Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "But let's actually come up with a formal definition for continuity, and then see if it feels intuitive for us. So the formal definition of continuity, let's start here, we'll start with continuity at a point. So we could say the function F is continuous, continuous at X equals C, continuous at X equals C, if and only if, I'll draw this two-way arrow to show if and only if, the two-sided limit of F of X as X approaches C is equal to F of C. So this seems very technical, but let's just think about what it's saying. It's saying, look, if the limit as we approach C from the left and the right of F of X, if that's actually the value of our function there, then we are continuous at that point. So let's look at three examples. Let's look at one example where, by our picking up the pencil idea, it feels like we are continuous at a point, and then let's think about a couple of examples where it doesn't seem like we're continuous at a point, and see how this more rigorous definition applies. So let's say that my function, so let's say this right over here is, Y is equal to F of X, and we care about the behavior right over here when X is equal to C. This is my X axis, that's my Y axis."}, {"video_title": "Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "It's saying, look, if the limit as we approach C from the left and the right of F of X, if that's actually the value of our function there, then we are continuous at that point. So let's look at three examples. Let's look at one example where, by our picking up the pencil idea, it feels like we are continuous at a point, and then let's think about a couple of examples where it doesn't seem like we're continuous at a point, and see how this more rigorous definition applies. So let's say that my function, so let's say this right over here is, Y is equal to F of X, and we care about the behavior right over here when X is equal to C. This is my X axis, that's my Y axis. So we care about the behavior when X is equal to C. And so notice, from our first intuitive sense, I can definitely draw this function as we go through X equals C without picking up my pencil, so it feels continuous there. There's no jumps or discontinuities that we can tell. It just kind of keeps on going."}, {"video_title": "Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say that my function, so let's say this right over here is, Y is equal to F of X, and we care about the behavior right over here when X is equal to C. This is my X axis, that's my Y axis. So we care about the behavior when X is equal to C. And so notice, from our first intuitive sense, I can definitely draw this function as we go through X equals C without picking up my pencil, so it feels continuous there. There's no jumps or discontinuities that we can tell. It just kind of keeps on going. It seems all connected is one way to think about it. But let's think about this definition. Well, the limit as X approaches C from the left, it is, as we approach from the left, it looks like it is approaching, it looks like it is approaching F of C. So this is the value F of C right over here."}, {"video_title": "Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "It just kind of keeps on going. It seems all connected is one way to think about it. But let's think about this definition. Well, the limit as X approaches C from the left, it is, as we approach from the left, it looks like it is approaching, it looks like it is approaching F of C. So this is the value F of C right over here. And as we approach from the right, as we approach from the right, it also looks like it's approaching F of C. And we are defined right at X equals C, and it is the value that we are approaching from both the left or the right. So this seems good in this scenario. So now let's look at some scenarios that we would have to pick up the pencil as we draw the function through that point, through that, when X is equal to C. So let's look at a scenario."}, {"video_title": "Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the limit as X approaches C from the left, it is, as we approach from the left, it looks like it is approaching, it looks like it is approaching F of C. So this is the value F of C right over here. And as we approach from the right, as we approach from the right, it also looks like it's approaching F of C. And we are defined right at X equals C, and it is the value that we are approaching from both the left or the right. So this seems good in this scenario. So now let's look at some scenarios that we would have to pick up the pencil as we draw the function through that point, through that, when X is equal to C. So let's look at a scenario. Let's look at a scenario where we have what's often called a point discontinuity, although you don't have to know at this point, not no pun intended, the formal terminology for it. So let's say we have a function that, let's see, this is C. And let's say our function looks something like this. So we go like this, and at C, let's say it's equal to that."}, {"video_title": "Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So now let's look at some scenarios that we would have to pick up the pencil as we draw the function through that point, through that, when X is equal to C. So let's look at a scenario. Let's look at a scenario where we have what's often called a point discontinuity, although you don't have to know at this point, not no pun intended, the formal terminology for it. So let's say we have a function that, let's see, this is C. And let's say our function looks something like this. So we go like this, and at C, let's say it's equal to that. So F of C is right over here. Right over here. F of C would be that value."}, {"video_title": "Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So we go like this, and at C, let's say it's equal to that. So F of C is right over here. Right over here. F of C would be that value. But what's the limit as X approaches C? So the limit as X approaches C, and this would be a two-sided limit of F of X, well, this is, as we approach from the left, it looks like we are approaching this value right over here. And from the right, it looks like we are approaching that same value, and so we could call that L. And L is different than F of C. And so in this case, by our formal definition, we will not be continuous at, F will not be continuous for X, or at the point X, or when X is equal to C. And you can see that there."}, {"video_title": "Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "F of C would be that value. But what's the limit as X approaches C? So the limit as X approaches C, and this would be a two-sided limit of F of X, well, this is, as we approach from the left, it looks like we are approaching this value right over here. And from the right, it looks like we are approaching that same value, and so we could call that L. And L is different than F of C. And so in this case, by our formal definition, we will not be continuous at, F will not be continuous for X, or at the point X, or when X is equal to C. And you can see that there. If we tried to draw this, okay, my pencil's touching the paper, touching the paper, touching the paper. Uh-oh, if I needed to keep drawing this function, I'd have to pick up my pencil, move it over here, then pick it up again, and then jump right back down. But this rigorous definition is giving us the same conclusion."}, {"video_title": "Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And from the right, it looks like we are approaching that same value, and so we could call that L. And L is different than F of C. And so in this case, by our formal definition, we will not be continuous at, F will not be continuous for X, or at the point X, or when X is equal to C. And you can see that there. If we tried to draw this, okay, my pencil's touching the paper, touching the paper, touching the paper. Uh-oh, if I needed to keep drawing this function, I'd have to pick up my pencil, move it over here, then pick it up again, and then jump right back down. But this rigorous definition is giving us the same conclusion. The limit as we approach X equals C from the left and the right, it's a different value than F of C, and so this is not continuous. Not, not continuous. And let's think about another scenario."}, {"video_title": "Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "But this rigorous definition is giving us the same conclusion. The limit as we approach X equals C from the left and the right, it's a different value than F of C, and so this is not continuous. Not, not continuous. And let's think about another scenario. Let's think about a scenario, and actually, maybe let's think about a scenario where the limit, the two-sided limit doesn't even exist. So, there are my axes, X and Y, and let's say it's doing something like this. Let's say it's doing something like this, and then it does something like this and goes like that."}, {"video_title": "Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And let's think about another scenario. Let's think about a scenario, and actually, maybe let's think about a scenario where the limit, the two-sided limit doesn't even exist. So, there are my axes, X and Y, and let's say it's doing something like this. Let's say it's doing something like this, and then it does something like this and goes like that. And let's say that this right over here is our C. And so, let's see, this is F of C right over here. That is, let me draw it a little bit neater. That is F of C, and it does look like the limit as X approaches C from the left."}, {"video_title": "Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say it's doing something like this, and then it does something like this and goes like that. And let's say that this right over here is our C. And so, let's see, this is F of C right over here. That is, let me draw it a little bit neater. That is F of C, and it does look like the limit as X approaches C from the left. So, from values less than C, it does look like that is approaching F of C. But if we look at the limit as X approaches C from the right, that looks like it's approaching some other value. That looks like it's approaching this value right over here, let's call it L. That's approaching L, and L does not equal F of C. And so, in this situation, the two-sided limit doesn't even exist. We're approaching two different values when we approach from the left and from the right."}, {"video_title": "Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "That is F of C, and it does look like the limit as X approaches C from the left. So, from values less than C, it does look like that is approaching F of C. But if we look at the limit as X approaches C from the right, that looks like it's approaching some other value. That looks like it's approaching this value right over here, let's call it L. That's approaching L, and L does not equal F of C. And so, in this situation, the two-sided limit doesn't even exist. We're approaching two different values when we approach from the left and from the right. And so, the limit doesn't even exist at C. This is definitely not going to be continuous. And this matches up to our expectations with our little, do we have to pick up the pencil test? If I have to draw this, I can leave my pencil, it's on the paper, it's on the paper, it's on the paper, it's on the paper."}, {"video_title": "Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "We're approaching two different values when we approach from the left and from the right. And so, the limit doesn't even exist at C. This is definitely not going to be continuous. And this matches up to our expectations with our little, do we have to pick up the pencil test? If I have to draw this, I can leave my pencil, it's on the paper, it's on the paper, it's on the paper, it's on the paper. How am I going to continue to draw this function, this graph of the function, without picking up my pencil? Pick it up, put it back down, and then keep drawing it. So, once again, this right over here is not continuous, both intuitively by our pick up the pencil definition, and also by this more rigorous definition, where, in this case, the limit, the two-sided limit at X equals C doesn't even exist."}, {"video_title": "Continuity at a point Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "If I have to draw this, I can leave my pencil, it's on the paper, it's on the paper, it's on the paper, it's on the paper. How am I going to continue to draw this function, this graph of the function, without picking up my pencil? Pick it up, put it back down, and then keep drawing it. So, once again, this right over here is not continuous, both intuitively by our pick up the pencil definition, and also by this more rigorous definition, where, in this case, the limit, the two-sided limit at X equals C doesn't even exist. So, we're definitely not gonna be continuous. But even when the two-sided limit does exist, but the limit is a different value than the value of the function, that will also not be continuous. The only situation that it's going to be continuous is if the two-sided limit approaches the same value as the value of the function."}, {"video_title": "Worked example Approximation with local linearity AP Calculus AB Khan Academy.mp3", "Sentence": "We're told the function f is twice differentiable with f of two equals one, f prime of two is equal to four, and f prime prime of two is equal to three. What is the value of the approximation of f of 1.9 using the line tangent to the graph of f at x equals two? So pause this video and see if you can figure this out. This is an actual question from a past AP calculus exam. All right, now let's do this together. And if I was actually doing this on exam, I would just cut to the chase and I would figure out the equation of the tangent line at x equals two, go through the point two comma one, and then I would figure out, okay, when x is equal to 1.9, what is the value of y? And that would be my approximation."}, {"video_title": "Worked example Approximation with local linearity AP Calculus AB Khan Academy.mp3", "Sentence": "This is an actual question from a past AP calculus exam. All right, now let's do this together. And if I was actually doing this on exam, I would just cut to the chase and I would figure out the equation of the tangent line at x equals two, go through the point two comma one, and then I would figure out, okay, when x is equal to 1.9, what is the value of y? And that would be my approximation. But for the sake of learning and getting the intuition here, let's just make sure we understand what's happening. So let me graph this. So let's say that's my y-axis, and then this is my x-axis, and this is x equals one, this is x equals two, this is y equals one."}, {"video_title": "Worked example Approximation with local linearity AP Calculus AB Khan Academy.mp3", "Sentence": "And that would be my approximation. But for the sake of learning and getting the intuition here, let's just make sure we understand what's happening. So let me graph this. So let's say that's my y-axis, and then this is my x-axis, and this is x equals one, this is x equals two, this is y equals one. We know that the point two comma one is on the graph of y is equal to f of x. So we know that point right over there is there. And we also know the slope of the tangent line."}, {"video_title": "Worked example Approximation with local linearity AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say that's my y-axis, and then this is my x-axis, and this is x equals one, this is x equals two, this is y equals one. We know that the point two comma one is on the graph of y is equal to f of x. So we know that point right over there is there. And we also know the slope of the tangent line. The slope of the tangent line is four. So it's gonna look something like this. It's gonna probably even be a little steeper than that."}, {"video_title": "Worked example Approximation with local linearity AP Calculus AB Khan Academy.mp3", "Sentence": "And we also know the slope of the tangent line. The slope of the tangent line is four. So it's gonna look something like this. It's gonna probably even be a little steeper than that. The slope of the tangent line is gonna look something like that. And we don't know much more about it. We know the second derivative here."}, {"video_title": "Worked example Approximation with local linearity AP Calculus AB Khan Academy.mp3", "Sentence": "It's gonna probably even be a little steeper than that. The slope of the tangent line is gonna look something like that. And we don't know much more about it. We know the second derivative here. But what they're asking us to do is without knowing what the function actually looks like, the function might look something like this. Let me just draw something. So the function might look, might look something like this."}, {"video_title": "Worked example Approximation with local linearity AP Calculus AB Khan Academy.mp3", "Sentence": "We know the second derivative here. But what they're asking us to do is without knowing what the function actually looks like, the function might look something like this. Let me just draw something. So the function might look, might look something like this. We're trying to figure out what f of 1.9 is. So if x is 1.9, f of 1.9, if that's the way the function actually looked, might be this value right over here. But we don't know for sure because we don't know much more about the function."}, {"video_title": "Worked example Approximation with local linearity AP Calculus AB Khan Academy.mp3", "Sentence": "So the function might look, might look something like this. We're trying to figure out what f of 1.9 is. So if x is 1.9, f of 1.9, if that's the way the function actually looked, might be this value right over here. But we don't know for sure because we don't know much more about the function. But what they're suggesting for us to do is use this tangent line. If we know the equation of this tangent line here, we could say, well, what does that tangent line equal when x equals 1.9? When x equals 1.9, it equals that point right over there."}, {"video_title": "Worked example Approximation with local linearity AP Calculus AB Khan Academy.mp3", "Sentence": "But we don't know for sure because we don't know much more about the function. But what they're suggesting for us to do is use this tangent line. If we know the equation of this tangent line here, we could say, well, what does that tangent line equal when x equals 1.9? When x equals 1.9, it equals that point right over there. And then we could use that as our approximation for f of 1.9. Well, to do that, we need to know the equation of the tangent line. And we could do that in point-slope form."}, {"video_title": "Worked example Approximation with local linearity AP Calculus AB Khan Academy.mp3", "Sentence": "When x equals 1.9, it equals that point right over there. And then we could use that as our approximation for f of 1.9. Well, to do that, we need to know the equation of the tangent line. And we could do that in point-slope form. We would just have to say y minus the y value that we know is on that line, the point two comma one we know is on that line. So y minus one is going to be equal to the slope of our tangent line, which we know is going to be equal to four, times x minus the x value that corresponds to that y value, so x minus two. So now we just have to substitute x equals 1.9 to get our approximation for f of 1.9."}, {"video_title": "Worked example Approximation with local linearity AP Calculus AB Khan Academy.mp3", "Sentence": "And we could do that in point-slope form. We would just have to say y minus the y value that we know is on that line, the point two comma one we know is on that line. So y minus one is going to be equal to the slope of our tangent line, which we know is going to be equal to four, times x minus the x value that corresponds to that y value, so x minus two. So now we just have to substitute x equals 1.9 to get our approximation for f of 1.9. So we'd say y minus one is equal to four times 1.9 minus two. 1.9 minus two is negative 0.1. And let's see, four times negative 0.1, this all simplifies to negative 0.4."}, {"video_title": "Worked example Approximation with local linearity AP Calculus AB Khan Academy.mp3", "Sentence": "So now we just have to substitute x equals 1.9 to get our approximation for f of 1.9. So we'd say y minus one is equal to four times 1.9 minus two. 1.9 minus two is negative 0.1. And let's see, four times negative 0.1, this all simplifies to negative 0.4. Now you add one to both sides, you get y is equal to, if you add one here, you're gonna get 0.6. So this, I didn't draw it quite to scale, 0.6 might be something closer to right around there. But there you go, that is our approximation for f of 1.9, which is choice B."}, {"video_title": "Worked example Approximation with local linearity AP Calculus AB Khan Academy.mp3", "Sentence": "And let's see, four times negative 0.1, this all simplifies to negative 0.4. Now you add one to both sides, you get y is equal to, if you add one here, you're gonna get 0.6. So this, I didn't draw it quite to scale, 0.6 might be something closer to right around there. But there you go, that is our approximation for f of 1.9, which is choice B. And we're done. And one interesting thing to note is we didn't have to use all the information they gave us. We did not have to use this information about the second derivative in order to solve the problem."}, {"video_title": "Worked example identifying separable equations AP Calculus AB Khan Academy.mp3", "Sentence": "And I encourage you to pause this video and see which of these are actually separable. Now, the way that I approach this is I try to solve for the derivative, and if when I solve for the derivative, if I get dy dx is equal to some function of y times some other function of x, then I say, okay, this is separable, because I could rewrite this as, I could divide both sides by g of y, and I get one over g of y, which is itself a function of y, times dy is equal to h of x dx. You would go from this first equation to the second equation just by dividing both sides by g of y and multiplying both sides by dx, and then it's clear you have a separable equation, you can integrate both sides. But the key is let's solve for the derivative and see if we can put this in a form where we have the product of a function of y times a function of x. So let's do it with this first one here. So let's see, if I subtract y from both sides, I'm just trying to solve for the derivative of y with respect to x, I'm gonna get x times, I'll write y prime as the derivative of y with respect to x is equal to three minus y. So I subtracted y from both sides."}, {"video_title": "Worked example identifying separable equations AP Calculus AB Khan Academy.mp3", "Sentence": "But the key is let's solve for the derivative and see if we can put this in a form where we have the product of a function of y times a function of x. So let's do it with this first one here. So let's see, if I subtract y from both sides, I'm just trying to solve for the derivative of y with respect to x, I'm gonna get x times, I'll write y prime as the derivative of y with respect to x is equal to three minus y. So I subtracted y from both sides. Now, let's see, if I divide both sides by x, I'm gonna get the derivative of y with respect to x is equal to, actually, I'm gonna write it this way, I'm gonna write it three minus y times one over x. And so it's clear, I'm able to write the derivative as the product of a function of y and a function of x. So this indeed is separable."}, {"video_title": "Worked example identifying separable equations AP Calculus AB Khan Academy.mp3", "Sentence": "So I subtracted y from both sides. Now, let's see, if I divide both sides by x, I'm gonna get the derivative of y with respect to x is equal to, actually, I'm gonna write it this way, I'm gonna write it three minus y times one over x. And so it's clear, I'm able to write the derivative as the product of a function of y and a function of x. So this indeed is separable. And I could show you, I can multiply both sides by dx, and I can divide both sides by three minus y now, and I would get one over three minus y dy is equal to one over x dx. So clearly, this one right over here is separable. Now let's do the second one."}, {"video_title": "Worked example identifying separable equations AP Calculus AB Khan Academy.mp3", "Sentence": "So this indeed is separable. And I could show you, I can multiply both sides by dx, and I can divide both sides by three minus y now, and I would get one over three minus y dy is equal to one over x dx. So clearly, this one right over here is separable. Now let's do the second one. And I'm gonna just do the same technique, I'll do it in a different color so we don't get all of our math all jumbled together. So in this second one, let's see, if I subtract the two x, the two y from both sides, so actually, let me just do, whoops, let me do a couple things at once. I'm gonna subtract two x from both sides."}, {"video_title": "Worked example identifying separable equations AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's do the second one. And I'm gonna just do the same technique, I'll do it in a different color so we don't get all of our math all jumbled together. So in this second one, let's see, if I subtract the two x, the two y from both sides, so actually, let me just do, whoops, let me do a couple things at once. I'm gonna subtract two x from both sides. I am going to subtract two y from both sides. So I'm gonna subtract two y from both sides. I'm gonna add one to both sides."}, {"video_title": "Worked example identifying separable equations AP Calculus AB Khan Academy.mp3", "Sentence": "I'm gonna subtract two x from both sides. I am going to subtract two y from both sides. So I'm gonna subtract two y from both sides. I'm gonna add one to both sides. So I'm gonna add one to both sides. And then what am I going to get if I do that? This is gonna be zero, this is gonna be zero, this is gonna be zero."}, {"video_title": "Worked example identifying separable equations AP Calculus AB Khan Academy.mp3", "Sentence": "I'm gonna add one to both sides. So I'm gonna add one to both sides. And then what am I going to get if I do that? This is gonna be zero, this is gonna be zero, this is gonna be zero. I'm gonna have two times the derivative of y with respect to x is equal to negative two x minus two y plus one. And now let's see, I can divide everything by two. I would get the derivative of y with respect to x is equal to, and actually, yeah, I would get, I'm just gonna divide by two."}, {"video_title": "Worked example identifying separable equations AP Calculus AB Khan Academy.mp3", "Sentence": "This is gonna be zero, this is gonna be zero, this is gonna be zero. I'm gonna have two times the derivative of y with respect to x is equal to negative two x minus two y plus one. And now let's see, I can divide everything by two. I would get the derivative of y with respect to x is equal to, and actually, yeah, I would get, I'm just gonna divide by two. So I'm gonna get negative x minus y and then I'm going to get plus 1 1 2. So it's not obvious to me how I can write this as a product of a function of x and a function of y. So this one does not feel, this one right over here is not separable."}, {"video_title": "Worked example identifying separable equations AP Calculus AB Khan Academy.mp3", "Sentence": "I would get the derivative of y with respect to x is equal to, and actually, yeah, I would get, I'm just gonna divide by two. So I'm gonna get negative x minus y and then I'm going to get plus 1 1 2. So it's not obvious to me how I can write this as a product of a function of x and a function of y. So this one does not feel, this one right over here is not separable. I don't know how to write this as a function of x times a function of y. So this one I'm gonna say is not separable. Now this one, they've already written it for us as a function of x times a function of y."}, {"video_title": "Worked example identifying separable equations AP Calculus AB Khan Academy.mp3", "Sentence": "So this one does not feel, this one right over here is not separable. I don't know how to write this as a function of x times a function of y. So this one I'm gonna say is not separable. Now this one, they've already written it for us as a function of x times a function of y. So this one is clearly separable right over here and if you want me to do the separating, I can rewrite this as, well, this is dy dx. If I multiply both sides by dx and divide both sides by this right over here, I would get one over y squared plus y. Dy is equal to x squared plus x dx. So clearly separable."}, {"video_title": "Worked example identifying separable equations AP Calculus AB Khan Academy.mp3", "Sentence": "Now this one, they've already written it for us as a function of x times a function of y. So this one is clearly separable right over here and if you want me to do the separating, I can rewrite this as, well, this is dy dx. If I multiply both sides by dx and divide both sides by this right over here, I would get one over y squared plus y. Dy is equal to x squared plus x dx. So clearly separable. Alright, now this last choice, this is interesting. They've essentially distributed the derivative right over here. So let's see, if we were to unfactor the derivative, I'm just gonna solve for dy dx."}, {"video_title": "Worked example identifying separable equations AP Calculus AB Khan Academy.mp3", "Sentence": "So clearly separable. Alright, now this last choice, this is interesting. They've essentially distributed the derivative right over here. So let's see, if we were to unfactor the derivative, I'm just gonna solve for dy dx. So I'm gonna factor it out. I'm gonna get dy dx times x plus y, x plus y is equal to x. Now if I were to divide both sides by x plus y, I'm gonna get dy dx is equal to x over x plus y."}, {"video_title": "Worked example identifying separable equations AP Calculus AB Khan Academy.mp3", "Sentence": "So let's see, if we were to unfactor the derivative, I'm just gonna solve for dy dx. So I'm gonna factor it out. I'm gonna get dy dx times x plus y, x plus y is equal to x. Now if I were to divide both sides by x plus y, I'm gonna get dy dx is equal to x over x plus y. And here, my algebraic toolkit of how do I separate x and y so I can write this as a function of x times a function of y, not obvious to me here. So this one is not separable. There's only the first one and the third one."}, {"video_title": "Mean value theorem example square root function AP Calculus AB Khan Academy.mp3", "Sentence": "So let's just remind ourselves what it means for c to be the number that satisfies the mean value theorem for f. This means that over this interval, c is a point, x equals c where the slope of the tangent line at x equals c so I could write f prime of c, so that is the slope of the tangent line when x is equal to c, this is equal to the slope of the secant line that connects these two points so this is going to be equal to, see the slope of the secant line that connects the points three f of three and one f of one so it's going to be f of three minus f of one over three minus one and if you wanted to think about what this means visually it would look something like this. So this is our x axis and this is one, two actually let me spread it out a little bit more, one, two and three and so you have one comma f of one right over there, so that is at the point one comma f of one and we could evaluate that actually what, that's one comma one, right? So that's going to be the point one comma one and then you have the point three comma, let's see, you're going to have four times three is twelve minus three is nine, so it's going to be three comma three three comma three, so maybe it's right over there three comma three and it might look, the curve might look something like this, so it might look something like that. So if you think about the slope of the line that connects these two points, so this line that connects those two points all the mean value theorem, I'm going to do that in a different color, all the mean value theorem tells us is that there's a point between one and three where the slope of the tangent line has the exact same slope, so if I were to eyeball it, it looks like it's right around there although we are actually going to solve for it, so some point where the slope of the tangent line is equal to the slope of the line that connects these two end points and their corresponding function values, so that is c, that would be c right over there. So really we just have to solve this, so let's first just find out what f prime of x is and then we could substitute a c in there and then we can evaluate this on the right hand side. So I'm going to rewrite f of x f of x is equal to and I'm going to write it as four x to the minus three to the one half power, it makes it a little bit more obvious that we can apply the power rule and the chain rule here, so f prime of x is going to be the derivative of 4 x minus 3 to the one half with respect to 4 x minus 3, so that is going to be one half times 4 x minus 3 to the negative one half and then we are going to multiply that times the derivative of 4 x minus 3 with respect to x. Well the derivative of 4 x with respect to x is just 4 and the derivative of negative 3 with respect to x well, that's going to be zero, so the derivative of 4 x minus 3 with respect to x is four, so times four."}, {"video_title": "Mean value theorem example square root function AP Calculus AB Khan Academy.mp3", "Sentence": "So if you think about the slope of the line that connects these two points, so this line that connects those two points all the mean value theorem, I'm going to do that in a different color, all the mean value theorem tells us is that there's a point between one and three where the slope of the tangent line has the exact same slope, so if I were to eyeball it, it looks like it's right around there although we are actually going to solve for it, so some point where the slope of the tangent line is equal to the slope of the line that connects these two end points and their corresponding function values, so that is c, that would be c right over there. So really we just have to solve this, so let's first just find out what f prime of x is and then we could substitute a c in there and then we can evaluate this on the right hand side. So I'm going to rewrite f of x f of x is equal to and I'm going to write it as four x to the minus three to the one half power, it makes it a little bit more obvious that we can apply the power rule and the chain rule here, so f prime of x is going to be the derivative of 4 x minus 3 to the one half with respect to 4 x minus 3, so that is going to be one half times 4 x minus 3 to the negative one half and then we are going to multiply that times the derivative of 4 x minus 3 with respect to x. Well the derivative of 4 x with respect to x is just 4 and the derivative of negative 3 with respect to x well, that's going to be zero, so the derivative of 4 x minus 3 with respect to x is four, so times four. So f prime of x, f prime of x is equal to four times 1 1\u20442, which is two over the square root of four x minus three. Four x minus three to the 1 1\u20442 would just be the square root of four x minus three, but it's the negative 1 1\u20442, so we're gonna put it in the denominator right over here. And so f prime of c, we could rewrite this as two over the square root of four c minus three."}, {"video_title": "Mean value theorem example square root function AP Calculus AB Khan Academy.mp3", "Sentence": "Well the derivative of 4 x with respect to x is just 4 and the derivative of negative 3 with respect to x well, that's going to be zero, so the derivative of 4 x minus 3 with respect to x is four, so times four. So f prime of x, f prime of x is equal to four times 1 1\u20442, which is two over the square root of four x minus three. Four x minus three to the 1 1\u20442 would just be the square root of four x minus three, but it's the negative 1 1\u20442, so we're gonna put it in the denominator right over here. And so f prime of c, we could rewrite this as two over the square root of four c minus three. And what is that going to be equal to? That is going to be equal to, let's see, f of three we already figured out is three. F of one we already figured out is one."}, {"video_title": "Mean value theorem example square root function AP Calculus AB Khan Academy.mp3", "Sentence": "And so f prime of c, we could rewrite this as two over the square root of four c minus three. And what is that going to be equal to? That is going to be equal to, let's see, f of three we already figured out is three. F of one we already figured out is one. And so we get three minus one over three minus one. Well, that's gonna be two over two, which is equal to one. So there's some point between one and three where the derivative at that point, the slope of the tangent line, is equal to one."}, {"video_title": "Mean value theorem example square root function AP Calculus AB Khan Academy.mp3", "Sentence": "F of one we already figured out is one. And so we get three minus one over three minus one. Well, that's gonna be two over two, which is equal to one. So there's some point between one and three where the derivative at that point, the slope of the tangent line, is equal to one. So let's see if we can solve this thing right over here. Well, we can multiply both sides of this by the square root of four c minus three. And so then we are going to get, we're going to get two is equal to the square root of four c minus three."}, {"video_title": "Mean value theorem example square root function AP Calculus AB Khan Academy.mp3", "Sentence": "So there's some point between one and three where the derivative at that point, the slope of the tangent line, is equal to one. So let's see if we can solve this thing right over here. Well, we can multiply both sides of this by the square root of four c minus three. And so then we are going to get, we're going to get two is equal to the square root of four c minus three. All I did is multiply both sides of this by square root of four c minus three to get rid of this in the denominator. And so, let's see, now to get rid of the radical, we can square both sides. And so, actually let me just show that."}, {"video_title": "Mean value theorem example square root function AP Calculus AB Khan Academy.mp3", "Sentence": "And so then we are going to get, we're going to get two is equal to the square root of four c minus three. All I did is multiply both sides of this by square root of four c minus three to get rid of this in the denominator. And so, let's see, now to get rid of the radical, we can square both sides. And so, actually let me just show that. So now we can square both sides. So we get four is equal to four c minus three. Add three to both sides."}, {"video_title": "Mean value theorem example square root function AP Calculus AB Khan Academy.mp3", "Sentence": "And so, actually let me just show that. So now we can square both sides. So we get four is equal to four c minus three. Add three to both sides. Seven is equal to four c. And then divide both sides by four. I'll go right here to do it. You're going to get c is equal to 7 4ths."}, {"video_title": "Mean value theorem example square root function AP Calculus AB Khan Academy.mp3", "Sentence": "Add three to both sides. Seven is equal to four c. And then divide both sides by four. I'll go right here to do it. You're going to get c is equal to 7 4ths. C is equal to, is equal to seven over four, which is equal to 1 3 4ths. Or we could view this as 1.75. So actually, the c value is a little bit closer."}, {"video_title": "Mean value theorem example square root function AP Calculus AB Khan Academy.mp3", "Sentence": "You're going to get c is equal to 7 4ths. C is equal to, is equal to seven over four, which is equal to 1 3 4ths. Or we could view this as 1.75. So actually, the c value is a little bit closer. I hand drew this. It's closer to about right over there on our diagram. And actually that looks pretty, pretty close."}, {"video_title": "Mean value theorem example square root function AP Calculus AB Khan Academy.mp3", "Sentence": "So actually, the c value is a little bit closer. I hand drew this. It's closer to about right over there on our diagram. And actually that looks pretty, pretty close. That actually looks pretty good. I just hand drew this curve, so it's definitely not exact. But anyway, hopefully that gives you a sense of what's going on here."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "And as we'll see, differential equations are super useful for modeling and simulating phenomena and understanding how they operate. But we'll get into that later. For now, let's just think about, or at least look at, what a differential equation actually is. So if I were to write, so here's an example of a differential equation. If I were to write that the second derivative of y plus two times the first derivative of y is equal to three times y, this right over here is a differential equation. Another way we could write it, if we said that y is a function of x, we could write this in function notation. We could write the second derivative of our function with respect to x plus two times the first derivative of our function is equal to three times our function."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "So if I were to write, so here's an example of a differential equation. If I were to write that the second derivative of y plus two times the first derivative of y is equal to three times y, this right over here is a differential equation. Another way we could write it, if we said that y is a function of x, we could write this in function notation. We could write the second derivative of our function with respect to x plus two times the first derivative of our function is equal to three times our function. Or if we wanted to use the Leibniz notation, we could also write, we could also write the second derivative of y with respect to x plus two times the first derivative of y with respect to x is equal to three times y. All three of these, I guess, equations are really representing the same thing. They're saying, okay, can I find functions where the second derivative of the function plus two times the first derivative of the function is equal to three times the function itself?"}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "We could write the second derivative of our function with respect to x plus two times the first derivative of our function is equal to three times our function. Or if we wanted to use the Leibniz notation, we could also write, we could also write the second derivative of y with respect to x plus two times the first derivative of y with respect to x is equal to three times y. All three of these, I guess, equations are really representing the same thing. They're saying, okay, can I find functions where the second derivative of the function plus two times the first derivative of the function is equal to three times the function itself? So just to be clear, these are all essentially saying the same thing. And you might have just caught from how I described it that the solution to a differential equation is a function or a class of functions. It's not just a value or a set of values."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "They're saying, okay, can I find functions where the second derivative of the function plus two times the first derivative of the function is equal to three times the function itself? So just to be clear, these are all essentially saying the same thing. And you might have just caught from how I described it that the solution to a differential equation is a function or a class of functions. It's not just a value or a set of values. So the solution here, so the solution to a differential equation is a function, or a set of functions, or a class of functions. And it's important to contrast this relative to a traditional equation. So let me write that down."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "It's not just a value or a set of values. So the solution here, so the solution to a differential equation is a function, or a set of functions, or a class of functions. And it's important to contrast this relative to a traditional equation. So let me write that down. So a traditional equation, I guess I could say, maybe I shouldn't say traditional equation. Differential equations have been around for a while. So let me write this as a, maybe an algebraic equation that you're familiar with."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "So let me write that down. So a traditional equation, I guess I could say, maybe I shouldn't say traditional equation. Differential equations have been around for a while. So let me write this as a, maybe an algebraic equation that you're familiar with. Algebraic, an algebraic equation might look something like, and I'll just write a simple quadratic, say x squared, x squared plus three x plus two is equal to zero. The solutions to this algebraic equation are going to be numbers or a set of numbers. We can solve this as going to be x plus two times x plus one is equal to zero."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "So let me write this as a, maybe an algebraic equation that you're familiar with. Algebraic, an algebraic equation might look something like, and I'll just write a simple quadratic, say x squared, x squared plus three x plus two is equal to zero. The solutions to this algebraic equation are going to be numbers or a set of numbers. We can solve this as going to be x plus two times x plus one is equal to zero. So x could be equal to negative two or x could be equal to negative one. The solutions here are numbers or a set of values or set of values that satisfy the equation. Here it's a relationship between a function and its derivatives and so the solutions or the solution is going to be a function or a set of functions."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "We can solve this as going to be x plus two times x plus one is equal to zero. So x could be equal to negative two or x could be equal to negative one. The solutions here are numbers or a set of values or set of values that satisfy the equation. Here it's a relationship between a function and its derivatives and so the solutions or the solution is going to be a function or a set of functions. Now let's make that a little bit more tangible. What would a solution to something like any of these three which really represent the same thing, what would a solution actually look like? And actually let me move over to the, let me move this over a little bit."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "Here it's a relationship between a function and its derivatives and so the solutions or the solution is going to be a function or a set of functions. Now let's make that a little bit more tangible. What would a solution to something like any of these three which really represent the same thing, what would a solution actually look like? And actually let me move over to the, let me move this over a little bit. Move this a little bit so that we can take a look at what some of these solutions could look like. Let me erase this little stuff that I have right over here. So I'm just going to give you examples of solutions here."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "And actually let me move over to the, let me move this over a little bit. Move this a little bit so that we can take a look at what some of these solutions could look like. Let me erase this little stuff that I have right over here. So I'm just going to give you examples of solutions here. We'll verify that these indeed are solutions for, I guess this is really just one differential equation represented in different ways, but hopefully appreciate what a solution to a differential equation looks like and that there is often more than one solution or that there's a whole class of functions that could be a solution. So one solution to this differential equation, and I'll just write it as our first one, so one solution, I'll call it y one, and I could even write it as y one of x to make it explicit that it is a function of x. One solution is y one of x is equal to e to the negative 3x."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "So I'm just going to give you examples of solutions here. We'll verify that these indeed are solutions for, I guess this is really just one differential equation represented in different ways, but hopefully appreciate what a solution to a differential equation looks like and that there is often more than one solution or that there's a whole class of functions that could be a solution. So one solution to this differential equation, and I'll just write it as our first one, so one solution, I'll call it y one, and I could even write it as y one of x to make it explicit that it is a function of x. One solution is y one of x is equal to e to the negative 3x. And I encourage you to pause this video right now and find the first derivative of y one or the second derivative of y one and verify that it does indeed satisfy this differential equation. So I'm assuming you've had a go at it, so let's work through this together. So that's y one."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "One solution is y one of x is equal to e to the negative 3x. And I encourage you to pause this video right now and find the first derivative of y one or the second derivative of y one and verify that it does indeed satisfy this differential equation. So I'm assuming you've had a go at it, so let's work through this together. So that's y one. So the first derivative of y one, well, this is going to be, let's see, we just have to do the chain rule here, derivative of negative 3x with respect to x is just negative three, and the derivative of e to the negative 3x with respect to negative 3x is just e to the negative 3x. And if we take the second derivative of y one, this is equal to, same exact idea, derivative of this is negative three times negative three, it's going to be nine e to the negative 3x. And now we can just substitute these values into the differential equation or these expressions into the differential equation to verify that this is indeed going to be true for this function."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "So that's y one. So the first derivative of y one, well, this is going to be, let's see, we just have to do the chain rule here, derivative of negative 3x with respect to x is just negative three, and the derivative of e to the negative 3x with respect to negative 3x is just e to the negative 3x. And if we take the second derivative of y one, this is equal to, same exact idea, derivative of this is negative three times negative three, it's going to be nine e to the negative 3x. And now we can just substitute these values into the differential equation or these expressions into the differential equation to verify that this is indeed going to be true for this function. So let's verify that. So let me, so we first have the second derivative of y, so that's that term right over there. So we have nine e to the negative 3x plus two times the first derivative."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "And now we can just substitute these values into the differential equation or these expressions into the differential equation to verify that this is indeed going to be true for this function. So let's verify that. So let me, so we first have the second derivative of y, so that's that term right over there. So we have nine e to the negative 3x plus two times the first derivative. So that's going to be two times this right over here. So it's going to be minus six, I'll just write plus negative six e to the negative 3x. Notice I just took this two times the first derivative."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "So we have nine e to the negative 3x plus two times the first derivative. So that's going to be two times this right over here. So it's going to be minus six, I'll just write plus negative six e to the negative 3x. Notice I just took this two times the first derivative. Two times the first derivative is going to be equal to, or needs to be equal to, if this indeed does satisfy, if y one does indeed satisfy the differential equation, this needs to be equal to three times y. Well three times y is three times e to the negative 3x. Three e to the negative 3x."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "Notice I just took this two times the first derivative. Two times the first derivative is going to be equal to, or needs to be equal to, if this indeed does satisfy, if y one does indeed satisfy the differential equation, this needs to be equal to three times y. Well three times y is three times e to the negative 3x. Three e to the negative 3x. And let's see if that indeed is true. So these two terms right over here, nine e to the negative 3x, essentially minus six e to the negative 3x, that's going to be three e to the negative 3x, which is indeed equal to three e to the negative 3x. So y one is indeed a solution to this differential equation."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "Three e to the negative 3x. And let's see if that indeed is true. So these two terms right over here, nine e to the negative 3x, essentially minus six e to the negative 3x, that's going to be three e to the negative 3x, which is indeed equal to three e to the negative 3x. So y one is indeed a solution to this differential equation. But as we'll see, it is not the only solution to this differential equation. For example, for example, let's say y two is equal to e to the x is also a solution to this differential equation. And I encourage you to pause the video again and verify that it's a solution."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "So y one is indeed a solution to this differential equation. But as we'll see, it is not the only solution to this differential equation. For example, for example, let's say y two is equal to e to the x is also a solution to this differential equation. And I encourage you to pause the video again and verify that it's a solution. So assuming you've had a go at it, so the first derivative of this, this is pretty straightforward, is e to the x. Second derivative is one of the profound things of the exponential function. The second derivative here is also e to the x."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "And I encourage you to pause the video again and verify that it's a solution. So assuming you've had a go at it, so the first derivative of this, this is pretty straightforward, is e to the x. Second derivative is one of the profound things of the exponential function. The second derivative here is also e to the x. And so if I have, so the second derivative, let me just do it in those same colors. So the second derivative is going to be e to the x, e to the x plus two times e to the x, plus two times e to the x, is indeed going to be equal to three times e to the x, three times e to the x. This is absolutely going to be true."}, {"video_title": "Differential equation introduction First order differential equations Khan Academy.mp3", "Sentence": "The second derivative here is also e to the x. And so if I have, so the second derivative, let me just do it in those same colors. So the second derivative is going to be e to the x, e to the x plus two times e to the x, plus two times e to the x, is indeed going to be equal to three times e to the x, three times e to the x. This is absolutely going to be true. E to the x plus two to the x is three e to the x. So y two is also a solution to this differential equation. So that's a start."}, {"video_title": "Infinite series as limit of partial sums Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let's say that we have an infinite series S, so it's the sum from n equals one to infinity of a sub n. So we could write it out, a sub one plus a sub two, and we're just gonna go on and on and on for infinity. We're gonna go on and on and on forever. So let's say, and I've written it in very general terms, let's say we have a formula for the partial sums of S. So we know that S sub n is equal to two n to the third over n plus one times n plus two. Now my question to you is, based on what I've just told you, I've, you know, S is this, I've in a very general way written this infinite series, but I have the partial sum, the sum of the first n terms of S is given by this formula right over here. Does this series converge or diverge? Does this thing converge to some finite value or is it unbounded and does it diverge? Well, one way to think about this, one way to think about this is the idea that our infinite series S is just the limit as n approaches infinity of our partial sums."}, {"video_title": "Infinite series as limit of partial sums Series AP Calculus BC Khan Academy.mp3", "Sentence": "Now my question to you is, based on what I've just told you, I've, you know, S is this, I've in a very general way written this infinite series, but I have the partial sum, the sum of the first n terms of S is given by this formula right over here. Does this series converge or diverge? Does this thing converge to some finite value or is it unbounded and does it diverge? Well, one way to think about this, one way to think about this is the idea that our infinite series S is just the limit as n approaches infinity of our partial sums. So what do we mean by that? Well, you could imagine a sequence of partial sums here. So you have S sub one, S sub two, S sub three, and you keep going, so this would be the sum of the first term, this would be the sum of the first two terms, this would be the sum of the first three terms."}, {"video_title": "Infinite series as limit of partial sums Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, one way to think about this, one way to think about this is the idea that our infinite series S is just the limit as n approaches infinity of our partial sums. So what do we mean by that? Well, you could imagine a sequence of partial sums here. So you have S sub one, S sub two, S sub three, and you keep going, so this would be the sum of the first term, this would be the sum of the first two terms, this would be the sum of the first three terms. And just think about, well, what happens to this sequence as you go, as n, as n right over here approaches infinity? Because that's what the series is. It's the sum of the first, I guess you could say the first infinite terms."}, {"video_title": "Infinite series as limit of partial sums Series AP Calculus BC Khan Academy.mp3", "Sentence": "So you have S sub one, S sub two, S sub three, and you keep going, so this would be the sum of the first term, this would be the sum of the first two terms, this would be the sum of the first three terms. And just think about, well, what happens to this sequence as you go, as n, as n right over here approaches infinity? Because that's what the series is. It's the sum of the first, I guess you could say the first infinite terms. It's the sum of all, you have an infinite number of terms here. Well, let's think about what this is. The limit as n approaches infinity of S sub n, well, that's just going to be, that's going to be the limit as n approaches infinity of this business right over here."}, {"video_title": "Infinite series as limit of partial sums Series AP Calculus BC Khan Academy.mp3", "Sentence": "It's the sum of the first, I guess you could say the first infinite terms. It's the sum of all, you have an infinite number of terms here. Well, let's think about what this is. The limit as n approaches infinity of S sub n, well, that's just going to be, that's going to be the limit as n approaches infinity of this business right over here. Two n to the third power over n plus one over n, or times n plus two. And there's several ways that you could evaluate this. One way is you could just realize, hey, look, in the bottom, this is going to be a second degree polynomial."}, {"video_title": "Infinite series as limit of partial sums Series AP Calculus BC Khan Academy.mp3", "Sentence": "The limit as n approaches infinity of S sub n, well, that's just going to be, that's going to be the limit as n approaches infinity of this business right over here. Two n to the third power over n plus one over n, or times n plus two. And there's several ways that you could evaluate this. One way is you could just realize, hey, look, in the bottom, this is going to be a second degree polynomial. And up here, you have a third degree, so the numerator's gonna grow faster than the denominator, so this is going to be unbounded. So that will immediately tell you, well, this is gonna approach infinity, so S is going to diverge. But if you wanna do it a little bit less hand-wavy than that, we can actually do a little bit more algebra."}, {"video_title": "Infinite series as limit of partial sums Series AP Calculus BC Khan Academy.mp3", "Sentence": "One way is you could just realize, hey, look, in the bottom, this is going to be a second degree polynomial. And up here, you have a third degree, so the numerator's gonna grow faster than the denominator, so this is going to be unbounded. So that will immediately tell you, well, this is gonna approach infinity, so S is going to diverge. But if you wanna do it a little bit less hand-wavy than that, we can actually do a little bit more algebra. Limit as n approaches infinity, two n to the third power over, let's multiply this out, n squared plus three n plus two. And let's see, we can divide the numerator and the denominator by n squared, so this is going to be the limit as n approaches infinity of, if we divide the numerator by n squared, you're going to have, actually, let's divide the numerator, and well, yeah, let's divide it by, yeah, n squared. So if we divide the numerator by n squared, we're gonna have two n, and then the denominator divided by n squared, you're gonna have one plus three over n plus two over n squared."}, {"video_title": "Infinite series as limit of partial sums Series AP Calculus BC Khan Academy.mp3", "Sentence": "But if you wanna do it a little bit less hand-wavy than that, we can actually do a little bit more algebra. Limit as n approaches infinity, two n to the third power over, let's multiply this out, n squared plus three n plus two. And let's see, we can divide the numerator and the denominator by n squared, so this is going to be the limit as n approaches infinity of, if we divide the numerator by n squared, you're going to have, actually, let's divide the numerator, and well, yeah, let's divide it by, yeah, n squared. So if we divide the numerator by n squared, we're gonna have two n, and then the denominator divided by n squared, you're gonna have one plus three over n plus two over n squared. And now when you look at it like this, it becomes pretty clear. This thing, as n approaches infinity, this thing is gonna go towards infinity, but this thing down here, the denominator, this is gonna go towards zero, this is gonna go towards zero, so the denominator's gonna go towards one. So this whole thing, this whole thing is, the limit is gonna go to infinity."}, {"video_title": "Infinite series as limit of partial sums Series AP Calculus BC Khan Academy.mp3", "Sentence": "So if we divide the numerator by n squared, we're gonna have two n, and then the denominator divided by n squared, you're gonna have one plus three over n plus two over n squared. And now when you look at it like this, it becomes pretty clear. This thing, as n approaches infinity, this thing is gonna go towards infinity, but this thing down here, the denominator, this is gonna go towards zero, this is gonna go towards zero, so the denominator's gonna go towards one. So this whole thing, this whole thing is, the limit is gonna go to infinity. And since the limit of the partial sums goes to infinity, that means that this infinite series is not going to be a finite value. It's just going to diverge. So this character right over here is going to diverge."}, {"video_title": "Infinite series as limit of partial sums Series AP Calculus BC Khan Academy.mp3", "Sentence": "So this whole thing, this whole thing is, the limit is gonna go to infinity. And since the limit of the partial sums goes to infinity, that means that this infinite series is not going to be a finite value. It's just going to diverge. So this character right over here is going to diverge. In order for it to have converged, this thing should have come, this limit should have been some finite value. So hopefully that makes sense. All we said is look, infinite series, we had a formula for the partial sum of the first n terms, and then we said, oh look, the series itself, the infinite series, you could view it as a limit of, as n approaches infinity, of the partial sum S sub n. We said, hey, that approaches infinity, this thing is diverging."}, {"video_title": "Introduction to l'H\u00f4pital's rule Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "It's a slope around the point as we take the limit of points closer and closer to the point in question. You've seen that many, many, many times over. In this video, we're going to do it in the opposite direction. We're going to use derivatives to figure out limits. In particular, limits that end up in indeterminate form. When I say by indeterminate form, I mean that when we just take the limit as it is, we end up with something like 0 over 0 or infinity over infinity or negative infinity over infinity. All of these are indeterminate, undefined forms."}, {"video_title": "Introduction to l'H\u00f4pital's rule Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "We're going to use derivatives to figure out limits. In particular, limits that end up in indeterminate form. When I say by indeterminate form, I mean that when we just take the limit as it is, we end up with something like 0 over 0 or infinity over infinity or negative infinity over infinity. All of these are indeterminate, undefined forms. To do that, we're going to use L'Hopital's Rule. In this video, I'm just going to show you what L'Hopital's Rule says and how to apply it. It's fairly straightforward."}, {"video_title": "Introduction to l'H\u00f4pital's rule Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "All of these are indeterminate, undefined forms. To do that, we're going to use L'Hopital's Rule. In this video, I'm just going to show you what L'Hopital's Rule says and how to apply it. It's fairly straightforward. It's actually a very useful tool sometimes if you're in some type of a math competition. They ask you to find a difficult limit that when you just plug the numbers in, you get something like this. L'Hopital's Rule is normally what they are testing you for."}, {"video_title": "Introduction to l'H\u00f4pital's rule Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "It's fairly straightforward. It's actually a very useful tool sometimes if you're in some type of a math competition. They ask you to find a difficult limit that when you just plug the numbers in, you get something like this. L'Hopital's Rule is normally what they are testing you for. In a future video, I might prove it, but that gets a little bit more involved. The application is actually reasonably straightforward. What L'Hopital's Rule tells us is that if we have, and I'll do it in abstract form first, but I think when I show you the example, it will all be made clear."}, {"video_title": "Introduction to l'H\u00f4pital's rule Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "L'Hopital's Rule is normally what they are testing you for. In a future video, I might prove it, but that gets a little bit more involved. The application is actually reasonably straightforward. What L'Hopital's Rule tells us is that if we have, and I'll do it in abstract form first, but I think when I show you the example, it will all be made clear. If the limit as x approaches c of f of x is equal to 0, and the limit as x approaches c of g of x is equal to 0, and the limit as x approaches c of f prime of x over g prime of x exists and it equals L, then all of these conditions have to be met. This is the indeterminate form of 0 over 0, so this is the first case. Then we can say that the limit as x approaches c of f of x over g of x is also going to be equal to L. This might seem a little bit bizarre to you right now, and I'm actually going to write the other case, and then I'll do an example."}, {"video_title": "Introduction to l'H\u00f4pital's rule Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "What L'Hopital's Rule tells us is that if we have, and I'll do it in abstract form first, but I think when I show you the example, it will all be made clear. If the limit as x approaches c of f of x is equal to 0, and the limit as x approaches c of g of x is equal to 0, and the limit as x approaches c of f prime of x over g prime of x exists and it equals L, then all of these conditions have to be met. This is the indeterminate form of 0 over 0, so this is the first case. Then we can say that the limit as x approaches c of f of x over g of x is also going to be equal to L. This might seem a little bit bizarre to you right now, and I'm actually going to write the other case, and then I'll do an example. We'll do multiple examples, and the examples are going to make it all clear. This is the first case, and the example we're going to do is actually going to be an example of this case. The other case is if the limit as x approaches c of f of x is equal to positive or negative infinity, and the limit as x approaches c of g of x is equal to positive or negative infinity, and the limit of, I guess you could say the quotient of the derivatives exists, and the limit as x approaches c of f prime of x over g prime of x is equal to L, then we can make this same statement again."}, {"video_title": "Introduction to l'H\u00f4pital's rule Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Then we can say that the limit as x approaches c of f of x over g of x is also going to be equal to L. This might seem a little bit bizarre to you right now, and I'm actually going to write the other case, and then I'll do an example. We'll do multiple examples, and the examples are going to make it all clear. This is the first case, and the example we're going to do is actually going to be an example of this case. The other case is if the limit as x approaches c of f of x is equal to positive or negative infinity, and the limit as x approaches c of g of x is equal to positive or negative infinity, and the limit of, I guess you could say the quotient of the derivatives exists, and the limit as x approaches c of f prime of x over g prime of x is equal to L, then we can make this same statement again. Then we can make this exact same statement. Let me just copy that out. Then this again."}, {"video_title": "Introduction to l'H\u00f4pital's rule Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "The other case is if the limit as x approaches c of f of x is equal to positive or negative infinity, and the limit as x approaches c of g of x is equal to positive or negative infinity, and the limit of, I guess you could say the quotient of the derivatives exists, and the limit as x approaches c of f prime of x over g prime of x is equal to L, then we can make this same statement again. Then we can make this exact same statement. Let me just copy that out. Then this again. Edit, copy, and then let me paste it. In either of these two situations, just to make sure you understand what you're looking at, this is a situation where if you just tried to evaluate this limit right here, you're going to get f of c, which is 0, or the limit as x approaches c of f of x over the limit as x approaches c of g of x, and that's going to give you 0 over 0. You say, hey, I don't know what that limit is, but this says, well, look, if this limit exists, I could take the derivative of each of these functions and then try to evaluate that limit."}, {"video_title": "Introduction to l'H\u00f4pital's rule Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Then this again. Edit, copy, and then let me paste it. In either of these two situations, just to make sure you understand what you're looking at, this is a situation where if you just tried to evaluate this limit right here, you're going to get f of c, which is 0, or the limit as x approaches c of f of x over the limit as x approaches c of g of x, and that's going to give you 0 over 0. You say, hey, I don't know what that limit is, but this says, well, look, if this limit exists, I could take the derivative of each of these functions and then try to evaluate that limit. If I get a number, if that exists, then they're going to be the same limit. This is a situation where when we take the limit, we get infinity over infinity, or negative infinity, or positive infinity over positive or negative infinity. These are the two indeterminate forms."}, {"video_title": "Introduction to l'H\u00f4pital's rule Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "You say, hey, I don't know what that limit is, but this says, well, look, if this limit exists, I could take the derivative of each of these functions and then try to evaluate that limit. If I get a number, if that exists, then they're going to be the same limit. This is a situation where when we take the limit, we get infinity over infinity, or negative infinity, or positive infinity over positive or negative infinity. These are the two indeterminate forms. To make it all clear, let me just show you an example, because I think this will make things a lot more clear. Let's say we are trying to find the limit. Notice in a new color."}, {"video_title": "Introduction to l'H\u00f4pital's rule Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "These are the two indeterminate forms. To make it all clear, let me just show you an example, because I think this will make things a lot more clear. Let's say we are trying to find the limit. Notice in a new color. Let me do it in this purplish color. Let's say we wanted to find the limit as x approaches 0 of sine of x over x. Now, if we just view this, if we just try to evaluate it at 0 or take the limit as we approach 0 in each of these functions, we're going to get something that looks like 0 over 0."}, {"video_title": "Introduction to l'H\u00f4pital's rule Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Notice in a new color. Let me do it in this purplish color. Let's say we wanted to find the limit as x approaches 0 of sine of x over x. Now, if we just view this, if we just try to evaluate it at 0 or take the limit as we approach 0 in each of these functions, we're going to get something that looks like 0 over 0. Sine of 0 is 0, or the limit as x approaches 0 of sine of x is 0, and obviously as x approaches 0 of x, that's also going to be 0. This is our indeterminate form. If you want to think about it, this is our f of x."}, {"video_title": "Introduction to l'H\u00f4pital's rule Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Now, if we just view this, if we just try to evaluate it at 0 or take the limit as we approach 0 in each of these functions, we're going to get something that looks like 0 over 0. Sine of 0 is 0, or the limit as x approaches 0 of sine of x is 0, and obviously as x approaches 0 of x, that's also going to be 0. This is our indeterminate form. If you want to think about it, this is our f of x. That f of x right there is sine of x. Our g of x, this g of x right there for this first case, is the x. g of x is equal to x, and f of x is equal to sine of x. Notice, we definitely know that this meets the first two constraints."}, {"video_title": "Introduction to l'H\u00f4pital's rule Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "If you want to think about it, this is our f of x. That f of x right there is sine of x. Our g of x, this g of x right there for this first case, is the x. g of x is equal to x, and f of x is equal to sine of x. Notice, we definitely know that this meets the first two constraints. The limit as x, and in this case, c is 0. The limit as x approaches 0 of sine of x is 0, and the limit as x approaches 0 of x is also equal to 0. We get our indeterminate form."}, {"video_title": "Introduction to l'H\u00f4pital's rule Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Notice, we definitely know that this meets the first two constraints. The limit as x, and in this case, c is 0. The limit as x approaches 0 of sine of x is 0, and the limit as x approaches 0 of x is also equal to 0. We get our indeterminate form. Let's see at least whether this limit even exists. If we take the derivative of f of x and we put that over the derivative of g of x and take that the limit as x approaches 0 in this case, that's our c. Let's see if this limit exists. I'll do that in the blue."}, {"video_title": "Introduction to l'H\u00f4pital's rule Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "We get our indeterminate form. Let's see at least whether this limit even exists. If we take the derivative of f of x and we put that over the derivative of g of x and take that the limit as x approaches 0 in this case, that's our c. Let's see if this limit exists. I'll do that in the blue. Let me write the derivatives of the two functions. f prime of x, if f of x is sine of x, what's f prime of x? It's just cosine of x."}, {"video_title": "Introduction to l'H\u00f4pital's rule Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "I'll do that in the blue. Let me write the derivatives of the two functions. f prime of x, if f of x is sine of x, what's f prime of x? It's just cosine of x. We've learned that many times. If g of x is x, what is g prime of x? That's super easy."}, {"video_title": "Introduction to l'H\u00f4pital's rule Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "It's just cosine of x. We've learned that many times. If g of x is x, what is g prime of x? That's super easy. The derivative of x is just 1. Let's try to take the limit as x approaches 0 of f prime of x over g prime of x, over their derivatives. That's going to be the limit as x approaches 0 of cosine of x over 1."}, {"video_title": "Introduction to l'H\u00f4pital's rule Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "That's super easy. The derivative of x is just 1. Let's try to take the limit as x approaches 0 of f prime of x over g prime of x, over their derivatives. That's going to be the limit as x approaches 0 of cosine of x over 1. I wrote that 1 a little strange. Over 1. This is pretty straightforward."}, {"video_title": "Introduction to l'H\u00f4pital's rule Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "That's going to be the limit as x approaches 0 of cosine of x over 1. I wrote that 1 a little strange. Over 1. This is pretty straightforward. What is this going to be? As x approaches 0 of cosine of x, that's going to be equal to 1. That's equal to 1."}, {"video_title": "Introduction to l'H\u00f4pital's rule Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "This is pretty straightforward. What is this going to be? As x approaches 0 of cosine of x, that's going to be equal to 1. That's equal to 1. Obviously, the limit as x approaches 0 of 1, that's also going to be equal to 1. In this situation, we just saw that the limit as x approaches, our c in this case is 0, as x approaches 0 of f prime of x over g prime of x is equal to 1. This limit exists and it equals 1."}, {"video_title": "Introduction to l'H\u00f4pital's rule Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "That's equal to 1. Obviously, the limit as x approaches 0 of 1, that's also going to be equal to 1. In this situation, we just saw that the limit as x approaches, our c in this case is 0, as x approaches 0 of f prime of x over g prime of x is equal to 1. This limit exists and it equals 1. We've met all of the conditions. This is the case we're dealing with. Limit as x approaches 0 of sine of x is equal to 0."}, {"video_title": "Introduction to l'H\u00f4pital's rule Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "This limit exists and it equals 1. We've met all of the conditions. This is the case we're dealing with. Limit as x approaches 0 of sine of x is equal to 0. Limit as x approaches 0 of x is also equal to 0. The limit of the derivative of sine of x over the derivative of x, which is cosine of x over 1, we found this to be equal to 1. All of these top conditions are met."}, {"video_title": "Limit comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let's remind ourselves, give ourselves a review of the comparison test, see where it can be useful, and maybe see where it might not be so useful, but luckily we'll also see the limit comparison test, which can be applicable in a broader category of situations. So we've already seen this. We want to prove that the infinite series from n equals one to infinity of one over two to the n plus one converges. How can we do that? Well, each of these terms are greater than or equal to zero, and we can construct another series where each of the corresponding terms are greater than each of these corresponding terms, and that other series, the one that jumps out at, that will likely jump out at most folks, would be one over two to the n. One over two to the n is greater than, is greater than, and all we have to really say is greater than or equal to, but we can actually explicitly say it's, well, I'll just write it's greater than or equal to one over two to the n plus one for n is equal to one to all the way to infinity. Why? Because this denominator is always going to be greater by one if your denominator is greater, the overall expression is going to be less."}, {"video_title": "Limit comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "How can we do that? Well, each of these terms are greater than or equal to zero, and we can construct another series where each of the corresponding terms are greater than each of these corresponding terms, and that other series, the one that jumps out at, that will likely jump out at most folks, would be one over two to the n. One over two to the n is greater than, is greater than, and all we have to really say is greater than or equal to, but we can actually explicitly say it's, well, I'll just write it's greater than or equal to one over two to the n plus one for n is equal to one to all the way to infinity. Why? Because this denominator is always going to be greater by one if your denominator is greater, the overall expression is going to be less. And because of that, because each of these terms are, they're all positive, this one, each corresponding term is greater than that one, and by the comparison test, because this one converges, this kind of provides an upper bound, because this series we already know converges, we can say, so because this one converges, we can say that this one converges. Now let's see, let's see if we can apply a similar logic to a slightly different series. Let's say we have the series, the sum from n equals one to infinity of one over two to the n minus one."}, {"video_title": "Limit comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Because this denominator is always going to be greater by one if your denominator is greater, the overall expression is going to be less. And because of that, because each of these terms are, they're all positive, this one, each corresponding term is greater than that one, and by the comparison test, because this one converges, this kind of provides an upper bound, because this series we already know converges, we can say, so because this one converges, we can say that this one converges. Now let's see, let's see if we can apply a similar logic to a slightly different series. Let's say we have the series, the sum from n equals one to infinity of one over two to the n minus one. In this situation, can we do just the straight up comparison test? Well, no, because you cannot say that one over two to the n is greater than or equal to one over two to the n minus one. Here, the denominator is lower, means the expression is greater, which means that this can't, each of these terms can't provide an upper bound on this one."}, {"video_title": "Limit comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let's say we have the series, the sum from n equals one to infinity of one over two to the n minus one. In this situation, can we do just the straight up comparison test? Well, no, because you cannot say that one over two to the n is greater than or equal to one over two to the n minus one. Here, the denominator is lower, means the expression is greater, which means that this can't, each of these terms can't provide an upper bound on this one. This one is a little bit larger. But the other hand, you're like, okay, I get that. But look, as n gets large, the two to the n is gonna really dominate the minus one or the plus one or the, or this one has nothing there, just has two to the n. The two to the n is really gonna describe the behavior of what this thing does."}, {"video_title": "Limit comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Here, the denominator is lower, means the expression is greater, which means that this can't, each of these terms can't provide an upper bound on this one. This one is a little bit larger. But the other hand, you're like, okay, I get that. But look, as n gets large, the two to the n is gonna really dominate the minus one or the plus one or the, or this one has nothing there, just has two to the n. The two to the n is really gonna describe the behavior of what this thing does. And I would agree with you, but we just haven't proven that it actually works. And that's where the limit comparison test comes in helpful. So let me write that down."}, {"video_title": "Limit comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "But look, as n gets large, the two to the n is gonna really dominate the minus one or the plus one or the, or this one has nothing there, just has two to the n. The two to the n is really gonna describe the behavior of what this thing does. And I would agree with you, but we just haven't proven that it actually works. And that's where the limit comparison test comes in helpful. So let me write that down. Limit, limit comparison test. Limit comparison test. And I'll write it down a little bit formally, but then we'll apply it to this infinite series right over here."}, {"video_title": "Limit comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let me write that down. Limit, limit comparison test. Limit comparison test. And I'll write it down a little bit formally, but then we'll apply it to this infinite series right over here. So the limit comparison test tells us that if I have two infinite series, so let's say this is going from n equals k to infinity of a sub n, I'm not gonna prove it here. We'll just learn to apply it first. And this goes from n equals k to infinity of b sub n. And we know that each of the terms, a sub n, are greater than or equal to zero."}, {"video_title": "Limit comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And I'll write it down a little bit formally, but then we'll apply it to this infinite series right over here. So the limit comparison test tells us that if I have two infinite series, so let's say this is going from n equals k to infinity of a sub n, I'm not gonna prove it here. We'll just learn to apply it first. And this goes from n equals k to infinity of b sub n. And we know that each of the terms, a sub n, are greater than or equal to zero. And we know each of the terms b sub n are, actually we're just gonna say greater than zero. It's actually gonna show up in the denominator of an expression, so we don't want it to be equal to zero. For all the n's that we care about."}, {"video_title": "Limit comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And this goes from n equals k to infinity of b sub n. And we know that each of the terms, a sub n, are greater than or equal to zero. And we know each of the terms b sub n are, actually we're just gonna say greater than zero. It's actually gonna show up in the denominator of an expression, so we don't want it to be equal to zero. For all the n's that we care about. So for all n equal to k, k plus one, k plus two, on and on and on and on. And, and this is the key, this is where the limit of the limit comparison test comes into play. And if the limit, the limit as n approaches infinity of a sub n over b sub n, b sub n, is positive and finite, is positive and finite, then either both series converge or both series diverge."}, {"video_title": "Limit comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "For all the n's that we care about. So for all n equal to k, k plus one, k plus two, on and on and on and on. And, and this is the key, this is where the limit of the limit comparison test comes into play. And if the limit, the limit as n approaches infinity of a sub n over b sub n, b sub n, is positive and finite, is positive and finite, then either both series converge or both series diverge. So let me write that. So then that tells us that either, either both converge or both diverge. Which is really, really useful."}, {"video_title": "Limit comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And if the limit, the limit as n approaches infinity of a sub n over b sub n, b sub n, is positive and finite, is positive and finite, then either both series converge or both series diverge. So let me write that. So then that tells us that either, either both converge or both diverge. Which is really, really useful. It's kind of a more formal way of saying that, hey look, as n approaches infinity, if these have similar behaviors, then they're either going to converge or they're both going to diverge. Let's apply that right over here. Well if we say that our b sub n is one over two to the n, just like we did up there, one over two to the n, so we're going to compare, so these two series right over here, notice it satisfies all of these constraints."}, {"video_title": "Limit comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Which is really, really useful. It's kind of a more formal way of saying that, hey look, as n approaches infinity, if these have similar behaviors, then they're either going to converge or they're both going to diverge. Let's apply that right over here. Well if we say that our b sub n is one over two to the n, just like we did up there, one over two to the n, so we're going to compare, so these two series right over here, notice it satisfies all of these constraints. So let's take the limit, the limit as n approaches infinity of a sub n over b sub n. So it's going to be one over two to the n minus one over, over one over two to the n. And what's that going to be equal to? Well that's going to be equal to the limit as n approaches infinity of twos, if you divide by one over two to the n, that's just going to be the same thing as multiplying by two to the n. So it's going to be two to the n over, over two to the n minus one. Over two to the n minus one."}, {"video_title": "Limit comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well if we say that our b sub n is one over two to the n, just like we did up there, one over two to the n, so we're going to compare, so these two series right over here, notice it satisfies all of these constraints. So let's take the limit, the limit as n approaches infinity of a sub n over b sub n. So it's going to be one over two to the n minus one over, over one over two to the n. And what's that going to be equal to? Well that's going to be equal to the limit as n approaches infinity of twos, if you divide by one over two to the n, that's just going to be the same thing as multiplying by two to the n. So it's going to be two to the n over, over two to the n minus one. Over two to the n minus one. And this clearly, what's happening in the numerator and the denominator, these are approaching the same quantity, and actually we can even write it like, we can even write it like this, divide the numerator and the denominator by two to the n if you want. Although it's probably going to jump out at you at this point. So limit as n approaches infinity, let me scroll over to the right a little bit."}, {"video_title": "Limit comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Over two to the n minus one. And this clearly, what's happening in the numerator and the denominator, these are approaching the same quantity, and actually we can even write it like, we can even write it like this, divide the numerator and the denominator by two to the n if you want. Although it's probably going to jump out at you at this point. So limit as n approaches infinity, let me scroll over to the right a little bit. If I divide the numerator by two to the n, I'm just going to have one. If I divide the denominator by two to the n, I'm going to have one minus two, I can just write this one over two to the n. And now it becomes clear. This thing right over here is just going to go to zero, and you're going to have one over one."}, {"video_title": "Limit comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So limit as n approaches infinity, let me scroll over to the right a little bit. If I divide the numerator by two to the n, I'm just going to have one. If I divide the denominator by two to the n, I'm going to have one minus two, I can just write this one over two to the n. And now it becomes clear. This thing right over here is just going to go to zero, and you're going to have one over one. The important thing is that this limit is positive and finite. Because this thing is, so this thing right over here is positive and finite, the limit is one, is positive and finite, if this thing converges and this thing converges, if this thing diverges and this thing diverges, well we already know this thing converges. It's a geometric series where the common ratio is less than one."}, {"video_title": "Worked example forming a slope field AP Calculus AB Khan Academy.mp3", "Sentence": "In drawing the slope field for the differential equation, the derivative of y with respect to x is equal to y minus two x, I would place short line segments at select points on the xy plane. Complete the sentences. At the point negative one comma one, I would draw a short segment of slope blank. And like always, pause this video and see if you can fill out these three blanks. Well, when you're, the short segments that you're trying to draw to construct this slope field, you figure out their slope based on the differential equation. So you're saying when x is equal to negative one and y is equal to one, what is the derivative of y with respect to x? And that's what this differential equation tells us."}, {"video_title": "Worked example forming a slope field AP Calculus AB Khan Academy.mp3", "Sentence": "And like always, pause this video and see if you can fill out these three blanks. Well, when you're, the short segments that you're trying to draw to construct this slope field, you figure out their slope based on the differential equation. So you're saying when x is equal to negative one and y is equal to one, what is the derivative of y with respect to x? And that's what this differential equation tells us. So for this first case, the derivative of y with respect to x is going to be equal to y, which is one, minus two times x. X is negative one. So this is gonna be negative two, but you're subtracting it, so it's gonna be plus two. So the derivative of y with respect to x at this point is going to be three."}, {"video_title": "Worked example forming a slope field AP Calculus AB Khan Academy.mp3", "Sentence": "And that's what this differential equation tells us. So for this first case, the derivative of y with respect to x is going to be equal to y, which is one, minus two times x. X is negative one. So this is gonna be negative two, but you're subtracting it, so it's gonna be plus two. So the derivative of y with respect to x at this point is going to be three. So I would draw a short line segment or a short segment of slope three. And we keep going at the point zero comma two. Well, let's see, when x is zero and y is two, the derivative of y with respect to x is going to be equal to y, which is two, minus two times zero."}, {"video_title": "Worked example forming a slope field AP Calculus AB Khan Academy.mp3", "Sentence": "So the derivative of y with respect to x at this point is going to be three. So I would draw a short line segment or a short segment of slope three. And we keep going at the point zero comma two. Well, let's see, when x is zero and y is two, the derivative of y with respect to x is going to be equal to y, which is two, minus two times zero. Well, that's just going to be two. And then last but not least, for this third point, the derivative of y with respect to x is going to be equal to y, which is three, minus two times x. X here is two. Two times two, three minus four is equal to four."}, {"video_title": "Worked example forming a slope field AP Calculus AB Khan Academy.mp3", "Sentence": "Well, let's see, when x is zero and y is two, the derivative of y with respect to x is going to be equal to y, which is two, minus two times zero. Well, that's just going to be two. And then last but not least, for this third point, the derivative of y with respect to x is going to be equal to y, which is three, minus two times x. X here is two. Two times two, three minus four is equal to four. Three minus four is equal to negative one. And that's all that problem asks us to do. Now, if we actually had to do it, it would look something like, I'll try to draw it real fast."}, {"video_title": "Worked example forming a slope field AP Calculus AB Khan Academy.mp3", "Sentence": "Two times two, three minus four is equal to four. Three minus four is equal to negative one. And that's all that problem asks us to do. Now, if we actually had to do it, it would look something like, I'll try to draw it real fast. So let's see, let me make sure I have space for all of these points here. So that's my coordinate axes. And I want to get the point zero comma two."}, {"video_title": "Worked example forming a slope field AP Calculus AB Khan Academy.mp3", "Sentence": "Now, if we actually had to do it, it would look something like, I'll try to draw it real fast. So let's see, let me make sure I have space for all of these points here. So that's my coordinate axes. And I want to get the point zero comma two. So that's zero comma two. Actually, I want to go all the way to two comma three, so let me get some space here. So one, two, three, and then one, two, three, and then we have to go negative one comma one, so we might go right over here."}, {"video_title": "Worked example forming a slope field AP Calculus AB Khan Academy.mp3", "Sentence": "And I want to get the point zero comma two. So that's zero comma two. Actually, I want to go all the way to two comma three, so let me get some space here. So one, two, three, and then one, two, three, and then we have to go negative one comma one, so we might go right over here. And so for this first one, and this exercise isn't asking us to do it, but I'm just making it very clear how we would construct the slope field. So the point negative one comma one, negative one comma one, a short segment of slope three. So slope three would look something like that."}, {"video_title": "Worked example forming a slope field AP Calculus AB Khan Academy.mp3", "Sentence": "So one, two, three, and then one, two, three, and then we have to go negative one comma one, so we might go right over here. And so for this first one, and this exercise isn't asking us to do it, but I'm just making it very clear how we would construct the slope field. So the point negative one comma one, negative one comma one, a short segment of slope three. So slope three would look something like that. Then at the point zero comma two, a slope of two. Zero comma two, the slope is going to be two, which looks something like that. And then at the point two comma three, at two comma three, a short segment of slope negative one."}, {"video_title": "Worked example forming a slope field AP Calculus AB Khan Academy.mp3", "Sentence": "So slope three would look something like that. Then at the point zero comma two, a slope of two. Zero comma two, the slope is going to be two, which looks something like that. And then at the point two comma three, at two comma three, a short segment of slope negative one. So two comma three, a segment of slope negative one. It would look something like that. And you would keep doing this at more and more points."}, {"video_title": "Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "I guess we could call it a relationship. And if we were to graph all of the points x and y that satisfy this relationship, we get a unit circle like this. And what I'm curious about in this video is how we can figure out the slope of the tangent line at any point of this unit circle. And what immediately might be jumping out in your brain is well, a circle defined this way, this isn't a function. It's not y explicitly defined as a function of x. For any x value, you actually have two possible y's that satisfy this relationship right over here. So you might be tempted to maybe split this up into two separate functions of x."}, {"video_title": "Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And what immediately might be jumping out in your brain is well, a circle defined this way, this isn't a function. It's not y explicitly defined as a function of x. For any x value, you actually have two possible y's that satisfy this relationship right over here. So you might be tempted to maybe split this up into two separate functions of x. You could say y is equal to the positive square root of one minus x squared. And you could say y is equal to the negative square root of one minus x squared. Take the derivatives of each of these separately, and you would be able to find the derivative for any x, or the derivative of the slope of the tangent line at any point."}, {"video_title": "Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So you might be tempted to maybe split this up into two separate functions of x. You could say y is equal to the positive square root of one minus x squared. And you could say y is equal to the negative square root of one minus x squared. Take the derivatives of each of these separately, and you would be able to find the derivative for any x, or the derivative of the slope of the tangent line at any point. But what I wanna do in this video is literally leverage the chain rule to take the derivative implicitly so that I don't have to explicitly define y as a function of x either way. And the way we do that is literally just apply the derivative operator to both sides of this equation and then apply what we know about the chain rule. Because we are not explicitly defining y as a function of x and explicitly getting y is equal to f prime of x, they call this, which is really just an application of the chain rule, we call it implicit differentiation."}, {"video_title": "Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Take the derivatives of each of these separately, and you would be able to find the derivative for any x, or the derivative of the slope of the tangent line at any point. But what I wanna do in this video is literally leverage the chain rule to take the derivative implicitly so that I don't have to explicitly define y as a function of x either way. And the way we do that is literally just apply the derivative operator to both sides of this equation and then apply what we know about the chain rule. Because we are not explicitly defining y as a function of x and explicitly getting y is equal to f prime of x, they call this, which is really just an application of the chain rule, we call it implicit differentiation. Implicit differentiation. And what I want you to keep in the back of your mind the entire time is that it's just an application of the chain rule. So let's apply the derivative operator to both sides of this."}, {"video_title": "Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Because we are not explicitly defining y as a function of x and explicitly getting y is equal to f prime of x, they call this, which is really just an application of the chain rule, we call it implicit differentiation. Implicit differentiation. And what I want you to keep in the back of your mind the entire time is that it's just an application of the chain rule. So let's apply the derivative operator to both sides of this. So it's the derivative with respect to x of x squared plus y squared, x squared plus y squared on the left-hand side of our equation, and then that's going to be equal to the derivative with respect to x on the right-hand side. I'm just doing the same exact thing to both sides of this equation. Now, if I take the derivative of the sum of two terms, that's the same thing as taking the sum of the derivative."}, {"video_title": "Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let's apply the derivative operator to both sides of this. So it's the derivative with respect to x of x squared plus y squared, x squared plus y squared on the left-hand side of our equation, and then that's going to be equal to the derivative with respect to x on the right-hand side. I'm just doing the same exact thing to both sides of this equation. Now, if I take the derivative of the sum of two terms, that's the same thing as taking the sum of the derivative. So this is going to be the same thing as the derivative with respect to x of x squared plus the derivative with respect to x of y squared. I'm writing all my orange stuff first. So let's see, this is going to be x squared, it's gonna be y squared, and then this is going to be equal to the derivative with respect to x of a constant."}, {"video_title": "Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Now, if I take the derivative of the sum of two terms, that's the same thing as taking the sum of the derivative. So this is going to be the same thing as the derivative with respect to x of x squared plus the derivative with respect to x of y squared. I'm writing all my orange stuff first. So let's see, this is going to be x squared, it's gonna be y squared, and then this is going to be equal to the derivative with respect to x of a constant. This isn't changing with respect to x. So we just get zero. Now, this first term right over here, we have done many, many, many, many, many times."}, {"video_title": "Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let's see, this is going to be x squared, it's gonna be y squared, and then this is going to be equal to the derivative with respect to x of a constant. This isn't changing with respect to x. So we just get zero. Now, this first term right over here, we have done many, many, many, many, many times. The derivative with respect to x of x squared is just the power rule here. It's going to be two times x to the first power, or we could just say two x. Now, what's interesting is what we're doing right over here."}, {"video_title": "Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Now, this first term right over here, we have done many, many, many, many, many times. The derivative with respect to x of x squared is just the power rule here. It's going to be two times x to the first power, or we could just say two x. Now, what's interesting is what we're doing right over here. The derivative with respect to x of y squared. And the realization here is to just apply the chain rule. If we're taking the derivative with respect to x of this something, we just have to take the derivative, let me make it clear, we're just gonna take the derivative of our something, the derivative of y squared, that's what we're taking, you can kind of view that as a function, with respect to y, with respect to y, and then multiply that times the derivative of y, the derivative of y with respect to x."}, {"video_title": "Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Now, what's interesting is what we're doing right over here. The derivative with respect to x of y squared. And the realization here is to just apply the chain rule. If we're taking the derivative with respect to x of this something, we just have to take the derivative, let me make it clear, we're just gonna take the derivative of our something, the derivative of y squared, that's what we're taking, you can kind of view that as a function, with respect to y, with respect to y, and then multiply that times the derivative of y, the derivative of y with respect to x. We're assuming that y does change with respect to x. Y is not some type of a constant that we're writing just in abstract terms. So we're taking the derivative of this whole thing with respect to y. Once again, just the chain rule."}, {"video_title": "Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "If we're taking the derivative with respect to x of this something, we just have to take the derivative, let me make it clear, we're just gonna take the derivative of our something, the derivative of y squared, that's what we're taking, you can kind of view that as a function, with respect to y, with respect to y, and then multiply that times the derivative of y, the derivative of y with respect to x. We're assuming that y does change with respect to x. Y is not some type of a constant that we're writing just in abstract terms. So we're taking the derivative of this whole thing with respect to y. Once again, just the chain rule. And then we're taking the derivative of y with respect to x. It might be a little bit clearer if you kind of thought of it as the derivative with respect to x of y as a function of x. Y as a function of x. This might be, or y as a function of x squared, which is essentially another way of writing what we had here."}, {"video_title": "Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Once again, just the chain rule. And then we're taking the derivative of y with respect to x. It might be a little bit clearer if you kind of thought of it as the derivative with respect to x of y as a function of x. Y as a function of x. This might be, or y as a function of x squared, which is essentially another way of writing what we had here. This might be a little bit clearer in terms of the chain rule. The derivative of y as a function of x squared with respect to y of x, so the derivative of something squared, the derivative of something squared with respect to that something, times the derivative of that something with respect to x. This is just the chain rule."}, {"video_title": "Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "This might be, or y as a function of x squared, which is essentially another way of writing what we had here. This might be a little bit clearer in terms of the chain rule. The derivative of y as a function of x squared with respect to y of x, so the derivative of something squared, the derivative of something squared with respect to that something, times the derivative of that something with respect to x. This is just the chain rule. I wanna say it over and over again. This is just the chain rule. So let's do that."}, {"video_title": "Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "This is just the chain rule. I wanna say it over and over again. This is just the chain rule. So let's do that. What do we get on the right-hand side over here? And I'll write it over here as well. This would be equal to the derivative of y squared with respect to y is just going to be two times y, two times y, just an application of the chain rule."}, {"video_title": "Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let's do that. What do we get on the right-hand side over here? And I'll write it over here as well. This would be equal to the derivative of y squared with respect to y is just going to be two times y, two times y, just an application of the chain rule. And the derivative of y with respect to x, well, we don't know what that is, so we're just gonna leave that as times the derivative of y times the derivative of y with respect to x. So let's just write this down over here. So what we have is two x, two x plus, two x plus derivative of something squared with respect to that something is two times the something."}, {"video_title": "Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "This would be equal to the derivative of y squared with respect to y is just going to be two times y, two times y, just an application of the chain rule. And the derivative of y with respect to x, well, we don't know what that is, so we're just gonna leave that as times the derivative of y times the derivative of y with respect to x. So let's just write this down over here. So what we have is two x, two x plus, two x plus derivative of something squared with respect to that something is two times the something. In this case, the something is y, so two times y, and then times the derivative of y, the derivative of y with respect to x. And this is all going to be equal to, all going to be equal to zero. Now that was interesting."}, {"video_title": "Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So what we have is two x, two x plus, two x plus derivative of something squared with respect to that something is two times the something. In this case, the something is y, so two times y, and then times the derivative of y, the derivative of y with respect to x. And this is all going to be equal to, all going to be equal to zero. Now that was interesting. Now we have an equation that has the derivative of y with respect to x in it. And this is what we essentially want to solve for. This is the slope of the tangent line at any point."}, {"video_title": "Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Now that was interesting. Now we have an equation that has the derivative of y with respect to x in it. And this is what we essentially want to solve for. This is the slope of the tangent line at any point. So all we have to do at this point is solve for the derivative of y with respect to x, solve this equation. So let's do that. And actually, just so we all, so we can do this whole thing on the same page so we can see where we started, let me copy and paste this up here."}, {"video_title": "Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "This is the slope of the tangent line at any point. So all we have to do at this point is solve for the derivative of y with respect to x, solve this equation. So let's do that. And actually, just so we all, so we can do this whole thing on the same page so we can see where we started, let me copy and paste this up here. This is where we left off, and let's continue there. So let's say, let's subtract two x from both sides. So we're left with two y times the derivative of y with respect to x is equal to, we're subtracting two x from both sides, so it's equal to negative two x."}, {"video_title": "Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And actually, just so we all, so we can do this whole thing on the same page so we can see where we started, let me copy and paste this up here. This is where we left off, and let's continue there. So let's say, let's subtract two x from both sides. So we're left with two y times the derivative of y with respect to x is equal to, we're subtracting two x from both sides, so it's equal to negative two x. And then if we really want to solve for the derivative of y with respect to x, we can just divide both sides by two y. We just divide both sides by two y. And we are left with the derivative of y with respect to x."}, {"video_title": "Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So we're left with two y times the derivative of y with respect to x is equal to, we're subtracting two x from both sides, so it's equal to negative two x. And then if we really want to solve for the derivative of y with respect to x, we can just divide both sides by two y. We just divide both sides by two y. And we are left with the derivative of y with respect to x. Let's scroll down a little bit. The derivative of y with respect to x is equal to, is equal to, well, the twos cancel out. We are left with negative x, negative x over, over y."}, {"video_title": "Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And we are left with the derivative of y with respect to x. Let's scroll down a little bit. The derivative of y with respect to x is equal to, is equal to, well, the twos cancel out. We are left with negative x, negative x over, over y. So this is interesting. We didn't have to explicitly define y as a function of x here, but we got our derivative in terms of an x and a y, not just only in terms of an x. But what does this mean?"}, {"video_title": "Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We are left with negative x, negative x over, over y. So this is interesting. We didn't have to explicitly define y as a function of x here, but we got our derivative in terms of an x and a y, not just only in terms of an x. But what does this mean? Well, if we wanted to find, let's say we wanted to find the derivative at this point, this point right over here, which if you're familiar with the unit circle, so if this was a 45 degree angle, this would be the square root of two over two, comma, the square root of two over two. What is the slope of the tangent line there? Well, we figured it out."}, {"video_title": "Implicit differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "But what does this mean? Well, if we wanted to find, let's say we wanted to find the derivative at this point, this point right over here, which if you're familiar with the unit circle, so if this was a 45 degree angle, this would be the square root of two over two, comma, the square root of two over two. What is the slope of the tangent line there? Well, we figured it out. It's going to be negative x over y. So the slope of the tangent line here, the slope of the tangent line right over here, the slope is going to be equal to negative x, so negative square root of two over two over y, over square root of two over two, which is equal to negative one. And that looks just about right."}, {"video_title": "Analyzing related rates problems equations (trig) AP Calculus AB Khan Academy.mp3", "Sentence": "At a certain instant, t sub zero, the top of the ladder is a distance y of t sub zero of 15 meters from the ground. What is the rate of change of the angle, theta of t between the ground and the ladder at that instant? What I'm gonna do is draw this out, and really the first step is to think about, well, what equation will be helpful for us to solve this problem? Then we might just go ahead and actually solve the problem. A 20 meter ladder is leaning against a wall. Let me draw ourselves a wall here. That is my wall."}, {"video_title": "Analyzing related rates problems equations (trig) AP Calculus AB Khan Academy.mp3", "Sentence": "Then we might just go ahead and actually solve the problem. A 20 meter ladder is leaning against a wall. Let me draw ourselves a wall here. That is my wall. Now let me draw our 20 meter ladder. Maybe it looks something like that. That is 20 meters."}, {"video_title": "Analyzing related rates problems equations (trig) AP Calculus AB Khan Academy.mp3", "Sentence": "That is my wall. Now let me draw our 20 meter ladder. Maybe it looks something like that. That is 20 meters. They say the distance x of t between the bottom of the ladder and the wall, so it's this distance right over here, this distance right over here is x of t. They say it's increasing at a rate of three meters per minute. We know that we could either say x prime of t, which is the same thing as dx dt, is equal to three meters. I'll write it out, because it's hard if I just said m per m. It might not be that clear."}, {"video_title": "Analyzing related rates problems equations (trig) AP Calculus AB Khan Academy.mp3", "Sentence": "That is 20 meters. They say the distance x of t between the bottom of the ladder and the wall, so it's this distance right over here, this distance right over here is x of t. They say it's increasing at a rate of three meters per minute. We know that we could either say x prime of t, which is the same thing as dx dt, is equal to three meters. I'll write it out, because it's hard if I just said m per m. It might not be that clear. Meters per minute, so they give us that piece of information. The rate of change of x with respect to time, they give us that. At a certain instant, t sub zero, the top of the ladder is a distance of 15 meters."}, {"video_title": "Analyzing related rates problems equations (trig) AP Calculus AB Khan Academy.mp3", "Sentence": "I'll write it out, because it's hard if I just said m per m. It might not be that clear. Meters per minute, so they give us that piece of information. The rate of change of x with respect to time, they give us that. At a certain instant, t sub zero, the top of the ladder is a distance of 15 meters. The top of the ladder, so let's make this very clear. This distance right over here is y of t, y of t. They say at time t sub zero, y of t is 15 meters. Let me just write it here."}, {"video_title": "Analyzing related rates problems equations (trig) AP Calculus AB Khan Academy.mp3", "Sentence": "At a certain instant, t sub zero, the top of the ladder is a distance of 15 meters. The top of the ladder, so let's make this very clear. This distance right over here is y of t, y of t. They say at time t sub zero, y of t is 15 meters. Let me just write it here. Y of t sub zero is equal to 15 meters. Let me write this right over here. This is y of t sub zero."}, {"video_title": "Analyzing related rates problems equations (trig) AP Calculus AB Khan Academy.mp3", "Sentence": "Let me just write it here. Y of t sub zero is equal to 15 meters. Let me write this right over here. This is y of t sub zero. Let's just assume that we're drawing it at that moment, t sub zero, because I think that's going to be important. Y at t sub zero is equal to 15 meters. They want to know what is the rate of change of the angle theta between the ground and the ladder."}, {"video_title": "Analyzing related rates problems equations (trig) AP Calculus AB Khan Academy.mp3", "Sentence": "This is y of t sub zero. Let's just assume that we're drawing it at that moment, t sub zero, because I think that's going to be important. Y at t sub zero is equal to 15 meters. They want to know what is the rate of change of the angle theta between the ground and the ladder. This is the same. Theta is also going to change with respect to time. It's going to be a function of time between the ground and the ladder at that instant."}, {"video_title": "Analyzing related rates problems equations (trig) AP Calculus AB Khan Academy.mp3", "Sentence": "They want to know what is the rate of change of the angle theta between the ground and the ladder. This is the same. Theta is also going to change with respect to time. It's going to be a function of time between the ground and the ladder at that instant. Theta, let me get a new color here. Theta is this angle right over here. This is theta, and it's also going to be a function of time."}, {"video_title": "Analyzing related rates problems equations (trig) AP Calculus AB Khan Academy.mp3", "Sentence": "It's going to be a function of time between the ground and the ladder at that instant. Theta, let me get a new color here. Theta is this angle right over here. This is theta, and it's also going to be a function of time. What we'll always want to do in these related rates problems is we want to set up an equation, and really an algebraic equation, maybe a little bit of trigonometry involved, that relates the things that we care about, and then we're likely to have to take the derivative of both sides of that in order to relate the related rates. Let's see. We want to know the rate of change of the angle between the ground and the ladder at that instant."}, {"video_title": "Analyzing related rates problems equations (trig) AP Calculus AB Khan Academy.mp3", "Sentence": "This is theta, and it's also going to be a function of time. What we'll always want to do in these related rates problems is we want to set up an equation, and really an algebraic equation, maybe a little bit of trigonometry involved, that relates the things that we care about, and then we're likely to have to take the derivative of both sides of that in order to relate the related rates. Let's see. We want to know the rate of change of the angle between the ground and the ladder at that instant. What we need to figure out, we want to figure out theta prime at t sub zero. This is what we want to figure out. They've given us some interesting things."}, {"video_title": "Analyzing related rates problems equations (trig) AP Calculus AB Khan Academy.mp3", "Sentence": "We want to know the rate of change of the angle between the ground and the ladder at that instant. What we need to figure out, we want to figure out theta prime at t sub zero. This is what we want to figure out. They've given us some interesting things. They've given us, I guess, our rate of change of x with respect to time is constant at three meters per minute, and we know what y is at that moment. Let's see. Can we create a relationship?"}, {"video_title": "Analyzing related rates problems equations (trig) AP Calculus AB Khan Academy.mp3", "Sentence": "They've given us some interesting things. They've given us, I guess, our rate of change of x with respect to time is constant at three meters per minute, and we know what y is at that moment. Let's see. Can we create a relationship? Because they gave us dx dt, it'll actually be more useful to find a relationship between x and theta, and then take the derivative of both sides, and then use this information, possibly, to figure out what the appropriate value of x or theta is at that moment. Let's do that. How does x relate to theta?"}, {"video_title": "Analyzing related rates problems equations (trig) AP Calculus AB Khan Academy.mp3", "Sentence": "Can we create a relationship? Because they gave us dx dt, it'll actually be more useful to find a relationship between x and theta, and then take the derivative of both sides, and then use this information, possibly, to figure out what the appropriate value of x or theta is at that moment. Let's do that. How does x relate to theta? Well, we use a little bit of trigonometry right over here. If you took the hypotenuse times the cosine of theta, you would get x. Let me write this right over here."}, {"video_title": "Analyzing related rates problems equations (trig) AP Calculus AB Khan Academy.mp3", "Sentence": "How does x relate to theta? Well, we use a little bit of trigonometry right over here. If you took the hypotenuse times the cosine of theta, you would get x. Let me write this right over here. x of t, x of t, is equal to the hypotenuse, 20 meters, that's the length of the ladder, times the cosine, cosine of theta, and I could say the cosine of theta of t just to make it clear that this is a function of time. This comes straight out of trigonometry, actually our basic trigonometric function definitions. Now, why is this useful?"}, {"video_title": "Analyzing related rates problems equations (trig) AP Calculus AB Khan Academy.mp3", "Sentence": "Let me write this right over here. x of t, x of t, is equal to the hypotenuse, 20 meters, that's the length of the ladder, times the cosine, cosine of theta, and I could say the cosine of theta of t just to make it clear that this is a function of time. This comes straight out of trigonometry, actually our basic trigonometric function definitions. Now, why is this useful? Why do I think this is useful? Well, let's think about what happens when I take the derivative of both sides using the chain rule. On the left-hand side, I am going to have an x prime of t, and then that's going to be equal to, what do I end up on the right-hand side?"}, {"video_title": "Analyzing related rates problems equations (trig) AP Calculus AB Khan Academy.mp3", "Sentence": "Now, why is this useful? Why do I think this is useful? Well, let's think about what happens when I take the derivative of both sides using the chain rule. On the left-hand side, I am going to have an x prime of t, and then that's going to be equal to, what do I end up on the right-hand side? Well, using the chain rule, first I'll take the derivative with respect to theta, and so that's just going to be negative 20 sine of theta of t, and then I need to multiply that times theta prime of t. So, what I could do is say, hey, look, at t sub zero, I know what x prime of t is, I could try to figure out what sine of theta of t is, and then I'll just solve for this right over there. So, let's do that. So, at t sub zero, so at t is equal to t sub zero, t sub zero, what we're gonna have, x prime of t, well, that's at every time, it's three meters per minute, we'll assume that our rates are in meters per minute, and just our values are in meters when we're talking about distance, and our angles are in radians."}, {"video_title": "Analyzing related rates problems equations (trig) AP Calculus AB Khan Academy.mp3", "Sentence": "On the left-hand side, I am going to have an x prime of t, and then that's going to be equal to, what do I end up on the right-hand side? Well, using the chain rule, first I'll take the derivative with respect to theta, and so that's just going to be negative 20 sine of theta of t, and then I need to multiply that times theta prime of t. So, what I could do is say, hey, look, at t sub zero, I know what x prime of t is, I could try to figure out what sine of theta of t is, and then I'll just solve for this right over there. So, let's do that. So, at t sub zero, so at t is equal to t sub zero, t sub zero, what we're gonna have, x prime of t, well, that's at every time, it's three meters per minute, we'll assume that our rates are in meters per minute, and just our values are in meters when we're talking about distance, and our angles are in radians. So, this is going to be equal to three is equal to negative 20 times sine of theta of t times the derivative of theta with respect to time. So, how do we figure out what sine of theta of t is going to be? Well, let's just use that other information they gave us, and I'm gonna scroll down a little bit, get a little bit more real estate."}, {"video_title": "Analyzing related rates problems equations (trig) AP Calculus AB Khan Academy.mp3", "Sentence": "So, at t sub zero, so at t is equal to t sub zero, t sub zero, what we're gonna have, x prime of t, well, that's at every time, it's three meters per minute, we'll assume that our rates are in meters per minute, and just our values are in meters when we're talking about distance, and our angles are in radians. So, this is going to be equal to three is equal to negative 20 times sine of theta of t times the derivative of theta with respect to time. So, how do we figure out what sine of theta of t is going to be? Well, let's just use that other information they gave us, and I'm gonna scroll down a little bit, get a little bit more real estate. So, sine of theta, let me write it over here, sine of theta at time t sub naught, that's what we care about, t is equal to t sub naught, what's that going to be? Well, sine is opposite over hypotenuse, so that's going to be y at t sub naught over our hypotenuse of 20 meters. Well, that's going to be equal to, that's going to be equal to, they tell us y of t sub naught is 15 meters over 20 meters, which is the same thing as 3 4ths."}, {"video_title": "Analyzing related rates problems equations (trig) AP Calculus AB Khan Academy.mp3", "Sentence": "Well, let's just use that other information they gave us, and I'm gonna scroll down a little bit, get a little bit more real estate. So, sine of theta, let me write it over here, sine of theta at time t sub naught, that's what we care about, t is equal to t sub naught, what's that going to be? Well, sine is opposite over hypotenuse, so that's going to be y at t sub naught over our hypotenuse of 20 meters. Well, that's going to be equal to, that's going to be equal to, they tell us y of t sub naught is 15 meters over 20 meters, which is the same thing as 3 4ths. So, by this yellow information, they actually told us that this right over here is going to be equal to 3 4ths. So, this times 3 4ths times the rate of change of theta with respect to t. And so, now we just solve for this, and we're done. So, this is going to be, what's negative 20 times 3 4ths?"}, {"video_title": "Analyzing related rates problems equations (trig) AP Calculus AB Khan Academy.mp3", "Sentence": "Well, that's going to be equal to, that's going to be equal to, they tell us y of t sub naught is 15 meters over 20 meters, which is the same thing as 3 4ths. So, by this yellow information, they actually told us that this right over here is going to be equal to 3 4ths. So, this times 3 4ths times the rate of change of theta with respect to t. And so, now we just solve for this, and we're done. So, this is going to be, what's negative 20 times 3 4ths? That is negative 15, that is negative 15. If we divide both sides by negative 15, we get theta prime of t is equal to three over negative 15, three over negative 15, three over negative 15, which is the same thing as being equal to negative 1 5th, and the units here would be in radians per minute, because our rates are all in per minute. So, if I wanted to, I could write radians per minute."}, {"video_title": "One-sided limits from graphs Limits Differential Calculus Khan Academy.mp3", "Sentence": "So if we were to ask ourselves, what is the value of our function approaching as we approach x equals 2 from values less than x equals 2. So as you imagine, as we approach x equals 2, so x equals 1, x equals 1.5, x equals 1.9, x equals 1.999, x equals 1.99999999, what is f of x approaching? We see that f of x seems to be approaching this value, seems to be approaching this value right over here, it seems to be approaching 5. And so the way we would denote that is the limit of f of x as x approaches 2, and we're going to specify the direction, as x approaches 2 from the negative direction, we put the negative as a superscript after the 2 to denote the direction that we're approaching, this is not a negative 2, we're approaching 2 from the negative direction. We're approaching 2 from values less than 2, we're getting closer and closer to 2, but from below, 1.9, 1.99, 1.999999, as x gets closer and closer to those values, what is f of x approaching? And we see here that it is approaching 5. But what if we were asked a different question, or I guess the natural other question, what is the limit of f of x as x approaches 2 from values greater than 2?"}, {"video_title": "One-sided limits from graphs Limits Differential Calculus Khan Academy.mp3", "Sentence": "And so the way we would denote that is the limit of f of x as x approaches 2, and we're going to specify the direction, as x approaches 2 from the negative direction, we put the negative as a superscript after the 2 to denote the direction that we're approaching, this is not a negative 2, we're approaching 2 from the negative direction. We're approaching 2 from values less than 2, we're getting closer and closer to 2, but from below, 1.9, 1.99, 1.999999, as x gets closer and closer to those values, what is f of x approaching? And we see here that it is approaching 5. But what if we were asked a different question, or I guess the natural other question, what is the limit of f of x as x approaches 2 from values greater than 2? So this is a little superscript positive right over here. So now we're going to approach x equals 2, but we're going to approach it from this direction. x equals 3, x equals 2.5, x equals 2.1, x equals 2.01, x equals 2.0001, we're going to get closer and closer to 2, but we're coming from values that are larger than 2."}, {"video_title": "One-sided limits from graphs Limits Differential Calculus Khan Academy.mp3", "Sentence": "But what if we were asked a different question, or I guess the natural other question, what is the limit of f of x as x approaches 2 from values greater than 2? So this is a little superscript positive right over here. So now we're going to approach x equals 2, but we're going to approach it from this direction. x equals 3, x equals 2.5, x equals 2.1, x equals 2.01, x equals 2.0001, we're going to get closer and closer to 2, but we're coming from values that are larger than 2. So here, when x equals 3, f of x is here, when x equals 2.5, f of x is here, when x equals 2.01, f of x looks like it's right over here. So in this situation, we're getting closer and closer to f of x equaling 1. It never does quite equal that, it actually then just has a jump discontinuity, but this seems to be the limiting value when we approach x from values greater than 2."}, {"video_title": "One-sided limits from graphs Limits Differential Calculus Khan Academy.mp3", "Sentence": "x equals 3, x equals 2.5, x equals 2.1, x equals 2.01, x equals 2.0001, we're going to get closer and closer to 2, but we're coming from values that are larger than 2. So here, when x equals 3, f of x is here, when x equals 2.5, f of x is here, when x equals 2.01, f of x looks like it's right over here. So in this situation, we're getting closer and closer to f of x equaling 1. It never does quite equal that, it actually then just has a jump discontinuity, but this seems to be the limiting value when we approach x from values greater than 2. So this right over here is equal to 1. And so when we think about limits in general, the only way that a limit at 2 will actually exist is if both of these one-sided limits are actually the same thing. In this situation, they aren't."}, {"video_title": "One-sided limits from graphs Limits Differential Calculus Khan Academy.mp3", "Sentence": "It never does quite equal that, it actually then just has a jump discontinuity, but this seems to be the limiting value when we approach x from values greater than 2. So this right over here is equal to 1. And so when we think about limits in general, the only way that a limit at 2 will actually exist is if both of these one-sided limits are actually the same thing. In this situation, they aren't. As we approach 2 from values below 2, the function seems to be approaching 5, and as we approach 2 from values above 2, the function seems to be approaching 1. So in this case, the limit of f of x as x approaches 2 from the negative direction does not equal the limit of f of x as x approaches 2 from the positive direction. And since this is the case that they're not equal, the limit does not exist."}, {"video_title": "One-sided limits from graphs Limits Differential Calculus Khan Academy.mp3", "Sentence": "In this situation, they aren't. As we approach 2 from values below 2, the function seems to be approaching 5, and as we approach 2 from values above 2, the function seems to be approaching 1. So in this case, the limit of f of x as x approaches 2 from the negative direction does not equal the limit of f of x as x approaches 2 from the positive direction. And since this is the case that they're not equal, the limit does not exist. The limit as x approaches 2 in general of f of x does not exist. In order for it to have existed, these two things would have had to be equal to each other. For example, if someone were to say, what is the limit of f of x as x approaches 4?"}, {"video_title": "One-sided limits from graphs Limits Differential Calculus Khan Academy.mp3", "Sentence": "And since this is the case that they're not equal, the limit does not exist. The limit as x approaches 2 in general of f of x does not exist. In order for it to have existed, these two things would have had to be equal to each other. For example, if someone were to say, what is the limit of f of x as x approaches 4? Well, then we could think about the two one-sided limits, the one-sided limit from below and the one-sided limit from above. So we could say, well, let's see. The limit of f of x as x approaches 4 from below."}, {"video_title": "One-sided limits from graphs Limits Differential Calculus Khan Academy.mp3", "Sentence": "For example, if someone were to say, what is the limit of f of x as x approaches 4? Well, then we could think about the two one-sided limits, the one-sided limit from below and the one-sided limit from above. So we could say, well, let's see. The limit of f of x as x approaches 4 from below. So let me draw that. We care about x equals 4. As x equals 4 from below, so when x equals 3, we're here where f of 3 is negative 2. f of 3.5 seems to be right over here."}, {"video_title": "One-sided limits from graphs Limits Differential Calculus Khan Academy.mp3", "Sentence": "The limit of f of x as x approaches 4 from below. So let me draw that. We care about x equals 4. As x equals 4 from below, so when x equals 3, we're here where f of 3 is negative 2. f of 3.5 seems to be right over here. f of 3.9 seems to be right over here. f of 3.999, we're getting closer and closer to our function equaling negative 5. So this, the limit as we approach 4 from below, this one-sided limit from the left, we could say, this is going to be equal to negative 5."}, {"video_title": "One-sided limits from graphs Limits Differential Calculus Khan Academy.mp3", "Sentence": "As x equals 4 from below, so when x equals 3, we're here where f of 3 is negative 2. f of 3.5 seems to be right over here. f of 3.9 seems to be right over here. f of 3.999, we're getting closer and closer to our function equaling negative 5. So this, the limit as we approach 4 from below, this one-sided limit from the left, we could say, this is going to be equal to negative 5. And if we were to ask ourselves the limit of f of x as x approaches 4 from the right, from values larger than 4, x approaches 4 from the right, well, same exercise. f of 5 gets us here. f of 4.5 seems right around here."}, {"video_title": "One-sided limits from graphs Limits Differential Calculus Khan Academy.mp3", "Sentence": "So this, the limit as we approach 4 from below, this one-sided limit from the left, we could say, this is going to be equal to negative 5. And if we were to ask ourselves the limit of f of x as x approaches 4 from the right, from values larger than 4, x approaches 4 from the right, well, same exercise. f of 5 gets us here. f of 4.5 seems right around here. f of 4.1 seems right about here. f of 4.01 seems right around here. And even f of 4 is actually defined, but we're getting closer and closer to it."}, {"video_title": "One-sided limits from graphs Limits Differential Calculus Khan Academy.mp3", "Sentence": "f of 4.5 seems right around here. f of 4.1 seems right about here. f of 4.01 seems right around here. And even f of 4 is actually defined, but we're getting closer and closer to it. And we see once again we are approaching 5. Even if f of 4 was not defined, on either side we would be approaching 5. Sorry, we would be approaching negative 5."}, {"video_title": "One-sided limits from graphs Limits Differential Calculus Khan Academy.mp3", "Sentence": "And even f of 4 is actually defined, but we're getting closer and closer to it. And we see once again we are approaching 5. Even if f of 4 was not defined, on either side we would be approaching 5. Sorry, we would be approaching negative 5. So this is also approaching negative 5. And since the limit from the left-hand side is equal to the limit from the right-hand side, we can say, so these two things are equal, and because these two things are equal, we know that the limit of f of x as x approaches 4 is equal to 5. Let's look at a few more examples."}, {"video_title": "One-sided limits from graphs Limits Differential Calculus Khan Academy.mp3", "Sentence": "Sorry, we would be approaching negative 5. So this is also approaching negative 5. And since the limit from the left-hand side is equal to the limit from the right-hand side, we can say, so these two things are equal, and because these two things are equal, we know that the limit of f of x as x approaches 4 is equal to 5. Let's look at a few more examples. So let's ask ourselves the limit of f of x. Now this is our new f of x depicted here. As x approaches 8, and let's approach 8 from the left."}, {"video_title": "One-sided limits from graphs Limits Differential Calculus Khan Academy.mp3", "Sentence": "Let's look at a few more examples. So let's ask ourselves the limit of f of x. Now this is our new f of x depicted here. As x approaches 8, and let's approach 8 from the left. As x approaches 8 from values less than 8. So what's this going to be equal to? And I encourage you to pause the video to try to figure it out yourself."}, {"video_title": "One-sided limits from graphs Limits Differential Calculus Khan Academy.mp3", "Sentence": "As x approaches 8, and let's approach 8 from the left. As x approaches 8 from values less than 8. So what's this going to be equal to? And I encourage you to pause the video to try to figure it out yourself. So x is getting closer and closer to 8. So if x is 7, f of 7 is here. If x is 7.5, f of 7.5 is here."}, {"video_title": "One-sided limits from graphs Limits Differential Calculus Khan Academy.mp3", "Sentence": "And I encourage you to pause the video to try to figure it out yourself. So x is getting closer and closer to 8. So if x is 7, f of 7 is here. If x is 7.5, f of 7.5 is here. So it looks like our value of f of x is getting closer and closer and closer to 3. So it looks like the limit of f of x as x approaches 8 from the negative side is equal to 3. What about from the positive side?"}, {"video_title": "One-sided limits from graphs Limits Differential Calculus Khan Academy.mp3", "Sentence": "If x is 7.5, f of 7.5 is here. So it looks like our value of f of x is getting closer and closer and closer to 3. So it looks like the limit of f of x as x approaches 8 from the negative side is equal to 3. What about from the positive side? What about the limit of f of x as x approaches 8 from the positive direction, or from the right side? Well here we see as x is 9, this is our f of x, as x is 8.5, this is our f of 8.5. It seems like we're approaching f of x equaling 1."}, {"video_title": "One-sided limits from graphs Limits Differential Calculus Khan Academy.mp3", "Sentence": "What about from the positive side? What about the limit of f of x as x approaches 8 from the positive direction, or from the right side? Well here we see as x is 9, this is our f of x, as x is 8.5, this is our f of 8.5. It seems like we're approaching f of x equaling 1. So notice, these two limits are different. So the non-one-sided limit, or the two-sided limit, does not exist at f of x, or as we approach 8. So let me write that down."}, {"video_title": "One-sided limits from graphs Limits Differential Calculus Khan Academy.mp3", "Sentence": "It seems like we're approaching f of x equaling 1. So notice, these two limits are different. So the non-one-sided limit, or the two-sided limit, does not exist at f of x, or as we approach 8. So let me write that down. The limit of f of x as x approaches 8, because these two things are not the same value, this does not exist. Let's do one more example. Here they're actually asking us a question."}, {"video_title": "One-sided limits from graphs Limits Differential Calculus Khan Academy.mp3", "Sentence": "So let me write that down. The limit of f of x as x approaches 8, because these two things are not the same value, this does not exist. Let's do one more example. Here they're actually asking us a question. The function f is graphed below. What appears to be the value of the one-sided limit, the limit of f of x, this is f of x, as x approaches negative 2 from the negative direction. So this is the negative 2 from the negative direction."}, {"video_title": "One-sided limits from graphs Limits Differential Calculus Khan Academy.mp3", "Sentence": "Here they're actually asking us a question. The function f is graphed below. What appears to be the value of the one-sided limit, the limit of f of x, this is f of x, as x approaches negative 2 from the negative direction. So this is the negative 2 from the negative direction. So we care what happens as x approaches negative 2. We see f of x is actually undefined right over there. But let's see what happens as we approach from the negative direction, or as we approach from values less than negative 2, or as we approach from the left."}, {"video_title": "One-sided limits from graphs Limits Differential Calculus Khan Academy.mp3", "Sentence": "So this is the negative 2 from the negative direction. So we care what happens as x approaches negative 2. We see f of x is actually undefined right over there. But let's see what happens as we approach from the negative direction, or as we approach from values less than negative 2, or as we approach from the left. As we approach from the left, f of negative 4 is right over here. So this is f of negative 4. f of negative 3 is right over here. f of negative 3 is right over there."}, {"video_title": "One-sided limits from graphs Limits Differential Calculus Khan Academy.mp3", "Sentence": "But let's see what happens as we approach from the negative direction, or as we approach from values less than negative 2, or as we approach from the left. As we approach from the left, f of negative 4 is right over here. So this is f of negative 4. f of negative 3 is right over here. f of negative 3 is right over there. f of negative 2.5 seems to be right over here. We seem to be getting closer and closer to f of x being equal to 4, at least visually. So I would say that it looks, at least graphically, the limit of f of x as x approaches 2 from the negative direction is equal to 4."}, {"video_title": "One-sided limits from graphs Limits Differential Calculus Khan Academy.mp3", "Sentence": "f of negative 3 is right over there. f of negative 2.5 seems to be right over here. We seem to be getting closer and closer to f of x being equal to 4, at least visually. So I would say that it looks, at least graphically, the limit of f of x as x approaches 2 from the negative direction is equal to 4. Now, if we also ask ourselves the limit of f of x as x approaches negative 2 from the positive direction, we would get a similar result. Now we're going to approach from when x is 0, f of x seems to be right over here. When x is 1, f of x is right over here."}, {"video_title": "One-sided limits from graphs Limits Differential Calculus Khan Academy.mp3", "Sentence": "So I would say that it looks, at least graphically, the limit of f of x as x approaches 2 from the negative direction is equal to 4. Now, if we also ask ourselves the limit of f of x as x approaches negative 2 from the positive direction, we would get a similar result. Now we're going to approach from when x is 0, f of x seems to be right over here. When x is 1, f of x is right over here. When x is negative 1.99, so sorry, this was when x is negative 1, f of x is there. When x is negative 1.9, f of x seems to be right over here. So once again, we seem to be getting closer and closer to 4."}, {"video_title": "Limits of piecewise functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Pause this video and see if you can figure out what these various limits would be. Some of them are one-sided, and some of them are regular limits or two-sided limits. All right, let's start with this first one. The limit as x approaches four from values larger than equaling four. So that's what that plus tells us. And so when x is greater than four, our f of x is equal to square root of x. So as we are approaching four from the right, we are really thinking about this part of the function."}, {"video_title": "Limits of piecewise functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "The limit as x approaches four from values larger than equaling four. So that's what that plus tells us. And so when x is greater than four, our f of x is equal to square root of x. So as we are approaching four from the right, we are really thinking about this part of the function. And so this is going to be equal to the square root of four. Even though right at four, our f of x is equal to this, we are approaching from values greater than four. We're approaching from the right."}, {"video_title": "Limits of piecewise functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So as we are approaching four from the right, we are really thinking about this part of the function. And so this is going to be equal to the square root of four. Even though right at four, our f of x is equal to this, we are approaching from values greater than four. We're approaching from the right. So we would use this part of our function definition. And so this is going to be equal to two. Now what about our limit of f of x as we approach four from the left?"}, {"video_title": "Limits of piecewise functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "We're approaching from the right. So we would use this part of our function definition. And so this is going to be equal to two. Now what about our limit of f of x as we approach four from the left? Well, then we would use this part of our function definition. And so this is going to be equal to four plus two over four minus one, which is equal to six over three, which is equal to two. And so if we wanna say what is the limit of f of x as x approaches four?"}, {"video_title": "Limits of piecewise functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Now what about our limit of f of x as we approach four from the left? Well, then we would use this part of our function definition. And so this is going to be equal to four plus two over four minus one, which is equal to six over three, which is equal to two. And so if we wanna say what is the limit of f of x as x approaches four? Well, this is a good scenario here because from both the left and the right, as we approach x equals four, we're approaching the same value. And we know that in order for the two-sided limit to have a limit, you have to be approaching the same thing from the right and the left. And we are, and so this is going to be equal to two."}, {"video_title": "Limits of piecewise functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And so if we wanna say what is the limit of f of x as x approaches four? Well, this is a good scenario here because from both the left and the right, as we approach x equals four, we're approaching the same value. And we know that in order for the two-sided limit to have a limit, you have to be approaching the same thing from the right and the left. And we are, and so this is going to be equal to two. Now what's the limit as x approaches two of f of x? Well, as x approaches two, we are going to be completely in this scenario right over here. Now interesting things do happen at x equals one here."}, {"video_title": "Limits of piecewise functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And we are, and so this is going to be equal to two. Now what's the limit as x approaches two of f of x? Well, as x approaches two, we are going to be completely in this scenario right over here. Now interesting things do happen at x equals one here. Our denominator goes to zero. But at x equals two, this part of the curve is gonna be continuous. So we can just substitute the value."}, {"video_title": "Limits of piecewise functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Now interesting things do happen at x equals one here. Our denominator goes to zero. But at x equals two, this part of the curve is gonna be continuous. So we can just substitute the value. It's going to be two plus two over two minus one, which is four over one, which is equal to four. Let's do another example. So we have another piecewise function."}, {"video_title": "Limits of piecewise functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So we can just substitute the value. It's going to be two plus two over two minus one, which is four over one, which is equal to four. Let's do another example. So we have another piecewise function. And so let's pause our video and figure out these things. All right, now let's do this together. So what's the limit as x approaches negative one from the right?"}, {"video_title": "Limits of piecewise functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So we have another piecewise function. And so let's pause our video and figure out these things. All right, now let's do this together. So what's the limit as x approaches negative one from the right? So if we're approaching from the right, when we are greater than or equal to negative one, we are in this part of our piecewise function. And so we would say this is going to approach, this is gonna be two to the negative one power, which is equal to 1 1\u20442. What about if we're approaching from the left?"}, {"video_title": "Limits of piecewise functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So what's the limit as x approaches negative one from the right? So if we're approaching from the right, when we are greater than or equal to negative one, we are in this part of our piecewise function. And so we would say this is going to approach, this is gonna be two to the negative one power, which is equal to 1 1\u20442. What about if we're approaching from the left? Well, if we're approaching from the left, we're in this scenario right over here. We're to the left of x equals negative one. And so this is going to be equal to the sine, because we're in this case for our piecewise function, of negative one plus one, which is a sine of zero, which is equal to zero."}, {"video_title": "Limits of piecewise functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "What about if we're approaching from the left? Well, if we're approaching from the left, we're in this scenario right over here. We're to the left of x equals negative one. And so this is going to be equal to the sine, because we're in this case for our piecewise function, of negative one plus one, which is a sine of zero, which is equal to zero. Now what's the two-sided limit as x approaches negative one of g of x? Well, we're approaching two different values as we approach from the right and as we approach from the left. And if our one-sided limits aren't approaching the same value, well, then this limit does not exist."}, {"video_title": "Limits of piecewise functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And so this is going to be equal to the sine, because we're in this case for our piecewise function, of negative one plus one, which is a sine of zero, which is equal to zero. Now what's the two-sided limit as x approaches negative one of g of x? Well, we're approaching two different values as we approach from the right and as we approach from the left. And if our one-sided limits aren't approaching the same value, well, then this limit does not exist. Does not exist. And what's the limit of g of x as x approaches zero from the right? Well, if we're talking about approaching zero from the right, we are going to be in this case right over here."}, {"video_title": "Limits of piecewise functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And if our one-sided limits aren't approaching the same value, well, then this limit does not exist. Does not exist. And what's the limit of g of x as x approaches zero from the right? Well, if we're talking about approaching zero from the right, we are going to be in this case right over here. Zero is definitely in this interval. And over this interval, this right over here is going to be continuous. And so we can just substitute x equals zero there."}, {"video_title": "Area between a curve and the x-axis AP Calculus AB Khan Academy.mp3", "Sentence": "So we've got the function f of x is equal to x squared. And what I'm concerned with is finding the area under the curve y is equal to f of x, so that's my y axis. This is my x axis. And let me draw my function. My function looks like this, at least in the first quadrant. That's where I'll graph it for now. I could also graph it, obviously, in the second quadrant."}, {"video_title": "Area between a curve and the x-axis AP Calculus AB Khan Academy.mp3", "Sentence": "And let me draw my function. My function looks like this, at least in the first quadrant. That's where I'll graph it for now. I could also graph it, obviously, in the second quadrant. But what I care about is the area under this curve and above the positive x axis between x equals 1 and x equals 4. And I'm tired of approximating areas. I want to find the exact area under this curve, above the x axis."}, {"video_title": "Area between a curve and the x-axis AP Calculus AB Khan Academy.mp3", "Sentence": "I could also graph it, obviously, in the second quadrant. But what I care about is the area under this curve and above the positive x axis between x equals 1 and x equals 4. And I'm tired of approximating areas. I want to find the exact area under this curve, above the x axis. And the way we denote the exact area under the curve, this little brown shaded area, is using the definite integral. It's a definite integral from 1 to 4 of f of x dx. And the way I conceptualize where this notation comes from is we imagine an infinite number of infinitely thin rectangles that we sum up to find this area."}, {"video_title": "Area between a curve and the x-axis AP Calculus AB Khan Academy.mp3", "Sentence": "I want to find the exact area under this curve, above the x axis. And the way we denote the exact area under the curve, this little brown shaded area, is using the definite integral. It's a definite integral from 1 to 4 of f of x dx. And the way I conceptualize where this notation comes from is we imagine an infinite number of infinitely thin rectangles that we sum up to find this area. And let me draw one of those infinitely thin rectangles, maybe not so infinitely thin. So let me draw it like this. So that would be one of the rectangles."}, {"video_title": "Area between a curve and the x-axis AP Calculus AB Khan Academy.mp3", "Sentence": "And the way I conceptualize where this notation comes from is we imagine an infinite number of infinitely thin rectangles that we sum up to find this area. And let me draw one of those infinitely thin rectangles, maybe not so infinitely thin. So let me draw it like this. So that would be one of the rectangles. That would be another rectangle. This should be reminiscent of a Riemann sum. In fact, that's where the Riemann integral comes from, taking a Riemann sum where you have an infinite number of these rectangles where the width of each of the rectangles, this is how I conceptualize it, is dx."}, {"video_title": "Area between a curve and the x-axis AP Calculus AB Khan Academy.mp3", "Sentence": "So that would be one of the rectangles. That would be another rectangle. This should be reminiscent of a Riemann sum. In fact, that's where the Riemann integral comes from, taking a Riemann sum where you have an infinite number of these rectangles where the width of each of the rectangles, this is how I conceptualize it, is dx. And the height of this rectangle is the function evaluated at an x that's within this interval right over here. And so this part right over here is the area of one of those rectangles and we're summing them all up. And this is kind of an elongated s reminiscent of a sigma for summing."}, {"video_title": "Area between a curve and the x-axis AP Calculus AB Khan Academy.mp3", "Sentence": "In fact, that's where the Riemann integral comes from, taking a Riemann sum where you have an infinite number of these rectangles where the width of each of the rectangles, this is how I conceptualize it, is dx. And the height of this rectangle is the function evaluated at an x that's within this interval right over here. And so this part right over here is the area of one of those rectangles and we're summing them all up. And this is kind of an elongated s reminiscent of a sigma for summing. We're summing up the infinite number of those infinitely thin rectangles or the areas of those infinitely thin rectangles between 1 and 4. So that's where the notation of the definite integral comes from. But we still haven't done anything."}, {"video_title": "Area between a curve and the x-axis AP Calculus AB Khan Academy.mp3", "Sentence": "And this is kind of an elongated s reminiscent of a sigma for summing. We're summing up the infinite number of those infinitely thin rectangles or the areas of those infinitely thin rectangles between 1 and 4. So that's where the notation of the definite integral comes from. But we still haven't done anything. We've just written some notation that says the exact area between 1 and 4 under the curve f of x and above the x-axis. In order to actually do anything really productive with this, we have to turn to the second fundamental theorem of calculus, sometimes called part 2 of the fundamental theorem of calculus, which tells us that if f has an antiderivative, so if we have the antiderivative of f, so f of x is derivative of some function capital F of x, or another way of saying it is capital F of x is the antiderivative of lowercase f of x, then I can evaluate this thing, and we do a whole video on conceptually understanding why this makes sense, we can evaluate this by evaluating the antiderivative of f or an antiderivative of f at 4 and from that subtract the antiderivative evaluated at 1. So let's do it for this particular case right over here."}, {"video_title": "Area between a curve and the x-axis AP Calculus AB Khan Academy.mp3", "Sentence": "But we still haven't done anything. We've just written some notation that says the exact area between 1 and 4 under the curve f of x and above the x-axis. In order to actually do anything really productive with this, we have to turn to the second fundamental theorem of calculus, sometimes called part 2 of the fundamental theorem of calculus, which tells us that if f has an antiderivative, so if we have the antiderivative of f, so f of x is derivative of some function capital F of x, or another way of saying it is capital F of x is the antiderivative of lowercase f of x, then I can evaluate this thing, and we do a whole video on conceptually understanding why this makes sense, we can evaluate this by evaluating the antiderivative of f or an antiderivative of f at 4 and from that subtract the antiderivative evaluated at 1. So let's do it for this particular case right over here. So we are taking, I'll just rewrite this statement, instead of writing f of x I'll write x squared, so the definite integral from 1 to 4 of x squared dx, well, we're just going to have to figure out what the antiderivative is. So if f of x is equal to x squared, what is capital F of x equal to? What is the antiderivative?"}, {"video_title": "Area between a curve and the x-axis AP Calculus AB Khan Academy.mp3", "Sentence": "So let's do it for this particular case right over here. So we are taking, I'll just rewrite this statement, instead of writing f of x I'll write x squared, so the definite integral from 1 to 4 of x squared dx, well, we're just going to have to figure out what the antiderivative is. So if f of x is equal to x squared, what is capital F of x equal to? What is the antiderivative? Well, you might remember from your power rules that if you take the derivative with respect to x of x to the third, you are going to get 3x squared, which is pretty darn close to x squared except for this factor of 3, so let's divide both sides by 3. And you get the derivative of x to the third divided by 3 is indeed x squared, or you could say that this is the same thing as the derivative with respect to x of x to the third over 3. So what is the derivative of this?"}, {"video_title": "Area between a curve and the x-axis AP Calculus AB Khan Academy.mp3", "Sentence": "What is the antiderivative? Well, you might remember from your power rules that if you take the derivative with respect to x of x to the third, you are going to get 3x squared, which is pretty darn close to x squared except for this factor of 3, so let's divide both sides by 3. And you get the derivative of x to the third divided by 3 is indeed x squared, or you could say that this is the same thing as the derivative with respect to x of x to the third over 3. So what is the derivative of this? It will be 3 times 1 third, and then you'll decrement the power, it will just be x squared. So this right over here, once again, is x squared. It's just equal to x squared."}, {"video_title": "Area between a curve and the x-axis AP Calculus AB Khan Academy.mp3", "Sentence": "So what is the derivative of this? It will be 3 times 1 third, and then you'll decrement the power, it will just be x squared. So this right over here, once again, is x squared. It's just equal to x squared. So in this case, our capital F of x, our antiderivative, is x to the third over 3. And so we just have to evaluate that at 4 and at 1. And sometimes the notation we'll use is, we'll say that the antiderivative is x to the third over 3, and we're going to evaluate it, the one I always like to just write the numbers up here, at 4 and from that subtract it, evaluate it at 1."}, {"video_title": "Area between a curve and the x-axis AP Calculus AB Khan Academy.mp3", "Sentence": "It's just equal to x squared. So in this case, our capital F of x, our antiderivative, is x to the third over 3. And so we just have to evaluate that at 4 and at 1. And sometimes the notation we'll use is, we'll say that the antiderivative is x to the third over 3, and we're going to evaluate it, the one I always like to just write the numbers up here, at 4 and from that subtract it, evaluate it at 1. Sometimes you'll see people write a little line here too, where they'll say we're going to evaluate it at 4 and then at 1, but I'll just do it without the line. So we're going to evaluate this thing at 4, and from that subtract it, evaluate it at 1. So this is going to be equal to 4 to the third power of 64, so it's going to be 64 over 3."}, {"video_title": "Area between a curve and the x-axis AP Calculus AB Khan Academy.mp3", "Sentence": "And sometimes the notation we'll use is, we'll say that the antiderivative is x to the third over 3, and we're going to evaluate it, the one I always like to just write the numbers up here, at 4 and from that subtract it, evaluate it at 1. Sometimes you'll see people write a little line here too, where they'll say we're going to evaluate it at 4 and then at 1, but I'll just do it without the line. So we're going to evaluate this thing at 4, and from that subtract it, evaluate it at 1. So this is going to be equal to 4 to the third power of 64, so it's going to be 64 over 3. Let me color code it. This right over here is this right over there, and then from that we're going to subtract this business evaluated at 1. Well, when you evaluate it at 1, you get 1 to the third is 1 over 3."}, {"video_title": "Area between a curve and the x-axis AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be equal to 4 to the third power of 64, so it's going to be 64 over 3. Let me color code it. This right over here is this right over there, and then from that we're going to subtract this business evaluated at 1. Well, when you evaluate it at 1, you get 1 to the third is 1 over 3. You get 1 third. So just to be clear, this is this right over there. And then we are ready to just subtract these fractions, 64 over 3 minus 1 third is equal to 63 over 3, and 3 goes into 63 exactly 21 times."}, {"video_title": "Interpreting definite integral as net change AP Calculus AB Khan Academy.mp3", "Sentence": "So for example, this rate curve, this might represent the speed of a car and how the speed of a car is changing with respect to time. And so this shows us that our rate is actually changing. This isn't distance as a function of time, this is rate as a function of time. So this looks like the car is accelerating. At time one, it is going 10 meters per second. And at time five, let's assume that all of these are in seconds. So at five seconds, it is going 20 meters per second."}, {"video_title": "Interpreting definite integral as net change AP Calculus AB Khan Academy.mp3", "Sentence": "So this looks like the car is accelerating. At time one, it is going 10 meters per second. And at time five, let's assume that all of these are in seconds. So at five seconds, it is going 20 meters per second. So it is accelerating. Now, the relationship between the rate function and the area is that if we're able to figure out this area, then that is the change in distance of the car. So rate or speed in this case is distance per unit time."}, {"video_title": "Interpreting definite integral as net change AP Calculus AB Khan Academy.mp3", "Sentence": "So at five seconds, it is going 20 meters per second. So it is accelerating. Now, the relationship between the rate function and the area is that if we're able to figure out this area, then that is the change in distance of the car. So rate or speed in this case is distance per unit time. If we're able to figure out the area under that curve, it will actually give us our change in distance from time one to time five. It won't tell us our total distance because we won't know what happened before time one if we're not concerned with that area. And the intuition for that, it's a little bit easier if you were dealing with rectangles."}, {"video_title": "Interpreting definite integral as net change AP Calculus AB Khan Academy.mp3", "Sentence": "So rate or speed in this case is distance per unit time. If we're able to figure out the area under that curve, it will actually give us our change in distance from time one to time five. It won't tell us our total distance because we won't know what happened before time one if we're not concerned with that area. And the intuition for that, it's a little bit easier if you were dealing with rectangles. But just think about this. Let's make a rectangle that looks like a pretty good approximation for the area, let's say from time one to time two right over here. Well, what does this area of this rectangle represent?"}, {"video_title": "Interpreting definite integral as net change AP Calculus AB Khan Academy.mp3", "Sentence": "And the intuition for that, it's a little bit easier if you were dealing with rectangles. But just think about this. Let's make a rectangle that looks like a pretty good approximation for the area, let's say from time one to time two right over here. Well, what does this area of this rectangle represent? We would, to figure out the area, we would multiply one second, that would be the width here, times roughly, it looks like about 10 meters per second. And so the units here would be 10 meters per second times one second, or 10 meters. And we know from early physics or even before that if you multiply a rate times time or a speed times a time, you're gonna get a distance."}, {"video_title": "Interpreting definite integral as net change AP Calculus AB Khan Academy.mp3", "Sentence": "Well, what does this area of this rectangle represent? We would, to figure out the area, we would multiply one second, that would be the width here, times roughly, it looks like about 10 meters per second. And so the units here would be 10 meters per second times one second, or 10 meters. And we know from early physics or even before that if you multiply a rate times time or a speed times a time, you're gonna get a distance. And so the unit here is in distance. And as you can see, this area is going to represent, it's gonna be an approximation for the distance traveled. And so if you wanted to get an exact version or an exact number for the distance traveled, you would get the exact area under the curve."}, {"video_title": "Interpreting definite integral as net change AP Calculus AB Khan Academy.mp3", "Sentence": "And we know from early physics or even before that if you multiply a rate times time or a speed times a time, you're gonna get a distance. And so the unit here is in distance. And as you can see, this area is going to represent, it's gonna be an approximation for the distance traveled. And so if you wanted to get an exact version or an exact number for the distance traveled, you would get the exact area under the curve. And we have a notation for that. If you want the exact area under the curve right over here, we use definite integral notation. This area right over here, we could denote as a definite integral from one to five of r of t dt."}, {"video_title": "Interpreting definite integral as net change AP Calculus AB Khan Academy.mp3", "Sentence": "And so if you wanted to get an exact version or an exact number for the distance traveled, you would get the exact area under the curve. And we have a notation for that. If you want the exact area under the curve right over here, we use definite integral notation. This area right over here, we could denote as a definite integral from one to five of r of t dt. And once again, what does this represent? In this case, when our rate is speed, this represents, this whole expression represents our change in distance from t is equal to one to t is equal to five. Now with that context, let's actually try to do an example problem, the type that you might see on Khan Academy."}, {"video_title": "Interpreting definite integral as net change AP Calculus AB Khan Academy.mp3", "Sentence": "This area right over here, we could denote as a definite integral from one to five of r of t dt. And once again, what does this represent? In this case, when our rate is speed, this represents, this whole expression represents our change in distance from t is equal to one to t is equal to five. Now with that context, let's actually try to do an example problem, the type that you might see on Khan Academy. So this right over here tells us, Eden walked at a rate of r of t kilometers per hour, where t is the time in hours. Okay, so now t is in hours. What does the integral from the definite integral from two to three of r of t dt equals six mean?"}, {"video_title": "Interpreting definite integral as net change AP Calculus AB Khan Academy.mp3", "Sentence": "Now with that context, let's actually try to do an example problem, the type that you might see on Khan Academy. So this right over here tells us, Eden walked at a rate of r of t kilometers per hour, where t is the time in hours. Okay, so now t is in hours. What does the integral from the definite integral from two to three of r of t dt equals six mean? So before I even look at these choices, this is saying, so this is going from t equals two hours to t equals three hours, and it's essentially the area under the rate curve. And here the rate is, we're talking about a speed. Eden is walking at a certain number of kilometers per hour."}, {"video_title": "Interpreting definite integral as net change AP Calculus AB Khan Academy.mp3", "Sentence": "What does the integral from the definite integral from two to three of r of t dt equals six mean? So before I even look at these choices, this is saying, so this is going from t equals two hours to t equals three hours, and it's essentially the area under the rate curve. And here the rate is, we're talking about a speed. Eden is walking at a certain number of kilometers per hour. So what this means is that from time two hours to time three hours, Eden walked an extra six kilometers. So let's see which of these choices match that. Eden walked six kilometers each hour."}, {"video_title": "Interpreting definite integral as net change AP Calculus AB Khan Academy.mp3", "Sentence": "Eden is walking at a certain number of kilometers per hour. So what this means is that from time two hours to time three hours, Eden walked an extra six kilometers. So let's see which of these choices match that. Eden walked six kilometers each hour. It does tell us that from time two to three, Eden walked six kilometers, but doesn't mean, but we don't know what happened from time zero to time one or from time one to time two. So I would rule this out. Eden walked six kilometers in three hours."}, {"video_title": "Interpreting definite integral as net change AP Calculus AB Khan Academy.mp3", "Sentence": "Eden walked six kilometers each hour. It does tell us that from time two to three, Eden walked six kilometers, but doesn't mean, but we don't know what happened from time zero to time one or from time one to time two. So I would rule this out. Eden walked six kilometers in three hours. So this is a common misconception. People will look at the top bound and say, okay, this area represented by the definite integral, this tells us how far in total we have walked up until that point. That is not what this represents."}, {"video_title": "Interpreting definite integral as net change AP Calculus AB Khan Academy.mp3", "Sentence": "Eden walked six kilometers in three hours. So this is a common misconception. People will look at the top bound and say, okay, this area represented by the definite integral, this tells us how far in total we have walked up until that point. That is not what this represents. This represents the change in distance from time two to time three. So I'll rule that out. Eden walked six kilometers during the third hour."}, {"video_title": "Interpreting definite integral as net change AP Calculus AB Khan Academy.mp3", "Sentence": "That is not what this represents. This represents the change in distance from time two to time three. So I'll rule that out. Eden walked six kilometers during the third hour. Yes, that's what we've been talking about. From time equal two hours to time equal three hours, Eden walked six kilometers, and you could view that as the third hour going from time two to time three. Eden's rate increased by six kilometers per hour between hours two and three."}, {"video_title": "Interpreting definite integral as net change AP Calculus AB Khan Academy.mp3", "Sentence": "Eden walked six kilometers during the third hour. Yes, that's what we've been talking about. From time equal two hours to time equal three hours, Eden walked six kilometers, and you could view that as the third hour going from time two to time three. Eden's rate increased by six kilometers per hour between hours two and three. So let's be very clear. This right over here, this isn't a rate. This is the area under the rate curve."}, {"video_title": "Interpreting definite integral as net change AP Calculus AB Khan Academy.mp3", "Sentence": "Eden's rate increased by six kilometers per hour between hours two and three. So let's be very clear. This right over here, this isn't a rate. This is the area under the rate curve. That's what this definite integral is representing. And so this isn't telling us about our rate changing. This is telling us how does the thing that the rate is measuring the change of, how does that change from time two to time three?"}, {"video_title": "Motion problems with integrals displacement vs. distance AP Calculus AB Khan Academy.mp3", "Sentence": "And to get our bearings there, I'm going to introduce a few ideas. So the first idea is that of displacement. So you might use that word in everyday language, and it literally means your change in position. Your change in position. Now a related idea that sometimes gets confused with displacement is the notion of distance traveled. And you might say, well, isn't that just the same thing as change in position? And you will see shortly, no, it isn't always the same thing."}, {"video_title": "Motion problems with integrals displacement vs. distance AP Calculus AB Khan Academy.mp3", "Sentence": "Your change in position. Now a related idea that sometimes gets confused with displacement is the notion of distance traveled. And you might say, well, isn't that just the same thing as change in position? And you will see shortly, no, it isn't always the same thing. The distance traveled, this is the total length of path. Total length of path. So what are we talking about?"}, {"video_title": "Motion problems with integrals displacement vs. distance AP Calculus AB Khan Academy.mp3", "Sentence": "And you will see shortly, no, it isn't always the same thing. The distance traveled, this is the total length of path. Total length of path. So what are we talking about? Well, let's say, and we're going to introduce a little bit of calculus now. Let's say that we have a particle's velocity function. And so let's say our velocity as a function of time is equal to five minus t. Now this is a one-dimensional velocity function."}, {"video_title": "Motion problems with integrals displacement vs. distance AP Calculus AB Khan Academy.mp3", "Sentence": "So what are we talking about? Well, let's say, and we're going to introduce a little bit of calculus now. Let's say that we have a particle's velocity function. And so let's say our velocity as a function of time is equal to five minus t. Now this is a one-dimensional velocity function. Let's say it's just telling us our velocity in the horizontal direction. And oftentimes when something's one dimension, people forget, well, that too can be a vector quantity. In fact, this velocity is a vector quantity because you could think of it, if it's positive, it's moving to the right, and if it's negative, it's moving to the left."}, {"video_title": "Motion problems with integrals displacement vs. distance AP Calculus AB Khan Academy.mp3", "Sentence": "And so let's say our velocity as a function of time is equal to five minus t. Now this is a one-dimensional velocity function. Let's say it's just telling us our velocity in the horizontal direction. And oftentimes when something's one dimension, people forget, well, that too can be a vector quantity. In fact, this velocity is a vector quantity because you could think of it, if it's positive, it's moving to the right, and if it's negative, it's moving to the left. So it has a direction. And so sometimes you will see a vector quantity like this have a little arrow on it. Or you will see it bolded, or you will see it bolded like that."}, {"video_title": "Motion problems with integrals displacement vs. distance AP Calculus AB Khan Academy.mp3", "Sentence": "In fact, this velocity is a vector quantity because you could think of it, if it's positive, it's moving to the right, and if it's negative, it's moving to the left. So it has a direction. And so sometimes you will see a vector quantity like this have a little arrow on it. Or you will see it bolded, or you will see it bolded like that. I like to write an arrow in it, although that's not always the convention used in different classes. Now let's plot what this velocity function actually looks like. And I did that ahead of time."}, {"video_title": "Motion problems with integrals displacement vs. distance AP Calculus AB Khan Academy.mp3", "Sentence": "Or you will see it bolded, or you will see it bolded like that. I like to write an arrow in it, although that's not always the convention used in different classes. Now let's plot what this velocity function actually looks like. And I did that ahead of time. So you can see here, time equals zero. Let's say time is in seconds. And our velocity is in meters per second."}, {"video_title": "Motion problems with integrals displacement vs. distance AP Calculus AB Khan Academy.mp3", "Sentence": "And I did that ahead of time. So you can see here, time equals zero. Let's say time is in seconds. And our velocity is in meters per second. So this is meters per second right over here, and this is seconds in this axis. At exactly time zero, this object is traveling at five meters per second. And we could say to the right."}, {"video_title": "Motion problems with integrals displacement vs. distance AP Calculus AB Khan Academy.mp3", "Sentence": "And our velocity is in meters per second. So this is meters per second right over here, and this is seconds in this axis. At exactly time zero, this object is traveling at five meters per second. And we could say to the right. It has a velocity of positive five meters per second. But then it keeps decelerating at a constant rate. So five seconds into it, right at five seconds, the particle has no velocity, and then it starts having negative velocity, which you could interpret as moving to the left."}, {"video_title": "Motion problems with integrals displacement vs. distance AP Calculus AB Khan Academy.mp3", "Sentence": "And we could say to the right. It has a velocity of positive five meters per second. But then it keeps decelerating at a constant rate. So five seconds into it, right at five seconds, the particle has no velocity, and then it starts having negative velocity, which you could interpret as moving to the left. So let's think about a few things. First, let's think about what is the displacement over the first five seconds? Over first five seconds."}, {"video_title": "Motion problems with integrals displacement vs. distance AP Calculus AB Khan Academy.mp3", "Sentence": "So five seconds into it, right at five seconds, the particle has no velocity, and then it starts having negative velocity, which you could interpret as moving to the left. So let's think about a few things. First, let's think about what is the displacement over the first five seconds? Over first five seconds. Well, we've seen already multiple times. If you wanna find the change in quantity, you can take the integral of the rate function of it. And so velocity is actually the rate of displacement is one way to think about it."}, {"video_title": "Motion problems with integrals displacement vs. distance AP Calculus AB Khan Academy.mp3", "Sentence": "Over first five seconds. Well, we've seen already multiple times. If you wanna find the change in quantity, you can take the integral of the rate function of it. And so velocity is actually the rate of displacement is one way to think about it. So displacement over the first five seconds, we could take the integral from zero to five, zero to five, of our velocity function, of our velocity function, just like that. And we can even calculate this really fast. That would just be this area right over here, which we could just use a little bit of geometry."}, {"video_title": "Motion problems with integrals displacement vs. distance AP Calculus AB Khan Academy.mp3", "Sentence": "And so velocity is actually the rate of displacement is one way to think about it. So displacement over the first five seconds, we could take the integral from zero to five, zero to five, of our velocity function, of our velocity function, just like that. And we can even calculate this really fast. That would just be this area right over here, which we could just use a little bit of geometry. This is a five by five triangle. So five times five is 25, times 1 1\u20442, remember, area of a triangle's 1 1\u20442 base times height. So this is going to be 12.5, and let's see, this is going to be meters per second times seconds, so 12.5 meters."}, {"video_title": "Motion problems with integrals displacement vs. distance AP Calculus AB Khan Academy.mp3", "Sentence": "That would just be this area right over here, which we could just use a little bit of geometry. This is a five by five triangle. So five times five is 25, times 1 1\u20442, remember, area of a triangle's 1 1\u20442 base times height. So this is going to be 12.5, and let's see, this is going to be meters per second times seconds, so 12.5 meters. So that's the change in position for that particle over the first five seconds. Wherever it started, it's now going to be 12.5 meters to the right of it, assuming that positive is to the right. Now what about over the first 10 seconds?"}, {"video_title": "Motion problems with integrals displacement vs. distance AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be 12.5, and let's see, this is going to be meters per second times seconds, so 12.5 meters. So that's the change in position for that particle over the first five seconds. Wherever it started, it's now going to be 12.5 meters to the right of it, assuming that positive is to the right. Now what about over the first 10 seconds? Now this gets interesting, and I encourage you to pause your video and think about it. What would be the displacement over the first 10 seconds? Well, we would just do the same thing, the integral from zero to 10 of our velocity function, our one-dimensional velocity function, dt."}, {"video_title": "Motion problems with integrals displacement vs. distance AP Calculus AB Khan Academy.mp3", "Sentence": "Now what about over the first 10 seconds? Now this gets interesting, and I encourage you to pause your video and think about it. What would be the displacement over the first 10 seconds? Well, we would just do the same thing, the integral from zero to 10 of our velocity function, our one-dimensional velocity function, dt. And so that would be the area from here all the way to right over there, so this entire area. But you might appreciate, when you're taking a definite integral, if we are below the t-axis and above the function like this, this is going to be negative area. And in fact, this area and this area are going to exactly cancel out, and you're going to get zero meters."}, {"video_title": "Motion problems with integrals displacement vs. distance AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we would just do the same thing, the integral from zero to 10 of our velocity function, our one-dimensional velocity function, dt. And so that would be the area from here all the way to right over there, so this entire area. But you might appreciate, when you're taking a definite integral, if we are below the t-axis and above the function like this, this is going to be negative area. And in fact, this area and this area are going to exactly cancel out, and you're going to get zero meters. Now you might be saying, how can that be? After 10 seconds, how do we, why is our displacement only zero meters? This particle's been moving the entire time."}, {"video_title": "Motion problems with integrals displacement vs. distance AP Calculus AB Khan Academy.mp3", "Sentence": "And in fact, this area and this area are going to exactly cancel out, and you're going to get zero meters. Now you might be saying, how can that be? After 10 seconds, how do we, why is our displacement only zero meters? This particle's been moving the entire time. Well, remember what's going on. The first five seconds, it's moving to the right, it's decelerating the whole time, and then right at five seconds, it has gone 12.5 meters to the right. But then it starts, its velocity starts becoming negative, and the particle starts moving to the left."}, {"video_title": "Motion problems with integrals displacement vs. distance AP Calculus AB Khan Academy.mp3", "Sentence": "This particle's been moving the entire time. Well, remember what's going on. The first five seconds, it's moving to the right, it's decelerating the whole time, and then right at five seconds, it has gone 12.5 meters to the right. But then it starts, its velocity starts becoming negative, and the particle starts moving to the left. And so over the next five seconds, it actually moves 12.5 meters to the left, and then these two things net out. And so the particle has gone over 10 seconds, 12.5 meters to the right, and then 12.5 meters to the left, and so its change in position is zero meters. It has not changed."}, {"video_title": "Motion problems with integrals displacement vs. distance AP Calculus AB Khan Academy.mp3", "Sentence": "But then it starts, its velocity starts becoming negative, and the particle starts moving to the left. And so over the next five seconds, it actually moves 12.5 meters to the left, and then these two things net out. And so the particle has gone over 10 seconds, 12.5 meters to the right, and then 12.5 meters to the left, and so its change in position is zero meters. It has not changed. Now you might start to be appreciating what the difference between displacement and distance traveled is. So distance, if you're talking about your total length of path, you don't care as much about direction. And so instead of thinking about velocity, what we would do is think about speed."}, {"video_title": "Motion problems with integrals displacement vs. distance AP Calculus AB Khan Academy.mp3", "Sentence": "It has not changed. Now you might start to be appreciating what the difference between displacement and distance traveled is. So distance, if you're talking about your total length of path, you don't care as much about direction. And so instead of thinking about velocity, what we would do is think about speed. And speed is, you could view in this case, especially in this one-dimensional case, this is equal to the absolute value of velocity. Later on, when we do multiple dimensions, it would be the magnitude of the velocity function, which is what the absolute value function does in one dimension. So what would this look like if we plotted it?"}, {"video_title": "Motion problems with integrals displacement vs. distance AP Calculus AB Khan Academy.mp3", "Sentence": "And so instead of thinking about velocity, what we would do is think about speed. And speed is, you could view in this case, especially in this one-dimensional case, this is equal to the absolute value of velocity. Later on, when we do multiple dimensions, it would be the magnitude of the velocity function, which is what the absolute value function does in one dimension. So what would this look like if we plotted it? Well, the absolute value of the velocity function would just look like that. And so if you want the distance, you would find the, the distance traveled, I should say, you would find the integral over the appropriate change in time of the speed function, right over here, which we have graphed. So notice, if we want the distance traveled, so I'll just say, I'll write it out, distance traveled over first five seconds, first five seconds, first five seconds, what would it be?"}, {"video_title": "Motion problems with integrals displacement vs. distance AP Calculus AB Khan Academy.mp3", "Sentence": "So what would this look like if we plotted it? Well, the absolute value of the velocity function would just look like that. And so if you want the distance, you would find the, the distance traveled, I should say, you would find the integral over the appropriate change in time of the speed function, right over here, which we have graphed. So notice, if we want the distance traveled, so I'll just say, I'll write it out, distance traveled over first five seconds, first five seconds, first five seconds, what would it be? Well, it would be the integral from zero to five of the absolute value of our velocity function, which is, you could just view it as our speed function right over here, dt. And so it would be this area, which we already know to be 12.5 meters. So for the first five seconds, your distance and displacement are consistent."}, {"video_title": "Motion problems with integrals displacement vs. distance AP Calculus AB Khan Academy.mp3", "Sentence": "So notice, if we want the distance traveled, so I'll just say, I'll write it out, distance traveled over first five seconds, first five seconds, first five seconds, what would it be? Well, it would be the integral from zero to five of the absolute value of our velocity function, which is, you could just view it as our speed function right over here, dt. And so it would be this area, which we already know to be 12.5 meters. So for the first five seconds, your distance and displacement are consistent. Well, that's because you have, in this case, the velocity function is positive, so the absolute value of it is still going to be positive. But if you think about over the first 10 seconds, your distance, 10 seconds, what is it going to be? Pause the video and try to think about it."}, {"video_title": "Motion problems with integrals displacement vs. distance AP Calculus AB Khan Academy.mp3", "Sentence": "So for the first five seconds, your distance and displacement are consistent. Well, that's because you have, in this case, the velocity function is positive, so the absolute value of it is still going to be positive. But if you think about over the first 10 seconds, your distance, 10 seconds, what is it going to be? Pause the video and try to think about it. Well, that's going to be the integral from zero to 10 of the absolute value of our velocity function, which is going to be equal to what? Well, it's going to be this area plus this area right over here. So plus this area right over here."}, {"video_title": "Motion problems with integrals displacement vs. distance AP Calculus AB Khan Academy.mp3", "Sentence": "Pause the video and try to think about it. Well, that's going to be the integral from zero to 10 of the absolute value of our velocity function, which is going to be equal to what? Well, it's going to be this area plus this area right over here. So plus this area right over here. And so this is going to be five times five times 1 1\u20442 plus five times five times 1 1\u20442, which is going to be 25 meters. The particle has gone 12.5 meters to the right, and then it goes back 12.5 meters to the left. Your displacement, your net change in positions is zero, but the total length of path traveled is 25 meters."}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And then they define h of x for us in terms of both f of x and g of x. And what we're curious about is what is the derivative with respect to x of h of x at x is equal to 9. And so I encourage you to pause this video and think about it on your own before I work through it. So let's think about it a little bit. So another way just to get familiar with the notation of writing this, the derivative of h of x with respect to x at x equals 9. This is equivalent to h. Let me do it in that blue color. It is equivalent to h prime."}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So let's think about it a little bit. So another way just to get familiar with the notation of writing this, the derivative of h of x with respect to x at x equals 9. This is equivalent to h. Let me do it in that blue color. It is equivalent to h prime. And the prime signifies that we're taking the derivative. h prime of x when x equals 9. So h prime of 9 is what this really is."}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "It is equivalent to h prime. And the prime signifies that we're taking the derivative. h prime of x when x equals 9. So h prime of 9 is what this really is. Actually, let me do this in a different color. So this is h prime of 9. So let's think about what that is."}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So h prime of 9 is what this really is. Actually, let me do this in a different color. So this is h prime of 9. So let's think about what that is. Let's take the derivative of both sides of this expression to figure out what the derivative with respect to x of h is. So we get derivative with respect to x of h of x is going to be equal to the derivative with respect to x of all of this business. So I could actually just rewrite it."}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So let's think about what that is. Let's take the derivative of both sides of this expression to figure out what the derivative with respect to x of h is. So we get derivative with respect to x of h of x is going to be equal to the derivative with respect to x of all of this business. So I could actually just rewrite it. 3 times f of x plus 2 times g of x. Now, this right over here, the derivative of the sum of two terms, that's going to be the same thing as the sum of the derivatives of each of the terms. So this is going to be the same thing as the derivative with respect to x of 3 times f of x plus the derivative with respect to x of 2 times g of x."}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So I could actually just rewrite it. 3 times f of x plus 2 times g of x. Now, this right over here, the derivative of the sum of two terms, that's going to be the same thing as the sum of the derivatives of each of the terms. So this is going to be the same thing as the derivative with respect to x of 3 times f of x plus the derivative with respect to x of 2 times g of x. Now, the derivative of a number, or I guess you could say a scaling factor, times a function. The derivative of a scalar times a function is the same thing as the scalar times the derivative of the function. What does that mean?"}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be the same thing as the derivative with respect to x of 3 times f of x plus the derivative with respect to x of 2 times g of x. Now, the derivative of a number, or I guess you could say a scaling factor, times a function. The derivative of a scalar times a function is the same thing as the scalar times the derivative of the function. What does that mean? Well, that just means that this first term right over here, that's going to be equivalent to 3 times the derivative with respect to x of f of x plus this part over here is the same thing as 2. Let me make sure I don't run out of space here. Plus 2 times the derivative with respect to x of g of x."}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "What does that mean? Well, that just means that this first term right over here, that's going to be equivalent to 3 times the derivative with respect to x of f of x plus this part over here is the same thing as 2. Let me make sure I don't run out of space here. Plus 2 times the derivative with respect to x of g of x. So the derivative of h with respect to x is equal to 3 times the derivative of f with respect to x plus 2 times the derivative of g with respect to x. And if we wanted to write it in this kind of prime notation here, we could rewrite it as h prime of x is equal to 3 times f prime of x. So this whole part right over here, that is the same thing as f prime of x."}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Plus 2 times the derivative with respect to x of g of x. So the derivative of h with respect to x is equal to 3 times the derivative of f with respect to x plus 2 times the derivative of g with respect to x. And if we wanted to write it in this kind of prime notation here, we could rewrite it as h prime of x is equal to 3 times f prime of x. So this whole part right over here, that is the same thing as f prime of x. So it's 3 times f prime of x plus 2 times g prime of x. And once you kind of are more fluent with this property, the derivative of the sum of two things is the sum of the derivatives. And the derivative of a scalar times something is the same thing as the scalar times the derivative of that something."}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So this whole part right over here, that is the same thing as f prime of x. So it's 3 times f prime of x plus 2 times g prime of x. And once you kind of are more fluent with this property, the derivative of the sum of two things is the sum of the derivatives. And the derivative of a scalar times something is the same thing as the scalar times the derivative of that something. You really could have gone straight from here to here pretty quickly. Now why is this interesting? Well, now we can evaluate this function when x is equal to 9."}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And the derivative of a scalar times something is the same thing as the scalar times the derivative of that something. You really could have gone straight from here to here pretty quickly. Now why is this interesting? Well, now we can evaluate this function when x is equal to 9. So h prime of 9 is the same thing as 3 times f prime of 9 plus 2 times g prime of 9. Now what is f prime of 9, the derivative of our function f when x is equal to 9? Well, they tell us when x is equal to 9, f of 9 is 1."}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, now we can evaluate this function when x is equal to 9. So h prime of 9 is the same thing as 3 times f prime of 9 plus 2 times g prime of 9. Now what is f prime of 9, the derivative of our function f when x is equal to 9? Well, they tell us when x is equal to 9, f of 9 is 1. But more importantly, f prime of 9 is 3. So this part right over here evaluates that part's 3. But what's g prime of 9?"}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, they tell us when x is equal to 9, f of 9 is 1. But more importantly, f prime of 9 is 3. So this part right over here evaluates that part's 3. But what's g prime of 9? So let's look at this function a little bit more closely. So there's a couple of ways we could think about it. Actually, let's try to graph it."}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "But what's g prime of 9? So let's look at this function a little bit more closely. So there's a couple of ways we could think about it. Actually, let's try to graph it. Now I think that could be interesting just to visualize what's going on here. So let's say that's our y-axis. And this right over here is our x-axis."}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Actually, let's try to graph it. Now I think that could be interesting just to visualize what's going on here. So let's say that's our y-axis. And this right over here is our x-axis. Now when does an absolute value function like this, when is this going to hit a minimum point? Well, the absolute value of something is always going to be non-negative. So it hits a minimum point when this thing is equal to 0."}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And this right over here is our x-axis. Now when does an absolute value function like this, when is this going to hit a minimum point? Well, the absolute value of something is always going to be non-negative. So it hits a minimum point when this thing is equal to 0. Well, when is this thing equal to 0? When x equals 1, this thing is equal to 0. So we hit a minimum point when x is equal to 1."}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So it hits a minimum point when this thing is equal to 0. Well, when is this thing equal to 0? When x equals 1, this thing is equal to 0. So we hit a minimum point when x is equal to 1. And when x equals 1, this term is 0, absolute value of 0 is 0. g of 1 is 1. So we have this point right over there. Now what happens after that?"}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So we hit a minimum point when x is equal to 1. And when x equals 1, this term is 0, absolute value of 0 is 0. g of 1 is 1. So we have this point right over there. Now what happens after that? What happens for x greater than 1? Actually, let me write this down. So g of x is equal to."}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Now what happens after that? What happens for x greater than 1? Actually, let me write this down. So g of x is equal to. And in general, whenever you have an absolute value or a relatively simple absolute value function like this, you could think of it, you could break it up into two functions. Or you could think about this function over different intervals, when the absolute value is non-negative and when the absolute value is negative. So when the absolute value is non-negative, that's when x is greater than or equal to 0."}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So g of x is equal to. And in general, whenever you have an absolute value or a relatively simple absolute value function like this, you could think of it, you could break it up into two functions. Or you could think about this function over different intervals, when the absolute value is non-negative and when the absolute value is negative. So when the absolute value is non-negative, that's when x is greater than or equal to 0. And when the absolute value is non-negative, if you're taking the absolute value of a non-negative number, then it's just going to be itself. The absolute value of 0 is 0. Absolute value of 1 is 1."}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So when the absolute value is non-negative, that's when x is greater than or equal to 0. And when the absolute value is non-negative, if you're taking the absolute value of a non-negative number, then it's just going to be itself. The absolute value of 0 is 0. Absolute value of 1 is 1. Absolute value of 100 is 100. So then you could ignore the absolute value for x is greater than or equal to 0. For x is greater than or equal to 1."}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Absolute value of 1 is 1. Absolute value of 100 is 100. So then you could ignore the absolute value for x is greater than or equal to 0. For x is greater than or equal to 1. This thing right over here is non-negative. And so it'll just evaluate to x minus 1. So this is going to be x minus 1 plus 1, which is the same thing as just x."}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "For x is greater than or equal to 1. This thing right over here is non-negative. And so it'll just evaluate to x minus 1. So this is going to be x minus 1 plus 1, which is the same thing as just x. Minus 1 plus 1, they just cancel out. Now, when this term right over here is negative, and that's going to happen for x is less than 1, well, then the absolute value is going to be the opposite of it. You give me the absolute value of a negative number, it's going to be the opposite."}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be x minus 1 plus 1, which is the same thing as just x. Minus 1 plus 1, they just cancel out. Now, when this term right over here is negative, and that's going to happen for x is less than 1, well, then the absolute value is going to be the opposite of it. You give me the absolute value of a negative number, it's going to be the opposite. The absolute value of negative 8 is positive 8. So it's going to be that the negative of x minus 1 is 1 minus x plus 1. Or we could say 2 minus x."}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "You give me the absolute value of a negative number, it's going to be the opposite. The absolute value of negative 8 is positive 8. So it's going to be that the negative of x minus 1 is 1 minus x plus 1. Or we could say 2 minus x. 2 minus x. Now, so for x is greater than or equal to 1, we would look at this expression. Now, what's the slope of that?"}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Or we could say 2 minus x. 2 minus x. Now, so for x is greater than or equal to 1, we would look at this expression. Now, what's the slope of that? Well, the slope of that is 1. So we're going to have a curve that looks like, or a line, I guess we could say, that looks like this. For all x is greater than or equal to 1."}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Now, what's the slope of that? Well, the slope of that is 1. So we're going to have a curve that looks like, or a line, I guess we could say, that looks like this. For all x is greater than or equal to 1. So the important thing, remember, we're going to think about the slope of the tangent line when we think about the derivative of g. So slope is equal to 1. And for x less than 1, well, our slope now, if we look right over here, our slope is negative. Our slope is negative 1."}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "For all x is greater than or equal to 1. So the important thing, remember, we're going to think about the slope of the tangent line when we think about the derivative of g. So slope is equal to 1. And for x less than 1, well, our slope now, if we look right over here, our slope is negative. Our slope is negative 1. So it's going to look like this. It's going to look like that. But for the point in question, if we're thinking about g prime of 9, so 9 is someplace out here."}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Our slope is negative 1. So it's going to look like this. It's going to look like that. But for the point in question, if we're thinking about g prime of 9, so 9 is someplace out here. So what is g prime of 9? So g prime of 9, let me make it clear, this graph right over here. This is the graph of g of x."}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "But for the point in question, if we're thinking about g prime of 9, so 9 is someplace out here. So what is g prime of 9? So g prime of 9, let me make it clear, this graph right over here. This is the graph of g of x. Or we could say y. This is the graph y equals g of x. y is equal to g of x. So what is g prime of 9?"}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "This is the graph of g of x. Or we could say y. This is the graph y equals g of x. y is equal to g of x. So what is g prime of 9? Well, that's the slope when x is equal to 9. Well, the slope is going to be equal to 1. So g prime of 9 is 1."}, {"video_title": "Basic derivative rules table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So what is g prime of 9? Well, that's the slope when x is equal to 9. Well, the slope is going to be equal to 1. So g prime of 9 is 1. So what does this evaluate to? This is going to be 3 times 3. So this part right over here is 9 plus 2 times 1, plus 2, which is equal to 11."}, {"video_title": "Rewriting before integrating AP Calculus AB Khan Academy.mp3", "Sentence": "Pause this video and see if you can evaluate this. So you might be saying, oh, what kind of fancy technique could I use? But you will see sometimes the fanciest or maybe the least fancy, but the best technique is to just simplify this algebraically. So in this situation, what happens if we distribute this x squared? Well, then we're going to get a polynomial here within the integral. So this is going to be equal to the integral of x squared times three x is three x to the third and then negative one times x squared is minus x squared and then that times dx. And now this is pretty straightforward to evaluate."}, {"video_title": "Rewriting before integrating AP Calculus AB Khan Academy.mp3", "Sentence": "So in this situation, what happens if we distribute this x squared? Well, then we're going to get a polynomial here within the integral. So this is going to be equal to the integral of x squared times three x is three x to the third and then negative one times x squared is minus x squared and then that times dx. And now this is pretty straightforward to evaluate. This is going to be equal to the antiderivative of x to the third is x to the fourth over four. So this is going to be three times x to the fourth over four I could write it that way or let me just write it, x to the fourth over four and then the antiderivative of x squared is x to the third over three. So minus x to the third over three and this is an indefinite integral."}, {"video_title": "Rewriting before integrating AP Calculus AB Khan Academy.mp3", "Sentence": "And now this is pretty straightforward to evaluate. This is going to be equal to the antiderivative of x to the third is x to the fourth over four. So this is going to be three times x to the fourth over four I could write it that way or let me just write it, x to the fourth over four and then the antiderivative of x squared is x to the third over three. So minus x to the third over three and this is an indefinite integral. There might be a constant there. So let me write that down and we're done. The big takeaway is you just have to do a little bit of distribution to get a form where it's easy to evaluate the antiderivative."}, {"video_title": "Rewriting before integrating AP Calculus AB Khan Academy.mp3", "Sentence": "So minus x to the third over three and this is an indefinite integral. There might be a constant there. So let me write that down and we're done. The big takeaway is you just have to do a little bit of distribution to get a form where it's easy to evaluate the antiderivative. Let's do another example. Let's say that we want to take the indefinite integral of, it's going to be a hairy expression, so x to the third plus three x squared minus five, all of that over x squared dx. What would this be?"}, {"video_title": "Rewriting before integrating AP Calculus AB Khan Academy.mp3", "Sentence": "The big takeaway is you just have to do a little bit of distribution to get a form where it's easy to evaluate the antiderivative. Let's do another example. Let's say that we want to take the indefinite integral of, it's going to be a hairy expression, so x to the third plus three x squared minus five, all of that over x squared dx. What would this be? Pause the video again and see if you can figure it out. So once again, your brain might want to try to do some fancy tricks or whatever else but the main insight here is to realize that you could just simplify it algebraically. What happens if you just divide each of these terms by x squared?"}, {"video_title": "Rewriting before integrating AP Calculus AB Khan Academy.mp3", "Sentence": "What would this be? Pause the video again and see if you can figure it out. So once again, your brain might want to try to do some fancy tricks or whatever else but the main insight here is to realize that you could just simplify it algebraically. What happens if you just divide each of these terms by x squared? Well then this thing is going to be equal to, put some parentheses here, x to the third divided by x squared is just going to be x. Three x squared divided by x squared is just three and then negative five divided by x squared, you could just write that as negative five times x to the negative two power. And so once again, we just need to use the reverse power rule here to take the antiderivative."}, {"video_title": "Rewriting before integrating AP Calculus AB Khan Academy.mp3", "Sentence": "What happens if you just divide each of these terms by x squared? Well then this thing is going to be equal to, put some parentheses here, x to the third divided by x squared is just going to be x. Three x squared divided by x squared is just three and then negative five divided by x squared, you could just write that as negative five times x to the negative two power. And so once again, we just need to use the reverse power rule here to take the antiderivative. This is going to be, let's see, the antiderivative of x is x squared over two, x squared over two plus the antiderivative of three and the antiderivative of three would just be three x. The antiderivative of negative five x to the negative two. So we would increment the exponent by one, positive one and then divide by that value."}, {"video_title": "Rewriting before integrating AP Calculus AB Khan Academy.mp3", "Sentence": "And so once again, we just need to use the reverse power rule here to take the antiderivative. This is going to be, let's see, the antiderivative of x is x squared over two, x squared over two plus the antiderivative of three and the antiderivative of three would just be three x. The antiderivative of negative five x to the negative two. So we would increment the exponent by one, positive one and then divide by that value. So it would be negative five x to the negative one. We're adding one to negative one. All of that divided by negative one which is the same, we could write it like that."}, {"video_title": "Rewriting before integrating AP Calculus AB Khan Academy.mp3", "Sentence": "So we would increment the exponent by one, positive one and then divide by that value. So it would be negative five x to the negative one. We're adding one to negative one. All of that divided by negative one which is the same, we could write it like that. Well these two would just, you'd have minus divided by negative one so it's really just going, you can rewrite it like this, plus five x to the negative one and you could take the derivative of this to verify that it would indeed give you that. And of course, we can't forget our plus c. Never forget that if you're taking an indefinite integral. All right, let's just do one more for good measure."}, {"video_title": "Rewriting before integrating AP Calculus AB Khan Academy.mp3", "Sentence": "All of that divided by negative one which is the same, we could write it like that. Well these two would just, you'd have minus divided by negative one so it's really just going, you can rewrite it like this, plus five x to the negative one and you could take the derivative of this to verify that it would indeed give you that. And of course, we can't forget our plus c. Never forget that if you're taking an indefinite integral. All right, let's just do one more for good measure. Let's say we're taking the indefinite integral of the cube root of x to the fifth dx. Cube root of x to the fifth dx. Pause the video and see if you can evaluate this."}, {"video_title": "Rewriting before integrating AP Calculus AB Khan Academy.mp3", "Sentence": "All right, let's just do one more for good measure. Let's say we're taking the indefinite integral of the cube root of x to the fifth dx. Cube root of x to the fifth dx. Pause the video and see if you can evaluate this. Try to write a little bit neater. X to the fifth dx. Pause the video and try to figure it out."}, {"video_title": "Rewriting before integrating AP Calculus AB Khan Academy.mp3", "Sentence": "Pause the video and see if you can evaluate this. Try to write a little bit neater. X to the fifth dx. Pause the video and try to figure it out. So here the realization is, well if you just rewrite all of this as one exponent, so this is equal to the indefinite integral of x to the fifth to the 1 3rd. I just rewrote the cube root as the 1 3rd power dx which is the same thing as the integral of x to the, if I raise something to a power and then raise that to a power, I can multiply those two exponents. That's just exponent properties."}, {"video_title": "Rewriting before integrating AP Calculus AB Khan Academy.mp3", "Sentence": "Pause the video and try to figure it out. So here the realization is, well if you just rewrite all of this as one exponent, so this is equal to the indefinite integral of x to the fifth to the 1 3rd. I just rewrote the cube root as the 1 3rd power dx which is the same thing as the integral of x to the, if I raise something to a power and then raise that to a power, I can multiply those two exponents. That's just exponent properties. X to the 5 3rd dx. Many of you might have just gone straight to this step right over here. And then once again, we just have to use the reverse power rule."}, {"video_title": "Rewriting before integrating AP Calculus AB Khan Academy.mp3", "Sentence": "That's just exponent properties. X to the 5 3rd dx. Many of you might have just gone straight to this step right over here. And then once again, we just have to use the reverse power rule. This is going to be x to the, we increment this 5 3rds by one or we can add 3 3rds to it. So it's x to the 8 3rds and then we divide by 8 3rds and multiply by its reciprocal. So we could just say 3 8ths times x to the 8 3rds and of course we have r plus c. And verify this."}, {"video_title": "Rewriting before integrating AP Calculus AB Khan Academy.mp3", "Sentence": "And then once again, we just have to use the reverse power rule. This is going to be x to the, we increment this 5 3rds by one or we can add 3 3rds to it. So it's x to the 8 3rds and then we divide by 8 3rds and multiply by its reciprocal. So we could just say 3 8ths times x to the 8 3rds and of course we have r plus c. And verify this. If you use the power rule here, you'd have 8 3rds times the 3 8ths which just gives you a coefficient of one. And then you decrement this by 3 3rds or one, you get to 5 3rds which is exactly what we originally had. So the big takeaway of this video, many times the most powerful integration technique is literally just algebraic simplification first."}, {"video_title": "Intermediate value theorem example Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "So before I even look at this, what do we know about the intermediate value theorem? Well, it applies here. It's a continuous function on this closed interval. We know what the value of the function is at negative two. It's three, so let me write that. F of negative two is equal to three and f of one, they tell us right over here, is equal to six. And all the intermediate value theorem tells us, and if this is completely unfamiliar to you, I encourage you to watch the video on the intermediate value theorem, is that if we have a continuous function on some closed interval, then the function must take on every value between the values at the end points of the interval."}, {"video_title": "Intermediate value theorem example Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "We know what the value of the function is at negative two. It's three, so let me write that. F of negative two is equal to three and f of one, they tell us right over here, is equal to six. And all the intermediate value theorem tells us, and if this is completely unfamiliar to you, I encourage you to watch the video on the intermediate value theorem, is that if we have a continuous function on some closed interval, then the function must take on every value between the values at the end points of the interval. Or another way to say it is, for any L between three and six, there is at least one C, there is at least one C, one C between, or I could say one C in the interval from negative two to one, the closed interval, such that f of C is equal to L. This comes straight out of the intermediate value theorem. And just saying it in everyday language is, look, this is a continuous function. Actually, I'll draw it visually in a few seconds, but it makes sense that if it's continuous, if I were to draw the graph, I can't pick up my pencil, well, then it makes sense that I would have to take on every value between three and six, or there's at least one point in this interval where I take on any given value between three and six."}, {"video_title": "Intermediate value theorem example Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "And all the intermediate value theorem tells us, and if this is completely unfamiliar to you, I encourage you to watch the video on the intermediate value theorem, is that if we have a continuous function on some closed interval, then the function must take on every value between the values at the end points of the interval. Or another way to say it is, for any L between three and six, there is at least one C, there is at least one C, one C between, or I could say one C in the interval from negative two to one, the closed interval, such that f of C is equal to L. This comes straight out of the intermediate value theorem. And just saying it in everyday language is, look, this is a continuous function. Actually, I'll draw it visually in a few seconds, but it makes sense that if it's continuous, if I were to draw the graph, I can't pick up my pencil, well, then it makes sense that I would have to take on every value between three and six, or there's at least one point in this interval where I take on any given value between three and six. So let's see which of these answers are consistent with that. We only pick one. So f of C equals four, so that would be a case where L is equal to four."}, {"video_title": "Intermediate value theorem example Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "Actually, I'll draw it visually in a few seconds, but it makes sense that if it's continuous, if I were to draw the graph, I can't pick up my pencil, well, then it makes sense that I would have to take on every value between three and six, or there's at least one point in this interval where I take on any given value between three and six. So let's see which of these answers are consistent with that. We only pick one. So f of C equals four, so that would be a case where L is equal to four. So there's at least one C in this interval such that f of C is equal to four. We could say that, but that's not exactly what they're saying here. F of C could be four for at least one C, not in this interval."}, {"video_title": "Intermediate value theorem example Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "So f of C equals four, so that would be a case where L is equal to four. So there's at least one C in this interval such that f of C is equal to four. We could say that, but that's not exactly what they're saying here. F of C could be four for at least one C, not in this interval. Remember, the C is our X. This is our X right over here. So the C is going to be in this interval."}, {"video_title": "Intermediate value theorem example Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "F of C could be four for at least one C, not in this interval. Remember, the C is our X. This is our X right over here. So the C is going to be in this interval. I'll take a look at it visually in a second so that we can validate that. We're not saying for at least one C between three and six, f of C is equal to four. We're saying for at least one C in this interval, f of C is going to be equal to four."}, {"video_title": "Intermediate value theorem example Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "So the C is going to be in this interval. I'll take a look at it visually in a second so that we can validate that. We're not saying for at least one C between three and six, f of C is equal to four. We're saying for at least one C in this interval, f of C is going to be equal to four. It's important that four is between three and six because that's the value of our function, and the C needs to be in our closed interval along the X axis. So I'm gonna rule this out. They're trying to confuse us."}, {"video_title": "Intermediate value theorem example Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "We're saying for at least one C in this interval, f of C is going to be equal to four. It's important that four is between three and six because that's the value of our function, and the C needs to be in our closed interval along the X axis. So I'm gonna rule this out. They're trying to confuse us. All right, f of C equals zero for at least one C between negative two and one. Well, here they got the interval along the X axis right. That's where the C would be between, but it's not guaranteed by the intermediate value theorem that f of C is going to be equal to zero because zero is not between three and six."}, {"video_title": "Intermediate value theorem example Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "They're trying to confuse us. All right, f of C equals zero for at least one C between negative two and one. Well, here they got the interval along the X axis right. That's where the C would be between, but it's not guaranteed by the intermediate value theorem that f of C is going to be equal to zero because zero is not between three and six. So I'm gonna rule that one out. I'm going to rule this one out, saying f of C equals zero. And let's see, we're only left with this one, so I hope it works."}, {"video_title": "Intermediate value theorem example Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "That's where the C would be between, but it's not guaranteed by the intermediate value theorem that f of C is going to be equal to zero because zero is not between three and six. So I'm gonna rule that one out. I'm going to rule this one out, saying f of C equals zero. And let's see, we're only left with this one, so I hope it works. So f of C is equal to four. Well, that seems reasonable because four is between three and six for at least one C between negative two and one. Well, yeah, because that's in this interval right over here."}, {"video_title": "Intermediate value theorem example Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "And let's see, we're only left with this one, so I hope it works. So f of C is equal to four. Well, that seems reasonable because four is between three and six for at least one C between negative two and one. Well, yeah, because that's in this interval right over here. So I am feeling good about that. And we could think about this visually as well. The intermediate value theorem, when you think about it visually, makes a lot of sense."}, {"video_title": "Intermediate value theorem example Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "Well, yeah, because that's in this interval right over here. So I am feeling good about that. And we could think about this visually as well. The intermediate value theorem, when you think about it visually, makes a lot of sense. So let me draw the X axis first, actually, and then let me draw my Y axis, and I'm gonna draw them at different scales because my Y axis, well, let's see, if this is six, this is three, that's my Y axis, this is one, this is negative one, this is negative two. And so we're continuous on the closed interval from negative two to one, and f of negative two is equal to three. So let me plot that."}, {"video_title": "Intermediate value theorem example Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "The intermediate value theorem, when you think about it visually, makes a lot of sense. So let me draw the X axis first, actually, and then let me draw my Y axis, and I'm gonna draw them at different scales because my Y axis, well, let's see, if this is six, this is three, that's my Y axis, this is one, this is negative one, this is negative two. And so we're continuous on the closed interval from negative two to one, and f of negative two is equal to three. So let me plot that. F of negative two is equal to three. So that's right over there. And f of one is equal to six."}, {"video_title": "Intermediate value theorem example Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "So let me plot that. F of negative two is equal to three. So that's right over there. And f of one is equal to six. So that's right over there. And so let's try to draw a continuous function. So a continuous function includes these points, and it's continuous, so an intuitive way to think about it is I can't pick up my pencil if I'm drawing the graph of the function, which contains these two points."}, {"video_title": "Intermediate value theorem example Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "And f of one is equal to six. So that's right over there. And so let's try to draw a continuous function. So a continuous function includes these points, and it's continuous, so an intuitive way to think about it is I can't pick up my pencil if I'm drawing the graph of the function, which contains these two points. So I can't pick up my, I can't do that. That would be picking up my pencil. So it is a continuous function."}, {"video_title": "Intermediate value theorem example Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "So a continuous function includes these points, and it's continuous, so an intuitive way to think about it is I can't pick up my pencil if I'm drawing the graph of the function, which contains these two points. So I can't pick up my, I can't do that. That would be picking up my pencil. So it is a continuous function. So it takes on every value, as we can see, it definitely does that. It takes on every value between three and six. It might take on other values, but we know for sure it has to take on every value between three and six."}, {"video_title": "Intermediate value theorem example Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "So it is a continuous function. So it takes on every value, as we can see, it definitely does that. It takes on every value between three and six. It might take on other values, but we know for sure it has to take on every value between three and six. And so if we think about four, four is right over here. The way I drew it, it actually looks like it's almost taking on that value right at the Y axis. I forgot to label my X axis here."}, {"video_title": "Intermediate value theorem example Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "It might take on other values, but we know for sure it has to take on every value between three and six. And so if we think about four, four is right over here. The way I drew it, it actually looks like it's almost taking on that value right at the Y axis. I forgot to label my X axis here. But you can see, it took on that value in for a C, in this case, between negative two and one. And I could have drawn that graph multiple different ways. I could have drawn it something like, I could have done it, and actually it takes on, there's multiple times it takes on the value four here."}, {"video_title": "Intermediate value theorem example Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "I forgot to label my X axis here. But you can see, it took on that value in for a C, in this case, between negative two and one. And I could have drawn that graph multiple different ways. I could have drawn it something like, I could have done it, and actually it takes on, there's multiple times it takes on the value four here. So this could be our C, but once again, it's between the interval negative two and one. This could be our C, once again, in the interval between negative two and one. Or this could be our C, in between the interval of negative two and one."}, {"video_title": "Intermediate value theorem example Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "I could have drawn it something like, I could have done it, and actually it takes on, there's multiple times it takes on the value four here. So this could be our C, but once again, it's between the interval negative two and one. This could be our C, once again, in the interval between negative two and one. Or this could be our C, in between the interval of negative two and one. That's just the way I happened to draw it. I could have drawn this thing as just a straight line. I could have drawn it like this."}, {"video_title": "Intermediate value theorem example Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "Or this could be our C, in between the interval of negative two and one. That's just the way I happened to draw it. I could have drawn this thing as just a straight line. I could have drawn it like this. And then it looks like it's taking on four, only once, and it's doing it right around there. This isn't necessarily true, that you take on, you take on, that you become four for at least one C between three and six. Three and six aren't even on our graph here."}, {"video_title": "Intermediate value theorem example Existence theorems AP Calculus AB Khan Academy.mp3", "Sentence": "I could have drawn it like this. And then it looks like it's taking on four, only once, and it's doing it right around there. This isn't necessarily true, that you take on, you take on, that you become four for at least one C between three and six. Three and six aren't even on our graph here. I would have to go all the way to, so two, three. There's no guarantee that our function, that our function takes on four for one C between three and six. We don't even know what the function does when X is between three and six."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable) Khan Academy.mp3", "Sentence": "So let's do step by step. So first let's think about continuity. So for continuity, for g to be continuous at x equals one, that means that g of one, that means that g of one must be equal to the limit as x approaches one of g, of g of x. Well g of one, what is that going to be? G of one, we're gonna fall into this case, one minus one squared is going to be zero. So if we can show that the limit of g of x as x approaches one is the same as g of one, is equal to zero, then we know we're continuous there. Let's do the left and right-handed limits here."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable) Khan Academy.mp3", "Sentence": "Well g of one, what is that going to be? G of one, we're gonna fall into this case, one minus one squared is going to be zero. So if we can show that the limit of g of x as x approaches one is the same as g of one, is equal to zero, then we know we're continuous there. Let's do the left and right-handed limits here. So if we do the left-handed limit, and that's especially useful because we're in these different clauses here as we approach from the left and the right-hand side. So as x approaches one from the left-hand side of g of x, well we're gonna be falling into this situation here as we approach from the left, as x is less than one. So this is going to be the same thing as that."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable) Khan Academy.mp3", "Sentence": "Let's do the left and right-handed limits here. So if we do the left-handed limit, and that's especially useful because we're in these different clauses here as we approach from the left and the right-hand side. So as x approaches one from the left-hand side of g of x, well we're gonna be falling into this situation here as we approach from the left, as x is less than one. So this is going to be the same thing as that. That's what g of x is equal to when we are less than one as we're approaching from the left. Well this thing is defined and it's continuous for all real numbers, so we can just substitute one in for x and we get this is equal to zero. So so far so good, let's do one more of these."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable) Khan Academy.mp3", "Sentence": "So this is going to be the same thing as that. That's what g of x is equal to when we are less than one as we're approaching from the left. Well this thing is defined and it's continuous for all real numbers, so we can just substitute one in for x and we get this is equal to zero. So so far so good, let's do one more of these. Let's approach from the right-hand side as x approaches one from the right-hand side of g of x. Well now we're falling into this case. So g of x, if we're to the right of one, if our value's greater than or equal to one, it's gonna be x minus one squared."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable) Khan Academy.mp3", "Sentence": "So so far so good, let's do one more of these. Let's approach from the right-hand side as x approaches one from the right-hand side of g of x. Well now we're falling into this case. So g of x, if we're to the right of one, if our value's greater than or equal to one, it's gonna be x minus one squared. Well once again, x minus one squared, that is defined for all real numbers, it's continuous for all real numbers, so we could just pop that one in there, you get one minus one squared, well that's just zero again. So the left-hand limit, the right-hand limit are both equal zero, which means that the limit is equal, the limit of g of x as x approaches one is equal to zero, which is the same thing as g of one, so we are good with continuity. So we can rule out all the ones that are saying that it's not continuous."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable) Khan Academy.mp3", "Sentence": "So g of x, if we're to the right of one, if our value's greater than or equal to one, it's gonna be x minus one squared. Well once again, x minus one squared, that is defined for all real numbers, it's continuous for all real numbers, so we could just pop that one in there, you get one minus one squared, well that's just zero again. So the left-hand limit, the right-hand limit are both equal zero, which means that the limit is equal, the limit of g of x as x approaches one is equal to zero, which is the same thing as g of one, so we are good with continuity. So we can rule out all the ones that are saying that it's not continuous. So we could rule out that one, and we can rule out that one right over there. So now let's think about whether it is differentiable. So differentiability, so differentiability, I'll write differentiability, ability."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable) Khan Academy.mp3", "Sentence": "So we can rule out all the ones that are saying that it's not continuous. So we could rule out that one, and we can rule out that one right over there. So now let's think about whether it is differentiable. So differentiability, so differentiability, I'll write differentiability, ability. Did I, let's see, that's a long word, differentiability. All right, differentiability, what needs to be true here? Well we have to have a defined limit as x approaches one for f of x minus f of one over, oh let me be careful, it's not f, it's g. So we need to have a defined limit for g of x minus g of one over x minus one."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable) Khan Academy.mp3", "Sentence": "So differentiability, so differentiability, I'll write differentiability, ability. Did I, let's see, that's a long word, differentiability. All right, differentiability, what needs to be true here? Well we have to have a defined limit as x approaches one for f of x minus f of one over, oh let me be careful, it's not f, it's g. So we need to have a defined limit for g of x minus g of one over x minus one. And so let's just try to evaluate this limit from the left and right-hand sides. And we could simplify it, we already know that g of one is zero, so that's just going to be zero. So we just need to find the limit as x approaches one of g of x over x minus one, or see if we can find a limit."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable) Khan Academy.mp3", "Sentence": "Well we have to have a defined limit as x approaches one for f of x minus f of one over, oh let me be careful, it's not f, it's g. So we need to have a defined limit for g of x minus g of one over x minus one. And so let's just try to evaluate this limit from the left and right-hand sides. And we could simplify it, we already know that g of one is zero, so that's just going to be zero. So we just need to find the limit as x approaches one of g of x over x minus one, or see if we can find a limit. So let's first think about the limit as we approach from the left-hand side of g of x over x minus, g of x over x minus one. Well as we approach from the left-hand side, g of x is that right over there. So we could write this, instead of writing g of x, we could write this as x minus one, x minus one over x minus one."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable) Khan Academy.mp3", "Sentence": "So we just need to find the limit as x approaches one of g of x over x minus one, or see if we can find a limit. So let's first think about the limit as we approach from the left-hand side of g of x over x minus, g of x over x minus one. Well as we approach from the left-hand side, g of x is that right over there. So we could write this, instead of writing g of x, we could write this as x minus one, x minus one over x minus one. And as long as we aren't equal to one, this thing is going to be equal, as long as x does not equal one, x minus one over x minus one is just going to be one. So this limit is going to be one. So that one worked out."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable) Khan Academy.mp3", "Sentence": "So we could write this, instead of writing g of x, we could write this as x minus one, x minus one over x minus one. And as long as we aren't equal to one, this thing is going to be equal, as long as x does not equal one, x minus one over x minus one is just going to be one. So this limit is going to be one. So that one worked out. Now let's think about the limit as x approaches one from the right-hand side. Of, once again, I could write g of x minus g of one, but g of one is just zero, so I'll just write g of x over x minus one. Well what's g of x now?"}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable) Khan Academy.mp3", "Sentence": "So that one worked out. Now let's think about the limit as x approaches one from the right-hand side. Of, once again, I could write g of x minus g of one, but g of one is just zero, so I'll just write g of x over x minus one. Well what's g of x now? Well it's x minus one squared. So instead of writing g of x, I could write this as x minus one squared over x minus one. And so as long as x does not equal one, and we're just doing the limit, we're saying as we approach one from the right-hand side, well this expression right over here, you have x minus one squared divided by x minus one, well that's just going to give us x minus one."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable) Khan Academy.mp3", "Sentence": "Well what's g of x now? Well it's x minus one squared. So instead of writing g of x, I could write this as x minus one squared over x minus one. And so as long as x does not equal one, and we're just doing the limit, we're saying as we approach one from the right-hand side, well this expression right over here, you have x minus one squared divided by x minus one, well that's just going to give us x minus one. X minus one squared divided by x minus one is just going to be x minus one. And this limit, well this expression right over here is going to be continuous and defined for sure all x's that are not equaling one. Actually let me, well it was before it was this, x minus one squared over x minus one."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable) Khan Academy.mp3", "Sentence": "And so as long as x does not equal one, and we're just doing the limit, we're saying as we approach one from the right-hand side, well this expression right over here, you have x minus one squared divided by x minus one, well that's just going to give us x minus one. X minus one squared divided by x minus one is just going to be x minus one. And this limit, well this expression right over here is going to be continuous and defined for sure all x's that are not equaling one. Actually let me, well it was before it was this, x minus one squared over x minus one. This thing right over here, as I said, it's not defined for x equals one, but it is defined for anything not, for x does not equal one, and we're just approaching one. And if we wanted to simplify this expression, it would get, this would just be, I think I just did this, but I'm just making sure I'm doing it right. This is going to be the same thing as that for x not being equal to one."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable) Khan Academy.mp3", "Sentence": "Actually let me, well it was before it was this, x minus one squared over x minus one. This thing right over here, as I said, it's not defined for x equals one, but it is defined for anything not, for x does not equal one, and we're just approaching one. And if we wanted to simplify this expression, it would get, this would just be, I think I just did this, but I'm just making sure I'm doing it right. This is going to be the same thing as that for x not being equal to one. Well this is just going to be zero. You could just evaluate when x is equal to one here, this is going to be equal to zero. And so notice, you get a different limit for this definition of derivative as we approach from the left-hand side or the right-hand side."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable) Khan Academy.mp3", "Sentence": "This is going to be the same thing as that for x not being equal to one. Well this is just going to be zero. You could just evaluate when x is equal to one here, this is going to be equal to zero. And so notice, you get a different limit for this definition of derivative as we approach from the left-hand side or the right-hand side. And that makes sense. This graph is going to look something like, we have a slope of one, so it's going to look something like this. And then right when x is equal to one and the value of our function is zero, it looks something like this."}, {"video_title": "Differentiability at a point algebraic (function isn't differentiable) Khan Academy.mp3", "Sentence": "And so notice, you get a different limit for this definition of derivative as we approach from the left-hand side or the right-hand side. And that makes sense. This graph is going to look something like, we have a slope of one, so it's going to look something like this. And then right when x is equal to one and the value of our function is zero, it looks something like this. It looks something like this. And so the graph is continuous. The graph for sure is continuous, but our slope coming into that point is one and our slope right when we leave that point is zero."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "What we're going to do in this video is prove that the limit as theta approaches zero of sine of theta over theta is equal to one. So let's start with a little bit of a geometric or trigonometric construction that I have here. So this white circle, this is a unit circle. Let me label it as such. So it has radius one, unit circle. So what does the length of this salmon colored line represent? Well, the height of this line would be the y coordinate of where this radius intersects the unit circle."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Let me label it as such. So it has radius one, unit circle. So what does the length of this salmon colored line represent? Well, the height of this line would be the y coordinate of where this radius intersects the unit circle. And so by definition, by the unit circle definition of trig functions, the length of this line is going to be sine of theta. If we wanted to make sure that it also worked for thetas that end up in the fourth quadrant, that will be useful, we can just ensure that it's the absolute value of the sine of theta. Now what about this blue line over here?"}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the height of this line would be the y coordinate of where this radius intersects the unit circle. And so by definition, by the unit circle definition of trig functions, the length of this line is going to be sine of theta. If we wanted to make sure that it also worked for thetas that end up in the fourth quadrant, that will be useful, we can just ensure that it's the absolute value of the sine of theta. Now what about this blue line over here? Can I express that in terms of a trigonometric function? Well, let's think about it. What would tangent of theta be?"}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Now what about this blue line over here? Can I express that in terms of a trigonometric function? Well, let's think about it. What would tangent of theta be? Let me write it over here. Tangent of theta is equal to opposite over adjacent. So if we look at this broader triangle right over here, this is our angle theta in radians."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "What would tangent of theta be? Let me write it over here. Tangent of theta is equal to opposite over adjacent. So if we look at this broader triangle right over here, this is our angle theta in radians. This is the opposite side. The adjacent side down here, this just has length one. Remember, this is a unit circle."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So if we look at this broader triangle right over here, this is our angle theta in radians. This is the opposite side. The adjacent side down here, this just has length one. Remember, this is a unit circle. So this just has length one. So the tangent of theta is the opposite side. The opposite side is equal to the tangent of theta."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Remember, this is a unit circle. So this just has length one. So the tangent of theta is the opposite side. The opposite side is equal to the tangent of theta. And just like before, this is going to be a positive value if we're sitting here in the first quadrant, but I want things to work in both the first and the fourth quadrant for the sake of our proof, so I'm just gonna put an absolute value here. So now that we've done that, I'm gonna think about some triangles and their respective areas. So first, I'm gonna draw a triangle that sits in this wedge, in this pie piece, this pie slice within this circle."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "The opposite side is equal to the tangent of theta. And just like before, this is going to be a positive value if we're sitting here in the first quadrant, but I want things to work in both the first and the fourth quadrant for the sake of our proof, so I'm just gonna put an absolute value here. So now that we've done that, I'm gonna think about some triangles and their respective areas. So first, I'm gonna draw a triangle that sits in this wedge, in this pie piece, this pie slice within this circle. So I can construct this triangle. And so let's think about the area of what I am shading in right over here. How can I express that area?"}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So first, I'm gonna draw a triangle that sits in this wedge, in this pie piece, this pie slice within this circle. So I can construct this triangle. And so let's think about the area of what I am shading in right over here. How can I express that area? Well, it's a triangle. We know that the area of a triangle is 1 1 2 base times height. We know the height is the absolute value of the sine of theta, and we know that the base is equal to one."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "How can I express that area? Well, it's a triangle. We know that the area of a triangle is 1 1 2 base times height. We know the height is the absolute value of the sine of theta, and we know that the base is equal to one. So the area here is going to be equal to 1 1 2 times our base, which is one, times our height, which is the absolute value of the sine of theta. I'll rewrite it over here. I could just rewrite that as the absolute value of the sine of theta over two."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "We know the height is the absolute value of the sine of theta, and we know that the base is equal to one. So the area here is going to be equal to 1 1 2 times our base, which is one, times our height, which is the absolute value of the sine of theta. I'll rewrite it over here. I could just rewrite that as the absolute value of the sine of theta over two. Now let's think about the area of this wedge that I am highlighting in this yellow color. So what fraction of the entire circle is this going to be? If I were to go all the way around the circle, it would be two pi radians."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "I could just rewrite that as the absolute value of the sine of theta over two. Now let's think about the area of this wedge that I am highlighting in this yellow color. So what fraction of the entire circle is this going to be? If I were to go all the way around the circle, it would be two pi radians. So this is theta over two piths of the entire circle, and we know the area of the circle. This is a unit circle. It has a radius one."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "If I were to go all the way around the circle, it would be two pi radians. So this is theta over two piths of the entire circle, and we know the area of the circle. This is a unit circle. It has a radius one. So it would be times the area of the circle, which would be pi times the radius squared. The radius is one, so it's just gonna be times pi. And so the area of this wedge right over here, theta over two."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "It has a radius one. So it would be times the area of the circle, which would be pi times the radius squared. The radius is one, so it's just gonna be times pi. And so the area of this wedge right over here, theta over two. And if we wanted to make this work for thetas in the fourth quadrant, we could just write an absolute value sign right over there because we're talking about positive area. And now let's think about this larger triangle in this blue color. And this is pretty straightforward."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And so the area of this wedge right over here, theta over two. And if we wanted to make this work for thetas in the fourth quadrant, we could just write an absolute value sign right over there because we're talking about positive area. And now let's think about this larger triangle in this blue color. And this is pretty straightforward. The area here is gonna be 1 1\u20442 times base times height. So the area, and once again, this is this entire area, that's going to be 1 1\u20442 times our base, which is one, times our height, which is our absolute value of tangent of theta. And so I can just write that down as the absolute value of the tangent of theta over two."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And this is pretty straightforward. The area here is gonna be 1 1\u20442 times base times height. So the area, and once again, this is this entire area, that's going to be 1 1\u20442 times our base, which is one, times our height, which is our absolute value of tangent of theta. And so I can just write that down as the absolute value of the tangent of theta over two. Now how would you compare the areas of this pink or this salmon-colored triangle, which sits inside of this wedge, and how would you compare that area of the wedge to the bigger triangle? Well, it's clear that the area of the salmon triangle is less than or equal to the area of the wedge, and the area of the wedge is less than or equal to the area of the big blue triangle. The wedge includes the salmon triangle plus this area right over here."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And so I can just write that down as the absolute value of the tangent of theta over two. Now how would you compare the areas of this pink or this salmon-colored triangle, which sits inside of this wedge, and how would you compare that area of the wedge to the bigger triangle? Well, it's clear that the area of the salmon triangle is less than or equal to the area of the wedge, and the area of the wedge is less than or equal to the area of the big blue triangle. The wedge includes the salmon triangle plus this area right over here. And then the blue triangle includes the wedge plus it has this area right over here. So I think we can feel good visually that this statement right over here is true. And now I'm just going to do a little bit of algebraic manipulation."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "The wedge includes the salmon triangle plus this area right over here. And then the blue triangle includes the wedge plus it has this area right over here. So I think we can feel good visually that this statement right over here is true. And now I'm just going to do a little bit of algebraic manipulation. Let me multiply everything by two. So I can rewrite that the absolute value of sine of theta is less than or equal to the absolute value of theta, which is less than or equal to the absolute value of tangent of theta. And let's see, actually, instead of writing the absolute value of tangent of theta, I'm gonna rewrite that as the absolute value of sine of theta over the absolute value of cosine of theta."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And now I'm just going to do a little bit of algebraic manipulation. Let me multiply everything by two. So I can rewrite that the absolute value of sine of theta is less than or equal to the absolute value of theta, which is less than or equal to the absolute value of tangent of theta. And let's see, actually, instead of writing the absolute value of tangent of theta, I'm gonna rewrite that as the absolute value of sine of theta over the absolute value of cosine of theta. That's going to be the same thing as the absolute value of tangent of theta. And the reason why I did that is we can now divide everything by the absolute value of sine of theta. Since we're dividing by a positive quantity, it's not going to change the direction of the inequalities."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And let's see, actually, instead of writing the absolute value of tangent of theta, I'm gonna rewrite that as the absolute value of sine of theta over the absolute value of cosine of theta. That's going to be the same thing as the absolute value of tangent of theta. And the reason why I did that is we can now divide everything by the absolute value of sine of theta. Since we're dividing by a positive quantity, it's not going to change the direction of the inequalities. So let's do that. I'm gonna divide this by an absolute value of sine of theta. I'm gonna divide this by an absolute value of the sine of theta."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Since we're dividing by a positive quantity, it's not going to change the direction of the inequalities. So let's do that. I'm gonna divide this by an absolute value of sine of theta. I'm gonna divide this by an absolute value of the sine of theta. And then I'm gonna divide this by an absolute value of the sine of theta. And what do I get? Well, over here, I get a one."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "I'm gonna divide this by an absolute value of the sine of theta. And then I'm gonna divide this by an absolute value of the sine of theta. And what do I get? Well, over here, I get a one. And on the right-hand side, I get a one over the absolute value of cosine theta. These two cancel out. So the next step I'm gonna do is take the reciprocal of everything."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, over here, I get a one. And on the right-hand side, I get a one over the absolute value of cosine theta. These two cancel out. So the next step I'm gonna do is take the reciprocal of everything. And so when I take the reciprocal of everything, that actually will switch the inequalities. The reciprocal of one is still going to be one. But now, since I'm taking the reciprocal of this here, it's going to be greater than or equal to the absolute value of the sine of theta over the absolute value of theta."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So the next step I'm gonna do is take the reciprocal of everything. And so when I take the reciprocal of everything, that actually will switch the inequalities. The reciprocal of one is still going to be one. But now, since I'm taking the reciprocal of this here, it's going to be greater than or equal to the absolute value of the sine of theta over the absolute value of theta. And that's going to be greater than or equal to the reciprocal of one over the absolute value of cosine of theta is the absolute value of cosine of theta. We really just care about the first and fourth quadrants. You can think about this theta approaching zero from that direction or from that direction there."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "But now, since I'm taking the reciprocal of this here, it's going to be greater than or equal to the absolute value of the sine of theta over the absolute value of theta. And that's going to be greater than or equal to the reciprocal of one over the absolute value of cosine of theta is the absolute value of cosine of theta. We really just care about the first and fourth quadrants. You can think about this theta approaching zero from that direction or from that direction there. So that would be the first and fourth quadrants. So if we're in the first quadrant and theta is positive, sine of theta is gonna be positive as well. And if we're in the fourth quadrant and theta's negative, well, sine of theta's gonna have the same sign."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "You can think about this theta approaching zero from that direction or from that direction there. So that would be the first and fourth quadrants. So if we're in the first quadrant and theta is positive, sine of theta is gonna be positive as well. And if we're in the fourth quadrant and theta's negative, well, sine of theta's gonna have the same sign. It's going to be negative as well. And so these absolute value signs aren't necessary. In the first quadrant, sine of theta and theta are both positive."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And if we're in the fourth quadrant and theta's negative, well, sine of theta's gonna have the same sign. It's going to be negative as well. And so these absolute value signs aren't necessary. In the first quadrant, sine of theta and theta are both positive. In the fourth quadrant, they're both negative, but when you divide them, you're going to get a positive value. So I can erase those. If we're in the first or fourth quadrant, our x value is not negative, and so cosine of theta, which is the x coordinate on our unit circle, is not going to be negative."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "In the first quadrant, sine of theta and theta are both positive. In the fourth quadrant, they're both negative, but when you divide them, you're going to get a positive value. So I can erase those. If we're in the first or fourth quadrant, our x value is not negative, and so cosine of theta, which is the x coordinate on our unit circle, is not going to be negative. And so we don't need the absolute value signs over there. Now, we should pause a second because we're actually almost done. We have just set up three functions."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "If we're in the first or fourth quadrant, our x value is not negative, and so cosine of theta, which is the x coordinate on our unit circle, is not going to be negative. And so we don't need the absolute value signs over there. Now, we should pause a second because we're actually almost done. We have just set up three functions. You could think of this as f of x is equal to, you could view this as f of theta is equal to one, g of theta is equal to this, and h of theta is equal to that. And over the interval that we care about, we could say four, negative pi over two is less than theta, is less than pi over two. But over this interval, this is true for any theta over which these functions are defined."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "We have just set up three functions. You could think of this as f of x is equal to, you could view this as f of theta is equal to one, g of theta is equal to this, and h of theta is equal to that. And over the interval that we care about, we could say four, negative pi over two is less than theta, is less than pi over two. But over this interval, this is true for any theta over which these functions are defined. Sine of theta over theta is defined over this interval except where theta is equal to zero. But since we're defined everywhere else, we can now find the limit. So what we can say is, well, by the squeeze theorem or by the sandwich theorem, if this is true over the interval, then we also know that the following is true."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "But over this interval, this is true for any theta over which these functions are defined. Sine of theta over theta is defined over this interval except where theta is equal to zero. But since we're defined everywhere else, we can now find the limit. So what we can say is, well, by the squeeze theorem or by the sandwich theorem, if this is true over the interval, then we also know that the following is true. And this we deserve a little bit of a drum roll. The limit as theta approaches zero of this is going to be greater than or equal to the limit as theta approaches zero of this, which is the one that we care about, sine of theta over theta, which is going to be greater than or equal to the limit as theta approaches zero of this. Now, this is clearly going to be just equal to one."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So what we can say is, well, by the squeeze theorem or by the sandwich theorem, if this is true over the interval, then we also know that the following is true. And this we deserve a little bit of a drum roll. The limit as theta approaches zero of this is going to be greater than or equal to the limit as theta approaches zero of this, which is the one that we care about, sine of theta over theta, which is going to be greater than or equal to the limit as theta approaches zero of this. Now, this is clearly going to be just equal to one. This is what we care about. And this, what's the limit as theta approaches zero of cosine of theta? Well, cosine of zero is just one, and it's a continuous function, so this is just going to be one."}, {"video_title": "Limit of sin(x) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Now, this is clearly going to be just equal to one. This is what we care about. And this, what's the limit as theta approaches zero of cosine of theta? Well, cosine of zero is just one, and it's a continuous function, so this is just going to be one. So let's see. This limit is going to be less than or equal to one, and it's going to be greater than or equal to one. So this must be equal to one, and we are done."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "We're going to assume we know that the limit as x approaches zero of sine of x over x, that this is equal to one. And I'm not gonna reprove this in this video, but we have a whole other video dedicated to proving this famous limit. And we do it using the squeeze or the sandwich theorem. So let's see if we can work this out. So the first thing we're going to do is algebraically manipulate this expression a little bit. What I'm going to do is I'm gonna multiply both the numerator and the denominator by one plus cosine of x. So times the denominator, I have to do the same thing, one plus cosine of x. I'm not changing the value of the expression."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "So let's see if we can work this out. So the first thing we're going to do is algebraically manipulate this expression a little bit. What I'm going to do is I'm gonna multiply both the numerator and the denominator by one plus cosine of x. So times the denominator, I have to do the same thing, one plus cosine of x. I'm not changing the value of the expression. This is just multiplying it by one. But what does that do for us? Well, I can rewrite the whole thing as the limit as x approaches zero."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "So times the denominator, I have to do the same thing, one plus cosine of x. I'm not changing the value of the expression. This is just multiplying it by one. But what does that do for us? Well, I can rewrite the whole thing as the limit as x approaches zero. So one minus cosine of x times one plus cosine of x. Well, that is just going to be, let me do this in another color. That is going to be one squared, which is just one, minus cosine squared of x. Cosine squared of x, difference of squares."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "Well, I can rewrite the whole thing as the limit as x approaches zero. So one minus cosine of x times one plus cosine of x. Well, that is just going to be, let me do this in another color. That is going to be one squared, which is just one, minus cosine squared of x. Cosine squared of x, difference of squares. And then in the denominator, I am going to have these, which is just x times one plus cosine of x. Now what is one minus cosine squared of x? Well, this comes straight out of the Pythagorean identity, trig identity."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "That is going to be one squared, which is just one, minus cosine squared of x. Cosine squared of x, difference of squares. And then in the denominator, I am going to have these, which is just x times one plus cosine of x. Now what is one minus cosine squared of x? Well, this comes straight out of the Pythagorean identity, trig identity. This is the same thing as sine squared of x. So sine squared of x. And so I can rewrite all of this as being equal to the limit as x approaches zero."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "Well, this comes straight out of the Pythagorean identity, trig identity. This is the same thing as sine squared of x. So sine squared of x. And so I can rewrite all of this as being equal to the limit as x approaches zero. And let me rewrite this as, instead of sine squared of x, that's the same thing as sine of x times sine of x. Let me write it that way. Sine x times sine x."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "And so I can rewrite all of this as being equal to the limit as x approaches zero. And let me rewrite this as, instead of sine squared of x, that's the same thing as sine of x times sine of x. Let me write it that way. Sine x times sine x. So I'll take the first sine of x. So I'll do, I'll take this one right over here and put it over this x. So sine of x over x times the second sine of x, let's say this one, over one plus cosine of x."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "Sine x times sine x. So I'll take the first sine of x. So I'll do, I'll take this one right over here and put it over this x. So sine of x over x times the second sine of x, let's say this one, over one plus cosine of x. Times sine of x over one plus cosine of x. All I've done is I've leveraged a trigonometric identity and I've done a little bit of algebraic manipulation. Well here, the limit of the product of these two expressions is going to be the same thing as the product of the limits."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "So sine of x over x times the second sine of x, let's say this one, over one plus cosine of x. Times sine of x over one plus cosine of x. All I've done is I've leveraged a trigonometric identity and I've done a little bit of algebraic manipulation. Well here, the limit of the product of these two expressions is going to be the same thing as the product of the limits. So I can rewrite this as being equal to the limit as x approaches zero of sine of x over x times the limit as x approaches zero of sine of x over one plus cosine of x. Now, we said going into this video that we're going to assume that we know what this is. We've proven it in other videos."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "Well here, the limit of the product of these two expressions is going to be the same thing as the product of the limits. So I can rewrite this as being equal to the limit as x approaches zero of sine of x over x times the limit as x approaches zero of sine of x over one plus cosine of x. Now, we said going into this video that we're going to assume that we know what this is. We've proven it in other videos. What is the limit as x approaches zero of sine of x over x? Well, that is equal to one. So this whole limit is just going to be dependent on whatever this is equal to."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "We've proven it in other videos. What is the limit as x approaches zero of sine of x over x? Well, that is equal to one. So this whole limit is just going to be dependent on whatever this is equal to. Well, this is pretty straightforward here. As x approaches zero, the numerator's approaching zero, sine of zero is zero. The denominator is approaching, cosine of zero is one, so the denominator is approaching two."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "So this whole limit is just going to be dependent on whatever this is equal to. Well, this is pretty straightforward here. As x approaches zero, the numerator's approaching zero, sine of zero is zero. The denominator is approaching, cosine of zero is one, so the denominator is approaching two. So this is approaching zero over two or just zero. So that's approaching zero. One times zero, well, this is just going to be equal to zero and we're done."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy (2).mp3", "Sentence": "The denominator is approaching, cosine of zero is one, so the denominator is approaching two. So this is approaching zero over two or just zero. So that's approaching zero. One times zero, well, this is just going to be equal to zero and we're done. Using that fact and a little bit of trig identities and a little bit of algebraic manipulation, we were able to show that our original limit, the limit as x approaches zero of one minus cosine of x over x is equal to zero. I encourage you to graph it. You will see that that makes sense from a graphical point of view as well."}, {"video_title": "Worked example Product rule with table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "When x is equal to three, the value of the function is six, f of three is six, you could view it that way, h of three is zero, f prime of three is six, and h prime of three is four. And now they want us to evaluate the derivative with respect to x of the product of f of x and h of x when x is equal to three. So one way you could view this is, if we viewed some function, if we viewed some function g of x, g of x as being equal to the product of f of x and h of x, this expression is the derivative of g of x. So we could write g prime of x is equal to the derivative with respect to x of f of x times h of x, which is what we see right here which is what we want to evaluate at x equals three. So we essentially want to evaluate g prime of three. This is what they're asking us to do. Well, to do that, let's go first up here."}, {"video_title": "Worked example Product rule with table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So we could write g prime of x is equal to the derivative with respect to x of f of x times h of x, which is what we see right here which is what we want to evaluate at x equals three. So we essentially want to evaluate g prime of three. This is what they're asking us to do. Well, to do that, let's go first up here. Let's just think about what they're doing. They're asking us to take the derivative with respect to x of the product of two functions that we have some information about. Well, if we're taking the derivative of the product of two functions, you could imagine that the product rule could prove useful here."}, {"video_title": "Worked example Product rule with table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, to do that, let's go first up here. Let's just think about what they're doing. They're asking us to take the derivative with respect to x of the product of two functions that we have some information about. Well, if we're taking the derivative of the product of two functions, you could imagine that the product rule could prove useful here. So I'm just gonna restate the product rule. This is going to be equal to the derivative of the first function, f prime of x, times the second function, not taking its derivative, plus the first function, not taking its derivative, f of x, times the derivative of the second function, h prime of x. So if you're trying to find g prime of three, well, that's just going to be f prime of three times h of three plus f of three times h prime of three."}, {"video_title": "Worked example Product rule with table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, if we're taking the derivative of the product of two functions, you could imagine that the product rule could prove useful here. So I'm just gonna restate the product rule. This is going to be equal to the derivative of the first function, f prime of x, times the second function, not taking its derivative, plus the first function, not taking its derivative, f of x, times the derivative of the second function, h prime of x. So if you're trying to find g prime of three, well, that's just going to be f prime of three times h of three plus f of three times h prime of three. And lucky for us, they give us what all of these things evaluate to. f prime of three, right over here, they tell us. f prime, when x is equal to three, is equal to six."}, {"video_title": "Worked example Product rule with table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So if you're trying to find g prime of three, well, that's just going to be f prime of three times h of three plus f of three times h prime of three. And lucky for us, they give us what all of these things evaluate to. f prime of three, right over here, they tell us. f prime, when x is equal to three, is equal to six. So this right over here is six. H of three, they give us that too. H of three, when x is three, the value of our function is zero."}, {"video_title": "Worked example Product rule with table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "f prime, when x is equal to three, is equal to six. So this right over here is six. H of three, they give us that too. H of three, when x is three, the value of our function is zero. So this is zero. So this first term, you just get six times zero, which is gonna be zero, but we'll get to that. Now, f of three."}, {"video_title": "Worked example Product rule with table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "H of three, when x is three, the value of our function is zero. So this is zero. So this first term, you just get six times zero, which is gonna be zero, but we'll get to that. Now, f of three. F of three, well, the function, when x is equal to three, y, or the f of three is equal to six. So that is six. And then finally, the h prime evaluated at three, h prime of x, when x is equal to three, h prime of x is equal to four."}, {"video_title": "Worked example Product rule with table Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Now, f of three. F of three, well, the function, when x is equal to three, y, or the f of three is equal to six. So that is six. And then finally, the h prime evaluated at three, h prime of x, when x is equal to three, h prime of x is equal to four. Or you could say this is h prime of three. So this is four. And so there you have it."}, {"video_title": "Fundamental theorem of calculus (Part 2) AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say we've got some function f that is continuous over the interval between c and d. And the reason why I'm using c and d instead of a and b is so I can use a and b for later. And let's say we set up some function capital F of x, which is defined as the area under the curve between c and some value x, where x is in this interval where f is continuous, under the curve, so it's the area under the curve between c and x. So if this is x right over here, under the curve, f of t, dt. So this right over here, f of x, is that area. That right over there is what f of x is. Now the fundamental theorem of calculus tells us that if f is continuous over this interval, then f of x is differentiable at every x in the interval and the derivative, the derivative of capital F of x. And let me be clear."}, {"video_title": "Fundamental theorem of calculus (Part 2) AP Calculus AB Khan Academy.mp3", "Sentence": "So this right over here, f of x, is that area. That right over there is what f of x is. Now the fundamental theorem of calculus tells us that if f is continuous over this interval, then f of x is differentiable at every x in the interval and the derivative, the derivative of capital F of x. And let me be clear. Capital F of x is differentiable at every possible x between c and d, and the derivative of capital F of x is going to be equal to lowercase f of x. Fair enough. Now what I wanna do in this video is connect the first fundamental theorem of calculus to the second part, or the second fundamental theorem of calculus, which we tend to use to actually evaluate definite integrals."}, {"video_title": "Fundamental theorem of calculus (Part 2) AP Calculus AB Khan Academy.mp3", "Sentence": "And let me be clear. Capital F of x is differentiable at every possible x between c and d, and the derivative of capital F of x is going to be equal to lowercase f of x. Fair enough. Now what I wanna do in this video is connect the first fundamental theorem of calculus to the second part, or the second fundamental theorem of calculus, which we tend to use to actually evaluate definite integrals. So let's think about, let's think about what f of, f of b, what f of b minus f of a is, what this is, where both b and a are also in this interval. So f of b, and we're gonna assume that b is larger than a. So let's say that b is this right over here."}, {"video_title": "Fundamental theorem of calculus (Part 2) AP Calculus AB Khan Academy.mp3", "Sentence": "Now what I wanna do in this video is connect the first fundamental theorem of calculus to the second part, or the second fundamental theorem of calculus, which we tend to use to actually evaluate definite integrals. So let's think about, let's think about what f of, f of b, what f of b minus f of a is, what this is, where both b and a are also in this interval. So f of b, and we're gonna assume that b is larger than a. So let's say that b is this right over here. Let me do that in the same color. So let's say that b is right over here. F of b is going to be equal to, we just literally replaced the b where you see the x, it's going to be equal to the definite integral between c and b of f of t, dt, which is just another way of saying, it's just another way of saying is the area under the curve between c and b."}, {"video_title": "Fundamental theorem of calculus (Part 2) AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say that b is this right over here. Let me do that in the same color. So let's say that b is right over here. F of b is going to be equal to, we just literally replaced the b where you see the x, it's going to be equal to the definite integral between c and b of f of t, dt, which is just another way of saying, it's just another way of saying is the area under the curve between c and b. So this f of b, capital F of b, is all of this business, all of this business right over here. And from that, we are going to want to subtract, we are going to subtract capital F of a, which is just the integral between c and lowercase a of lowercase f of t, dt. So let's say that this is, let's say that this is a right over here."}, {"video_title": "Fundamental theorem of calculus (Part 2) AP Calculus AB Khan Academy.mp3", "Sentence": "F of b is going to be equal to, we just literally replaced the b where you see the x, it's going to be equal to the definite integral between c and b of f of t, dt, which is just another way of saying, it's just another way of saying is the area under the curve between c and b. So this f of b, capital F of b, is all of this business, all of this business right over here. And from that, we are going to want to subtract, we are going to subtract capital F of a, which is just the integral between c and lowercase a of lowercase f of t, dt. So let's say that this is, let's say that this is a right over here. Capital F of a is just literally the area between c and a under the curve, lowercase f of t. So it's this right over here. It's all of this business, all of this right over here. So what is, if you have this blue area, which is all of this, and you subtract out this red, this magenta area, what are you left with?"}, {"video_title": "Fundamental theorem of calculus (Part 2) AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say that this is, let's say that this is a right over here. Capital F of a is just literally the area between c and a under the curve, lowercase f of t. So it's this right over here. It's all of this business, all of this right over here. So what is, if you have this blue area, which is all of this, and you subtract out this red, this magenta area, what are you left with? Well, you're left with this green area. You're left with this green area right over here. And how would we represent that?"}, {"video_title": "Fundamental theorem of calculus (Part 2) AP Calculus AB Khan Academy.mp3", "Sentence": "So what is, if you have this blue area, which is all of this, and you subtract out this red, this magenta area, what are you left with? Well, you're left with this green area. You're left with this green area right over here. And how would we represent that? How would we denote that? Well, we could denote that as the definite integral between a and b of f of t, of f of t, dt. And there you have it."}, {"video_title": "Fundamental theorem of calculus (Part 2) AP Calculus AB Khan Academy.mp3", "Sentence": "And how would we represent that? How would we denote that? Well, we could denote that as the definite integral between a and b of f of t, of f of t, dt. And there you have it. This right over here is the second fundamental theorem of calculus. It tells us that this, if f is continuous on the interval, that this is going to be equal to the antiderivative or an antiderivative of f. And we see right over here that capital F is the antiderivative of f. So we could view this as capital F antiderivative. This is how we define capital F. The antiderivative, or we didn't define it that way, but the fundamental theorem of calculus tells us that capital F is an antiderivative of lowercase f. So right over here, this tells you, if you have a definite integral like this, it completely equivalent to an antiderivative of it evaluated at f and from that, or evaluated at b, and from that, you subtract it, evaluated at a."}, {"video_title": "Fundamental theorem of calculus (Part 2) AP Calculus AB Khan Academy.mp3", "Sentence": "And there you have it. This right over here is the second fundamental theorem of calculus. It tells us that this, if f is continuous on the interval, that this is going to be equal to the antiderivative or an antiderivative of f. And we see right over here that capital F is the antiderivative of f. So we could view this as capital F antiderivative. This is how we define capital F. The antiderivative, or we didn't define it that way, but the fundamental theorem of calculus tells us that capital F is an antiderivative of lowercase f. So right over here, this tells you, if you have a definite integral like this, it completely equivalent to an antiderivative of it evaluated at f and from that, or evaluated at b, and from that, you subtract it, evaluated at a. So normally it looks like this. I just switched the order. The definite integral from a to b of f of t dt is equal to an antiderivative of f, so capital F evaluated at b, and from that, the antiderivative, from that, subtract the antiderivative evaluated at a."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "So we're going to rotate it around the vertical line x is equal to 2. We're going to rotate it right around like that. And we will get, if we rotate it like that, we will get this shape, so this strange-looking shape. It's hollowed out in the middle. You can kind of use the y equals x squared part kind of hollows out the middle, and it's really the stuff in between that forms the wall of this kind of rotated shape. So let's think about how we can figure out the volume. And we're going to do it using the disk method, sometimes called the ring method."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "It's hollowed out in the middle. You can kind of use the y equals x squared part kind of hollows out the middle, and it's really the stuff in between that forms the wall of this kind of rotated shape. So let's think about how we can figure out the volume. And we're going to do it using the disk method, sometimes called the ring method. Actually, it's going to be more of the washer method to do this one right over here. So we're rotating around a vertical line. We want to use our disk or ring or washer method."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "And we're going to do it using the disk method, sometimes called the ring method. Actually, it's going to be more of the washer method to do this one right over here. So we're rotating around a vertical line. We want to use our disk or ring or washer method. And so it will be really helpful to have a bunch of rings stacked up in the y direction. So we're probably going to want to integrate with respect to y. Let me make it clear what I'm talking about."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "We want to use our disk or ring or washer method. And so it will be really helpful to have a bunch of rings stacked up in the y direction. So we're probably going to want to integrate with respect to y. Let me make it clear what I'm talking about. Each of our rings are going to look something like this. So let me do my best attempt to draw. So that's the inner radius of the ring defined by y is equal to x squared."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "Let me make it clear what I'm talking about. Each of our rings are going to look something like this. So let me do my best attempt to draw. So that's the inner radius of the ring defined by y is equal to x squared. That's the inner radius. And then the outside radius of the ring might look something like this. So the outside radius, my best attempt to draw it reasonably, the outside radius of the ring might look something like that."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "So that's the inner radius of the ring defined by y is equal to x squared. That's the inner radius. And then the outside radius of the ring might look something like this. So the outside radius, my best attempt to draw it reasonably, the outside radius of the ring might look something like that. And the depth of the ring would be a little dy, just like that. So let me draw the depth. So let me see, right over here."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "So the outside radius, my best attempt to draw it reasonably, the outside radius of the ring might look something like that. And the depth of the ring would be a little dy, just like that. So let me draw the depth. So let me see, right over here. You can kind of view it as the height of the ring because we are thinking in a vertical direction. So we have a ring for a given y, for say this y right over here. It's essentially the ring generated if you were to take this rectangle, this rectangle of height dy and rotate it around the line x equals 2."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "So let me see, right over here. You can kind of view it as the height of the ring because we are thinking in a vertical direction. So we have a ring for a given y, for say this y right over here. It's essentially the ring generated if you were to take this rectangle, this rectangle of height dy and rotate it around the line x equals 2. And we're going to construct a ring like that for each of the y's in our interval. So you can imagine stacking up a whole set of rings just like this and then taking the sum of the volumes of all of those rings and the limit of that sum as you have an infinite number of rings with, or close to infinite, or really infinite number of rings with infinitesimally small height or depth or dy. So let's figure out how we would do this."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "It's essentially the ring generated if you were to take this rectangle, this rectangle of height dy and rotate it around the line x equals 2. And we're going to construct a ring like that for each of the y's in our interval. So you can imagine stacking up a whole set of rings just like this and then taking the sum of the volumes of all of those rings and the limit of that sum as you have an infinite number of rings with, or close to infinite, or really infinite number of rings with infinitesimally small height or depth or dy. So let's figure out how we would do this. Well, we know that all we have to do is figure out what the volume of one of these rings are for a given y as a function of y and then integrate along all of them, sum them all up. So let's figure out what the volume of one of these rings are. And to do it, we're going to express these functions as functions of y."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "So let's figure out how we would do this. Well, we know that all we have to do is figure out what the volume of one of these rings are for a given y as a function of y and then integrate along all of them, sum them all up. So let's figure out what the volume of one of these rings are. And to do it, we're going to express these functions as functions of y. So our purple function, y is equal to square root of x. If we square both sides, we would get y squared is equal to x. Let me swap the sides so it makes it clear that now x is a function of y. x is equal to y squared."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "And to do it, we're going to express these functions as functions of y. So our purple function, y is equal to square root of x. If we square both sides, we would get y squared is equal to x. Let me swap the sides so it makes it clear that now x is a function of y. x is equal to y squared. That's our top function, the way we've drawn it, the kind of outer shell for our figure. And then y is equal to x squared. If you take the principal root of both sides of that, it all works out because we're operating in the first quadrant here."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "Let me swap the sides so it makes it clear that now x is a function of y. x is equal to y squared. That's our top function, the way we've drawn it, the kind of outer shell for our figure. And then y is equal to x squared. If you take the principal root of both sides of that, it all works out because we're operating in the first quadrant here. That's the part of it that we care about. So you're going to get x is equal to the square root of y. x is equal to square root of y. That is our yellow function right over there."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "If you take the principal root of both sides of that, it all works out because we're operating in the first quadrant here. That's the part of it that we care about. So you're going to get x is equal to the square root of y. x is equal to square root of y. That is our yellow function right over there. Now, how do we figure out the area of the surface of one of these rings or one of these washers? Well, the area, let me do this in orange because I drew that ring in orange. So the area of the surface right over here in orange as a function of y is going to be equal to the area of the circle if I just consider the outer radius."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "That is our yellow function right over there. Now, how do we figure out the area of the surface of one of these rings or one of these washers? Well, the area, let me do this in orange because I drew that ring in orange. So the area of the surface right over here in orange as a function of y is going to be equal to the area of the circle if I just consider the outer radius. And then I subtract out the area of the circle constructed by the inner radius. Just kind of subtract it out. So the outer circle radius, so it's going to be pi times outer radius squared minus pi times inner radius squared."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "So the area of the surface right over here in orange as a function of y is going to be equal to the area of the circle if I just consider the outer radius. And then I subtract out the area of the circle constructed by the inner radius. Just kind of subtract it out. So the outer circle radius, so it's going to be pi times outer radius squared minus pi times inner radius squared. And we want this all to be a function of y. So the outer radius as a function of y is going to be what? Well, it might be easier to visualize."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "So the outer circle radius, so it's going to be pi times outer radius squared minus pi times inner radius squared. And we want this all to be a function of y. So the outer radius as a function of y is going to be what? Well, it might be easier to visualize. Actually, I'll try it both places. So it's this entire distance right over here. Essentially, the distance between the vertical line, the horizontal distance between our vertical line and our outer function, the horizontal distance between our vertical line and our outer function."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "Well, it might be easier to visualize. Actually, I'll try it both places. So it's this entire distance right over here. Essentially, the distance between the vertical line, the horizontal distance between our vertical line and our outer function, the horizontal distance between our vertical line and our outer function. So if you think about it in terms of x, it's going to be 2 minus whatever the x value is right over here. So the x value right over here is going to be y squared. Remember, we want this as a function of y."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "Essentially, the distance between the vertical line, the horizontal distance between our vertical line and our outer function, the horizontal distance between our vertical line and our outer function. So if you think about it in terms of x, it's going to be 2 minus whatever the x value is right over here. So the x value right over here is going to be y squared. Remember, we want this as a function of y. So our outer radius, this whole distance is going to be 2 minus the x value here as a function of y. That x value is y squared. So the outer radius is 2 minus y squared."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "Remember, we want this as a function of y. So our outer radius, this whole distance is going to be 2 minus the x value here as a function of y. That x value is y squared. So the outer radius is 2 minus y squared. We want it as a function of y. 2 minus y squared. And then our inner radius is going to be equal to what?"}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "So the outer radius is 2 minus y squared. We want it as a function of y. 2 minus y squared. And then our inner radius is going to be equal to what? Well, that's going to be the difference, the horizontal distance between this vertical line and our inner function, our inner boundary. So it's going to be the horizontal distance between 2 and whatever x value this is. But this x value as a function of y is just the square root of y."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "And then our inner radius is going to be equal to what? Well, that's going to be the difference, the horizontal distance between this vertical line and our inner function, our inner boundary. So it's going to be the horizontal distance between 2 and whatever x value this is. But this x value as a function of y is just the square root of y. So it's going to be 2 minus the square root of y. And so now we can come up with an expression for area. It's going to be, and I'll just factor out, I'll leave the pi there."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "But this x value as a function of y is just the square root of y. So it's going to be 2 minus the square root of y. And so now we can come up with an expression for area. It's going to be, and I'll just factor out, I'll leave the pi there. So it's going to be pi, I'll write it over here. It's going to be pi times outer radius squared. Well, the outer radius is 2 minus y squared."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "It's going to be, and I'll just factor out, I'll leave the pi there. So it's going to be pi, I'll write it over here. It's going to be pi times outer radius squared. Well, the outer radius is 2 minus y squared. And let me just, well, I'll just write it, 2 minus y squared. And we're going to square that. Squared minus pi times the inner radius squared."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the outer radius is 2 minus y squared. And let me just, well, I'll just write it, 2 minus y squared. And we're going to square that. Squared minus pi times the inner radius squared. Well, we already figured that out. The inner radius is 2 minus square root of y. And we're going to square that one too."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "Squared minus pi times the inner radius squared. Well, we already figured that out. The inner radius is 2 minus square root of y. And we're going to square that one too. So this gives us the area of one of our rings as a function of y, the top of the ring where I shaded in orange. And now if we want the volume of one of those rings, we have to multiply it by its depth or its height, the way we've drawn it right over here. And its height, we've done this multiple times already, right over here, is an infinitesimal change in y."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "And we're going to square that one too. So this gives us the area of one of our rings as a function of y, the top of the ring where I shaded in orange. And now if we want the volume of one of those rings, we have to multiply it by its depth or its height, the way we've drawn it right over here. And its height, we've done this multiple times already, right over here, is an infinitesimal change in y. So we're going to multiply all that business times dy. This is the volume of one of our rings. And then we want to sum up all of the rings over our interval."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "And its height, we've done this multiple times already, right over here, is an infinitesimal change in y. So we're going to multiply all that business times dy. This is the volume of one of our rings. And then we want to sum up all of the rings over our interval. So we're going to sum up all of the rings over the interval. And when you take the integral sign, it's a sum where you're taking the limit as you have an infinite number of rings that become infinitesimally small in height or depth, depending on how you view it. And what's our interval?"}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "And then we want to sum up all of the rings over our interval. So we're going to sum up all of the rings over the interval. And when you take the integral sign, it's a sum where you're taking the limit as you have an infinite number of rings that become infinitesimally small in height or depth, depending on how you view it. And what's our interval? So we've looked at this multiple times. These two graphs, you could do it by inspection. You could try to solve it in some way, but it's pretty obvious that they intersect at, remember, we care about our y interval."}, {"video_title": "Washer method rotating around vertical line (not y-axis), part 1 AP Calculus AB Khan Academy.mp3", "Sentence": "And what's our interval? So we've looked at this multiple times. These two graphs, you could do it by inspection. You could try to solve it in some way, but it's pretty obvious that they intersect at, remember, we care about our y interval. They intersect at y is equal to 0 and y is equal to 1. Intersect at y equals 0 and y equals 1. And there you have it."}, {"video_title": "Formal and alternate form of the derivative example 1 Differential Calculus Khan Academy.mp3", "Sentence": "The alternate form of the derivative of the function f at a number a denoted by f prime of a is given by this stuff. Now, this might look a little strange to you, but if you really think about what it's saying, it's really just taking the slope of the tangent line between a comma f of a. So let's imagine some arbitrary function like this. Let's say that that is, well, I'll just write, that's our function f. That's our function f. And so you could have the point when x is equal to a. This is our x-axis. When x is equal to a, this is the point a f of a. You notice a f of a."}, {"video_title": "Formal and alternate form of the derivative example 1 Differential Calculus Khan Academy.mp3", "Sentence": "Let's say that that is, well, I'll just write, that's our function f. That's our function f. And so you could have the point when x is equal to a. This is our x-axis. When x is equal to a, this is the point a f of a. You notice a f of a. And then we could take the slope between that and some arbitrary point. Let's call that x. So this is the point x f of x."}, {"video_title": "Formal and alternate form of the derivative example 1 Differential Calculus Khan Academy.mp3", "Sentence": "You notice a f of a. And then we could take the slope between that and some arbitrary point. Let's call that x. So this is the point x f of x. And notice, this top, the numerator right here, this is just our change in the value of our function. Or you could view that as a change in the vertical axis. So that would give you this distance right over here."}, {"video_title": "Formal and alternate form of the derivative example 1 Differential Calculus Khan Academy.mp3", "Sentence": "So this is the point x f of x. And notice, this top, the numerator right here, this is just our change in the value of our function. Or you could view that as a change in the vertical axis. So that would give you this distance right over here. That's what we're doing up here in the numerator. And then in the denominator, we're finding the change in our horizontal values, horizontal coordinates. Let me do that in a different color."}, {"video_title": "Formal and alternate form of the derivative example 1 Differential Calculus Khan Academy.mp3", "Sentence": "So that would give you this distance right over here. That's what we're doing up here in the numerator. And then in the denominator, we're finding the change in our horizontal values, horizontal coordinates. Let me do that in a different color. So the change in the horizontal, that's this right over here. And then they're trying to find the limit as x approaches a. So as x gets closer and closer and closer and closer to a, what's going to happen is that when x is out here, we have this secant line."}, {"video_title": "Formal and alternate form of the derivative example 1 Differential Calculus Khan Academy.mp3", "Sentence": "Let me do that in a different color. So the change in the horizontal, that's this right over here. And then they're trying to find the limit as x approaches a. So as x gets closer and closer and closer and closer to a, what's going to happen is that when x is out here, we have this secant line. We're finding the slope of this secant line. But as x gets closer and closer, the secant lines better and better and better approximate the slope of the tangent line. Where the limit as x approaches a, but doesn't quite equal a, is going to be, this is actually our definition of our derivative, or I guess the alternate form of the derivative definition."}, {"video_title": "Formal and alternate form of the derivative example 1 Differential Calculus Khan Academy.mp3", "Sentence": "So as x gets closer and closer and closer and closer to a, what's going to happen is that when x is out here, we have this secant line. We're finding the slope of this secant line. But as x gets closer and closer, the secant lines better and better and better approximate the slope of the tangent line. Where the limit as x approaches a, but doesn't quite equal a, is going to be, this is actually our definition of our derivative, or I guess the alternate form of the derivative definition. And this would be the slope of the tangent line, if it exists. So with all that out of the way, let's try to answer their question. With the alternative form of the derivative as an aid, make sense of the following limit expression by identifying the function f and the number a."}, {"video_title": "Formal and alternate form of the derivative example 1 Differential Calculus Khan Academy.mp3", "Sentence": "Where the limit as x approaches a, but doesn't quite equal a, is going to be, this is actually our definition of our derivative, or I guess the alternate form of the derivative definition. And this would be the slope of the tangent line, if it exists. So with all that out of the way, let's try to answer their question. With the alternative form of the derivative as an aid, make sense of the following limit expression by identifying the function f and the number a. So right here, they want to find the slope of the tangent line at 5. Here, they wanted to find the slope of the tangent line at a. So it's pretty clear that a is equal to 5 and that f of a is equal to 125."}, {"video_title": "Formal and alternate form of the derivative example 1 Differential Calculus Khan Academy.mp3", "Sentence": "With the alternative form of the derivative as an aid, make sense of the following limit expression by identifying the function f and the number a. So right here, they want to find the slope of the tangent line at 5. Here, they wanted to find the slope of the tangent line at a. So it's pretty clear that a is equal to 5 and that f of a is equal to 125. Now, what about f of x? Well, here, it's the limit of f of x minus f of a. Well, here's the limit as x to the third minus 125."}, {"video_title": "Formal and alternate form of the derivative example 1 Differential Calculus Khan Academy.mp3", "Sentence": "So it's pretty clear that a is equal to 5 and that f of a is equal to 125. Now, what about f of x? Well, here, it's the limit of f of x minus f of a. Well, here's the limit as x to the third minus 125. And this makes sense. If f of x is equal to x to the third, then it makes sense that f of 5 is going to be 5 to the third is going to be 125. And we're also taking up here, the limit as x approaches a."}, {"video_title": "Formal and alternate form of the derivative example 1 Differential Calculus Khan Academy.mp3", "Sentence": "Well, here's the limit as x to the third minus 125. And this makes sense. If f of x is equal to x to the third, then it makes sense that f of 5 is going to be 5 to the third is going to be 125. And we're also taking up here, the limit as x approaches a. Here, we're taking the limit as x approaches 5. So this is the derivative of the function f of x is equal to x to the third. Let me write that down in the green color."}, {"video_title": "Formal and alternate form of the derivative example 1 Differential Calculus Khan Academy.mp3", "Sentence": "And we're also taking up here, the limit as x approaches a. Here, we're taking the limit as x approaches 5. So this is the derivative of the function f of x is equal to x to the third. Let me write that down in the green color. x to the third at the number a is equal to 5. And so we can imagine this. Let's try to actually graph it just so we can imagine it."}, {"video_title": "Formal and alternate form of the derivative example 1 Differential Calculus Khan Academy.mp3", "Sentence": "Let me write that down in the green color. x to the third at the number a is equal to 5. And so we can imagine this. Let's try to actually graph it just so we can imagine it. Actually, I'll do it out here where I have a little bit better contrast with the colors. So let's say that is my y-axis. Let's say that this is my x-axis."}, {"video_title": "Formal and alternate form of the derivative example 1 Differential Calculus Khan Academy.mp3", "Sentence": "Let's try to actually graph it just so we can imagine it. Actually, I'll do it out here where I have a little bit better contrast with the colors. So let's say that is my y-axis. Let's say that this is my x-axis. I'm not going to quite draw it to scale. Let's say this right over here is a point 125 or y. This is when y equals 125."}, {"video_title": "Formal and alternate form of the derivative example 1 Differential Calculus Khan Academy.mp3", "Sentence": "Let's say that this is my x-axis. I'm not going to quite draw it to scale. Let's say this right over here is a point 125 or y. This is when y equals 125. This is when x is equal to 5. So they're clearly not at the same scale. But the function is going to look something like this."}, {"video_title": "Formal and alternate form of the derivative example 1 Differential Calculus Khan Academy.mp3", "Sentence": "This is when y equals 125. This is when x is equal to 5. So they're clearly not at the same scale. But the function is going to look something like this. We know what x to the third looks like. It looks something like this. We know it's going to look something like this."}, {"video_title": "Formal and alternate form of the derivative example 1 Differential Calculus Khan Academy.mp3", "Sentence": "But the function is going to look something like this. We know what x to the third looks like. It looks something like this. We know it's going to look something like this. So here, our a is equal to 5. This point right over here is 5, 125. And then we're taking the slope between that point and an arbitrary x value, or I should say an arbitrary other point on the curve."}, {"video_title": "Formal and alternate form of the derivative example 1 Differential Calculus Khan Academy.mp3", "Sentence": "We know it's going to look something like this. So here, our a is equal to 5. This point right over here is 5, 125. And then we're taking the slope between that point and an arbitrary x value, or I should say an arbitrary other point on the curve. So this right over here would be the point. We could call that x comma x to the third. We know that f of x is equal to x to the third."}, {"video_title": "Formal and alternate form of the derivative example 1 Differential Calculus Khan Academy.mp3", "Sentence": "And then we're taking the slope between that point and an arbitrary x value, or I should say an arbitrary other point on the curve. So this right over here would be the point. We could call that x comma x to the third. We know that f of x is equal to x to the third. And let me make it clear. This is a graph of y is equal to x to the third. And so this expression right over here, all of this, this is the slope between these two points."}, {"video_title": "Formal and alternate form of the derivative example 1 Differential Calculus Khan Academy.mp3", "Sentence": "We know that f of x is equal to x to the third. And let me make it clear. This is a graph of y is equal to x to the third. And so this expression right over here, all of this, this is the slope between these two points. And as we take the limit as x approaches 5, so right now this is our x. As x gets closer and closer and closer to 5, these secant lines are going to better and better approximate the slope of the tangent line at x equals 5. So the slope of the tangent line would look something like that."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Maybe nothing crazy happens, so let's just try it out. If we try to do x equals 0, what happens? We get 2 sine of 0, which is 0, minus sine of 2 times 0. That's going to be sine of 0 again, which is 0. So our numerator is going to be equal to 0. Sine of 0, that's 0, and then we have another sine of 0 there, so all 0s. And our denominator, we're going to have a 0 minus sine of 0."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "That's going to be sine of 0 again, which is 0. So our numerator is going to be equal to 0. Sine of 0, that's 0, and then we have another sine of 0 there, so all 0s. And our denominator, we're going to have a 0 minus sine of 0. Well, that's also going to be 0, but we have that indeterminate form, we have that undefined 0 over 0 that we talked about in the last video. So maybe we can use L'Hopital's Rule here. In order to use L'Hopital's Rule, then the limit as x approaches 0 of the derivative of this function over the derivative of this function needs to exist."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And our denominator, we're going to have a 0 minus sine of 0. Well, that's also going to be 0, but we have that indeterminate form, we have that undefined 0 over 0 that we talked about in the last video. So maybe we can use L'Hopital's Rule here. In order to use L'Hopital's Rule, then the limit as x approaches 0 of the derivative of this function over the derivative of this function needs to exist. So let's just apply L'Hopital's Rule and let's just take the derivative of each of these and see if we can find the limit. If we can, then that's going to be the limit of this thing. So this thing, assuming that it exists, is going to be equal to the limit as x approaches 0 of the derivative of this numerator up here."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "In order to use L'Hopital's Rule, then the limit as x approaches 0 of the derivative of this function over the derivative of this function needs to exist. So let's just apply L'Hopital's Rule and let's just take the derivative of each of these and see if we can find the limit. If we can, then that's going to be the limit of this thing. So this thing, assuming that it exists, is going to be equal to the limit as x approaches 0 of the derivative of this numerator up here. And so what's the derivative of the numerator going to be? I'll do it in a new color. I'll do it in green."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So this thing, assuming that it exists, is going to be equal to the limit as x approaches 0 of the derivative of this numerator up here. And so what's the derivative of the numerator going to be? I'll do it in a new color. I'll do it in green. Well, the derivative of 2 sine of x is 2 cosine of x. And then minus, well, the derivative of sine of 2x is 2 cosine of 2x. So minus 2 cosine of 2x."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "I'll do it in green. Well, the derivative of 2 sine of x is 2 cosine of x. And then minus, well, the derivative of sine of 2x is 2 cosine of 2x. So minus 2 cosine of 2x. Just use the chain rule there. Derivative of the inside is just 2. That's the 2 out there."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So minus 2 cosine of 2x. Just use the chain rule there. Derivative of the inside is just 2. That's the 2 out there. Derivative of the outside is cosine of 2x. And we had that negative number out there. So that's the derivative of our numerator."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "That's the 2 out there. Derivative of the outside is cosine of 2x. And we had that negative number out there. So that's the derivative of our numerator. And what is the derivative of our denominator? Well, derivative of x is just 1. And derivative of sine of x is just cosine of x."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So that's the derivative of our numerator. And what is the derivative of our denominator? Well, derivative of x is just 1. And derivative of sine of x is just cosine of x. So 1 minus cosine of x. So let's try to evaluate this limit. What do we get?"}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And derivative of sine of x is just cosine of x. So 1 minus cosine of x. So let's try to evaluate this limit. What do we get? If we put a 0 up here, we're going to get 2 times cosine of 0, which is 2. Let me write it like this. So this is 2 times cosine of 0, which is 1."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "What do we get? If we put a 0 up here, we're going to get 2 times cosine of 0, which is 2. Let me write it like this. So this is 2 times cosine of 0, which is 1. So it's 2 minus 2 cosine of 2 times 0. Cosine of, let me write it this way. Actually, let me just do it this way."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So this is 2 times cosine of 0, which is 1. So it's 2 minus 2 cosine of 2 times 0. Cosine of, let me write it this way. Actually, let me just do it this way. If we just straight up evaluate the limit of the numerator and denominator, what are we going to get? We're going to get 2 cosine of 0, which is 2, minus 2 times cosine of, well, this 2 times 0 is still going to be 0. So minus 2 times cosine of 0, which is 2."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Actually, let me just do it this way. If we just straight up evaluate the limit of the numerator and denominator, what are we going to get? We're going to get 2 cosine of 0, which is 2, minus 2 times cosine of, well, this 2 times 0 is still going to be 0. So minus 2 times cosine of 0, which is 2. All of that over 1 minus the cosine of 0, which is 1. So once again, we get 0 over 0. So does this mean that the limit doesn't exist?"}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So minus 2 times cosine of 0, which is 2. All of that over 1 minus the cosine of 0, which is 1. So once again, we get 0 over 0. So does this mean that the limit doesn't exist? No, it still might exist. We might just want to do L'Hopital's Rule again. Let me take the derivative of that and put it over the derivative of that and then take the limit."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So does this mean that the limit doesn't exist? No, it still might exist. We might just want to do L'Hopital's Rule again. Let me take the derivative of that and put it over the derivative of that and then take the limit. And maybe L'Hopital's Rule will help us on the next stage. So let's see if it gets us anywhere. So this should be equal to the limit, if L'Hopital's Rule applies here, we're not 100% sure yet."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Let me take the derivative of that and put it over the derivative of that and then take the limit. And maybe L'Hopital's Rule will help us on the next stage. So let's see if it gets us anywhere. So this should be equal to the limit, if L'Hopital's Rule applies here, we're not 100% sure yet. This should be equal to the limit as x approaches 0 of the derivative of that thing over the derivative of that thing. So what's the derivative of 2 cosine of x? Well, the derivative of cosine of x is negative sine of x."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So this should be equal to the limit, if L'Hopital's Rule applies here, we're not 100% sure yet. This should be equal to the limit as x approaches 0 of the derivative of that thing over the derivative of that thing. So what's the derivative of 2 cosine of x? Well, the derivative of cosine of x is negative sine of x. So it's negative 2 sine of x. And then the derivative of cosine of 2x is negative 2 sine of 2x. So we're going to have this negative cancel out with the negative on the negative 2 and then a 2 times a 2."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Well, the derivative of cosine of x is negative sine of x. So it's negative 2 sine of x. And then the derivative of cosine of 2x is negative 2 sine of 2x. So we're going to have this negative cancel out with the negative on the negative 2 and then a 2 times a 2. So it's going to be plus 4 sine of 2x. Let me make sure I did that right. We have the minus 2 or the negative 2 on the outside."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So we're going to have this negative cancel out with the negative on the negative 2 and then a 2 times a 2. So it's going to be plus 4 sine of 2x. Let me make sure I did that right. We have the minus 2 or the negative 2 on the outside. Derivative of cosine of 2x is going to be 2 times negative sine of x. So the 2 times 2 is 4. The negative sine of x times the negative right there is a plus."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "We have the minus 2 or the negative 2 on the outside. Derivative of cosine of 2x is going to be 2 times negative sine of x. So the 2 times 2 is 4. The negative sine of x times the negative right there is a plus. The positive sign, so it's sine of 2x. So that's the numerator when you take the derivative. And the denominator, this is just an exercise in taking derivatives."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "The negative sine of x times the negative right there is a plus. The positive sign, so it's sine of 2x. So that's the numerator when you take the derivative. And the denominator, this is just an exercise in taking derivatives. What's the derivative of the denominator? Derivative of 1 is 0. And derivative of negative cosine of x is just sine of x."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And the denominator, this is just an exercise in taking derivatives. What's the derivative of the denominator? Derivative of 1 is 0. And derivative of negative cosine of x is just sine of x. So let's take this limit. So this is going to be equal to, well immediately if I take x is equal to 0 in the denominator, I know that sine of 0 is just 0. Let's see what happens in the numerator."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And derivative of negative cosine of x is just sine of x. So let's take this limit. So this is going to be equal to, well immediately if I take x is equal to 0 in the denominator, I know that sine of 0 is just 0. Let's see what happens in the numerator. Negative 2 times sine of 0, that's going to be 0. And then plus 4 times sine of 2 times 0, that's still sine of 0. So that's still going to be 0."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Let's see what happens in the numerator. Negative 2 times sine of 0, that's going to be 0. And then plus 4 times sine of 2 times 0, that's still sine of 0. So that's still going to be 0. So once again we've got indeterminate form again. Are we done? Do we give up?"}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So that's still going to be 0. So once again we've got indeterminate form again. Are we done? Do we give up? Do we say that L'Hopital's rule didn't work? No. Because this could have been our first limit problem."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Do we give up? Do we say that L'Hopital's rule didn't work? No. Because this could have been our first limit problem. And if this is our first limit problem, we say hey, maybe we could use L'Hopital's rule here because we've got an indeterminate form where both the numerator and the denominator approach 0 as x approaches 0. So let's take the derivatives again. This will be equal to, if the limit exists, the limit as x approaches 0."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Because this could have been our first limit problem. And if this is our first limit problem, we say hey, maybe we could use L'Hopital's rule here because we've got an indeterminate form where both the numerator and the denominator approach 0 as x approaches 0. So let's take the derivatives again. This will be equal to, if the limit exists, the limit as x approaches 0. Let's take the derivative of the numerator. The derivative of negative 2 sine of x is negative 2 cosine of x. And then plus the derivative of 4 sine of 2x, well it's 2 times 4, which is 8 times cosine of 2x."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "This will be equal to, if the limit exists, the limit as x approaches 0. Let's take the derivative of the numerator. The derivative of negative 2 sine of x is negative 2 cosine of x. And then plus the derivative of 4 sine of 2x, well it's 2 times 4, which is 8 times cosine of 2x. Derivative of sine of 2x is 2 cosine of 2x. And that first 2 gets multiplied by the 4 to get the 8. And then the derivative of the denominator, derivative of sine of x is just cosine of x."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And then plus the derivative of 4 sine of 2x, well it's 2 times 4, which is 8 times cosine of 2x. Derivative of sine of 2x is 2 cosine of 2x. And that first 2 gets multiplied by the 4 to get the 8. And then the derivative of the denominator, derivative of sine of x is just cosine of x. So let's evaluate this character. It looks like we've made some headway, or maybe L'Hopital's rule will stop applying here because we take the limit as x approaches 0 of cosine of x. That is 1."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And then the derivative of the denominator, derivative of sine of x is just cosine of x. So let's evaluate this character. It looks like we've made some headway, or maybe L'Hopital's rule will stop applying here because we take the limit as x approaches 0 of cosine of x. That is 1. So we're definitely not going to get that indeterminate form, that 0 over 0 on this iteration. Let's see what happens to the numerator. We get negative 2 times cosine of 0."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "That is 1. So we're definitely not going to get that indeterminate form, that 0 over 0 on this iteration. Let's see what happens to the numerator. We get negative 2 times cosine of 0. Well, that's just negative 2 because cosine of 0 is 1. Plus 8 times cosine of 2x. Well, x is 0, so it's going to be cosine of 0, which is 1."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "We get negative 2 times cosine of 0. Well, that's just negative 2 because cosine of 0 is 1. Plus 8 times cosine of 2x. Well, x is 0, so it's going to be cosine of 0, which is 1. So it's just going to be an 8. So negative 2 plus 8. Well, this thing right here, negative 2 plus 8 is 6."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Well, x is 0, so it's going to be cosine of 0, which is 1. So it's just going to be an 8. So negative 2 plus 8. Well, this thing right here, negative 2 plus 8 is 6. 6 over 1. This whole thing is equal to 6. So L'Hopital's rule, it applies to this last step."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Well, this thing right here, negative 2 plus 8 is 6. 6 over 1. This whole thing is equal to 6. So L'Hopital's rule, it applies to this last step. If this was the problem we were given, we said, hey, when we tried to apply the limit, we get the limit as the numerator approaches 0 is 0. Limit as this denominator approaches 0 is 0. The limit as the derivative of the numerator over the derivative of the denominator, that exists and it equals 6."}, {"video_title": "L'H\u00f4pital's rule example 1 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So L'Hopital's rule, it applies to this last step. If this was the problem we were given, we said, hey, when we tried to apply the limit, we get the limit as the numerator approaches 0 is 0. Limit as this denominator approaches 0 is 0. The limit as the derivative of the numerator over the derivative of the denominator, that exists and it equals 6. So this limit must be equal to 6. Well, if this limit is equal to 6, by the same argument, this limit is also going to be equal to 6. And by the same argument, this limit has got to also be equal to 6."}, {"video_title": "Maclaurin series of cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And I had said at the end of the videos that a Maclaurin series is just a special case of a Taylor series. In the case of a Maclaurin series, we're approximating this function around x is equal to 0. In a Taylor series, and we'll talk about that in a future video, you can pick an arbitrary x value, or f of x value, we should say, around which to approximate the function. But with that said, let's just focus on the Maclaurin, because to some degree, it's a little bit simpler. And that by itself can lead us to some pretty profound conclusions about mathematics. And that's actually where I'm trying to get to. So let's take the Maclaurin series of some interesting functions."}, {"video_title": "Maclaurin series of cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "But with that said, let's just focus on the Maclaurin, because to some degree, it's a little bit simpler. And that by itself can lead us to some pretty profound conclusions about mathematics. And that's actually where I'm trying to get to. So let's take the Maclaurin series of some interesting functions. And I'm going to do functions where it's pretty easy to take the derivatives. And you can keep taking their derivatives over and over and over and over and over again. So let's take the Maclaurin series of cosine of x."}, {"video_title": "Maclaurin series of cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's take the Maclaurin series of some interesting functions. And I'm going to do functions where it's pretty easy to take the derivatives. And you can keep taking their derivatives over and over and over and over and over again. So let's take the Maclaurin series of cosine of x. So if f of x is equal to cosine of x, then before I even apply this formula that we somewhat derived in the last video, or at least got the intuitive for in the last video, let's take a bunch of derivatives of f of x just so that we have a good sense of it. So if we take the first derivative, derivative of cosine of x is negative sine of x. If we take the derivative of that, derivative of sine of x is cosine of x, but we have that negative there."}, {"video_title": "Maclaurin series of cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's take the Maclaurin series of cosine of x. So if f of x is equal to cosine of x, then before I even apply this formula that we somewhat derived in the last video, or at least got the intuitive for in the last video, let's take a bunch of derivatives of f of x just so that we have a good sense of it. So if we take the first derivative, derivative of cosine of x is negative sine of x. If we take the derivative of that, derivative of sine of x is cosine of x, but we have that negative there. So it's negative cosine of x. If we take the derivative of that, so this is the third derivative of cosine of x. Now it's just going to be positive sine of x."}, {"video_title": "Maclaurin series of cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "If we take the derivative of that, derivative of sine of x is cosine of x, but we have that negative there. So it's negative cosine of x. If we take the derivative of that, so this is the third derivative of cosine of x. Now it's just going to be positive sine of x. And if we take the derivative of that, we get cosine of x again. We get cosine of x again. So if we take the derivative of that, this is the fourth derivative."}, {"video_title": "Maclaurin series of cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Now it's just going to be positive sine of x. And if we take the derivative of that, we get cosine of x again. We get cosine of x again. So if we take the derivative of that, this is the fourth derivative. I shouldn't use this notation, but you get the idea. We'll get cosine of x again. And if you look at what we talked about in the last video, we want the function and we want its various derivatives evaluated at 0."}, {"video_title": "Maclaurin series of cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So if we take the derivative of that, this is the fourth derivative. I shouldn't use this notation, but you get the idea. We'll get cosine of x again. And if you look at what we talked about in the last video, we want the function and we want its various derivatives evaluated at 0. So let's evaluate them at 0. So f of 0, cosine of 0 is 1. Whether you're talking about 0 radians or 0 degrees, it doesn't matter."}, {"video_title": "Maclaurin series of cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And if you look at what we talked about in the last video, we want the function and we want its various derivatives evaluated at 0. So let's evaluate them at 0. So f of 0, cosine of 0 is 1. Whether you're talking about 0 radians or 0 degrees, it doesn't matter. Sine of 0 is 0. So this is f prime of 0 is 0. And then cosine of 0 is once again 1, but we have the negative out there."}, {"video_title": "Maclaurin series of cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Whether you're talking about 0 radians or 0 degrees, it doesn't matter. Sine of 0 is 0. So this is f prime of 0 is 0. And then cosine of 0 is once again 1, but we have the negative out there. So it becomes negative 1. So f, the second derivative evaluated at 0 is negative 1. Let's take the third derivative."}, {"video_title": "Maclaurin series of cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And then cosine of 0 is once again 1, but we have the negative out there. So it becomes negative 1. So f, the second derivative evaluated at 0 is negative 1. Let's take the third derivative. The third derivative evaluated at 0, well, sine of 0 is just 0. And then the fourth derivative evaluated at 0, cosine of 0 is 1. So f of prime, prime, prime at 0 is now equal to 1."}, {"video_title": "Maclaurin series of cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let's take the third derivative. The third derivative evaluated at 0, well, sine of 0 is just 0. And then the fourth derivative evaluated at 0, cosine of 0 is 1. So f of prime, prime, prime at 0 is now equal to 1. So you see an interesting pattern here. 1, 0, negative 1, 0, 1. Then you go to 0."}, {"video_title": "Maclaurin series of cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So f of prime, prime, prime at 0 is now equal to 1. So you see an interesting pattern here. 1, 0, negative 1, 0, 1. Then you go to 0. Then you go to negative 1, 0. And so if we were to apply this to find its Maclaurin representation, what would we get? Let me do my best attempt at this."}, {"video_title": "Maclaurin series of cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Then you go to 0. Then you go to negative 1, 0. And so if we were to apply this to find its Maclaurin representation, what would we get? Let me do my best attempt at this. So we would get our polynomial would be. So our polynomial approximation of cosine of x is going to be f of 0. f of 0 is 1. And then we have 1 plus f prime of 0 times x."}, {"video_title": "Maclaurin series of cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let me do my best attempt at this. So we would get our polynomial would be. So our polynomial approximation of cosine of x is going to be f of 0. f of 0 is 1. And then we have 1 plus f prime of 0 times x. But f prime of 0 is just 0. So we're not going to have this term over there. It's going to be 0 times x. I won't even take the trouble of writing it down."}, {"video_title": "Maclaurin series of cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And then we have 1 plus f prime of 0 times x. But f prime of 0 is just 0. So we're not going to have this term over there. It's going to be 0 times x. I won't even take the trouble of writing it down. It would be this 0 times x. Then plus f prime, prime, or its second derivative, which is negative 1. So I'll write negative 1 times x squared over 2 factorial, which in this case is just going to be 2."}, {"video_title": "Maclaurin series of cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "It's going to be 0 times x. I won't even take the trouble of writing it down. It would be this 0 times x. Then plus f prime, prime, or its second derivative, which is negative 1. So I'll write negative 1 times x squared over 2 factorial, which in this case is just going to be 2. But I'll just write it down here as 2 factorial. It'll make the pattern a little bit more obvious. And then we go to the next term, the third derivative evaluated at 0."}, {"video_title": "Maclaurin series of cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So I'll write negative 1 times x squared over 2 factorial, which in this case is just going to be 2. But I'll just write it down here as 2 factorial. It'll make the pattern a little bit more obvious. And then we go to the next term, the third derivative evaluated at 0. But the third derivative evaluated at 0 is just 0. So this term won't be there as well. Then you go to the fourth derivative."}, {"video_title": "Maclaurin series of cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And then we go to the next term, the third derivative evaluated at 0. But the third derivative evaluated at 0 is just 0. So this term won't be there as well. Then you go to the fourth derivative. The fourth derivative evaluated at 0 is positive 1. So this coefficient right here is going to be a 1. And so you're going to have 1 times x to the fourth over 4 factorial."}, {"video_title": "Maclaurin series of cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Then you go to the fourth derivative. The fourth derivative evaluated at 0 is positive 1. So this coefficient right here is going to be a 1. And so you're going to have 1 times x to the fourth over 4 factorial. So plus x to the fourth over 4 factorial. And I think you start seeing a pattern now. You have sign switches."}, {"video_title": "Maclaurin series of cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so you're going to have 1 times x to the fourth over 4 factorial. So plus x to the fourth over 4 factorial. And I think you start seeing a pattern now. You have sign switches. And you would see this if we kept going. So you can verify it for yourself if you don't believe me. So you have a positive sign, a negative sign, a positive sign, and then a positive sign."}, {"video_title": "Maclaurin series of cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "You have sign switches. And you would see this if we kept going. So you can verify it for yourself if you don't believe me. So you have a positive sign, a negative sign, a positive sign, and then a positive sign. And you're going to have a negative sign, so on and so forth. And this is 1 times x to the 0th power. Then you jump 2 to x to the squared, jump 2 to x to the fourth."}, {"video_title": "Maclaurin series of cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So you have a positive sign, a negative sign, a positive sign, and then a positive sign. And you're going to have a negative sign, so on and so forth. And this is 1 times x to the 0th power. Then you jump 2 to x to the squared, jump 2 to x to the fourth. And so if we kept that up, we had a positive sign. Now you have a negative sign. It would be x to the sixth over 6 factorial."}, {"video_title": "Maclaurin series of cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Then you jump 2 to x to the squared, jump 2 to x to the fourth. And so if we kept that up, we had a positive sign. Now you have a negative sign. It would be x to the sixth over 6 factorial. Then you have a positive sign, x to the eighth over 8 factorial. And then you'd have a negative sign, x to the 10th over 10 factorial. And you could just keep going that way."}, {"video_title": "Maclaurin series of cos(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "It would be x to the sixth over 6 factorial. Then you have a positive sign, x to the eighth over 8 factorial. And then you'd have a negative sign, x to the 10th over 10 factorial. And you could just keep going that way. And if you kept going with the series, this would be the polynomial representation of cosine of x. And it's frankly just kind of cool that it can be represented this way, that it's a pretty simple pattern here for a trigonometric function. Once again, it kind of tells you that all of this math is connected."}, {"video_title": "Analyzing unbounded limits mixed function Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So we're told that f of x is equal to x over one minus cosine of x minus two. And they ask us to select the correct description of the one-sided limits of f at x equals two. And we see that right at x equals two, if we try to evaluate f of two, we get two over one minus cosine of two minus two, which is the same thing as cosine of zero, and cosine of zero is just one, and so one minus one is zero, and so the function is not defined at x equals two, and that's why it might be interesting to find the limit as x approaches two, and especially the one-sided limits. And if the one-sided limits, well, I'll just leave it at that. So let's try to approach this. And there's actually a couple of ways you could do it. There's one way you could do this without a calculator, by just inspecting what's going on here and thinking about the properties of the cosine function."}, {"video_title": "Analyzing unbounded limits mixed function Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And if the one-sided limits, well, I'll just leave it at that. So let's try to approach this. And there's actually a couple of ways you could do it. There's one way you could do this without a calculator, by just inspecting what's going on here and thinking about the properties of the cosine function. And if that inspires you, pause the video and work it out, and I will do that at the end of this video. The other way, if you have a calculator, is to do it with a little bit of a table, like we've done in other example problems. So if we think about x approaching two from the positive direction, well, then we can make a little table here where you have x and then you have f of x."}, {"video_title": "Analyzing unbounded limits mixed function Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "There's one way you could do this without a calculator, by just inspecting what's going on here and thinking about the properties of the cosine function. And if that inspires you, pause the video and work it out, and I will do that at the end of this video. The other way, if you have a calculator, is to do it with a little bit of a table, like we've done in other example problems. So if we think about x approaching two from the positive direction, well, then we can make a little table here where you have x and then you have f of x. And so if we're approaching two from values greater than two you could have 2.1, 2.01. Now, the reason why I said calculators, these aren't trivial to evaluate because this would be, what, 2.1 over one minus cosine of 2.1 minus two is 0.1. I do not know what cosine of 0.1 is without a calculator."}, {"video_title": "Analyzing unbounded limits mixed function Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So if we think about x approaching two from the positive direction, well, then we can make a little table here where you have x and then you have f of x. And so if we're approaching two from values greater than two you could have 2.1, 2.01. Now, the reason why I said calculators, these aren't trivial to evaluate because this would be, what, 2.1 over one minus cosine of 2.1 minus two is 0.1. I do not know what cosine of 0.1 is without a calculator. I do know that cosine of zero is one, so this is very, very close to one without getting to one. And it's going to be less than one. Cosine is never going to be greater than one."}, {"video_title": "Analyzing unbounded limits mixed function Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "I do not know what cosine of 0.1 is without a calculator. I do know that cosine of zero is one, so this is very, very close to one without getting to one. And it's going to be less than one. Cosine is never going to be greater than one. The cosine function is bounded between negative one is less than cosine of x. I'll just write the x there, I don't need the parentheses, which is less than one. The cosine function just oscillates between these two values. So this thing is going to be approaching one, but it's going to be less than one."}, {"video_title": "Analyzing unbounded limits mixed function Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Cosine is never going to be greater than one. The cosine function is bounded between negative one is less than cosine of x. I'll just write the x there, I don't need the parentheses, which is less than one. The cosine function just oscillates between these two values. So this thing is going to be approaching one, but it's going to be less than one. It definitely cannot be greater than one. And that's actually a good hint for how you can just explore the structure here. And then you could say, all right, 2.01, well, that's going to be 2.01 over one minus cosine of 0.01."}, {"video_title": "Analyzing unbounded limits mixed function Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So this thing is going to be approaching one, but it's going to be less than one. It definitely cannot be greater than one. And that's actually a good hint for how you can just explore the structure here. And then you could say, all right, 2.01, well, that's going to be 2.01 over one minus cosine of 0.01. And this is going to be even closer to one without being one. So this could, you know, this is, but it's going to be less than one. No matter what, cosine of anything is going to be less than, it's going to be between negative one and one, and it could even be including those things."}, {"video_title": "Analyzing unbounded limits mixed function Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And then you could say, all right, 2.01, well, that's going to be 2.01 over one minus cosine of 0.01. And this is going to be even closer to one without being one. So this could, you know, this is, but it's going to be less than one. No matter what, cosine of anything is going to be less than, it's going to be between negative one and one, and it could even be including those things. But as we approach two, this thing is going to approach, it's going to approach one, I guess you could say approach one from below. And so you can start to make some intuitions here. If it's approaching one from below, this thing over here, this whole expression, is going to be positive."}, {"video_title": "Analyzing unbounded limits mixed function Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "No matter what, cosine of anything is going to be less than, it's going to be between negative one and one, and it could even be including those things. But as we approach two, this thing is going to approach, it's going to approach one, I guess you could say approach one from below. And so you can start to make some intuitions here. If it's approaching one from below, this thing over here, this whole expression, is going to be positive. And as we approach x equals two, well, the numerator is positive, it's approaching two, the denominator is positive. So this whole thing has to be approaching a positive value, or it could become unbounded in the positive direction. As we'll see, this is unbounded, because this thing is even closer to one than this thing."}, {"video_title": "Analyzing unbounded limits mixed function Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "If it's approaching one from below, this thing over here, this whole expression, is going to be positive. And as we approach x equals two, well, the numerator is positive, it's approaching two, the denominator is positive. So this whole thing has to be approaching a positive value, or it could become unbounded in the positive direction. As we'll see, this is unbounded, because this thing is even closer to one than this thing. And you would see that if you have a calculator. But needless to say, this is going to be unbounded in the positive direction, so we're going to be going towards positive infinity. So these two choices have that."}, {"video_title": "Analyzing unbounded limits mixed function Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "As we'll see, this is unbounded, because this thing is even closer to one than this thing. And you would see that if you have a calculator. But needless to say, this is going to be unbounded in the positive direction, so we're going to be going towards positive infinity. So these two choices have that. And we can make the exact same argument as we go, as we approach x in the negative, or from below, as we approach two from below, I should say. So that's x, and that is f of x. And once again, I don't have a calculator in front of me."}, {"video_title": "Analyzing unbounded limits mixed function Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So these two choices have that. And we can make the exact same argument as we go, as we approach x in the negative, or from below, as we approach two from below, I should say. So that's x, and that is f of x. And once again, I don't have a calculator in front of me. You could evaluate these things at a calculator, and it'll become very clear that these are positive, and as we get closer to two, they become even larger and larger positive values. And the same thing would happen if you did 1.9, and 1.9, and if you did 1.99. Because here you'd be 1.9 over one minus cosine."}, {"video_title": "Analyzing unbounded limits mixed function Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And once again, I don't have a calculator in front of me. You could evaluate these things at a calculator, and it'll become very clear that these are positive, and as we get closer to two, they become even larger and larger positive values. And the same thing would happen if you did 1.9, and 1.9, and if you did 1.99. Because here you'd be 1.9 over one minus cosine. Now here you'd have 1.9 minus two, so this would be negative 0.1. Let me scroll over a little bit. This second one would be 1.99 over one minus cosine of negative 0.01, 0.01."}, {"video_title": "Analyzing unbounded limits mixed function Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Because here you'd be 1.9 over one minus cosine. Now here you'd have 1.9 minus two, so this would be negative 0.1. Let me scroll over a little bit. This second one would be 1.99 over one minus cosine of negative 0.01, 0.01. And cosine of negative 0.1 is the same thing as cosine of 0.1. Cosine of negative 0.01 is the same thing as cosine of 0.01. So these two things, this is going to be equivalent to that, that is going to be equivalent to that."}, {"video_title": "Analyzing unbounded limits mixed function Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "This second one would be 1.99 over one minus cosine of negative 0.01, 0.01. And cosine of negative 0.1 is the same thing as cosine of 0.1. Cosine of negative 0.01 is the same thing as cosine of 0.01. So these two things, this is going to be equivalent to that, that is going to be equivalent to that. And once again, we're gonna be approaching positive infinity. So the only choice where all of that is true is this first one. Whether we approach two from the right-hand side or the left-hand side, we are approaching positive infinity."}, {"video_title": "Analyzing unbounded limits mixed function Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So these two things, this is going to be equivalent to that, that is going to be equivalent to that. And once again, we're gonna be approaching positive infinity. So the only choice where all of that is true is this first one. Whether we approach two from the right-hand side or the left-hand side, we are approaching positive infinity. Now the other way you could have deduced that is say, okay, as we approach two, the numerator is going to be positive, because two is positive. And then over here, as you approach two, cosine of anything can never be greater than one. It's going to approach one, but be less than one."}, {"video_title": "Analyzing unbounded limits mixed function Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Whether we approach two from the right-hand side or the left-hand side, we are approaching positive infinity. Now the other way you could have deduced that is say, okay, as we approach two, the numerator is going to be positive, because two is positive. And then over here, as you approach two, cosine of anything can never be greater than one. It's going to approach one, but be less than one. So if this is less than one, as x approaches two, it becomes one when x is equal to two. Well then this right over here, one minus something less than one is going to be positive. So you have a positive divided by a positive, so you're definitely going to get positive values as you approach two."}, {"video_title": "Worked example coefficient in Taylor polynomial Series AP Calculus BC Khan Academy.mp3", "Sentence": "We're given an f of x, and they say, what is the coefficient for the term containing x plus two to the fourth power in the Taylor polynomial centered at x equals negative two of f? So like always, see if you could take a stab at this video on your own before we work through it together. All right, now let's do this. So in general, our Taylor polynomial, p of x, it's going to have the form, and remember, we're centering at x equals negative two. So this means we're going to evaluate our function at where we're centering it. We are going to divide it by zero factorial, which is just one. I'm just gonna write them all out just so you see the pattern."}, {"video_title": "Worked example coefficient in Taylor polynomial Series AP Calculus BC Khan Academy.mp3", "Sentence": "So in general, our Taylor polynomial, p of x, it's going to have the form, and remember, we're centering at x equals negative two. So this means we're going to evaluate our function at where we're centering it. We are going to divide it by zero factorial, which is just one. I'm just gonna write them all out just so you see the pattern. And we could even say that's gonna be times x minus where we're centering it, but if we're subtracting a negative two, it's gonna be x plus two, and I could write to the zero power, but once again, that's just going to be one. So a lot of times, you won't see someone write this and this, but I'm writing it just to show that there's a consistent pattern. So then you're gonna have plus the first derivative evaluated at negative two divided by one factorial, which is still just one, times x plus two to the first power plus the second derivative evaluated at negative two over two factorial times x plus two squared."}, {"video_title": "Worked example coefficient in Taylor polynomial Series AP Calculus BC Khan Academy.mp3", "Sentence": "I'm just gonna write them all out just so you see the pattern. And we could even say that's gonna be times x minus where we're centering it, but if we're subtracting a negative two, it's gonna be x plus two, and I could write to the zero power, but once again, that's just going to be one. So a lot of times, you won't see someone write this and this, but I'm writing it just to show that there's a consistent pattern. So then you're gonna have plus the first derivative evaluated at negative two divided by one factorial, which is still just one, times x plus two to the first power plus the second derivative evaluated at negative two over two factorial times x plus two squared. I think you see where this is going. And really, all we care about is the one that has the fourth degree term. Well, actually, let me just write the third degree term too just so we get fluent in this."}, {"video_title": "Worked example coefficient in Taylor polynomial Series AP Calculus BC Khan Academy.mp3", "Sentence": "So then you're gonna have plus the first derivative evaluated at negative two divided by one factorial, which is still just one, times x plus two to the first power plus the second derivative evaluated at negative two over two factorial times x plus two squared. I think you see where this is going. And really, all we care about is the one that has the fourth degree term. Well, actually, let me just write the third degree term too just so we get fluent in this. So the third derivative evaluated at negative two over three factorial times x plus two to the third power, and now this is the part that we really care about, plus the fourth derivative. I could have just written a four there, but I think you get what I'm saying. And then evaluated at x equals negative two divided by four factorial times x plus two to the fourth power."}, {"video_title": "Worked example coefficient in Taylor polynomial Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, actually, let me just write the third degree term too just so we get fluent in this. So the third derivative evaluated at negative two over three factorial times x plus two to the third power, and now this is the part that we really care about, plus the fourth derivative. I could have just written a four there, but I think you get what I'm saying. And then evaluated at x equals negative two divided by four factorial times x plus two to the fourth power. So what's the coefficient here? Well, the coefficient is this business. So we'd have to take the fourth derivative of our original function, we had to take the fourth derivative of that original function, evaluate it at negative two, and divide it by four factorial."}, {"video_title": "Worked example coefficient in Taylor polynomial Series AP Calculus BC Khan Academy.mp3", "Sentence": "And then evaluated at x equals negative two divided by four factorial times x plus two to the fourth power. So what's the coefficient here? Well, the coefficient is this business. So we'd have to take the fourth derivative of our original function, we had to take the fourth derivative of that original function, evaluate it at negative two, and divide it by four factorial. So let's do that. So our function, so our first derivative, f prime of x, is just going to be, just gonna use the power rule a lot, six x to the fifth minus three x squared. Second derivative is going to be equal to, five times six is 30, x to the fourth, two times three minus six x to the first power."}, {"video_title": "Worked example coefficient in Taylor polynomial Series AP Calculus BC Khan Academy.mp3", "Sentence": "So we'd have to take the fourth derivative of our original function, we had to take the fourth derivative of that original function, evaluate it at negative two, and divide it by four factorial. So let's do that. So our function, so our first derivative, f prime of x, is just going to be, just gonna use the power rule a lot, six x to the fifth minus three x squared. Second derivative is going to be equal to, five times six is 30, x to the fourth, two times three minus six x to the first power. Third derivative, third derivative of x is going to be equal to four times 30 is 120, x to the third power, minus six. And then the fourth derivative, which is what we really care about, is going to be three times 120 is 360, x to the second power, and the derivative of a constant is just zero. So if we were to evaluate this at x equals negative two, so f, the fourth derivative, evaluated when x equals negative two is going to be 360 times, negative two squared is four."}, {"video_title": "Worked example coefficient in Taylor polynomial Series AP Calculus BC Khan Academy.mp3", "Sentence": "Second derivative is going to be equal to, five times six is 30, x to the fourth, two times three minus six x to the first power. Third derivative, third derivative of x is going to be equal to four times 30 is 120, x to the third power, minus six. And then the fourth derivative, which is what we really care about, is going to be three times 120 is 360, x to the second power, and the derivative of a constant is just zero. So if we were to evaluate this at x equals negative two, so f, the fourth derivative, evaluated when x equals negative two is going to be 360 times, negative two squared is four. I'm just gonna keep that as 360 times four. We can obviously evaluate that. But we're gonna have to divide it by four factorial."}, {"video_title": "Worked example coefficient in Taylor polynomial Series AP Calculus BC Khan Academy.mp3", "Sentence": "So if we were to evaluate this at x equals negative two, so f, the fourth derivative, evaluated when x equals negative two is going to be 360 times, negative two squared is four. I'm just gonna keep that as 360 times four. We can obviously evaluate that. But we're gonna have to divide it by four factorial. So the whole coefficient is going to be 360 times four, which is the numerator here, divided by four factorial, divided by four times three times two times one. Well, four divided by four, those is gonna be one. 360 divided by three, maybe I'll think of it this way, 360 divided by six is going to be 60."}, {"video_title": "Worked example coefficient in Taylor polynomial Series AP Calculus BC Khan Academy.mp3", "Sentence": "But we're gonna have to divide it by four factorial. So the whole coefficient is going to be 360 times four, which is the numerator here, divided by four factorial, divided by four times three times two times one. Well, four divided by four, those is gonna be one. 360 divided by three, maybe I'll think of it this way, 360 divided by six is going to be 60. And so that's all we have. We have 60 and then the denominator, we just have a one. So this is going to simplify to 60."}, {"video_title": "Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So it says, consider the differential equation. The derivative of y with respect to x is equal to three x minus two y. Let y is equal to g of x be a solution to the differential equation with the initial condition g of zero is equal to k, where k is constant. Euler's method, starting at x equals zero, with a step size of one, gives the approximation that g of two is approximately 4.5. Find the value of k. So once again, this is saying, hey look, we're gonna start with this initial condition. When x is equal to zero, y is equal to k, we're going to use Euler's method with a step size of one. So we're essentially going to use, we're gonna step once from zero to one, and then again from one to two, and then that approximation is going to give us 4.5."}, {"video_title": "Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "Euler's method, starting at x equals zero, with a step size of one, gives the approximation that g of two is approximately 4.5. Find the value of k. So once again, this is saying, hey look, we're gonna start with this initial condition. When x is equal to zero, y is equal to k, we're going to use Euler's method with a step size of one. So we're essentially going to use, we're gonna step once from zero to one, and then again from one to two, and then that approximation is going to give us 4.5. And so given that we started at k, we should be able to figure out what k was to get us to g of two being approximated as 4.5. So with that, I encourage you to pause the video and try to figure this out on your own. I am assuming you have tried to figure this out on your own."}, {"video_title": "Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So we're essentially going to use, we're gonna step once from zero to one, and then again from one to two, and then that approximation is going to give us 4.5. And so given that we started at k, we should be able to figure out what k was to get us to g of two being approximated as 4.5. So with that, I encourage you to pause the video and try to figure this out on your own. I am assuming you have tried to figure this out on your own. Now we can do it together. And I'll do the same thing that we did in the first video on Euler's method. I'll make a little table here."}, {"video_title": "Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "I am assuming you have tried to figure this out on your own. Now we can do it together. And I'll do the same thing that we did in the first video on Euler's method. I'll make a little table here. So let me make a little table. I could draw a straighter line than that. That's only marginally straighter, but it'll get the job done."}, {"video_title": "Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "I'll make a little table here. So let me make a little table. I could draw a straighter line than that. That's only marginally straighter, but it'll get the job done. So let's make this column x. I'm gonna give myself some space for y. I might do some calculation here. Y and then dy dx. Now, we can start at our initial condition."}, {"video_title": "Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "That's only marginally straighter, but it'll get the job done. So let's make this column x. I'm gonna give myself some space for y. I might do some calculation here. Y and then dy dx. Now, we can start at our initial condition. When x is equal to zero, y is equal to k. When x is equal to zero, y is equal to k. And so what's our derivative going to be at that point? Well, dy dx is equal to three x minus two y. So in this case, it's three times zero minus two times k, which is just equal to negative two k. And so now we can increment one more step."}, {"video_title": "Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "Now, we can start at our initial condition. When x is equal to zero, y is equal to k. When x is equal to zero, y is equal to k. And so what's our derivative going to be at that point? Well, dy dx is equal to three x minus two y. So in this case, it's three times zero minus two times k, which is just equal to negative two k. And so now we can increment one more step. We have a step size. I'll do this in a different color. We have a step size of one."}, {"video_title": "Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So in this case, it's three times zero minus two times k, which is just equal to negative two k. And so now we can increment one more step. We have a step size. I'll do this in a different color. We have a step size of one. So we're gonna, in each step, we're gonna increment, in each step, we're gonna increment x by one. And so we're now going to be at one. Now, what's our new y going to be?"}, {"video_title": "Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "We have a step size of one. So we're gonna, in each step, we're gonna increment, in each step, we're gonna increment x by one. And so we're now going to be at one. Now, what's our new y going to be? Well, if we increment x by one and our slope is negative two k, that means we're going to increment y by negative two k times one, or just negative two k. So negative two k. So k plus negative two k is negative k. So our approximation using Euler's method gets us the point one, negative k. And then what is going to be our slope starting at that point? So one, negative k. Our slope is going to be three times our x, which is one, minus two times our y, which is negative k now. And this is equal to three plus two k. Three plus two k. And now we'll do another step of one, because that's our step size."}, {"video_title": "Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "Now, what's our new y going to be? Well, if we increment x by one and our slope is negative two k, that means we're going to increment y by negative two k times one, or just negative two k. So negative two k. So k plus negative two k is negative k. So our approximation using Euler's method gets us the point one, negative k. And then what is going to be our slope starting at that point? So one, negative k. Our slope is going to be three times our x, which is one, minus two times our y, which is negative k now. And this is equal to three plus two k. Three plus two k. And now we'll do another step of one, because that's our step size. Another, whoops, now I'm going to get to two. And this is the one that we care about, right? Because we're trying to approximate g of two."}, {"video_title": "Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "And this is equal to three plus two k. Three plus two k. And now we'll do another step of one, because that's our step size. Another, whoops, now I'm going to get to two. And this is the one that we care about, right? Because we're trying to approximate g of two. So we have to say, oh, what does our approximation give us for y when x is equal to two? And we're going to have something expressed in k, but they're saying that's going to be 4.5, and then we could use that to solve for k. So what's this going to be? So if we increment by one in x, we should increment our y by one times three plus two k, which is just going to be, so we're going to increment by three plus two k. Three plus two k, or negative k plus three plus two k is just going to be three plus k. Three plus k. And they're telling us that our approximation gets that to be 4.5."}, {"video_title": "Worked example Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "Because we're trying to approximate g of two. So we have to say, oh, what does our approximation give us for y when x is equal to two? And we're going to have something expressed in k, but they're saying that's going to be 4.5, and then we could use that to solve for k. So what's this going to be? So if we increment by one in x, we should increment our y by one times three plus two k, which is just going to be, so we're going to increment by three plus two k. Three plus two k, or negative k plus three plus two k is just going to be three plus k. Three plus k. And they're telling us that our approximation gets that to be 4.5. So three plus k is equal to 4.5. So the k that we started with must have been, if we just subtract three from both sides, this is a decimal here, it must have been, k must be equal to 1.5. And you can verify that."}, {"video_title": "Integration by parts \u00c2\u00baln(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "The goal of this video is to try to figure out the antiderivative of the natural log of x. And it's not completely obvious how to approach this at first. Even if I were to tell you to use integration by parts, you'll say, wait, wait, integration by parts, you're looking for the antiderivative of something that can be expressed as a product of two functions. It looks like I only have one function right over here, the natural log of x. But it might become a little bit more obvious if I were to rewrite this as the integral of the natural log of x times 1 dx. Now, you do have the product of two functions. One is a function, a function of x."}, {"video_title": "Integration by parts \u00c2\u00baln(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "It looks like I only have one function right over here, the natural log of x. But it might become a little bit more obvious if I were to rewrite this as the integral of the natural log of x times 1 dx. Now, you do have the product of two functions. One is a function, a function of x. It's not actually dependent on x. It's always going to be 1. But you could have f of x is equal to 1."}, {"video_title": "Integration by parts \u00c2\u00baln(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "One is a function, a function of x. It's not actually dependent on x. It's always going to be 1. But you could have f of x is equal to 1. And now it might become a little bit more obvious to use integration by parts. Integration by parts tells us that if we have an integral that can be viewed as the product of one function and the derivative of another function and the derivative of another function, and this is really just the reverse product rule, and we've shown that multiple times already, this is going to be equal to the product of both functions, f of x times g of x minus the antiderivative of, instead of having f and g prime, you're going to have f prime and g. So f prime of x times g of x dx. And we've seen this multiple times."}, {"video_title": "Integration by parts \u00c2\u00baln(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "But you could have f of x is equal to 1. And now it might become a little bit more obvious to use integration by parts. Integration by parts tells us that if we have an integral that can be viewed as the product of one function and the derivative of another function and the derivative of another function, and this is really just the reverse product rule, and we've shown that multiple times already, this is going to be equal to the product of both functions, f of x times g of x minus the antiderivative of, instead of having f and g prime, you're going to have f prime and g. So f prime of x times g of x dx. And we've seen this multiple times. So when you figure out what should be f and what should be g, for f, you want to figure out something that's easy to take the derivative of, and it simplifies things, possibly, if you're taking the derivative of it. And for g prime of x, you want to find something where it's easy to take the antiderivative of it. So a good candidate for f of x is natural log of x."}, {"video_title": "Integration by parts \u00c2\u00baln(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "And we've seen this multiple times. So when you figure out what should be f and what should be g, for f, you want to figure out something that's easy to take the derivative of, and it simplifies things, possibly, if you're taking the derivative of it. And for g prime of x, you want to find something where it's easy to take the antiderivative of it. So a good candidate for f of x is natural log of x. If you were to take the derivative of it, it's 1 over x. Let me write this down. So let's say that f of x is equal to the natural log of x."}, {"video_title": "Integration by parts \u00c2\u00baln(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "So a good candidate for f of x is natural log of x. If you were to take the derivative of it, it's 1 over x. Let me write this down. So let's say that f of x is equal to the natural log of x. Then f prime of x is equal to 1 over x. And let's set g prime of x is equal to 1. So g prime of x is equal to 1."}, {"video_title": "Integration by parts \u00c2\u00baln(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "So let's say that f of x is equal to the natural log of x. Then f prime of x is equal to 1 over x. And let's set g prime of x is equal to 1. So g prime of x is equal to 1. That means that g of x could be equal to x. And so let's go back right over here. So this is going to be equal to f of x times g of x."}, {"video_title": "Integration by parts \u00c2\u00baln(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "So g prime of x is equal to 1. That means that g of x could be equal to x. And so let's go back right over here. So this is going to be equal to f of x times g of x. Well, f of x times g of x is x natural log of x. So g of x is x, and f of x is the natural log of x. I just like writing the x in front of the natural log of x to avoid ambiguity. So this is x natural log of x minus the antiderivative of f prime of x, which is 1 over x, times g of x, which is x, which is x dx."}, {"video_title": "Integration by parts \u00c2\u00baln(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "So this is going to be equal to f of x times g of x. Well, f of x times g of x is x natural log of x. So g of x is x, and f of x is the natural log of x. I just like writing the x in front of the natural log of x to avoid ambiguity. So this is x natural log of x minus the antiderivative of f prime of x, which is 1 over x, times g of x, which is x, which is x dx. Well, what's this going to be equal to? Well, what we have inside the integrand, this is just 1 over x times x, which is just equal to 1. So this simplifies quite nicely."}, {"video_title": "Integration by parts \u00c2\u00baln(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "So this is x natural log of x minus the antiderivative of f prime of x, which is 1 over x, times g of x, which is x, which is x dx. Well, what's this going to be equal to? Well, what we have inside the integrand, this is just 1 over x times x, which is just equal to 1. So this simplifies quite nicely. This is going to end up equaling x natural log of x minus the antiderivative of just dx, or the antiderivative of 1 dx, or the integral of 1 dx, I should say, or the antiderivative of 1, is just minus x. And this is just an antiderivative of this. If we want to write the entire class of antiderivatives, we just have to add a plus c here."}, {"video_title": "Integration by parts \u00c2\u00baln(x)dx AP Calculus BC Khan Academy.mp3", "Sentence": "So this simplifies quite nicely. This is going to end up equaling x natural log of x minus the antiderivative of just dx, or the antiderivative of 1 dx, or the integral of 1 dx, I should say, or the antiderivative of 1, is just minus x. And this is just an antiderivative of this. If we want to write the entire class of antiderivatives, we just have to add a plus c here. And we are done. We figured out the antiderivative of the natural log of x. I encourage you to take the derivative of this. For this part, you're going to use the product rule and verify that you do indeed get natural log of x when you take the derivative of this."}, {"video_title": "Geometric series as a function Infinite sequences and series AP Calculus BC Khan Academy.mp3", "Sentence": "So it's defined as an infinite series. And what I wanna explore in this video, is there another way to write this function? So it's not expressed as an infinite series. Well, some of you might be thinking, well, this looks like a geometric series on the right-hand side, an infinite geometric series. And we know what the sum of an infinite geometric series is if it converges. So maybe that's a way that we can express this. So let's try to do that."}, {"video_title": "Geometric series as a function Infinite sequences and series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, some of you might be thinking, well, this looks like a geometric series on the right-hand side, an infinite geometric series. And we know what the sum of an infinite geometric series is if it converges. So maybe that's a way that we can express this. So let's try to do that. So first let's just confirm that this is an infinite geometric series. And in order for it to be a geometric series, each successive term has to be some common ratio times the previous term. So to go from two to negative eight X squared, what do you have to multiply by?"}, {"video_title": "Geometric series as a function Infinite sequences and series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's try to do that. So first let's just confirm that this is an infinite geometric series. And in order for it to be a geometric series, each successive term has to be some common ratio times the previous term. So to go from two to negative eight X squared, what do you have to multiply by? Well, you have to multiply by negative four X squared. Now let's see, if you multiply negative eight X squared times negative four X squared, what do you get? Well, negative four times negative eight is positive 32."}, {"video_title": "Geometric series as a function Infinite sequences and series AP Calculus BC Khan Academy.mp3", "Sentence": "So to go from two to negative eight X squared, what do you have to multiply by? Well, you have to multiply by negative four X squared. Now let's see, if you multiply negative eight X squared times negative four X squared, what do you get? Well, negative four times negative eight is positive 32. X squared times X squared is X to the fourth. So that works. And then you multiply that times negative four X squared, and you indeed would get negative 128 X to the sixth."}, {"video_title": "Geometric series as a function Infinite sequences and series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, negative four times negative eight is positive 32. X squared times X squared is X to the fourth. So that works. And then you multiply that times negative four X squared, and you indeed would get negative 128 X to the sixth. So this indeed looks like an infinite geometric series on the right-hand side. In fact, we can rewrite F of X as being equal to the sum from N equals zero to infinity of, you have your first term, and then you have your common ratio, negative four X squared to the Nth power. Let's confirm that works."}, {"video_title": "Geometric series as a function Infinite sequences and series AP Calculus BC Khan Academy.mp3", "Sentence": "And then you multiply that times negative four X squared, and you indeed would get negative 128 X to the sixth. So this indeed looks like an infinite geometric series on the right-hand side. In fact, we can rewrite F of X as being equal to the sum from N equals zero to infinity of, you have your first term, and then you have your common ratio, negative four X squared to the Nth power. Let's confirm that works. When N equals zero, this is going to be one. So two times one is two, and that indeed is our first term there. And then to that, you're gonna add it to when N is equal to one."}, {"video_title": "Geometric series as a function Infinite sequences and series AP Calculus BC Khan Academy.mp3", "Sentence": "Let's confirm that works. When N equals zero, this is going to be one. So two times one is two, and that indeed is our first term there. And then to that, you're gonna add it to when N is equal to one. So that's just going to be two times negative four X squared, which is indeed this second term right over here. And so this looks like it works. Now, what is the sum of an infinite geometric series like this?"}, {"video_title": "Geometric series as a function Infinite sequences and series AP Calculus BC Khan Academy.mp3", "Sentence": "And then to that, you're gonna add it to when N is equal to one. So that's just going to be two times negative four X squared, which is indeed this second term right over here. And so this looks like it works. Now, what is the sum of an infinite geometric series like this? Well, it's going to be a finite value assuming the absolute value of your common ratio is less than one. So first of all, let's just think about under what conditions is the absolute value of our common ratio less than one. And then we could say, okay, that helps us to find a radius of convergence."}, {"video_title": "Geometric series as a function Infinite sequences and series AP Calculus BC Khan Academy.mp3", "Sentence": "Now, what is the sum of an infinite geometric series like this? Well, it's going to be a finite value assuming the absolute value of your common ratio is less than one. So first of all, let's just think about under what conditions is the absolute value of our common ratio less than one. And then we could say, okay, that helps us to find a radius of convergence. And then if X is in that zone, or if it's in that interval, then we can figure out a non-infinite geometric series way of expressing this function. So if we just think about under what circumstances will this converge, will it come out to a finite value? That's a situation in which the absolute value of your common ratio is less than one."}, {"video_title": "Geometric series as a function Infinite sequences and series AP Calculus BC Khan Academy.mp3", "Sentence": "And then we could say, okay, that helps us to find a radius of convergence. And then if X is in that zone, or if it's in that interval, then we can figure out a non-infinite geometric series way of expressing this function. So if we just think about under what circumstances will this converge, will it come out to a finite value? That's a situation in which the absolute value of your common ratio is less than one. And so let's see if we can simplify this a little bit. And no matter what X is, it's always going to be, X squared is always going to be non-negative. And so the only, so this entire expression is always going to be negative."}, {"video_title": "Geometric series as a function Infinite sequences and series AP Calculus BC Khan Academy.mp3", "Sentence": "That's a situation in which the absolute value of your common ratio is less than one. And so let's see if we can simplify this a little bit. And no matter what X is, it's always going to be, X squared is always going to be non-negative. And so the only, so this entire expression is always going to be negative. And so if you take the absolute value of it, this is going to be evaluate as four X squared, which is always going to be positive. So this is equivalent to four X squared, which needs to be less than one. Or we could say that X squared needs to be less than one fourth."}, {"video_title": "Geometric series as a function Infinite sequences and series AP Calculus BC Khan Academy.mp3", "Sentence": "And so the only, so this entire expression is always going to be negative. And so if you take the absolute value of it, this is going to be evaluate as four X squared, which is always going to be positive. So this is equivalent to four X squared, which needs to be less than one. Or we could say that X squared needs to be less than one fourth. Or we could say that X needs to be less than one half and greater than negative one half. One way to think about it is anywhere in this interval, if you square it, you're going to be less than one fourth. At one half, if you square it, it's equal to one fourth."}, {"video_title": "Geometric series as a function Infinite sequences and series AP Calculus BC Khan Academy.mp3", "Sentence": "Or we could say that X squared needs to be less than one fourth. Or we could say that X needs to be less than one half and greater than negative one half. One way to think about it is anywhere in this interval, if you square it, you're going to be less than one fourth. At one half, if you square it, it's equal to one fourth. And at negative one half, if you square it, it's equal to one fourth. But for lower absolute values, it's going to be less than one fourth. And so that's what this interval right here says."}, {"video_title": "Geometric series as a function Infinite sequences and series AP Calculus BC Khan Academy.mp3", "Sentence": "At one half, if you square it, it's equal to one fourth. And at negative one half, if you square it, it's equal to one fourth. But for lower absolute values, it's going to be less than one fourth. And so that's what this interval right here says. Another way to think about it is the absolute value of X needs to be less than one half. And so we've just defined an interval over which this infinite geometric series will converge. You could say this has a radius of convergence of, let me write it this way, radius of convergence, convergence of one half."}, {"video_title": "Geometric series as a function Infinite sequences and series AP Calculus BC Khan Academy.mp3", "Sentence": "And so that's what this interval right here says. Another way to think about it is the absolute value of X needs to be less than one half. And so we've just defined an interval over which this infinite geometric series will converge. You could say this has a radius of convergence of, let me write it this way, radius of convergence, convergence of one half. You can go one half above zero and one half below zero. But now that we've set the conditions under which this would converge, let's rewrite it. So this function is going to be equal to, we know what the sum of an infinite geometric series is."}, {"video_title": "Geometric series as a function Infinite sequences and series AP Calculus BC Khan Academy.mp3", "Sentence": "You could say this has a radius of convergence of, let me write it this way, radius of convergence, convergence of one half. You can go one half above zero and one half below zero. But now that we've set the conditions under which this would converge, let's rewrite it. So this function is going to be equal to, we know what the sum of an infinite geometric series is. It's going to be equal to the first term over one minus your common ratio, one minus negative four X squared. And so we can rewrite our function as F of X is equal to two over one subtract a negative one plus four X squared for the absolute value of X is less than one half. We have the interval over which we converge."}, {"video_title": "Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So I've been requested to do the proof of the derivative of the square root of x. So I thought I would do a quick video on the proof of the derivative of the square root of x. So we know from the definition of the derivative that the derivative of the function square root of x, that is equal to, let me switch colors just for variety, that's equal to the limit as delta x approaches 0. And some people say h approaches 0 or d approaches 0. I just use delta x. So the change in x approaches 0. And then we say f of x plus delta x."}, {"video_title": "Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And some people say h approaches 0 or d approaches 0. I just use delta x. So the change in x approaches 0. And then we say f of x plus delta x. So in this case, this is f of x. So it's the square root of x plus delta x minus f of x, in this case, the square root of x. All of that over the change in x, over delta x."}, {"video_title": "Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And then we say f of x plus delta x. So in this case, this is f of x. So it's the square root of x plus delta x minus f of x, in this case, the square root of x. All of that over the change in x, over delta x. So what I'm going to do, right now when I look at that, there's not much simplification I can do to make this come out with something meaningful. I'm going to multiply this fraction times, I'm going to multiply the numerator and the denominator by the conjugate of the numerator. So what do I mean by that?"}, {"video_title": "Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "All of that over the change in x, over delta x. So what I'm going to do, right now when I look at that, there's not much simplification I can do to make this come out with something meaningful. I'm going to multiply this fraction times, I'm going to multiply the numerator and the denominator by the conjugate of the numerator. So what do I mean by that? Let me rewrite it. Limit as delta x approaches 0. I'm just rewriting what I have here."}, {"video_title": "Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So what do I mean by that? Let me rewrite it. Limit as delta x approaches 0. I'm just rewriting what I have here. So I said the square root of x plus delta x minus square root of x, all of that over delta x. And I'm going to multiply that, after switching colors, times square root of x plus delta x plus the square root of x over the square root of x plus delta x plus the square root of x, right? This is just 1."}, {"video_title": "Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "I'm just rewriting what I have here. So I said the square root of x plus delta x minus square root of x, all of that over delta x. And I'm going to multiply that, after switching colors, times square root of x plus delta x plus the square root of x over the square root of x plus delta x plus the square root of x, right? This is just 1. So I could, of course, multiply that times, if we assume that x and delta x aren't both 0, this is a defined number. This would be 1. And we can do that."}, {"video_title": "Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "This is just 1. So I could, of course, multiply that times, if we assume that x and delta x aren't both 0, this is a defined number. This would be 1. And we can do that. This is 1 over 1. We're just multiplying it times this equation. And we get limit as delta x approaches 0."}, {"video_title": "Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And we can do that. This is 1 over 1. We're just multiplying it times this equation. And we get limit as delta x approaches 0. Well, if you view this as a minus b times a plus b, right? Let me do a little aside here. Let me say a plus b times a minus b is equal to a squared minus b squared, right?"}, {"video_title": "Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And we get limit as delta x approaches 0. Well, if you view this as a minus b times a plus b, right? Let me do a little aside here. Let me say a plus b times a minus b is equal to a squared minus b squared, right? So this is a plus b times a minus b. So it's going to be equal to a squared. So what's this quantity squared or this quantity squared, either one?"}, {"video_title": "Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Let me say a plus b times a minus b is equal to a squared minus b squared, right? So this is a plus b times a minus b. So it's going to be equal to a squared. So what's this quantity squared or this quantity squared, either one? These are my a's. Well, it's just going to be x plus delta x, right? So you get x plus delta x."}, {"video_title": "Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So what's this quantity squared or this quantity squared, either one? These are my a's. Well, it's just going to be x plus delta x, right? So you get x plus delta x. And then what's b squared? So minus square root of x is b in this analogy. So square root of x squared is just x."}, {"video_title": "Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So you get x plus delta x. And then what's b squared? So minus square root of x is b in this analogy. So square root of x squared is just x. And all of that over delta x times square root of x plus delta x plus the square root of x. Let's see what simplification we can do. Well, we have an x and then a minus x."}, {"video_title": "Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So square root of x squared is just x. And all of that over delta x times square root of x plus delta x plus the square root of x. Let's see what simplification we can do. Well, we have an x and then a minus x. So those cancel out. So we have delta x minus x. And then we're left in the numerator and the denominator."}, {"video_title": "Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Well, we have an x and then a minus x. So those cancel out. So we have delta x minus x. And then we're left in the numerator and the denominator. All we have is a delta x here and a delta x here. So let's divide the numerator and the denominator by delta x. So this goes to 1."}, {"video_title": "Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And then we're left in the numerator and the denominator. All we have is a delta x here and a delta x here. So let's divide the numerator and the denominator by delta x. So this goes to 1. And so this equals the limit. I'll write smaller because I'm running out of space. Limit as delta x approaches 0 of 1 over."}, {"video_title": "Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So this goes to 1. And so this equals the limit. I'll write smaller because I'm running out of space. Limit as delta x approaches 0 of 1 over. And of course, we can only do this assuming that delta, well, we're dividing by delta x to begin with. So we know it's not 0. It's just approaching 0."}, {"video_title": "Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Limit as delta x approaches 0 of 1 over. And of course, we can only do this assuming that delta, well, we're dividing by delta x to begin with. So we know it's not 0. It's just approaching 0. So we get square root of x plus delta x plus the square root of x. And now we can just directly take the limit as it approaches 0. We can just set delta x is equal to 0."}, {"video_title": "Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "It's just approaching 0. So we get square root of x plus delta x plus the square root of x. And now we can just directly take the limit as it approaches 0. We can just set delta x is equal to 0. That's what it's approaching. So that that equals 1 over the square root of x. Delta x is 0, so we can ignore that. We can take the limit all the way to 0."}, {"video_title": "Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "We can just set delta x is equal to 0. That's what it's approaching. So that that equals 1 over the square root of x. Delta x is 0, so we can ignore that. We can take the limit all the way to 0. And then this is, of course, just a square root of x here plus the square root of x. And that equals 1 over 2 square root of x. And that equals 1 half x to the negative 1 half."}, {"video_title": "Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "We can take the limit all the way to 0. And then this is, of course, just a square root of x here plus the square root of x. And that equals 1 over 2 square root of x. And that equals 1 half x to the negative 1 half. So we just proved that x to the 1 half power, the derivative of it, is 1 half x to the negative 1 half. And so it is consistent with the general property that the derivative of x to the n is equal to nx to the n minus 1. Even in this case where n was 1 half."}, {"video_title": "Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And that equals 1 half x to the negative 1 half. So we just proved that x to the 1 half power, the derivative of it, is 1 half x to the negative 1 half. And so it is consistent with the general property that the derivative of x to the n is equal to nx to the n minus 1. Even in this case where n was 1 half. Well, hopefully that's satisfying. I didn't prove it for all fractions, but this is a start. This is a common one you see."}, {"video_title": "Proof d dx(sqrt(x)) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Even in this case where n was 1 half. Well, hopefully that's satisfying. I didn't prove it for all fractions, but this is a start. This is a common one you see. Square root of x. And it's hopefully not too complicated of a proof. I will see you in future videos."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "We'd have to resort to numeric techniques to estimate the solutions. But let's go to what I would argue is the simplest form of differential equation to solve, and that's what's called a separable, separable differential equation. And we will see in a second why it is called a separable differential equation. So let's say that we have the derivative of y with respect to x is equal to negative x over y e to the x squared. So we have this differential equation, and we want to find the particular solution that goes through the point, that goes through the point zero comma one. And I encourage you to pause this video, and I'll give you a hint. If you can, on one side of this equation, through algebra, separate out the y's and the d y's, and on the other side, have all the x's and d x's, and then integrate, perhaps you can find the particular solution to this differential equation that contains this point."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "So let's say that we have the derivative of y with respect to x is equal to negative x over y e to the x squared. So we have this differential equation, and we want to find the particular solution that goes through the point, that goes through the point zero comma one. And I encourage you to pause this video, and I'll give you a hint. If you can, on one side of this equation, through algebra, separate out the y's and the d y's, and on the other side, have all the x's and d x's, and then integrate, perhaps you can find the particular solution to this differential equation that contains this point. Now if you can't do it, don't worry, because we're about to work through it. So as I said, let's use a little bit of algebra to get all the y's and d y's on one side, and all the x's and d x's on the other side. So one way, let's say I want to get all the y's and d y's on the left hand side, and all the x's and d x's on the right hand side."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "If you can, on one side of this equation, through algebra, separate out the y's and the d y's, and on the other side, have all the x's and d x's, and then integrate, perhaps you can find the particular solution to this differential equation that contains this point. Now if you can't do it, don't worry, because we're about to work through it. So as I said, let's use a little bit of algebra to get all the y's and d y's on one side, and all the x's and d x's on the other side. So one way, let's say I want to get all the y's and d y's on the left hand side, and all the x's and d x's on the right hand side. Well I can multiply both sides times y. So I can multiply both sides times y. That has the effect of putting the y's on the left hand side."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "So one way, let's say I want to get all the y's and d y's on the left hand side, and all the x's and d x's on the right hand side. Well I can multiply both sides times y. So I can multiply both sides times y. That has the effect of putting the y's on the left hand side. And then I can multiply both sides times d x. I can multiply both sides times d x. And we kind of treat, you can treat these differentials as you would treat a variable when you're manipulating it to essentially separate out the variables. And so this will cancel with that."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "That has the effect of putting the y's on the left hand side. And then I can multiply both sides times d x. I can multiply both sides times d x. And we kind of treat, you can treat these differentials as you would treat a variable when you're manipulating it to essentially separate out the variables. And so this will cancel with that. And so we are left with, we are left with y d y, y d y is equal to negative x, and actually let me write it this way. Let me write it as negative x e, actually I might want a little more space. So negative x e to the negative x squared d x."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "And so this will cancel with that. And so we are left with, we are left with y d y, y d y is equal to negative x, and actually let me write it this way. Let me write it as negative x e, actually I might want a little more space. So negative x e to the negative x squared d x. D x. Now why is this interesting? Because we can integrate both sides."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "So negative x e to the negative x squared d x. D x. Now why is this interesting? Because we can integrate both sides. And now this also highlights why we call this separable. You won't be able to do this with every differential equation. You won't be able to algebraically separate the y's and d y's on one side, and the x's and d x's on the other side."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "Because we can integrate both sides. And now this also highlights why we call this separable. You won't be able to do this with every differential equation. You won't be able to algebraically separate the y's and d y's on one side, and the x's and d x's on the other side. But this one we were able to. And so that's why this is called a separable differential equation. Differential, differential equation."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "You won't be able to algebraically separate the y's and d y's on one side, and the x's and d x's on the other side. But this one we were able to. And so that's why this is called a separable differential equation. Differential, differential equation. And it's usually the first technique that you should try. Hey, can I separate the y's and the x's? And as I said, this is not going to be true of many, if not most differential equations."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "Differential, differential equation. And it's usually the first technique that you should try. Hey, can I separate the y's and the x's? And as I said, this is not going to be true of many, if not most differential equations. But now that we did this, we can integrate both sides. So let's do that. So I'll find a nice color to integrate with."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "And as I said, this is not going to be true of many, if not most differential equations. But now that we did this, we can integrate both sides. So let's do that. So I'll find a nice color to integrate with. So I'm going to integrate, integrate both sides. Now if you integrate the left-hand side, what do you get? You get, and remember, we're integrating with respect to y here."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "So I'll find a nice color to integrate with. So I'm going to integrate, integrate both sides. Now if you integrate the left-hand side, what do you get? You get, and remember, we're integrating with respect to y here. So this is going to be y squared over two. And we could put some constant there. I could call that plus c one."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "You get, and remember, we're integrating with respect to y here. So this is going to be y squared over two. And we could put some constant there. I could call that plus c one. And if you're integrating, now that's going to be equal to, now the right-hand side we're integrating with respect to x. And let's see, you could do u substitution, or you could recognize that look, the derivative of negative x squared is going to be negative two x. So if that was a two there, and if you don't want to change the value of the integral, you put the 1 1 2 right over there."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "I could call that plus c one. And if you're integrating, now that's going to be equal to, now the right-hand side we're integrating with respect to x. And let's see, you could do u substitution, or you could recognize that look, the derivative of negative x squared is going to be negative two x. So if that was a two there, and if you don't want to change the value of the integral, you put the 1 1 2 right over there. So now you could either do u substitution explicitly, or you could do it in your head, where you said u is equal to negative x squared, and then du will be negative two x dx, or you can kind of do this in your head at this point. So I have something and its derivative, so I really could just integrate with respect to that something, with respect to that u. So this is going to be 1 1 2, this 1 1 2 right over here."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "So if that was a two there, and if you don't want to change the value of the integral, you put the 1 1 2 right over there. So now you could either do u substitution explicitly, or you could do it in your head, where you said u is equal to negative x squared, and then du will be negative two x dx, or you can kind of do this in your head at this point. So I have something and its derivative, so I really could just integrate with respect to that something, with respect to that u. So this is going to be 1 1 2, this 1 1 2 right over here. The antiderivative of this is e to the negative x squared, and then of course I might have some other constant. I'll just call that c two. And once again, if this part over here, what I just did seems strange, the u substitution, you might want to review that piece."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "So this is going to be 1 1 2, this 1 1 2 right over here. The antiderivative of this is e to the negative x squared, and then of course I might have some other constant. I'll just call that c two. And once again, if this part over here, what I just did seems strange, the u substitution, you might want to review that piece. Now, what can I do here? Well I have a constant on the left-hand side, it's an arbitrary constant, we don't know what it is. I haven't used this initial condition yet, we could call it."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "And once again, if this part over here, what I just did seems strange, the u substitution, you might want to review that piece. Now, what can I do here? Well I have a constant on the left-hand side, it's an arbitrary constant, we don't know what it is. I haven't used this initial condition yet, we could call it. So let me just subtract c one from both sides. So if I just subtract c one from both sides, I have an arbitrary, so this is going to cancel, and I have c two, sorry, let me, so this is c one, so these are going to cancel, and c two minus c one, these are both constants, arbitrary constants, we don't know what they are yet, and so we could just rewrite this as, on the left-hand side we have y squared over two is equal to, on the right-hand side, I'll write one half e, let me write that in blue, just because I wrote it in blue before, one half, one half e to the negative x squared, and I'll just say c two minus c one, let's just call that c. So if you take the sum of those two things, let's just call that c. And so now, this is kind of a general solution, we don't know what this constant is, and we haven't explicitly solved for y yet, but even in this form, we can now find a particular solution using this initial condition. Let me separate it out, this wasn't part, this wasn't part of this original expression right over here, but using this initial condition."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "I haven't used this initial condition yet, we could call it. So let me just subtract c one from both sides. So if I just subtract c one from both sides, I have an arbitrary, so this is going to cancel, and I have c two, sorry, let me, so this is c one, so these are going to cancel, and c two minus c one, these are both constants, arbitrary constants, we don't know what they are yet, and so we could just rewrite this as, on the left-hand side we have y squared over two is equal to, on the right-hand side, I'll write one half e, let me write that in blue, just because I wrote it in blue before, one half, one half e to the negative x squared, and I'll just say c two minus c one, let's just call that c. So if you take the sum of those two things, let's just call that c. And so now, this is kind of a general solution, we don't know what this constant is, and we haven't explicitly solved for y yet, but even in this form, we can now find a particular solution using this initial condition. Let me separate it out, this wasn't part, this wasn't part of this original expression right over here, but using this initial condition. So it tells us when x is zero, y needs to be equal to one. So we would have one squared, which is just one, over two is equal to one half, e to the negative zero squared, well that's just going to be, e to the zero is just one, so it's going to be one half plus c, and just like that, we're able to figure out if you subtract one half from both sides, c is equal to zero. So the relationship between y and x that goes through this point, we could just set c is equal to zero, so that's equal to zero, that's zero right over there, and so we are left with y squared over two is equal to e to the negative x squared over two."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "Let me separate it out, this wasn't part, this wasn't part of this original expression right over here, but using this initial condition. So it tells us when x is zero, y needs to be equal to one. So we would have one squared, which is just one, over two is equal to one half, e to the negative zero squared, well that's just going to be, e to the zero is just one, so it's going to be one half plus c, and just like that, we're able to figure out if you subtract one half from both sides, c is equal to zero. So the relationship between y and x that goes through this point, we could just set c is equal to zero, so that's equal to zero, that's zero right over there, and so we are left with y squared over two is equal to e to the negative x squared over two. Now we can multiply both sides by two, and we're going to get y squared, y squared, let me do that, so we're going to get y squared is equal to, is equal to e to the negative x squared. Now we can take the square root of both sides, and you could say, well look, y squared is equal to this, so y could be equal to the plus or minus square root of e to the negative x squared, of e to the negative x squared, but they gave us an initial condition where y is actually positive. So we're finding the particular solution that goes through this point, that means y is going to be the positive square root."}, {"video_title": "Separable differential equations introduction First order differential equations Khan Academy.mp3", "Sentence": "So the relationship between y and x that goes through this point, we could just set c is equal to zero, so that's equal to zero, that's zero right over there, and so we are left with y squared over two is equal to e to the negative x squared over two. Now we can multiply both sides by two, and we're going to get y squared, y squared, let me do that, so we're going to get y squared is equal to, is equal to e to the negative x squared. Now we can take the square root of both sides, and you could say, well look, y squared is equal to this, so y could be equal to the plus or minus square root of e to the negative x squared, of e to the negative x squared, but they gave us an initial condition where y is actually positive. So we're finding the particular solution that goes through this point, that means y is going to be the positive square root. If this was the point zero, negative one, then we would say y is the negative square root, but we know that y is the positive square root, it's the principal root right over there. Actually let me do that a little bit neater, so we can get rid of, whoops, I thought I was writing in black. So we can get rid of this right over here."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "What I have here in yellow is the graph of y equals f of x. Then here in this mauve color, I've graphed y is equal to the derivative of f, is f prime of x. And then here in blue, I've graphed y is equal to the second derivative of our function. So this is the derivative of this, of the first derivative right over there. And we've already seen examples of how can we identify minimum and maximum points. Obviously, if we have the graph in front of us, it's not hard for a human brain to identify this as a local maximum point. The function might take on higher values later on."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "So this is the derivative of this, of the first derivative right over there. And we've already seen examples of how can we identify minimum and maximum points. Obviously, if we have the graph in front of us, it's not hard for a human brain to identify this as a local maximum point. The function might take on higher values later on. And to identify this as a local minimum point, the function might take on lower values later on. But we saw, even if we don't have the graph in front of us, if we're able to take the derivative of the function, we might, or even if we're not able to take the derivative of the function, we might be able to identify these points as minimum or maximum. The way that we did it, we said, OK, what are the critical points for this function?"}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "The function might take on higher values later on. And to identify this as a local minimum point, the function might take on lower values later on. But we saw, even if we don't have the graph in front of us, if we're able to take the derivative of the function, we might, or even if we're not able to take the derivative of the function, we might be able to identify these points as minimum or maximum. The way that we did it, we said, OK, what are the critical points for this function? Well, critical points are where the function's derivative is either undefined or 0. This is the function's derivative. It is 0 here and here."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "The way that we did it, we said, OK, what are the critical points for this function? Well, critical points are where the function's derivative is either undefined or 0. This is the function's derivative. It is 0 here and here. So we would call those critical points. I don't see any undefined, any points at which the derivative is undefined just yet. So we would call here and here critical points."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "It is 0 here and here. So we would call those critical points. I don't see any undefined, any points at which the derivative is undefined just yet. So we would call here and here critical points. So these are candidate minimum, these are candidate points at which our function might take on a minimum or a maximum value. And the way that we figured out whether there was a minimum or maximum value is to look at the behavior of the derivative around that point. And over here, we saw the derivative is positive as we approach that point."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "So we would call here and here critical points. So these are candidate minimum, these are candidate points at which our function might take on a minimum or a maximum value. And the way that we figured out whether there was a minimum or maximum value is to look at the behavior of the derivative around that point. And over here, we saw the derivative is positive as we approach that point. And then it becomes negative. It goes from being positive to negative as we cross that point, which means that the function was increasing. If the derivative is positive, that means that the function was increasing as we approached that point, and then decreasing as we leave that point, which is a pretty good way to think about this being a maximum point."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "And over here, we saw the derivative is positive as we approach that point. And then it becomes negative. It goes from being positive to negative as we cross that point, which means that the function was increasing. If the derivative is positive, that means that the function was increasing as we approached that point, and then decreasing as we leave that point, which is a pretty good way to think about this being a maximum point. If we're increasing as we approach it and decreasing as we leave it, then this is definitely going to be a maximum point. Similarly, right over here, we see that the derivative is negative as we approach the point, which means that the function is decreasing. And we see that the derivative is positive as we exit that point."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "If the derivative is positive, that means that the function was increasing as we approached that point, and then decreasing as we leave that point, which is a pretty good way to think about this being a maximum point. If we're increasing as we approach it and decreasing as we leave it, then this is definitely going to be a maximum point. Similarly, right over here, we see that the derivative is negative as we approach the point, which means that the function is decreasing. And we see that the derivative is positive as we exit that point. We go from having a negative derivative to a positive derivative, which means the function goes from decreasing to increasing right around that point, which is a pretty good indication, or that is the indication, that this critical point is a point at which the function takes on a minimum value. What I want to do now is extend things by using the idea of concavity. And I know I'm mispronouncing it."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "And we see that the derivative is positive as we exit that point. We go from having a negative derivative to a positive derivative, which means the function goes from decreasing to increasing right around that point, which is a pretty good indication, or that is the indication, that this critical point is a point at which the function takes on a minimum value. What I want to do now is extend things by using the idea of concavity. And I know I'm mispronouncing it. Maybe it's con-cavity. But thinking about concavity, we can start to look at the second derivative rather than seeing just this transition to think about whether this is a minimum or a maximum point. So let's think about what's happening in this first region, this part of the curve up here where it looks like a arc, where it's opening downward, where it looks like an A without the cross beam or an upside down U."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "And I know I'm mispronouncing it. Maybe it's con-cavity. But thinking about concavity, we can start to look at the second derivative rather than seeing just this transition to think about whether this is a minimum or a maximum point. So let's think about what's happening in this first region, this part of the curve up here where it looks like a arc, where it's opening downward, where it looks like an A without the cross beam or an upside down U. And then we'll think about what's happening in this upward opening U part of the curve. So over this first interval right over here, if we start over here, the slope is very positive. Slope is very positive."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "So let's think about what's happening in this first region, this part of the curve up here where it looks like a arc, where it's opening downward, where it looks like an A without the cross beam or an upside down U. And then we'll think about what's happening in this upward opening U part of the curve. So over this first interval right over here, if we start over here, the slope is very positive. Slope is very positive. Then it becomes less positive. Then it becomes even less positive. It eventually gets to 0."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "Slope is very positive. Then it becomes less positive. Then it becomes even less positive. It eventually gets to 0. Then it keeps decreasing. Now it becomes slightly negative. Then it becomes even more negative."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "It eventually gets to 0. Then it keeps decreasing. Now it becomes slightly negative. Then it becomes even more negative. Then it becomes even more negative. And then it stops decreasing right around there. So the slope stops decreasing right around there."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "Then it becomes even more negative. Then it becomes even more negative. And then it stops decreasing right around there. So the slope stops decreasing right around there. And you see that in the derivative. The slope is decreasing, decreasing, decreasing, decreasing until that point. And then it starts to increase."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "So the slope stops decreasing right around there. And you see that in the derivative. The slope is decreasing, decreasing, decreasing, decreasing until that point. And then it starts to increase. So this entire section right over here, the slope is decreasing. The slope is decreasing. And you see it right over here when we take the derivative."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "And then it starts to increase. So this entire section right over here, the slope is decreasing. The slope is decreasing. And you see it right over here when we take the derivative. The derivative right over here over this entire interval is decreasing. And we also see that when we take the second derivative. If the derivative is decreasing, that means that the second derivative, the derivative of the derivative is negative."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "And you see it right over here when we take the derivative. The derivative right over here over this entire interval is decreasing. And we also see that when we take the second derivative. If the derivative is decreasing, that means that the second derivative, the derivative of the derivative is negative. And we see that that is indeed the case. over this entire interval, the second derivative is indeed negative. Now what happens as we start to transition to this upward opening u part of the curve?"}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "If the derivative is decreasing, that means that the second derivative, the derivative of the derivative is negative. And we see that that is indeed the case. over this entire interval, the second derivative is indeed negative. Now what happens as we start to transition to this upward opening u part of the curve? Well, here the derivative is reasonably negative. It's reasonably negative right there. But then it starts, it's still negative, but it becomes less negative and less negative and less negative, less negative and less negative and less negative."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "Now what happens as we start to transition to this upward opening u part of the curve? Well, here the derivative is reasonably negative. It's reasonably negative right there. But then it starts, it's still negative, but it becomes less negative and less negative and less negative, less negative and less negative and less negative. Then it becomes 0, it becomes 0 right over here. And then it becomes more and more and more positive. And you see that right over here."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "But then it starts, it's still negative, but it becomes less negative and less negative and less negative, less negative and less negative and less negative. Then it becomes 0, it becomes 0 right over here. And then it becomes more and more and more positive. And you see that right over here. So over this entire interval, the slope or the derivative is increasing. So the slope is in. The slope is increasing."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "And you see that right over here. So over this entire interval, the slope or the derivative is increasing. So the slope is in. The slope is increasing. And you see this over here. Over here the slope is 0. The slope of the derivative is 0."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "The slope is increasing. And you see this over here. Over here the slope is 0. The slope of the derivative is 0. The slope, the derivative itself isn't changing right at this moment. And then you see that the slope is increasing. And once again, we can visualize that on the second derivative."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "The slope of the derivative is 0. The slope, the derivative itself isn't changing right at this moment. And then you see that the slope is increasing. And once again, we can visualize that on the second derivative. The derivative of the derivative, if the derivative is increasing, that means the derivative of that must be positive. And it is indeed the case that the derivative is positive. And we have a word for this downward opening u and this upward opening u."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "And once again, we can visualize that on the second derivative. The derivative of the derivative, if the derivative is increasing, that means the derivative of that must be positive. And it is indeed the case that the derivative is positive. And we have a word for this downward opening u and this upward opening u. We call this concave downwards. Let me make this clear. Concave downwards, and we call this concave upwards."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "And we have a word for this downward opening u and this upward opening u. We call this concave downwards. Let me make this clear. Concave downwards, and we call this concave upwards. So let's review how we can identify concave downward intervals and concave upwards intervals. So if we're talking about concave downwards, we see several things. We see that the slope is decreasing, which is another way of saying that f prime of x is decreasing, which is another way of saying that the second derivative must be negative."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "Concave downwards, and we call this concave upwards. So let's review how we can identify concave downward intervals and concave upwards intervals. So if we're talking about concave downwards, we see several things. We see that the slope is decreasing, which is another way of saying that f prime of x is decreasing, which is another way of saying that the second derivative must be negative. If the first derivative is decreasing, the second derivative must be negative, which is another way of saying that the second derivative over that interval must be negative. So if you have a negative second derivative, then you are in a concave downward interval. Similarly, I have trouble saying that word."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "We see that the slope is decreasing, which is another way of saying that f prime of x is decreasing, which is another way of saying that the second derivative must be negative. If the first derivative is decreasing, the second derivative must be negative, which is another way of saying that the second derivative over that interval must be negative. So if you have a negative second derivative, then you are in a concave downward interval. Similarly, I have trouble saying that word. Let's think about concave upwards, where you have an upward opening u. Concave upwards, in these intervals, the slope is increasing. We have a negative slope, less negative, less negative, 0, positive, more positive, more positive, even more positive. So slope is increasing, which means that the derivative of the function is increasing."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "Similarly, I have trouble saying that word. Let's think about concave upwards, where you have an upward opening u. Concave upwards, in these intervals, the slope is increasing. We have a negative slope, less negative, less negative, 0, positive, more positive, more positive, even more positive. So slope is increasing, which means that the derivative of the function is increasing. And you see that right over here. This derivative is increasing in value, which means that the second derivative over an interval where we are concave upwards must be greater than 0. If the second derivative is greater than 0, that means that the first derivative is increasing, which means that the slope is increasing."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "So slope is increasing, which means that the derivative of the function is increasing. And you see that right over here. This derivative is increasing in value, which means that the second derivative over an interval where we are concave upwards must be greater than 0. If the second derivative is greater than 0, that means that the first derivative is increasing, which means that the slope is increasing. We are in a concave upward interval. Now, given all of these definitions that we've just given for concave downwards and concave upwards, can we come up with another way of identifying whether a critical point is a minimum point or a maximum point? Well, if you have a maximum point, if you have a critical point where the function is concave downwards, then you're going to be at a maximum point."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "If the second derivative is greater than 0, that means that the first derivative is increasing, which means that the slope is increasing. We are in a concave upward interval. Now, given all of these definitions that we've just given for concave downwards and concave upwards, can we come up with another way of identifying whether a critical point is a minimum point or a maximum point? Well, if you have a maximum point, if you have a critical point where the function is concave downwards, then you're going to be at a maximum point. Concave downwards, let's just be clear here, means that it's opening down like this. And when we're talking about a critical point, if we're assuming it's concave downwards over here, we're assuming differentiability over this interval. And so the critical point is going to be one where the slope is 0."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "Well, if you have a maximum point, if you have a critical point where the function is concave downwards, then you're going to be at a maximum point. Concave downwards, let's just be clear here, means that it's opening down like this. And when we're talking about a critical point, if we're assuming it's concave downwards over here, we're assuming differentiability over this interval. And so the critical point is going to be one where the slope is 0. So it's going to be that point right over there. So if you're concave downwards and you have a point where f prime of, let's say, a is equal to 0, then we have a maximum point at a. We have a maximum point at a."}, {"video_title": "Concavity introduction Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "And so the critical point is going to be one where the slope is 0. So it's going to be that point right over there. So if you're concave downwards and you have a point where f prime of, let's say, a is equal to 0, then we have a maximum point at a. We have a maximum point at a. And similarly, if we're concave upwards, that means that our function looks something like this. And if we found a point, obviously a critical point could also be where the function is not defined. But if we're assuming that our first derivative and second derivative is defined here, then the critical point is going to be one where the first derivative is going to be 0."}, {"video_title": "Evaluating definite integral with calculator AP Calculus BC Khan Academy.mp3", "Sentence": "In the last video, we tried to find the area of the region, I guess this combined area between the blue and this orange, the area, I guess the overlap between these two circles, and we came up with nine pi minus 18, all of that over eight. What I want to do in this video, so you could have also used a typical graphing calculator to come up with the same result, and it would have actually evaluated the definite integral. So let's see how you do that. And this, what I'm doing here, you could do this for a traditional, what if we're dealing with Cartesian coordinates, rectangular coordinates, or for polar coordinates, because it's really just about evaluating a definite integral. So we wanted to evaluate nine times the definite integral from zero to pi over four of sine squared theta d theta. So how do I do that? Well, I can go to second, Calculus, and I do the Fn, Int, that's definite integral."}, {"video_title": "Evaluating definite integral with calculator AP Calculus BC Khan Academy.mp3", "Sentence": "And this, what I'm doing here, you could do this for a traditional, what if we're dealing with Cartesian coordinates, rectangular coordinates, or for polar coordinates, because it's really just about evaluating a definite integral. So we wanted to evaluate nine times the definite integral from zero to pi over four of sine squared theta d theta. So how do I do that? Well, I can go to second, Calculus, and I do the Fn, Int, that's definite integral. So let's take, use that function, and the first thing you want to say, well, what are you taking the definite integral of? And we're taking the definite integral of, the definite integral of sine, actually I want the parentheses, sine, and I could use any variable here, as long as I'm consistent with what I am integrating with respect to. So I tend to use just the X button, because there is an X button, but we'll just assume that in this case, X is theta."}, {"video_title": "Evaluating definite integral with calculator AP Calculus BC Khan Academy.mp3", "Sentence": "Well, I can go to second, Calculus, and I do the Fn, Int, that's definite integral. So let's take, use that function, and the first thing you want to say, well, what are you taking the definite integral of? And we're taking the definite integral of, the definite integral of sine, actually I want the parentheses, sine, and I could use any variable here, as long as I'm consistent with what I am integrating with respect to. So I tend to use just the X button, because there is an X button, but we'll just assume that in this case, X is theta. So sine of X squared, instead of sine of theta squared, where once again we're assuming X is equal to theta. And then the next one, you specify, well, what's the variable you're taking the integral with respect to? In this case it's X."}, {"video_title": "Evaluating definite integral with calculator AP Calculus BC Khan Academy.mp3", "Sentence": "So I tend to use just the X button, because there is an X button, but we'll just assume that in this case, X is theta. So sine of X squared, instead of sine of theta squared, where once again we're assuming X is equal to theta. And then the next one, you specify, well, what's the variable you're taking the integral with respect to? In this case it's X. If we put in a theta here, then we would want to put a theta there as well. And then you want the bounds of integration, and you should assume that your calculator, or if you're doing this, if you're in radian mode, or if you're dealing with radians, you should assume you're in radian mode, I just did before I evaluated this. We're going between zero and pi over four."}, {"video_title": "Evaluating definite integral with calculator AP Calculus BC Khan Academy.mp3", "Sentence": "In this case it's X. If we put in a theta here, then we would want to put a theta there as well. And then you want the bounds of integration, and you should assume that your calculator, or if you're doing this, if you're in radian mode, or if you're dealing with radians, you should assume you're in radian mode, I just did before I evaluated this. We're going between zero and pi over four. Zero and pi over four. Pi over four. And then we get, so we get this number, and then we wanted to multiply it times nine."}, {"video_title": "Evaluating definite integral with calculator AP Calculus BC Khan Academy.mp3", "Sentence": "We're going between zero and pi over four. Zero and pi over four. Pi over four. And then we get, so we get this number, and then we wanted to multiply it times nine. So times, so my previous answer, times nine. If I just press times, it does this. Previous answer times nine is equal to this number, 1.28429."}, {"video_title": "Evaluating definite integral with calculator AP Calculus BC Khan Academy.mp3", "Sentence": "And then we get, so we get this number, and then we wanted to multiply it times nine. So times, so my previous answer, times nine. If I just press times, it does this. Previous answer times nine is equal to this number, 1.28429. So let's verify that that's the same exact value we got when we actually evaluated the integral by hand. So if we take nine, nine pi minus 18 divided by eight, divided by eight, what do we get? We get the exact same value."}, {"video_title": "Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And what I would like to do here is take the derivative of our function, which is essentially going to make us take a derivative of this polynomial expression. And we're gonna take the derivative with respect to x. So the first thing I'm gonna do is let's take the derivative of both sides. So we could say the derivative with respect to x of f of x, of f of x, is equal to the derivative with respect to x, the derivative with respect to x, of x to the fifth, x to the fifth, plus two, plus two x to the third, minus x squared. And so the notation, just to get familiar with it, you could view this as the derivative operator. This says, look, I wanna take the derivative of whatever is inside of the parentheses with respect to x. So the derivative of f with respect to x, we could use a notation that that is just f prime of, f prime of x."}, {"video_title": "Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So we could say the derivative with respect to x of f of x, of f of x, is equal to the derivative with respect to x, the derivative with respect to x, of x to the fifth, x to the fifth, plus two, plus two x to the third, minus x squared. And so the notation, just to get familiar with it, you could view this as the derivative operator. This says, look, I wanna take the derivative of whatever is inside of the parentheses with respect to x. So the derivative of f with respect to x, we could use a notation that that is just f prime of, f prime of x. And that is going to be equal to, now here we can use our derivative properties. The derivative of the sum or difference of a bunch of things is just the derivative of, is equal to the sum or the difference of the derivative of each of them. So this is equal to the derivative, let me just, it's the derivative with respect to x of each of these three things."}, {"video_title": "Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So the derivative of f with respect to x, we could use a notation that that is just f prime of, f prime of x. And that is going to be equal to, now here we can use our derivative properties. The derivative of the sum or difference of a bunch of things is just the derivative of, is equal to the sum or the difference of the derivative of each of them. So this is equal to the derivative, let me just, it's the derivative with respect to x of each of these three things. So the derivative with respect to x, actually let me just write it out like this, of that first term, plus the derivative with respect to x of that second term, minus the derivative with respect to x of that third term, of that third term. And I'll color code it here. So here I had an x to the fifth, so I'll put the x to the fifth there."}, {"video_title": "Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So this is equal to the derivative, let me just, it's the derivative with respect to x of each of these three things. So the derivative with respect to x, actually let me just write it out like this, of that first term, plus the derivative with respect to x of that second term, minus the derivative with respect to x of that third term, of that third term. And I'll color code it here. So here I had an x to the fifth, so I'll put the x to the fifth there. Here I had a two x, here I had a two x to the third, so I'll put the two x to the third there. And here I have a x squared, I'm subtracting x squared, so I'm subtracting the derivative with respect to x of x squared. So notice, all that's happening here is I'm taking the derivative individually of each of these terms, and then I'm adding or subtracting them the same way that the terms were added or subtracted."}, {"video_title": "Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So here I had an x to the fifth, so I'll put the x to the fifth there. Here I had a two x, here I had a two x to the third, so I'll put the two x to the third there. And here I have a x squared, I'm subtracting x squared, so I'm subtracting the derivative with respect to x of x squared. So notice, all that's happening here is I'm taking the derivative individually of each of these terms, and then I'm adding or subtracting them the same way that the terms were added or subtracted. And so what is this going to be equal to? Well, this is going to be equal to, for x to the fifth, we can just use the power rule. We can bring the five out front and decrement the exponent by one, so it becomes five x, we could say to the five minus one power, which of course is just four."}, {"video_title": "Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So notice, all that's happening here is I'm taking the derivative individually of each of these terms, and then I'm adding or subtracting them the same way that the terms were added or subtracted. And so what is this going to be equal to? Well, this is going to be equal to, for x to the fifth, we can just use the power rule. We can bring the five out front and decrement the exponent by one, so it becomes five x, we could say to the five minus one power, which of course is just four. And then for this second one, we could do it in a few steps. Actually, let me just write it out here. So I could write, I could write the derivative with respect to x of two x to the third power is the same thing, it's equal to, the same, we could bring the constant out, the derivative with, two times the derivative with respect to x of x to the third power."}, {"video_title": "Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "We can bring the five out front and decrement the exponent by one, so it becomes five x, we could say to the five minus one power, which of course is just four. And then for this second one, we could do it in a few steps. Actually, let me just write it out here. So I could write, I could write the derivative with respect to x of two x to the third power is the same thing, it's equal to, the same, we could bring the constant out, the derivative with, two times the derivative with respect to x of x to the third power. This is one of our, this is one of our derivative properties. The derivative of a constant times some expression is the same thing as the constant times the derivative of that expression. And what will the derivative with respect to x of, x to the third be?"}, {"video_title": "Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So I could write, I could write the derivative with respect to x of two x to the third power is the same thing, it's equal to, the same, we could bring the constant out, the derivative with, two times the derivative with respect to x of x to the third power. This is one of our, this is one of our derivative properties. The derivative of a constant times some expression is the same thing as the constant times the derivative of that expression. And what will the derivative with respect to x of, x to the third be? Well, we would bring the three out front and decrement the exponent, and so this would be equal to this two times the three times x to the three minus one power, which is of course the second power. So this would give us six x squared. So another way that you could have done it, I could just write, I could just write a six x squared here."}, {"video_title": "Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And what will the derivative with respect to x of, x to the third be? Well, we would bring the three out front and decrement the exponent, and so this would be equal to this two times the three times x to the three minus one power, which is of course the second power. So this would give us six x squared. So another way that you could have done it, I could just write, I could just write a six x squared here. So I could just, so this is going to be six x squared. And instead of going through all of this, you'll learn as you do more of these that you could have done this pretty much in your head, saying look, I have the three out here as an exponent. Let me multiply the three times this coefficient, because that's what we ended up doing anyway."}, {"video_title": "Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So another way that you could have done it, I could just write, I could just write a six x squared here. So I could just, so this is going to be six x squared. And instead of going through all of this, you'll learn as you do more of these that you could have done this pretty much in your head, saying look, I have the three out here as an exponent. Let me multiply the three times this coefficient, because that's what we ended up doing anyway. Three times the coefficient is six x, and then three minus one is two. So you didn't necessarily have to do this, but it's nice to see that this comes out of the derivative properties that we talk about in other videos. And then finally, we have minus, and we use the power rule right over here."}, {"video_title": "Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Let me multiply the three times this coefficient, because that's what we ended up doing anyway. Three times the coefficient is six x, and then three minus one is two. So you didn't necessarily have to do this, but it's nice to see that this comes out of the derivative properties that we talk about in other videos. And then finally, we have minus, and we use the power rule right over here. So bring the two out front and decrement the exponent. So it's going to be two. It's going to be two times x to the two minus one power, which is just one, which we could just write as two x."}, {"video_title": "Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And then finally, we have minus, and we use the power rule right over here. So bring the two out front and decrement the exponent. So it's going to be two. It's going to be two times x to the two minus one power, which is just one, which we could just write as two x. So just like that, we have been able to figure out the derivative of f. And you might say, well, what is this thing now? Well, now we have an expression that tells us the slope of the tangent line, or you could view it as the instantaneous rate of change with respect to x for any x value. So if I were to say, if I were now to say f prime, let's say f prime of two, this would tell me what is the slope of the tangent line of our function when x is equal to two."}, {"video_title": "Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "It's going to be two times x to the two minus one power, which is just one, which we could just write as two x. So just like that, we have been able to figure out the derivative of f. And you might say, well, what is this thing now? Well, now we have an expression that tells us the slope of the tangent line, or you could view it as the instantaneous rate of change with respect to x for any x value. So if I were to say, if I were now to say f prime, let's say f prime of two, this would tell me what is the slope of the tangent line of our function when x is equal to two. And I do that by using this expression. So this is going to be five times two to the fourth plus six times two squared, six times two squared, minus two times two, minus two times two. And this is going to be equal to, let's see, two to the fourth power is 16, 16 times five is 80, so that's 80."}, {"video_title": "Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So if I were to say, if I were now to say f prime, let's say f prime of two, this would tell me what is the slope of the tangent line of our function when x is equal to two. And I do that by using this expression. So this is going to be five times two to the fourth plus six times two squared, six times two squared, minus two times two, minus two times two. And this is going to be equal to, let's see, two to the fourth power is 16, 16 times five is 80, so that's 80. And then this is six times four, which is 24. And then we are going to subtract four. So this is 80 plus 24 is 104, minus four is equal to 100."}, {"video_title": "Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And this is going to be equal to, let's see, two to the fourth power is 16, 16 times five is 80, so that's 80. And then this is six times four, which is 24. And then we are going to subtract four. So this is 80 plus 24 is 104, minus four is equal to 100. So when x is equal to two, this curve is really steep. The slope is 100. If you were to graph the tangent line when x is equal to two, for every positive movement in the x direction by one, you're going to move up in the y direction by 100."}, {"video_title": "Differentiating polynomials example Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So this is 80 plus 24 is 104, minus four is equal to 100. So when x is equal to two, this curve is really steep. The slope is 100. If you were to graph the tangent line when x is equal to two, for every positive movement in the x direction by one, you're going to move up in the y direction by 100. So it's really steep there, and it makes sense. This is a pretty high degree. X to the fifth power, and then we're adding that to another high degree, x to the third power, and then we're subtracting a lower degree."}, {"video_title": "Writing a differential equation Differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "Its speed is inversely proportional to the square of the distance s it has traveled. Which equation describes this relationship? So I'm not gonna even look at these choices, and I'm just gonna try to parse this sentence up here and see if we can come up with an equation. So they tell us its speed is inversely proportional to what? To the square of the distance s it has traveled. So s is equal to distance. S is equal to distance."}, {"video_title": "Writing a differential equation Differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "So they tell us its speed is inversely proportional to what? To the square of the distance s it has traveled. So s is equal to distance. S is equal to distance. And how would we denote speed then, if s is distance? Well, speed is the rate of change of distance with respect to time. So our speed would be the rate of distance with respect to time."}, {"video_title": "Writing a differential equation Differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "S is equal to distance. And how would we denote speed then, if s is distance? Well, speed is the rate of change of distance with respect to time. So our speed would be the rate of distance with respect to time. The rate of change of distance with respect to time. So this is going to be our speed. So now that we got our notation, the s is the distance, the derivative of s with respect to time is speed, we can say the speed, which is d capital S, dt, is inversely proportional."}, {"video_title": "Writing a differential equation Differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "So our speed would be the rate of distance with respect to time. The rate of change of distance with respect to time. So this is going to be our speed. So now that we got our notation, the s is the distance, the derivative of s with respect to time is speed, we can say the speed, which is d capital S, dt, is inversely proportional. So it's inversely proportional. I'll write a proportionality constant over what? It's inversely proportional to what?"}, {"video_title": "Writing a differential equation Differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "So now that we got our notation, the s is the distance, the derivative of s with respect to time is speed, we can say the speed, which is d capital S, dt, is inversely proportional. So it's inversely proportional. I'll write a proportionality constant over what? It's inversely proportional to what? To the square of the distance. To the square of the distance it has traveled. So there you go."}, {"video_title": "Writing a differential equation Differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "It's inversely proportional to what? To the square of the distance. To the square of the distance it has traveled. So there you go. This is an equation that I think is describing a differential equation, really, that's describing what we have up here. Now let's see which of these choices match that. Well, actually, this one is exactly what we wrote."}, {"video_title": "Writing a differential equation Differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "So there you go. This is an equation that I think is describing a differential equation, really, that's describing what we have up here. Now let's see which of these choices match that. Well, actually, this one is exactly what we wrote. The speed, the rate of change of distance with respect to time is inversely proportional to the square of the distance. Now just to make sure we understand these other ones, let's just interpret them. This is saying that the distance, which is a function of time, is inversely proportional to the time squared."}, {"video_title": "Writing a differential equation Differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "Well, actually, this one is exactly what we wrote. The speed, the rate of change of distance with respect to time is inversely proportional to the square of the distance. Now just to make sure we understand these other ones, let's just interpret them. This is saying that the distance, which is a function of time, is inversely proportional to the time squared. That's not what they told us. This is saying that the distance is inversely proportional to the distance squared. That one is especially strange."}, {"video_title": "Writing a differential equation Differential equations AP Calculus AB Khan Academy.mp3", "Sentence": "This is saying that the distance, which is a function of time, is inversely proportional to the time squared. That's not what they told us. This is saying that the distance is inversely proportional to the distance squared. That one is especially strange. And this is saying that the distance with respect to time, the change in distance with respect to time, the derivative of the distance with respect to time, ds dt, or the speed, is inversely proportional to time squared. Well, that's not what they said. They said it's inversely proportional to the square of the distance it has traveled."}, {"video_title": "Finding function from power series by integrating Series AP Calculus BC Khan Academy.mp3", "Sentence": "And I say using this fact, find the function that corresponds to the following series. And like always, pause this video and see if you can work through it. Alright, so the first thing you might say is, well, how do we know that this expression is equal to this series? And you might recognize this series as a geometric series where the first term is negative two and then the common ratio, to get each successive term, we're multiplying by two x. We're multiplying by two x. And so for a geometric series like this, the sum is going to be the first term, negative two, over one minus the common ratio, which is exactly what we have over there. So that's why we know that."}, {"video_title": "Finding function from power series by integrating Series AP Calculus BC Khan Academy.mp3", "Sentence": "And you might recognize this series as a geometric series where the first term is negative two and then the common ratio, to get each successive term, we're multiplying by two x. We're multiplying by two x. And so for a geometric series like this, the sum is going to be the first term, negative two, over one minus the common ratio, which is exactly what we have over there. So that's why we know that. And this right over here, this gives us our radius of convergence, the x values for which this thing will actually converge. But now that we feel good about this first statement, let's try to answer their actual question. So we wanna find the function that corresponds to the following series."}, {"video_title": "Finding function from power series by integrating Series AP Calculus BC Khan Academy.mp3", "Sentence": "So that's why we know that. And this right over here, this gives us our radius of convergence, the x values for which this thing will actually converge. But now that we feel good about this first statement, let's try to answer their actual question. So we wanna find the function that corresponds to the following series. So my instinct is to say, well, how does this series relate to the series they gave us? Let's see, this first term is a negative two, this first term is a negative two x, this is a negative four x, this is a negative two x squared. So the thing that might jump out at you is that the second series they gave us is the antiderivative of this first one, or we could say this first series is the derivative of the second one."}, {"video_title": "Finding function from power series by integrating Series AP Calculus BC Khan Academy.mp3", "Sentence": "So we wanna find the function that corresponds to the following series. So my instinct is to say, well, how does this series relate to the series they gave us? Let's see, this first term is a negative two, this first term is a negative two x, this is a negative four x, this is a negative two x squared. So the thing that might jump out at you is that the second series they gave us is the antiderivative of this first one, or we could say this first series is the derivative of the second one. What's the derivative of negative two x with respect to x? Well, it's negative two. What's the derivative of negative two x squared with respect to x?"}, {"video_title": "Finding function from power series by integrating Series AP Calculus BC Khan Academy.mp3", "Sentence": "So the thing that might jump out at you is that the second series they gave us is the antiderivative of this first one, or we could say this first series is the derivative of the second one. What's the derivative of negative two x with respect to x? Well, it's negative two. What's the derivative of negative two x squared with respect to x? Well, it's negative four x, and so on and so forth. And so if we were to call this, let me call this thing right over here, actually let me just say that this is equal to g of x. Well, the way I can take this right-hand side and get g of x is by taking the antiderivative, taking the indefinite integral."}, {"video_title": "Finding function from power series by integrating Series AP Calculus BC Khan Academy.mp3", "Sentence": "What's the derivative of negative two x squared with respect to x? Well, it's negative four x, and so on and so forth. And so if we were to call this, let me call this thing right over here, actually let me just say that this is equal to g of x. Well, the way I can take this right-hand side and get g of x is by taking the antiderivative, taking the indefinite integral. And so what we can do, we can take the indefinite integral of both sides, dx. So dx. On the right-hand side, I'm gonna get g of x."}, {"video_title": "Finding function from power series by integrating Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, the way I can take this right-hand side and get g of x is by taking the antiderivative, taking the indefinite integral. And so what we can do, we can take the indefinite integral of both sides, dx. So dx. On the right-hand side, I'm gonna get g of x. And on the left-hand side, I am going to get, actually let me write it this way. So on the left-hand side, I'll just rewrite it, I'm gonna get the indefinite integral of, I'll write it as negative two dx over one minus two x. That's just another way of writing what I have on the left-hand side here."}, {"video_title": "Finding function from power series by integrating Series AP Calculus BC Khan Academy.mp3", "Sentence": "On the right-hand side, I'm gonna get g of x. And on the left-hand side, I am going to get, actually let me write it this way. So on the left-hand side, I'll just rewrite it, I'm gonna get the indefinite integral of, I'll write it as negative two dx over one minus two x. That's just another way of writing what I have on the left-hand side here. This is equal to g of x, right? If I take the antiderivative of this, or an antiderivative of this, is this g of x, where the constant would have been zero. So I'll say is equal to g of x."}, {"video_title": "Finding function from power series by integrating Series AP Calculus BC Khan Academy.mp3", "Sentence": "That's just another way of writing what I have on the left-hand side here. This is equal to g of x, right? If I take the antiderivative of this, or an antiderivative of this, is this g of x, where the constant would have been zero. So I'll say is equal to g of x. And so the key here is, well, what's the indefinite integral of this stuff? And you might immediately recognize, well, this, what I have here in the bottom, I have its derivative here on the top. If I consider this to be u, if I say u is equal to one minus two x, this is just u substitution, then du is equal to the derivative of this with respect to x, which is negative two dx."}, {"video_title": "Finding function from power series by integrating Series AP Calculus BC Khan Academy.mp3", "Sentence": "So I'll say is equal to g of x. And so the key here is, well, what's the indefinite integral of this stuff? And you might immediately recognize, well, this, what I have here in the bottom, I have its derivative here on the top. If I consider this to be u, if I say u is equal to one minus two x, this is just u substitution, then du is equal to the derivative of this with respect to x, which is negative two dx. And so I have the du right over here. So let me rewrite all of this. I can rewrite this as the integral of du over u du over u is equal to g of x."}, {"video_title": "Finding function from power series by integrating Series AP Calculus BC Khan Academy.mp3", "Sentence": "If I consider this to be u, if I say u is equal to one minus two x, this is just u substitution, then du is equal to the derivative of this with respect to x, which is negative two dx. And so I have the du right over here. So let me rewrite all of this. I can rewrite this as the integral of du over u du over u is equal to g of x. And this we can rewrite as the natural log of the absolute value of u plus c is equal to g of x. And then we can undo our u substitution. And so for u, I will undo, I will substitute back the one minus two x."}, {"video_title": "Finding function from power series by integrating Series AP Calculus BC Khan Academy.mp3", "Sentence": "I can rewrite this as the integral of du over u du over u is equal to g of x. And this we can rewrite as the natural log of the absolute value of u plus c is equal to g of x. And then we can undo our u substitution. And so for u, I will undo, I will substitute back the one minus two x. So I can write the natural log of the absolute value of one minus two x is equal to g, oh, plus c, don't want to forget that, plus c is equal to g of x. And so the next thing we want to do is, well, what is our c? And the easiest way to figure out c is let's substitute zero for x."}, {"video_title": "Finding function from power series by integrating Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so for u, I will undo, I will substitute back the one minus two x. So I can write the natural log of the absolute value of one minus two x is equal to g, oh, plus c, don't want to forget that, plus c is equal to g of x. And so the next thing we want to do is, well, what is our c? And the easiest way to figure out c is let's substitute zero for x. And so let's think about this a little bit. So if I put a zero here, if I put x equals, actually, let me just write it again. I'll have the natural log of, if I say x is zero, this is gonna be the natural log of the absolute value of one plus c is equal to g of zero."}, {"video_title": "Finding function from power series by integrating Series AP Calculus BC Khan Academy.mp3", "Sentence": "And the easiest way to figure out c is let's substitute zero for x. And so let's think about this a little bit. So if I put a zero here, if I put x equals, actually, let me just write it again. I'll have the natural log of, if I say x is zero, this is gonna be the natural log of the absolute value of one plus c is equal to g of zero. So what's g of zero? G of zero, every one of these terms is equal to zero. G of zero is equal to zero."}, {"video_title": "Finding function from power series by integrating Series AP Calculus BC Khan Academy.mp3", "Sentence": "I'll have the natural log of, if I say x is zero, this is gonna be the natural log of the absolute value of one plus c is equal to g of zero. So what's g of zero? G of zero, every one of these terms is equal to zero. G of zero is equal to zero. So is equal to zero. Well, natural log of one is just zero. So zero plus c is equal to zero."}, {"video_title": "Finding function from power series by integrating Series AP Calculus BC Khan Academy.mp3", "Sentence": "G of zero is equal to zero. So is equal to zero. Well, natural log of one is just zero. So zero plus c is equal to zero. C is equal to zero. So there you have it. I just took the antiderivative of both sides of this equation right over here."}, {"video_title": "Finding function from power series by integrating Series AP Calculus BC Khan Academy.mp3", "Sentence": "So zero plus c is equal to zero. C is equal to zero. So there you have it. I just took the antiderivative of both sides of this equation right over here. I figured out when I substituted zero for x, that okay, my constant here is going to be zero. And I get that this series, this g of x is equal to the natural log of the absolute value of one minus two x. Is equal to the natural log of the absolute value of one minus two x."}, {"video_title": "Finding function from power series by integrating Series AP Calculus BC Khan Academy.mp3", "Sentence": "I just took the antiderivative of both sides of this equation right over here. I figured out when I substituted zero for x, that okay, my constant here is going to be zero. And I get that this series, this g of x is equal to the natural log of the absolute value of one minus two x. Is equal to the natural log of the absolute value of one minus two x. Now, if we keep this restriction that we're in this open interval, then this is always going to be positive. We wouldn't have to write the absolute value, but we could be safe by writing the absolute value. But there you go."}, {"video_title": "Finding function from power series by integrating Series AP Calculus BC Khan Academy.mp3", "Sentence": "Is equal to the natural log of the absolute value of one minus two x. Now, if we keep this restriction that we're in this open interval, then this is always going to be positive. We wouldn't have to write the absolute value, but we could be safe by writing the absolute value. But there you go. Using the fact above, we found the function that corresponds to this following series. And I guess you could say the trick of it was recognizing that the series up here is a derivative of the series down here. And so this is going to be the derivative of the following series."}, {"video_title": "Worked example Area enclosed by cardioid AP Calculus BC Khan Academy.mp3", "Sentence": "Of course, we're dealing in polar coordinates here. And what I'm interested in is to see if we can figure out the area enclosed by this curve. And I encourage you to pause the video and try it on your own. All right, let's work through it together. So we've already seen, we've already given ourselves the intuition for the formula that the area enclosed by a polar graph is going to be equal to one half the definite integral from, I should say, our starting theta to our ending theta from alpha to beta of r of theta squared d theta. And so we essentially just, or we just have to apply this to this function right over here. So in this case, the area is going to be equal to one half the definite integral."}, {"video_title": "Worked example Area enclosed by cardioid AP Calculus BC Khan Academy.mp3", "Sentence": "All right, let's work through it together. So we've already seen, we've already given ourselves the intuition for the formula that the area enclosed by a polar graph is going to be equal to one half the definite integral from, I should say, our starting theta to our ending theta from alpha to beta of r of theta squared d theta. And so we essentially just, or we just have to apply this to this function right over here. So in this case, the area is going to be equal to one half the definite integral. Now what's our alpha and what's our beta? Well, we're going from theta is equal to zero radians, and we're essentially going all the way, when theta is equal to zero radians is one minus one, we're right over here. And then we go all the way around to theta is equal to two pi radians."}, {"video_title": "Worked example Area enclosed by cardioid AP Calculus BC Khan Academy.mp3", "Sentence": "So in this case, the area is going to be equal to one half the definite integral. Now what's our alpha and what's our beta? Well, we're going from theta is equal to zero radians, and we're essentially going all the way, when theta is equal to zero radians is one minus one, we're right over here. And then we go all the way around to theta is equal to two pi radians. Notice when we're back at two pi, cosine of two pi is one, one minus one is zero again, so we get back to that point. So we're going from theta is equal to zero radians to theta is equal to two pi radians. Now what's r of theta squared?"}, {"video_title": "Worked example Area enclosed by cardioid AP Calculus BC Khan Academy.mp3", "Sentence": "And then we go all the way around to theta is equal to two pi radians. Notice when we're back at two pi, cosine of two pi is one, one minus one is zero again, so we get back to that point. So we're going from theta is equal to zero radians to theta is equal to two pi radians. Now what's r of theta squared? Maybe I'll color code this a little bit. R of theta squared. Well, it's just going to be one minus cosine of theta."}, {"video_title": "Worked example Area enclosed by cardioid AP Calculus BC Khan Academy.mp3", "Sentence": "Now what's r of theta squared? Maybe I'll color code this a little bit. R of theta squared. Well, it's just going to be one minus cosine of theta. One minus cosine theta squared. And of course, we have our d theta. We have our d theta."}, {"video_title": "Worked example Area enclosed by cardioid AP Calculus BC Khan Academy.mp3", "Sentence": "Well, it's just going to be one minus cosine of theta. One minus cosine theta squared. And of course, we have our d theta. We have our d theta. And now we just have to evaluate this integral. So once again, at any point you feel inspired, try to evaluate this. So let's do this."}, {"video_title": "Worked example Area enclosed by cardioid AP Calculus BC Khan Academy.mp3", "Sentence": "We have our d theta. And now we just have to evaluate this integral. So once again, at any point you feel inspired, try to evaluate this. So let's do this. All right, so what I would do, so this is going to be equal to 1 1\u20442 times the definite integral from zero to two pi. And let me expand this out. This is going to be one minus two cosine theta cosine theta plus cosine squared theta, d theta, d theta."}, {"video_title": "Worked example Area enclosed by cardioid AP Calculus BC Khan Academy.mp3", "Sentence": "So let's do this. All right, so what I would do, so this is going to be equal to 1 1\u20442 times the definite integral from zero to two pi. And let me expand this out. This is going to be one minus two cosine theta cosine theta plus cosine squared theta, d theta, d theta. Now I know how to take the antiderivative of one. I know how to take the antiderivative of negative cosine of theta. But cosine squared theta, this is a little bit, it doesn't jump out at you that you can just use u substitution or something like that."}, {"video_title": "Worked example Area enclosed by cardioid AP Calculus BC Khan Academy.mp3", "Sentence": "This is going to be one minus two cosine theta cosine theta plus cosine squared theta, d theta, d theta. Now I know how to take the antiderivative of one. I know how to take the antiderivative of negative cosine of theta. But cosine squared theta, this is a little bit, it doesn't jump out at you that you can just use u substitution or something like that. But lucky for us, we have our trigonometric identities. And so we know that cosine squared of theta is just the same thing as 1 1\u20442 times one plus cosine of two theta. You learned this in trigonometry class."}, {"video_title": "Worked example Area enclosed by cardioid AP Calculus BC Khan Academy.mp3", "Sentence": "But cosine squared theta, this is a little bit, it doesn't jump out at you that you can just use u substitution or something like that. But lucky for us, we have our trigonometric identities. And so we know that cosine squared of theta is just the same thing as 1 1\u20442 times one plus cosine of two theta. You learned this in trigonometry class. If you didn't, well you learned it just now. And so that's why, well this is one of the more useful trigonometric identities if you're finding any type of antiderivative or if you're integrating anything. And so let's do that."}, {"video_title": "Worked example Area enclosed by cardioid AP Calculus BC Khan Academy.mp3", "Sentence": "You learned this in trigonometry class. If you didn't, well you learned it just now. And so that's why, well this is one of the more useful trigonometric identities if you're finding any type of antiderivative or if you're integrating anything. And so let's do that. Let's rewrite this right over here as 1 1\u20442 times one plus cosine of two theta. And let's see, maybe we could, well yeah, let's just do it like that. I guess we could if we want, well we'll just do it like that."}, {"video_title": "Worked example Area enclosed by cardioid AP Calculus BC Khan Academy.mp3", "Sentence": "And so let's do that. Let's rewrite this right over here as 1 1\u20442 times one plus cosine of two theta. And let's see, maybe we could, well yeah, let's just do it like that. I guess we could if we want, well we'll just do it like that. So this is going to be equal to, this is going to be equal to 1 1\u20442. And then we are going to, 1 1\u20442, now let's just start taking antiderivatives. 1 1\u20442."}, {"video_title": "Worked example Area enclosed by cardioid AP Calculus BC Khan Academy.mp3", "Sentence": "I guess we could if we want, well we'll just do it like that. So this is going to be equal to, this is going to be equal to 1 1\u20442. And then we are going to, 1 1\u20442, now let's just start taking antiderivatives. 1 1\u20442. Now the antiderivative of one with respect to theta is just going to be theta. The antiderivative of negative two cosine of theta, well that's just going to be negative two sine of theta. Negative two sine theta."}, {"video_title": "Worked example Area enclosed by cardioid AP Calculus BC Khan Academy.mp3", "Sentence": "1 1\u20442. Now the antiderivative of one with respect to theta is just going to be theta. The antiderivative of negative two cosine of theta, well that's just going to be negative two sine of theta. Negative two sine theta. You take the derivative of the derivative of sine is cosine. And the negative two, it'll just multiply it times the derivative of sine of theta, so it's negative two cosine of theta. And then we're going to have, let's see."}, {"video_title": "Worked example Area enclosed by cardioid AP Calculus BC Khan Academy.mp3", "Sentence": "Negative two sine theta. You take the derivative of the derivative of sine is cosine. And the negative two, it'll just multiply it times the derivative of sine of theta, so it's negative two cosine of theta. And then we're going to have, let's see. Actually let me distribute this. This is the same thing as 1 1\u20442 plus 1 1\u20442 cosine of two theta. So let's just assume, well it's this way."}, {"video_title": "Worked example Area enclosed by cardioid AP Calculus BC Khan Academy.mp3", "Sentence": "And then we're going to have, let's see. Actually let me distribute this. This is the same thing as 1 1\u20442 plus 1 1\u20442 cosine of two theta. So let's just assume, well it's this way. So the antiderivative of 1 1\u20442, so the antiderivative of 1 1\u20442, so I'm really looking at that right over there, is going to be 1 1\u20442 theta. 1 1\u20442 theta. And then the antiderivative of 1 1\u20442 cosine of two theta, let's see."}, {"video_title": "Worked example Area enclosed by cardioid AP Calculus BC Khan Academy.mp3", "Sentence": "So let's just assume, well it's this way. So the antiderivative of 1 1\u20442, so the antiderivative of 1 1\u20442, so I'm really looking at that right over there, is going to be 1 1\u20442 theta. 1 1\u20442 theta. And then the antiderivative of 1 1\u20442 cosine of two theta, let's see. Sine, the derivative of sine of two theta is two cosine of two theta. So this is going to, so the antiderivative of cosine of two theta, and you could do u-substitution if you like, but you might be able to do this in your head. The antiderivative of cosine of two theta is going to be 1 1\u20442 sine of two theta."}, {"video_title": "Worked example Area enclosed by cardioid AP Calculus BC Khan Academy.mp3", "Sentence": "And then the antiderivative of 1 1\u20442 cosine of two theta, let's see. Sine, the derivative of sine of two theta is two cosine of two theta. So this is going to, so the antiderivative of cosine of two theta, and you could do u-substitution if you like, but you might be able to do this in your head. The antiderivative of cosine of two theta is going to be 1 1\u20442 sine of two theta. And then you have this 1 1\u20442 right over here. So this is going to be, let me show you what I'm finding the antiderivative of, of that right over there, of this, and I guess this right over here. This is going to be plus 1\u20444 sine of two theta."}, {"video_title": "Worked example Area enclosed by cardioid AP Calculus BC Khan Academy.mp3", "Sentence": "The antiderivative of cosine of two theta is going to be 1 1\u20442 sine of two theta. And then you have this 1 1\u20442 right over here. So this is going to be, let me show you what I'm finding the antiderivative of, of that right over there, of this, and I guess this right over here. This is going to be plus 1\u20444 sine of two theta. And I encourage you to find the derivative here if that last part was a little bit confusing. Derivative of sine of two theta is two cosine of two theta. Two over 1\u20444 is 1 1\u20442."}, {"video_title": "Worked example Area enclosed by cardioid AP Calculus BC Khan Academy.mp3", "Sentence": "This is going to be plus 1\u20444 sine of two theta. And I encourage you to find the derivative here if that last part was a little bit confusing. Derivative of sine of two theta is two cosine of two theta. Two over 1\u20444 is 1 1\u20442. You get to 1 1\u20442 cosine of two theta. And we're going to evaluate that at two pi, at two pi and at zero. So when you evaluate it, well one thing that might jump out at you is when you evaluate this at zero, this whole thing is, everything, every term here is just going to be zero."}, {"video_title": "Worked example Area enclosed by cardioid AP Calculus BC Khan Academy.mp3", "Sentence": "Two over 1\u20444 is 1 1\u20442. You get to 1 1\u20442 cosine of two theta. And we're going to evaluate that at two pi, at two pi and at zero. So when you evaluate it, well one thing that might jump out at you is when you evaluate this at zero, this whole thing is, everything, every term here is just going to be zero. So that simplifies things nicely. So we really just have to take 1\u20442 of, it evaluated at two pi. So this is going to be 1\u20442 times two pi, two pi, and then sine of two pi is zero, so that's just going to be zero."}, {"video_title": "Worked example Area enclosed by cardioid AP Calculus BC Khan Academy.mp3", "Sentence": "So when you evaluate it, well one thing that might jump out at you is when you evaluate this at zero, this whole thing is, everything, every term here is just going to be zero. So that simplifies things nicely. So we really just have to take 1\u20442 of, it evaluated at two pi. So this is going to be 1\u20442 times two pi, two pi, and then sine of two pi is zero, so that's just going to be zero. And then plus 1\u20442 times two pi. So that's going to be plus pi. And then sine of two times two pi is sine of four pi."}, {"video_title": "Worked example Area enclosed by cardioid AP Calculus BC Khan Academy.mp3", "Sentence": "So this is going to be 1\u20442 times two pi, two pi, and then sine of two pi is zero, so that's just going to be zero. And then plus 1\u20442 times two pi. So that's going to be plus pi. And then sine of two times two pi is sine of four pi. That's still going to be zero. So this is going to be zero as well. And we are almost done."}, {"video_title": "Functions continuous on all real numbers Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So let's just remind ourselves what it means to be continuous, what a continuous function looks like. So a continuous function, let's say that's my y-axis, that is my x-axis, a function is going to be continuous over some interval if it just doesn't have any jumps or discontinuities or gaps over that interval. So if it's connected, and it for sure has to be defined over that interval without any gaps. So for example, a continuous function could look something like this. This function, let me make that line a little bit thicker. So this function right over here is continuous. It is connected over this interval, the interval that we can see."}, {"video_title": "Functions continuous on all real numbers Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So for example, a continuous function could look something like this. This function, let me make that line a little bit thicker. So this function right over here is continuous. It is connected over this interval, the interval that we can see. Now examples of discontinuous functions over interval or non-continuous functions, well, they would have gaps of some kind. They could have some type of an asymptotic discontinuity. So something like that, that makes it discontinuous."}, {"video_title": "Functions continuous on all real numbers Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "It is connected over this interval, the interval that we can see. Now examples of discontinuous functions over interval or non-continuous functions, well, they would have gaps of some kind. They could have some type of an asymptotic discontinuity. So something like that, that makes it discontinuous. They could have a jump discontinuity, something like that. They could just have a gap where they're not defined. So they could have a gap where they're not defined."}, {"video_title": "Functions continuous on all real numbers Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So something like that, that makes it discontinuous. They could have a jump discontinuity, something like that. They could just have a gap where they're not defined. So they could have a gap where they're not defined. Or maybe they actually are defined there, but it's removable discontinuity. So all of these are examples of discontinuous functions. Now if you want the more mathy understanding of that, and we've looked at this before, we say that a function, f, is continuous at some value x equals a if and only if, draw my little two-way arrows here, say if and only if the limit of f of x as x approaches a is equal to the value of the function at a."}, {"video_title": "Functions continuous on all real numbers Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So they could have a gap where they're not defined. Or maybe they actually are defined there, but it's removable discontinuity. So all of these are examples of discontinuous functions. Now if you want the more mathy understanding of that, and we've looked at this before, we say that a function, f, is continuous at some value x equals a if and only if, draw my little two-way arrows here, say if and only if the limit of f of x as x approaches a is equal to the value of the function at a. So once again, in order to be continuous there, you at least have to be defined there. Now when you look at these, the one thing that jumps out at me, in order to be continuous for all real numbers, you have to be defined for all real numbers, and g of x is not defined for all real numbers. It's not defined for negative values of x."}, {"video_title": "Functions continuous on all real numbers Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Now if you want the more mathy understanding of that, and we've looked at this before, we say that a function, f, is continuous at some value x equals a if and only if, draw my little two-way arrows here, say if and only if the limit of f of x as x approaches a is equal to the value of the function at a. So once again, in order to be continuous there, you at least have to be defined there. Now when you look at these, the one thing that jumps out at me, in order to be continuous for all real numbers, you have to be defined for all real numbers, and g of x is not defined for all real numbers. It's not defined for negative values of x. And so we would rule this one out. So let's think about f of x equals e to the x. It is defined for all real numbers, and as we'll see, most of the common functions that you've learned in math, they don't have these strange jumps or gaps or discontinuities."}, {"video_title": "Functions continuous on all real numbers Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "It's not defined for negative values of x. And so we would rule this one out. So let's think about f of x equals e to the x. It is defined for all real numbers, and as we'll see, most of the common functions that you've learned in math, they don't have these strange jumps or gaps or discontinuities. Some of them do, functions like one over x and things like that, but things like e to the x, it doesn't have any of those. We could graph e to the x. E to the x looks something like this. It's defined for all real numbers."}, {"video_title": "Functions continuous on all real numbers Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "It is defined for all real numbers, and as we'll see, most of the common functions that you've learned in math, they don't have these strange jumps or gaps or discontinuities. Some of them do, functions like one over x and things like that, but things like e to the x, it doesn't have any of those. We could graph e to the x. E to the x looks something like this. It's defined for all real numbers. There's no jumps or gaps of any kind, and so this f of x is continuous for all real numbers, and f only. Now I didn't do a very rigorous proof. You could if you like, but for the sake of this exercise, it's really more of getting this intuitive sense of like look, look, e to the x is defined for all real numbers, and so, and there's no jumps or gaps here, so it's reasonable to say that it's continuous, but you could do a more rigorous proof if you like as well."}, {"video_title": "Worked example convergent geometric series Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's see, to go from the first term to the second term, we multiply by 1 3rd, and then go to the next term, we are going to multiply by 1 3rd again, and we're going to keep doing that. So we can rewrite the series as eight plus eight times 1 3rd, eight times 1 3rd, plus, plus eight times 1 3rd squared, eight times 1 3rd squared. Each successive term, we multiply by 1 3rd again. And so when you look at it this way, you're like, okay, we could write this in sigma notation. This is going to be equal to, so this, the first thing we wrote is equal to this, which is equal to, this is equal to the sum, the sum, and we could start at zero or at one, depending on how we'd like to do it. We could say from k is equal to zero, and this is an infinite series right here. We're just gonna keep on going forever."}, {"video_title": "Worked example convergent geometric series Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so when you look at it this way, you're like, okay, we could write this in sigma notation. This is going to be equal to, so this, the first thing we wrote is equal to this, which is equal to, this is equal to the sum, the sum, and we could start at zero or at one, depending on how we'd like to do it. We could say from k is equal to zero, and this is an infinite series right here. We're just gonna keep on going forever. So two infinity of, well, what's our first term? Our first term is eight. So it's going to be eight times our common ratio, times our common ratio, 1 3rd, to the k, to the k power."}, {"video_title": "Worked example convergent geometric series Series AP Calculus BC Khan Academy.mp3", "Sentence": "We're just gonna keep on going forever. So two infinity of, well, what's our first term? Our first term is eight. So it's going to be eight times our common ratio, times our common ratio, 1 3rd, to the k, to the k power. And let me just verify that this indeed works, and I always do this just as a reality check, and I encourage you to do the same. So when k equals zero, that should be the first term right over here. You get eight times 1 3rd to the zero power, which is indeed eight."}, {"video_title": "Worked example convergent geometric series Series AP Calculus BC Khan Academy.mp3", "Sentence": "So it's going to be eight times our common ratio, times our common ratio, 1 3rd, to the k, to the k power. And let me just verify that this indeed works, and I always do this just as a reality check, and I encourage you to do the same. So when k equals zero, that should be the first term right over here. You get eight times 1 3rd to the zero power, which is indeed eight. When k is equal to one, that's gonna be our second term here. That's going to be eight times 1 3rd to the first power. That's what we have here."}, {"video_title": "Worked example convergent geometric series Series AP Calculus BC Khan Academy.mp3", "Sentence": "You get eight times 1 3rd to the zero power, which is indeed eight. When k is equal to one, that's gonna be our second term here. That's going to be eight times 1 3rd to the first power. That's what we have here. And so when k is equal to two, that is this term right over here. So these are all describing the same thing. So now that we've seen that we can write a geometric series in multiple ways, let's find the sum."}, {"video_title": "Worked example convergent geometric series Series AP Calculus BC Khan Academy.mp3", "Sentence": "That's what we have here. And so when k is equal to two, that is this term right over here. So these are all describing the same thing. So now that we've seen that we can write a geometric series in multiple ways, let's find the sum. Well, we've seen before, and we prove it in other videos, if you have a sum from k equals zero to infinity, and you have your first term, a times r to the k power, r to the k power, assuming this converges, so assuming that the absolute value of your common ratio is less than one, this is what needs to be true for convergence, this is going to be equal to, this is going to be equal to our first term, which is a over, over one minus our common ratio. One minus our common ratio. And if this looks unfamiliar to you, I encourage you to watch the video where we find the formula, we derive the formula for the sum of an infinite geometric series."}, {"video_title": "Worked example convergent geometric series Series AP Calculus BC Khan Academy.mp3", "Sentence": "So now that we've seen that we can write a geometric series in multiple ways, let's find the sum. Well, we've seen before, and we prove it in other videos, if you have a sum from k equals zero to infinity, and you have your first term, a times r to the k power, r to the k power, assuming this converges, so assuming that the absolute value of your common ratio is less than one, this is what needs to be true for convergence, this is going to be equal to, this is going to be equal to our first term, which is a over, over one minus our common ratio. One minus our common ratio. And if this looks unfamiliar to you, I encourage you to watch the video where we find the formula, we derive the formula for the sum of an infinite geometric series. But just applying that over here, we are going to get, this is going to be equal to our first term, which is eight, so that is eight over one minus, one minus our common ratio, over 1 3rd. And we know this is going to converge because our common ratio, the magnitude, the absolute value of 1 3rd is indeed less than one. And so this is all going to converge to, this is going to converge to eight over, one minus 1 3rd is 2 3rds, 2 3rds, which is the same thing as eight times 3 halves, which is, let's see, this could become, divide eight by two, that becomes four, and so this will become 12."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "Start with a slope field and figure out which differential equation is the slope field describing the solutions for. And so I encourage you to look at each of these options and think about which of these differential equations is being described by this slope field. I encourage you to pause the video right now and try it on your own. So I'm assuming you have had a go at it. So let's work through each of them. And the way I'm going to do it is I'm just gonna find some points that seem to be easy to do arithmetic with and we'll see if the slope described by the differential equation at that point is consistent with the slope depicted in the slope field. And I don't know, just for simplicity, maybe I'll do x equals one, y equals one for all of these."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "So I'm assuming you have had a go at it. So let's work through each of them. And the way I'm going to do it is I'm just gonna find some points that seem to be easy to do arithmetic with and we'll see if the slope described by the differential equation at that point is consistent with the slope depicted in the slope field. And I don't know, just for simplicity, maybe I'll do x equals one, y equals one for all of these. So I'm gonna see what, so when x equals one and y is equal to one. So this first differential equation right over here, if x is one and y is one, then dy dx would be negative one over one, or negative one. dy dx would be negative one."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "And I don't know, just for simplicity, maybe I'll do x equals one, y equals one for all of these. So I'm gonna see what, so when x equals one and y is equal to one. So this first differential equation right over here, if x is one and y is one, then dy dx would be negative one over one, or negative one. dy dx would be negative one. Now is that depicted here? When x is equal to one and y is equal to one, our slope isn't negative one, our slope here looks positive. So we can rule this one out."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "dy dx would be negative one. Now is that depicted here? When x is equal to one and y is equal to one, our slope isn't negative one, our slope here looks positive. So we can rule this one out. Now let's try the next one. So if x is equal to one and y is equal to one, well then dy dx would be equal to one minus one, or zero. And once again, I just picked x equals one and y equals one for convenience."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "So we can rule this one out. Now let's try the next one. So if x is equal to one and y is equal to one, well then dy dx would be equal to one minus one, or zero. And once again, I just picked x equals one and y equals one for convenience. I could have picked any other. I could have picked negative five and negative seven. This just makes the arithmetic a little easier."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "And once again, I just picked x equals one and y equals one for convenience. I could have picked any other. I could have picked negative five and negative seven. This just makes the arithmetic a little easier. Once again, when you look at that point that we've already looked at, our slope is clearly not zero. We have a positive slope here. So we could rule that out."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "This just makes the arithmetic a little easier. Once again, when you look at that point that we've already looked at, our slope is clearly not zero. We have a positive slope here. So we could rule that out. Once again, for this magenta differential equation, if x and y are both equal to one, then one minus one is once again going to be equal to zero. And we've already seen the slope is not zero here. So rule that one out."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "So we could rule that out. Once again, for this magenta differential equation, if x and y are both equal to one, then one minus one is once again going to be equal to zero. And we've already seen the slope is not zero here. So rule that one out. And now here we have x plus y. So when x is one, y is one, the derivative of y with respect to x is going to be one plus one, which is equal to two. Now this looks interesting."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "So rule that one out. And now here we have x plus y. So when x is one, y is one, the derivative of y with respect to x is going to be one plus one, which is equal to two. Now this looks interesting. It looks like this slope right over here could be two. This looks like one, this looks like two. I'd want to validate some other points, but this looks like a really, really good candidate."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "Now this looks interesting. It looks like this slope right over here could be two. This looks like one, this looks like two. I'd want to validate some other points, but this looks like a really, really good candidate. And you could also see what's happening here. When dy dx is equal to x plus y, you would expect that as x increases for a given y, your slope would increase. And as y increases for a given x, your slope increases."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "I'd want to validate some other points, but this looks like a really, really good candidate. And you could also see what's happening here. When dy dx is equal to x plus y, you would expect that as x increases for a given y, your slope would increase. And as y increases for a given x, your slope increases. And we see that. If we were to just follow, if we were to hold y constant one, but increase x along this line, we see that the slope is increasing. It's getting steeper."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "And as y increases for a given x, your slope increases. And we see that. If we were to just follow, if we were to hold y constant one, but increase x along this line, we see that the slope is increasing. It's getting steeper. And if we were to keep x constant and increase y across this line, we see that the slope increases. And in general, we see that the slope increases as we go to the top right. And we see that it decreases as we go to the bottom left and both x and y become much, much more negative."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "It's getting steeper. And if we were to keep x constant and increase y across this line, we see that the slope increases. And in general, we see that the slope increases as we go to the top right. And we see that it decreases as we go to the bottom left and both x and y become much, much more negative. So I'm feeling pretty good about this, especially if we can knock this one out here, if we can knock that one out. So dy dx is equal to x over y. Well then when x equals one, y equals one, dy dx would be equal to one."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "And we see that it decreases as we go to the bottom left and both x and y become much, much more negative. So I'm feeling pretty good about this, especially if we can knock this one out here, if we can knock that one out. So dy dx is equal to x over y. Well then when x equals one, y equals one, dy dx would be equal to one. And this slope looks larger than one. It looks like two, but since we're really just eyeballing it, let's see if we can find something that more clearly, where this more clearly falls apart. So let's look at the situation when, let's look at the situation when they both equal negative one."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "Well then when x equals one, y equals one, dy dx would be equal to one. And this slope looks larger than one. It looks like two, but since we're really just eyeballing it, let's see if we can find something that more clearly, where this more clearly falls apart. So let's look at the situation when, let's look at the situation when they both equal negative one. So x equals negative one and y is equal to negative one. Well in that case, dy dx should still be equal to one because you have negative one over one. Do we see that over here?"}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "So let's look at the situation when, let's look at the situation when they both equal negative one. So x equals negative one and y is equal to negative one. Well in that case, dy dx should still be equal to one because you have negative one over one. Do we see that over here? So when x is equal to negative one, y is equal to negative one, our derivative here looks negative. It looks like negative two, which is consistent with this yellow differential equation. The slope here is definitely not a positive one."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "Do we see that over here? So when x is equal to negative one, y is equal to negative one, our derivative here looks negative. It looks like negative two, which is consistent with this yellow differential equation. The slope here is definitely not a positive one. So we could rule this one out as well. And so we should feel pretty confident that this is the differential equation being described. And now that we've done it, we can actually think about, well okay, what are the solutions for this differential equation going to look like?"}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "The slope here is definitely not a positive one. So we could rule this one out as well. And so we should feel pretty confident that this is the differential equation being described. And now that we've done it, we can actually think about, well okay, what are the solutions for this differential equation going to look like? Well it depends where they start, if you have, or what points they contain. If you have a solution that contains that point, it looks like it might go, looks like it might do something, looks like it might do something like, something like this. If you had a solution that contained, I don't know, if you had a solution that contained this point, it might do something, it might do something like that."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "And now that we've done it, we can actually think about, well okay, what are the solutions for this differential equation going to look like? Well it depends where they start, if you have, or what points they contain. If you have a solution that contains that point, it looks like it might go, looks like it might do something, looks like it might do something like, something like this. If you had a solution that contained, I don't know, if you had a solution that contained this point, it might do something, it might do something like that. And of course it keeps going, it looks like it would asymptote towards y is equal to negative x, this downward sloping, this essentially is the line y is equal to negative x. Actually no, that's not the line y equals negative x, this is the line y is equal to negative x minus one. So that's this line right over here."}, {"video_title": "Differential equation from slope field First order differential equations Khan Academy.mp3", "Sentence": "If you had a solution that contained, I don't know, if you had a solution that contained this point, it might do something, it might do something like that. And of course it keeps going, it looks like it would asymptote towards y is equal to negative x, this downward sloping, this essentially is the line y is equal to negative x. Actually no, that's not the line y equals negative x, this is the line y is equal to negative x minus one. So that's this line right over here. And it looks like if you had, if you had your, in a solution, if you had a, if the solution contained, say this point right over here, you would actually, that would actually be a solution to the differential equation. Y is equal to, y is equal to negative x, whoops, y is equal to negative x minus one. And you can verify that."}, {"video_title": "Worked example divergent geometric series Series AP Calculus BC Khan Academy.mp3", "Sentence": "So we've got this infinite series here, and let's see, it looks like a geometric series. When you go from this first term to the second term, we are multiplying by negative three, and then to go to the next term, we're gonna multiply by negative three again. So it looks like we have a common ratio of negative three, so we could actually rewrite this series as being equal to negative 0.5, I could say, times negative three to the zero power. Negative three to the zero power, plus negative zero, or maybe I could just keep writing it this way, minus 0.5 times negative three to the first power, times negative three to the first power, minus 0.5, minus 0.5, times negative three to the second power, negative three to the second power, and we're just gonna keep going like that, and we could just say we're just gonna keep having minus 0.5 times negative three to higher and higher and higher powers. Or we could write this in sigma notation. This is equal to the same thing as the sum from, let's say, n equals zero to infinity, we're just gonna keep going on and on forever, and it's going to be this first, it's going to be, you can kind of think the thing we're multiplying by negative three to some power, so it's gonna be negative 0.5, actually let me just do that yellow color. So it's going to be negative 0.5 times negative three, negative, let me do that blue color, so times negative three to the nth power."}, {"video_title": "Worked example divergent geometric series Series AP Calculus BC Khan Academy.mp3", "Sentence": "Negative three to the zero power, plus negative zero, or maybe I could just keep writing it this way, minus 0.5 times negative three to the first power, times negative three to the first power, minus 0.5, minus 0.5, times negative three to the second power, negative three to the second power, and we're just gonna keep going like that, and we could just say we're just gonna keep having minus 0.5 times negative three to higher and higher and higher powers. Or we could write this in sigma notation. This is equal to the same thing as the sum from, let's say, n equals zero to infinity, we're just gonna keep going on and on forever, and it's going to be this first, it's going to be, you can kind of think the thing we're multiplying by negative three to some power, so it's gonna be negative 0.5, actually let me just do that yellow color. So it's going to be negative 0.5 times negative three, negative, let me do that blue color, so times negative three to the nth power. Here, this is when n is zero, here is n is one, here is n is equal to two. So we've been able to rewrite this in different ways, but let's actually see if we can evaluate this. And so we have a common ratio of negative three, so our r here is negative three."}, {"video_title": "Worked example divergent geometric series Series AP Calculus BC Khan Academy.mp3", "Sentence": "So it's going to be negative 0.5 times negative three, negative, let me do that blue color, so times negative three to the nth power. Here, this is when n is zero, here is n is one, here is n is equal to two. So we've been able to rewrite this in different ways, but let's actually see if we can evaluate this. And so we have a common ratio of negative three, so our r here is negative three. And the first thing that you should think about is, well, in order for this to converge, our common ratio, the magnitude of the common ratio, or the absolute value of the common ratio needs to be less than one for convergence. Convergence. And what is the absolute value of negative three?"}, {"video_title": "Worked example divergent geometric series Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so we have a common ratio of negative three, so our r here is negative three. And the first thing that you should think about is, well, in order for this to converge, our common ratio, the magnitude of the common ratio, or the absolute value of the common ratio needs to be less than one for convergence. Convergence. And what is the absolute value of negative three? Well, the absolute value of negative three is equal to three, which is definitely not less than one. So this thing will not converge. This thing will not converge."}, {"video_title": "Worked example divergent geometric series Series AP Calculus BC Khan Academy.mp3", "Sentence": "And what is the absolute value of negative three? Well, the absolute value of negative three is equal to three, which is definitely not less than one. So this thing will not converge. This thing will not converge. And even if you look at this, it makes sense, because the magnitudes of each of these terms are getting larger and larger and larger. We're flipping between adding and subtracting, but we're adding and subtracting larger and larger and larger and larger and larger values. Intuitively, when things converge, you're kind of, each successive term tends to get diminishingly small, or maybe it cancels out in some type of an interesting way."}, {"video_title": "Logarithmic functions differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "What is the derivative of y with respect to x going to be equal to? Now you might recognize immediately that this is a composite function. We're taking the log base four, not just of x, but we're taking that of another expression that involves x. So we could say, we could say this thing in blue, that's u of x. Let me do that in blue. So this thing in blue, that is u of x. U of x is equal to x squared plus x. And it's going to be useful later on to know what u prime of x is."}, {"video_title": "Logarithmic functions differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So we could say, we could say this thing in blue, that's u of x. Let me do that in blue. So this thing in blue, that is u of x. U of x is equal to x squared plus x. And it's going to be useful later on to know what u prime of x is. So that's going to be, just going to use the power rule here. So two x plus one, brought that two out front and decremented the exponent. Derivative with respect to x of x is one."}, {"video_title": "Logarithmic functions differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And it's going to be useful later on to know what u prime of x is. So that's going to be, just going to use the power rule here. So two x plus one, brought that two out front and decremented the exponent. Derivative with respect to x of x is one. And we could say the log base four of this stuff, well we could call that a function v. We could say v of, well if we said v of x, this would be log base four of x. And then we've shown in other videos that v prime of x is going to be very similar to if this was log base e or natural log, except we're going to scale it. So it's going to be one over, one over log base four, oh sorry, one over the natural log, the natural log of four times x."}, {"video_title": "Logarithmic functions differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Derivative with respect to x of x is one. And we could say the log base four of this stuff, well we could call that a function v. We could say v of, well if we said v of x, this would be log base four of x. And then we've shown in other videos that v prime of x is going to be very similar to if this was log base e or natural log, except we're going to scale it. So it's going to be one over, one over log base four, oh sorry, one over the natural log, the natural log of four times x. If this was v of x, if v of x was just natural log of x, our derivative would be one over x. But since it's log base four, and this comes straight out of the change of base formulas that you might have seen. And we have a video where we show this."}, {"video_title": "Logarithmic functions differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So it's going to be one over, one over log base four, oh sorry, one over the natural log, the natural log of four times x. If this was v of x, if v of x was just natural log of x, our derivative would be one over x. But since it's log base four, and this comes straight out of the change of base formulas that you might have seen. And we have a video where we show this. But we just scale it in the denominator with this natural log of four. Or you could think of scaling the whole expression by one over the natural log of four. But we can now use this information because y, this y can be viewed as v of, v of, remember v is the log base four of something."}, {"video_title": "Logarithmic functions differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And we have a video where we show this. But we just scale it in the denominator with this natural log of four. Or you could think of scaling the whole expression by one over the natural log of four. But we can now use this information because y, this y can be viewed as v of, v of, remember v is the log base four of something. But it's not v of x. We don't have just an x here. We have the whole expression that defines u of x."}, {"video_title": "Logarithmic functions differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "But we can now use this information because y, this y can be viewed as v of, v of, remember v is the log base four of something. But it's not v of x. We don't have just an x here. We have the whole expression that defines u of x. We have u of x right there. And let me draw a little line here so that we don't get those two sides confused. And so we know from the chain rule, the derivative of y with respect to x, this is going to be, this is going to be the derivative of v with respect to u."}, {"video_title": "Logarithmic functions differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We have the whole expression that defines u of x. We have u of x right there. And let me draw a little line here so that we don't get those two sides confused. And so we know from the chain rule, the derivative of y with respect to x, this is going to be, this is going to be the derivative of v with respect to u. Or we could call that v prime, v prime of u of x. v prime of u of x. Let me do the u of x in blue. v prime of u of x times u prime of x."}, {"video_title": "Logarithmic functions differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And so we know from the chain rule, the derivative of y with respect to x, this is going to be, this is going to be the derivative of v with respect to u. Or we could call that v prime, v prime of u of x. v prime of u of x. Let me do the u of x in blue. v prime of u of x times u prime of x. Well, what is v prime of u of x? We know what v prime of x is. If we wanted to do v prime of u of x, we would just replace wherever we see an x with the u of x."}, {"video_title": "Logarithmic functions differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "v prime of u of x times u prime of x. Well, what is v prime of u of x? We know what v prime of x is. If we wanted to do v prime of u of x, we would just replace wherever we see an x with the u of x. So this is going to be equal to v prime of u of x. And you just use it, you're taking the derivative of the green function with respect to the blue function. So it's going to be one over the natural log of four, natural log of four, times, instead of putting an x there, it would be times u of x."}, {"video_title": "Logarithmic functions differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "If we wanted to do v prime of u of x, we would just replace wherever we see an x with the u of x. So this is going to be equal to v prime of u of x. And you just use it, you're taking the derivative of the green function with respect to the blue function. So it's going to be one over the natural log of four, natural log of four, times, instead of putting an x there, it would be times u of x. Times u of x. And of course, that whole thing times u prime of x. And so, and I'm doing more steps, just hopefully so it's clearer what I'm doing here."}, {"video_title": "Logarithmic functions differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So it's going to be one over the natural log of four, natural log of four, times, instead of putting an x there, it would be times u of x. Times u of x. And of course, that whole thing times u prime of x. And so, and I'm doing more steps, just hopefully so it's clearer what I'm doing here. So this is one over the natural log of four. u of x is x squared plus x. So x squared plus x."}, {"video_title": "Logarithmic functions differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And so, and I'm doing more steps, just hopefully so it's clearer what I'm doing here. So this is one over the natural log of four. u of x is x squared plus x. So x squared plus x. And then we're going to multiply that times u prime of x. So times two x plus one. And so we can just rewrite this as two x plus one over, over, over the natural log of four, natural log of four, times x squared plus x."}, {"video_title": "Logarithmic functions differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So x squared plus x. And then we're going to multiply that times u prime of x. So times two x plus one. And so we can just rewrite this as two x plus one over, over, over the natural log of four, natural log of four, times x squared plus x. Times x squared plus x. And we're done. And we could distribute this natural log of four if we found that interesting, but we have just found the derivative of y with respect to x."}, {"video_title": "Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So to do that, I'm gonna have to figure out the critical points of our volume as a function of x, and to do that, I need to take the derivative of the volume, so let me do that, and before I even do that, it'll simplify things, so I don't have to use some product rule in some way and then have to simplify that. Let me just multiply this expression out. So let's rewrite volume as a function of x is equal to, and I'll write it all in yellow, so it's going to be x times, I'll multiply these two binomials first, so 20 times 30 is 600. Then I have 20 times negative two x, which is negative 40x. Then I have negative two x times 30, which is negative 60x, and then I have negative two x times negative two x, which is positive four x squared. So this part over here simplifies, and I can change the order to four x squared minus 100x plus 600. I just switched the order in which I'm writing them."}, {"video_title": "Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Then I have 20 times negative two x, which is negative 40x. Then I have negative two x times 30, which is negative 60x, and then I have negative two x times negative two x, which is positive four x squared. So this part over here simplifies, and I can change the order to four x squared minus 100x plus 600. I just switched the order in which I'm writing them. So that's that, and so I can rewrite the volume of x as being equal to x times all of this business, which is, let me make sure I have enough space, let me do it a little higher, which is equal to four x to the third power minus 100x squared plus 600x. Now it'll be pretty straightforward to take the derivative. So let's say that v prime of x, v prime of x is going to be equal to, I just have to use the power rule multiple times, so four times three is 12x to the three minus one power, 12x squared, minus 200 times x to the first power, which is just x, plus 600."}, {"video_title": "Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "I just switched the order in which I'm writing them. So that's that, and so I can rewrite the volume of x as being equal to x times all of this business, which is, let me make sure I have enough space, let me do it a little higher, which is equal to four x to the third power minus 100x squared plus 600x. Now it'll be pretty straightforward to take the derivative. So let's say that v prime of x, v prime of x is going to be equal to, I just have to use the power rule multiple times, so four times three is 12x to the three minus one power, 12x squared, minus 200 times x to the first power, which is just x, plus 600. And so now we just have to figure out when this is equal to zero. So we have to figure out when 12x squared minus 200x plus 600 is equal to zero. What x values gets my derivative to be equal to zero?"}, {"video_title": "Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say that v prime of x, v prime of x is going to be equal to, I just have to use the power rule multiple times, so four times three is 12x to the three minus one power, 12x squared, minus 200 times x to the first power, which is just x, plus 600. And so now we just have to figure out when this is equal to zero. So we have to figure out when 12x squared minus 200x plus 600 is equal to zero. What x values gets my derivative to be equal to zero? When is my slope equal to zero? I could also look for critical points where the derivative is undefined, but this derivative is defined, especially throughout my domain of x that I care about, between zero and 10. So I could try to factor this or try to simplify this a little bit, but I'm just gonna cut to the chase and try to use the quadratic formula here."}, {"video_title": "Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "What x values gets my derivative to be equal to zero? When is my slope equal to zero? I could also look for critical points where the derivative is undefined, but this derivative is defined, especially throughout my domain of x that I care about, between zero and 10. So I could try to factor this or try to simplify this a little bit, but I'm just gonna cut to the chase and try to use the quadratic formula here. So this is the x's that satisfy this. It's going to be, x is going to be equal to, so negative b, so it's 200, negative negative 200 is positive 200, plus or minus the square root of b squared, which is negative 200 squared, so I could just write that as 200 squared. Doesn't matter if it's negative 200 squared or 200 squared, I'm gonna get the same value."}, {"video_title": "Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So I could try to factor this or try to simplify this a little bit, but I'm just gonna cut to the chase and try to use the quadratic formula here. So this is the x's that satisfy this. It's going to be, x is going to be equal to, so negative b, so it's 200, negative negative 200 is positive 200, plus or minus the square root of b squared, which is negative 200 squared, so I could just write that as 200 squared. Doesn't matter if it's negative 200 squared or 200 squared, I'm gonna get the same value. So, let me give myself some more space. So negative 200 squared, well that's going to be four with four zeros. One, two, three, four."}, {"video_title": "Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Doesn't matter if it's negative 200 squared or 200 squared, I'm gonna get the same value. So, let me give myself some more space. So negative 200 squared, well that's going to be four with four zeros. One, two, three, four. So that's gonna be 40,000 minus four a c. So minus four times 12, 12 times 600, I still didn't give myself enough space, times 600, all of that over two times a. So all of that over 24. And I'll take out the calculator again to try to calculate this."}, {"video_title": "Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "One, two, three, four. So that's gonna be 40,000 minus four a c. So minus four times 12, 12 times 600, I still didn't give myself enough space, times 600, all of that over two times a. So all of that over 24. And I'll take out the calculator again to try to calculate this. So let me get out of graphing mode. All right, so first I'll try when I add the radical. So I'm going to get 200, 200 plus the square root of 40,000."}, {"video_title": "Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And I'll take out the calculator again to try to calculate this. So let me get out of graphing mode. All right, so first I'll try when I add the radical. So I'm going to get 200, 200 plus the square root of 40,000. I could have just written that, I could have just written that as 200 squared, but that's fine, 40,000 minus four times 12, times 600, times 600. And I get 305, which I then need to divide by 24. Which I'll divide by 24."}, {"video_title": "Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So I'm going to get 200, 200 plus the square root of 40,000. I could have just written that, I could have just written that as 200 squared, but that's fine, 40,000 minus four times 12, times 600, times 600. And I get 305, which I then need to divide by 24. Which I'll divide by 24. And I get 12.74. So one of my possible x's. So it equals 12.74."}, {"video_title": "Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Which I'll divide by 24. And I get 12.74. So one of my possible x's. So it equals 12.74. And now let me do the situation where I subtract what I had in the radical sign. So let me get my calculator back. And so now let me do 200."}, {"video_title": "Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So it equals 12.74. And now let me do the situation where I subtract what I had in the radical sign. So let me get my calculator back. And so now let me do 200. I probably could have done this slightly more efficiently, but this is fine. Minus the square root of 40,000, one, two, three. Minus four times 12, times 600."}, {"video_title": "Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And so now let me do 200. I probably could have done this slightly more efficiently, but this is fine. Minus the square root of 40,000, one, two, three. Minus four times 12, times 600. And I get, that's just the numerator. And then I'm gonna divide that by 24. Divide that by 24."}, {"video_title": "Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Minus four times 12, times 600. And I get, that's just the numerator. And then I'm gonna divide that by 24. Divide that by 24. And I get 3.92. Did I do that right? 200 minus 40,000 minus four times 12 times 600."}, {"video_title": "Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Divide that by 24. And I get 3.92. Did I do that right? 200 minus 40,000 minus four times 12 times 600. All of that divided by 24. My previous answer divided by 24. Gives me 3.92."}, {"video_title": "Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "200 minus 40,000 minus four times 12 times 600. All of that divided by 24. My previous answer divided by 24. Gives me 3.92. So it's 12.74 or 3.92. Now which of these can I use? Well, 12, x equals 12.74 is outside of our, outside of our valid values for x."}, {"video_title": "Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Gives me 3.92. So it's 12.74 or 3.92. Now which of these can I use? Well, 12, x equals 12.74 is outside of our, outside of our valid values for x. If x was equal to 12.74, we would cut past, we would completely cut past, the x's would start to overlap with each other. So x cannot be 12.74. So we get a critical point at x is equal to 3.92."}, {"video_title": "Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, 12, x equals 12.74 is outside of our, outside of our valid values for x. If x was equal to 12.74, we would cut past, we would completely cut past, the x's would start to overlap with each other. So x cannot be 12.74. So we get a critical point at x is equal to 3.92. And you could look at the graph, and you could say, oh look, that looks like a maximum value. But if you didn't have the graph at your disposal, you can then do the second derivative test and say, hey, are we concave upwards or concave downwards when x is equal to 3.92? Well, in order to do the second derivative test, you have to figure out what the second derivative is, so let's do that."}, {"video_title": "Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So we get a critical point at x is equal to 3.92. And you could look at the graph, and you could say, oh look, that looks like a maximum value. But if you didn't have the graph at your disposal, you can then do the second derivative test and say, hey, are we concave upwards or concave downwards when x is equal to 3.92? Well, in order to do the second derivative test, you have to figure out what the second derivative is, so let's do that. V prime prime of x is going to be equal to, is going to be equal to 24x, 24 times x to the first, minus 200. Minus, minus 200. And you can just look at inspection that this number right over here is less than four, so this thing right over here is going to be less than 100."}, {"video_title": "Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, in order to do the second derivative test, you have to figure out what the second derivative is, so let's do that. V prime prime of x is going to be equal to, is going to be equal to 24x, 24 times x to the first, minus 200. Minus, minus 200. And you can just look at inspection that this number right over here is less than four, so this thing right over here is going to be less than 100. You subtract, you subtract 200, so we can write the second derivative at 3.92 is going to be less than zero. You can figure out what the exact value is if you like. So because this is less than zero, we are concave downwards, concave downwards."}, {"video_title": "Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And you can just look at inspection that this number right over here is less than four, so this thing right over here is going to be less than 100. You subtract, you subtract 200, so we can write the second derivative at 3.92 is going to be less than zero. You can figure out what the exact value is if you like. So because this is less than zero, we are concave downwards, concave downwards. Another way of saying it is that the slope is, the slope is decreasing the entire time, concave downwards. When the slope is decreasing the entire time, our shape looks like that. The slope could start off high, lower, lower, gets to zero, even lower, lower, lower."}, {"video_title": "Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So because this is less than zero, we are concave downwards, concave downwards. Another way of saying it is that the slope is, the slope is decreasing the entire time, concave downwards. When the slope is decreasing the entire time, our shape looks like that. The slope could start off high, lower, lower, gets to zero, even lower, lower, lower. And we even saw that on the graph right over here. And since it's concave downwards, that implies that our critical point that sits where, during where the interval is concave downward, that critical point is a local maximum, is a maximum. So this is the x value at which our function attains our maximum."}, {"video_title": "Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "The slope could start off high, lower, lower, gets to zero, even lower, lower, lower. And we even saw that on the graph right over here. And since it's concave downwards, that implies that our critical point that sits where, during where the interval is concave downward, that critical point is a local maximum, is a maximum. So this is the x value at which our function attains our maximum. Now, what is that maximum value? Well, we could type that back in into our original expression for volume to figure what that is. So let's figure out what the volume, when we get to 3.92, is equal to."}, {"video_title": "Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So this is the x value at which our function attains our maximum. Now, what is that maximum value? Well, we could type that back in into our original expression for volume to figure what that is. So let's figure out what the volume, when we get to 3.92, is equal to. What is our maximum volume? So get the calculator back out. And I could use, it's obviously roughly 3.92, I could use this exact value."}, {"video_title": "Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let's figure out what the volume, when we get to 3.92, is equal to. What is our maximum volume? So get the calculator back out. And I could use, it's obviously roughly 3.92, I could use this exact value. Actually, let's, well, I'll just use 3.92 to get a rough sense of what our maximum value is, our maximum volume. So it'll be 3.92, 3.92. I'll just use this expression for the volume as a function of x."}, {"video_title": "Optimization box volume (Part 2) Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And I could use, it's obviously roughly 3.92, I could use this exact value. Actually, let's, well, I'll just use 3.92 to get a rough sense of what our maximum value is, our maximum volume. So it'll be 3.92, 3.92. I'll just use this expression for the volume as a function of x. 3.92 times 20 minus two times 3.92 times 30, 30 minus two times 3.92 gives us, and we deserve a drum roll now, gives us 1,056.3. So 1,056.3, which is a higher volume than we got when we just inspected it graphically. We probably could have gotten a little bit more precise if we zoomed in some, and then we would have gotten a little bit better of an answer."}, {"video_title": "Definite integral of piecewise function AP Calculus AB Khan Academy.mp3", "Sentence": "So we have an f of x right over here and it's defined piecewise. For x less than zero, f of x is x plus one. For x is greater than or equal to zero, f of x is cosine of pi x. And we want to evaluate the definite integral from negative one to one of f of x dx. And you might immediately say, well, which of these versions of f of x am I going to take the antiderivative from? Because from negative one to zero, I would think about x plus one, but then from zero to one, I would think about cosine pi x. And if you were thinking that, you're thinking in the right direction."}, {"video_title": "Definite integral of piecewise function AP Calculus AB Khan Academy.mp3", "Sentence": "And we want to evaluate the definite integral from negative one to one of f of x dx. And you might immediately say, well, which of these versions of f of x am I going to take the antiderivative from? Because from negative one to zero, I would think about x plus one, but then from zero to one, I would think about cosine pi x. And if you were thinking that, you're thinking in the right direction. And the way that we can make this a little bit more straightforward is to actually split up this definite integral. This is going to be equal to the definite integral from negative one to zero of f of x dx plus the integral from zero to one of f of x dx. Now, why was it useful for me to split it up this way?"}, {"video_title": "Definite integral of piecewise function AP Calculus AB Khan Academy.mp3", "Sentence": "And if you were thinking that, you're thinking in the right direction. And the way that we can make this a little bit more straightforward is to actually split up this definite integral. This is going to be equal to the definite integral from negative one to zero of f of x dx plus the integral from zero to one of f of x dx. Now, why was it useful for me to split it up this way? And in particular, to split it up, split the interval from negative one to one, split it into two intervals from negative one to zero and zero to one? Well, I did that because zero, x equals zero is where we switch, where f of x switches from being x plus one to cosine pi x. So if you look at the interval from negative one to zero, f of x is x plus one."}, {"video_title": "Definite integral of piecewise function AP Calculus AB Khan Academy.mp3", "Sentence": "Now, why was it useful for me to split it up this way? And in particular, to split it up, split the interval from negative one to one, split it into two intervals from negative one to zero and zero to one? Well, I did that because zero, x equals zero is where we switch, where f of x switches from being x plus one to cosine pi x. So if you look at the interval from negative one to zero, f of x is x plus one. So f of x here is x plus one. And then when you go from zero to one, f of x is cosine pi x. So cosine of pi x."}, {"video_title": "Definite integral of piecewise function AP Calculus AB Khan Academy.mp3", "Sentence": "So if you look at the interval from negative one to zero, f of x is x plus one. So f of x here is x plus one. And then when you go from zero to one, f of x is cosine pi x. So cosine of pi x. And so now we just have to evaluate each of these separately and add them together. So let's take the definite integral from negative one to zero of x plus one dx. Well, let's see, the antiderivative of x plus one is, antiderivative of x is x squared over two."}, {"video_title": "Definite integral of piecewise function AP Calculus AB Khan Academy.mp3", "Sentence": "So cosine of pi x. And so now we just have to evaluate each of these separately and add them together. So let's take the definite integral from negative one to zero of x plus one dx. Well, let's see, the antiderivative of x plus one is, antiderivative of x is x squared over two. I'm just incrementing the exponent and then dividing by that value. And then plus x, and you could view it as I'm doing the same thing. If this is x to the zero, it'll be x to the first, x to the first over one, which is just x."}, {"video_title": "Definite integral of piecewise function AP Calculus AB Khan Academy.mp3", "Sentence": "Well, let's see, the antiderivative of x plus one is, antiderivative of x is x squared over two. I'm just incrementing the exponent and then dividing by that value. And then plus x, and you could view it as I'm doing the same thing. If this is x to the zero, it'll be x to the first, x to the first over one, which is just x. And I'm gonna evaluate that at zero and subtract from that it evaluated at one, sorry, it evaluated at negative one. And so this is going to be equal to, if I evaluate it at zero, let me do this in another color. If I evaluate it at zero, it's going to be zero squared over two, which is, well, I'll just write it zero squared over two plus zero."}, {"video_title": "Definite integral of piecewise function AP Calculus AB Khan Academy.mp3", "Sentence": "If this is x to the zero, it'll be x to the first, x to the first over one, which is just x. And I'm gonna evaluate that at zero and subtract from that it evaluated at one, sorry, it evaluated at negative one. And so this is going to be equal to, if I evaluate it at zero, let me do this in another color. If I evaluate it at zero, it's going to be zero squared over two, which is, well, I'll just write it zero squared over two plus zero. Well, all of that's just gonna be equal to zero, minus it evaluated at, it evaluated at negative one. So minus negative one squared, negative one squared over two plus negative one. So negative one squared is just one."}, {"video_title": "Definite integral of piecewise function AP Calculus AB Khan Academy.mp3", "Sentence": "If I evaluate it at zero, it's going to be zero squared over two, which is, well, I'll just write it zero squared over two plus zero. Well, all of that's just gonna be equal to zero, minus it evaluated at, it evaluated at negative one. So minus negative one squared, negative one squared over two plus negative one. So negative one squared is just one. So it's 1 1 2 plus negative one, 1 1 2 plus negative one is, or 1 1 2 minus one is negative 1 1 2. So all of that is negative 1 1 2. But then we're subtracting negative 1 1 2."}, {"video_title": "Definite integral of piecewise function AP Calculus AB Khan Academy.mp3", "Sentence": "So negative one squared is just one. So it's 1 1 2 plus negative one, 1 1 2 plus negative one is, or 1 1 2 minus one is negative 1 1 2. So all of that is negative 1 1 2. But then we're subtracting negative 1 1 2. Zero minus negative 1 1 2 is going to be equal to positive 1 1 2. So this is going to be equal to positive 1 1 2. So this first part right over here is positive 1 1 2."}, {"video_title": "Definite integral of piecewise function AP Calculus AB Khan Academy.mp3", "Sentence": "But then we're subtracting negative 1 1 2. Zero minus negative 1 1 2 is going to be equal to positive 1 1 2. So this is going to be equal to positive 1 1 2. So this first part right over here is positive 1 1 2. And now let's evaluate the integral from zero to one of cosine pi, I don't need that first parenthesis, of cosine of pi x dx. What is this equal to? Now, if we were just trying to find the antiderivative of cosine of x, it's pretty straightforward."}, {"video_title": "Definite integral of piecewise function AP Calculus AB Khan Academy.mp3", "Sentence": "So this first part right over here is positive 1 1 2. And now let's evaluate the integral from zero to one of cosine pi, I don't need that first parenthesis, of cosine of pi x dx. What is this equal to? Now, if we were just trying to find the antiderivative of cosine of x, it's pretty straightforward. We know that the derivative with respect to x of sine of x is equal to cosine of x. Cosine of x. But that's not what we have here. We have cosine of pi x."}, {"video_title": "Definite integral of piecewise function AP Calculus AB Khan Academy.mp3", "Sentence": "Now, if we were just trying to find the antiderivative of cosine of x, it's pretty straightforward. We know that the derivative with respect to x of sine of x is equal to cosine of x. Cosine of x. But that's not what we have here. We have cosine of pi x. So there is a technique here, you could call it u substitution. You could say u is equal to pi x. If you don't know how to do that, you could still try to think about, think this through, where we could say, all right, well, maybe it involves sine of pi x somehow."}, {"video_title": "Definite integral of piecewise function AP Calculus AB Khan Academy.mp3", "Sentence": "We have cosine of pi x. So there is a technique here, you could call it u substitution. You could say u is equal to pi x. If you don't know how to do that, you could still try to think about, think this through, where we could say, all right, well, maybe it involves sine of pi x somehow. So the derivative with respect to x of sine of pi x would be what? Well, we would use the chain rule. It would be the derivative of the outside function with respect to the inside, or sine of pi x with respect to pi x, which would be cosine of pi x, and then times the derivative of the inside function with respect to x, so it would be times pi."}, {"video_title": "Definite integral of piecewise function AP Calculus AB Khan Academy.mp3", "Sentence": "If you don't know how to do that, you could still try to think about, think this through, where we could say, all right, well, maybe it involves sine of pi x somehow. So the derivative with respect to x of sine of pi x would be what? Well, we would use the chain rule. It would be the derivative of the outside function with respect to the inside, or sine of pi x with respect to pi x, which would be cosine of pi x, and then times the derivative of the inside function with respect to x, so it would be times pi. Or you could say the derivative of sine pi x is pi cosine of pi x. Now, we almost have that here, except we just need a pi. So what if we were to throw a pi right over here, but so we don't change the value, we also multiply by one over pi?"}, {"video_title": "Definite integral of piecewise function AP Calculus AB Khan Academy.mp3", "Sentence": "It would be the derivative of the outside function with respect to the inside, or sine of pi x with respect to pi x, which would be cosine of pi x, and then times the derivative of the inside function with respect to x, so it would be times pi. Or you could say the derivative of sine pi x is pi cosine of pi x. Now, we almost have that here, except we just need a pi. So what if we were to throw a pi right over here, but so we don't change the value, we also multiply by one over pi? So if you divide and multiply by the same number, you're not changing its value. One over pi times pi is just equal to one. But this is useful."}, {"video_title": "Definite integral of piecewise function AP Calculus AB Khan Academy.mp3", "Sentence": "So what if we were to throw a pi right over here, but so we don't change the value, we also multiply by one over pi? So if you divide and multiply by the same number, you're not changing its value. One over pi times pi is just equal to one. But this is useful. This is useful because we now know that pi cosine pi x is the derivative of sine pi x. So this is all going to be equal to, this is equal to one. Let me take that one over pi."}, {"video_title": "Definite integral of piecewise function AP Calculus AB Khan Academy.mp3", "Sentence": "But this is useful. This is useful because we now know that pi cosine pi x is the derivative of sine pi x. So this is all going to be equal to, this is equal to one. Let me take that one over pi. So this is equal to one over pi times, now we're going to evaluate, so the antiderivative here we just said is sine, sine of pi x, and we're going to evaluate that at one and at zero. So this is going to be equal to one over pi, one over pi, not pi, my hand is not listening to my mouth, one over pi times sine of pi, sine of pi minus sine of, minus sine of pi times zero, which is just zero. Well sine of pi, that's zero."}, {"video_title": "Definite integral of piecewise function AP Calculus AB Khan Academy.mp3", "Sentence": "Let me take that one over pi. So this is equal to one over pi times, now we're going to evaluate, so the antiderivative here we just said is sine, sine of pi x, and we're going to evaluate that at one and at zero. So this is going to be equal to one over pi, one over pi, not pi, my hand is not listening to my mouth, one over pi times sine of pi, sine of pi minus sine of, minus sine of pi times zero, which is just zero. Well sine of pi, that's zero. Sine of zero is zero. So you're going to have one over pi times zero minus zero. So this whole thing is just all going to be equal to zero."}, {"video_title": "Definite integral of piecewise function AP Calculus AB Khan Academy.mp3", "Sentence": "Well sine of pi, that's zero. Sine of zero is zero. So you're going to have one over pi times zero minus zero. So this whole thing is just all going to be equal to zero. So this first part was 1 1 2. This second part right over here is equal to zero. So the whole definite integral is going to be 1 1 2 plus zero which is equal to 1 1 2."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "So now we just have to evaluate it. And really the hardest part is going to be simplifying that and that right over there. So let's get to it. So what is this thing squared? Well, we're going to have to do a little bit of polynomial multiplication here. So I'm going to go into that same color I had. So we're going to have 4 minus x squared plus 2x times 4 minus."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "So what is this thing squared? Well, we're going to have to do a little bit of polynomial multiplication here. So I'm going to go into that same color I had. So we're going to have 4 minus x squared plus 2x times 4 minus. Actually, let me write that in the order of the terms or the degree of the terms. It's negative x squared plus 2x plus 4. I just switched the order of these things."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "So we're going to have 4 minus x squared plus 2x times 4 minus. Actually, let me write that in the order of the terms or the degree of the terms. It's negative x squared plus 2x plus 4. I just switched the order of these things. Times negative x squared plus 2x plus 4. So we're just going to multiply these two things. I shouldn't even write a multiplication symbol."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "I just switched the order of these things. Times negative x squared plus 2x plus 4. So we're just going to multiply these two things. I shouldn't even write a multiplication symbol. It looks too much like an x. So 4 times 4 is 16. 4 times 2x is 8x."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "I shouldn't even write a multiplication symbol. It looks too much like an x. So 4 times 4 is 16. 4 times 2x is 8x. 4 times negative x squared is negative 4x squared. 2x times 4 is 8x. 2x times 2x is 4x squared."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "4 times 2x is 8x. 4 times negative x squared is negative 4x squared. 2x times 4 is 8x. 2x times 2x is 4x squared. And then 2x times negative x squared is negative 2x to the third power. And now we just have to multiply negative x times all of this. So negative x squared times 4 is negative 4x squared."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "2x times 2x is 4x squared. And then 2x times negative x squared is negative 2x to the third power. And now we just have to multiply negative x times all of this. So negative x squared times 4 is negative 4x squared. Negative x squared times 2x is negative 2x to the third power. And then negative x squared times negative x squared is positive x to the fourth. So it's going to be positive x to the fourth."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "So negative x squared times 4 is negative 4x squared. Negative x squared times 2x is negative 2x to the third power. And then negative x squared times negative x squared is positive x to the fourth. So it's going to be positive x to the fourth. And now we just have to add up all of these terms. And we get x to the fourth. Add these two, minus 4x to the third."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "So it's going to be positive x to the fourth. And now we just have to add up all of these terms. And we get x to the fourth. Add these two, minus 4x to the third. And then this cancels with this, but we still have that. So minus 4x squared. You add these two right over here, you get plus 16x."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "Add these two, minus 4x to the third. And then this cancels with this, but we still have that. So minus 4x squared. You add these two right over here, you get plus 16x. And then you have plus 16. So that's this expanded out. And now if we want to x minus 4, or 4 minus x times 4 minus x."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "You add these two right over here, you get plus 16x. And then you have plus 16. So that's this expanded out. And now if we want to x minus 4, or 4 minus x times 4 minus x. So if we just have 4 minus x times 4 minus x. We could actually do it this way as well. But that's just going to be 4 times 4, which is 16."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "And now if we want to x minus 4, or 4 minus x times 4 minus x. So if we just have 4 minus x times 4 minus x. We could actually do it this way as well. But that's just going to be 4 times 4, which is 16. Plus 4 times negative x, which is negative 4x. Negative x times 4, another negative 4x. And then negative x times negative x is equal to plus x squared."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "But that's just going to be 4 times 4, which is 16. Plus 4 times negative x, which is negative 4x. Negative x times 4, another negative 4x. And then negative x times negative x is equal to plus x squared. So if we were to swap the order, we get x squared minus 8x plus 16. But we're going to subtract this. So if you have the negative sign out there, we're going to subtract this business."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "And then negative x times negative x is equal to plus x squared. So if we were to swap the order, we get x squared minus 8x plus 16. But we're going to subtract this. So if you have the negative sign out there, we're going to subtract this business. So let's just do it right over here, since we already have it set up. So we have this minus this. So we have this minus, we're going to subtract this."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "So if you have the negative sign out there, we're going to subtract this business. So let's just do it right over here, since we already have it set up. So we have this minus this. So we have this minus, we're going to subtract this. Or we could add the negative of it. So we'll put negative x squared plus 8x minus 16. So I'm just going to add the negative of this."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "So we have this minus, we're going to subtract this. Or we could add the negative of it. So we'll put negative x squared plus 8x minus 16. So I'm just going to add the negative of this. And we get, and I'll do this in a new color, let's see, we get x to the fourth minus 4x to the third power minus 5x squared plus 24x. And then these cancel out. So that's what we are left with."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "So I'm just going to add the negative of this. And we get, and I'll do this in a new color, let's see, we get x to the fourth minus 4x to the third power minus 5x squared plus 24x. And then these cancel out. So that's what we are left with. And so that's the inside of our integral. So we're going to take the integral of this thing, just so I don't have to keep rewriting this thing, from 0 until, if I remember correctly, 3. Yep, from 0 to 3 of this dx."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "So that's what we are left with. And so that's the inside of our integral. So we're going to take the integral of this thing, just so I don't have to keep rewriting this thing, from 0 until, if I remember correctly, 3. Yep, from 0 to 3 of this dx. And then we had a pi out front here. We had a pi out front here. So I'll just take that out of the integral."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "Yep, from 0 to 3 of this dx. And then we had a pi out front here. We had a pi out front here. So I'll just take that out of the integral. So times pi. Well, now we just have to take the antiderivative. And this is going to be equal to pi times, antiderivative of x to the fourth is x to the fifth over 5."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "So I'll just take that out of the integral. So times pi. Well, now we just have to take the antiderivative. And this is going to be equal to pi times, antiderivative of x to the fourth is x to the fifth over 5. Antiderivative of 4x to the third is actually x to the fourth. So this is going to be minus x to the fourth. You can verify that."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "And this is going to be equal to pi times, antiderivative of x to the fourth is x to the fifth over 5. Antiderivative of 4x to the third is actually x to the fourth. So this is going to be minus x to the fourth. You can verify that. Derivative is 4x. And then you decrement the exponent 4x to the third. So that works out."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "You can verify that. Derivative is 4x. And then you decrement the exponent 4x to the third. So that works out. And then the antiderivative of this is negative 5 thirds x to the third. Just incremented the exponent and divided by that. And then you have plus 24x squared over 2, or 12x squared."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "So that works out. And then the antiderivative of this is negative 5 thirds x to the third. Just incremented the exponent and divided by that. And then you have plus 24x squared over 2, or 12x squared. And we're going to evaluate that. I like to match colors for my opening and closing parentheses. We're going to evaluate that."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "And then you have plus 24x squared over 2, or 12x squared. And we're going to evaluate that. I like to match colors for my opening and closing parentheses. We're going to evaluate that. Actually, let me do it as brackets. We're going to evaluate that from 0 to 3. So this is going to be equal to pi times, let's evaluate all this business at 3."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "We're going to evaluate that. Actually, let me do it as brackets. We're going to evaluate that from 0 to 3. So this is going to be equal to pi times, let's evaluate all this business at 3. So we're going to get 3 to the fifth power. So let's see. 3 to the third is equal to 27."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be equal to pi times, let's evaluate all this business at 3. So we're going to get 3 to the fifth power. So let's see. 3 to the third is equal to 27. 3 to the fourth is equal to 81. 3 to the fifth is going to be equal to, this times 3 is 243. So it's going to be 243."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "3 to the third is equal to 27. 3 to the fourth is equal to 81. 3 to the fifth is going to be equal to, this times 3 is 243. So it's going to be 243. It's going to be some hairy math. 243 over 5 minus, well, 3 to the fourth is 81. Let's see."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "So it's going to be 243. It's going to be some hairy math. 243 over 5 minus, well, 3 to the fourth is 81. Let's see. We're going to have 3 to the third. So it's going to be minus 5 over 3 times 3 to the third power. So times 27."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "Let's see. We're going to have 3 to the third. So it's going to be minus 5 over 3 times 3 to the third power. So times 27. Well, 27 divided by 3 is just 9. 9 times 5 is 45. So let's just simplify that."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "So times 27. Well, 27 divided by 3 is just 9. 9 times 5 is 45. So let's just simplify that. So this is going to be equal to 45. Did I do that right? Yeah, it's essentially going to be like 3 squared times 5, because you're dividing by 3 here."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "So let's just simplify that. So this is going to be equal to 45. Did I do that right? Yeah, it's essentially going to be like 3 squared times 5, because you're dividing by 3 here. So it's going to have 45. And then finally, 3 squared is 9. 9 times 12 is 108."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "Yeah, it's essentially going to be like 3 squared times 5, because you're dividing by 3 here. So it's going to have 45. And then finally, 3 squared is 9. 9 times 12 is 108. So plus 108. These problems that involve hairy arithmetic are always the most stressful for me, but I'll try not to get too stressed. And then we're going to subtract out this whole thing evaluated at 0."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "9 times 12 is 108. So plus 108. These problems that involve hairy arithmetic are always the most stressful for me, but I'll try not to get too stressed. And then we're going to subtract out this whole thing evaluated at 0. But lucky for us, that's pretty simple. This evaluated at 0, 0, 0, 0, 0. So we're going to subtract out 0, which simplifies things a good bit."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "And then we're going to subtract out this whole thing evaluated at 0. But lucky for us, that's pretty simple. This evaluated at 0, 0, 0, 0, 0. So we're going to subtract out 0, which simplifies things a good bit. So now we just have to do some hairy fraction arithmetic. So let's do it. So what I'm going to do first is simplify all of this part."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "So we're going to subtract out 0, which simplifies things a good bit. So now we just have to do some hairy fraction arithmetic. So let's do it. So what I'm going to do first is simplify all of this part. And then I'm going to worry about putting it over a denominator of 5. So let's see what we have. We have negative 81 minus 45."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "So what I'm going to do first is simplify all of this part. And then I'm going to worry about putting it over a denominator of 5. So let's see what we have. We have negative 81 minus 45. So these two right over here become negative 126. And then negative 126 plus 108, well, that's just going to be the same thing as negative 26 plus 8, which is going to be negative 18. So this whole thing simplifies to negative 18."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "We have negative 81 minus 45. So these two right over here become negative 126. And then negative 126 plus 108, well, that's just going to be the same thing as negative 26 plus 8, which is going to be negative 18. So this whole thing simplifies to negative 18. Did I do that right? So this is negative 126. And then negative 126, yes, it will be negative 18."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "So this whole thing simplifies to negative 18. Did I do that right? So this is negative 126. And then negative 126, yes, it will be negative 18. So now we just have to write negative 18 over 5. Negative 18 over 5 is the same thing as negative, let's see, 5 times 10 is 50, plus 40. So that's going to be negative 90 over 5."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "And then negative 126, yes, it will be negative 18. So now we just have to write negative 18 over 5. Negative 18 over 5 is the same thing as negative, let's see, 5 times 10 is 50, plus 40. So that's going to be negative 90 over 5. So this whole thing has simplified to equal to pi times 243 over 5 minus 90 over 5, which is equal to pi. I deserve a drum roll now. This was some hairy mathematics."}, {"video_title": "Washer method rotating around horizontal line (not x-axis), part 2 AP Calculus AB Khan Academy.mp3", "Sentence": "So that's going to be negative 90 over 5. So this whole thing has simplified to equal to pi times 243 over 5 minus 90 over 5, which is equal to pi. I deserve a drum roll now. This was some hairy mathematics. So 243 minus 90 is going to be 153 over 5. Or we can write this as 153 pi over 5. And after all that math, you sometimes forget what we were doing in the first place."}, {"video_title": "Parametric curve arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "But we're going to think about, and I'm gonna talk about it in generalities in this video, but in future videos we're going to be dealing with more concrete examples. But we're gonna think about what is the path that is traced out from when t is equal to a. So this is where we are when t is equal to a. So in this case, this point would be x of a, comma y of a, that's this point. And then as we increase from t equals a to t is equal to b. So our curve might do something like this. So this is when t is equal to b. T is equal to b, so this point right over here is x of b, comma y of b."}, {"video_title": "Parametric curve arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So in this case, this point would be x of a, comma y of a, that's this point. And then as we increase from t equals a to t is equal to b. So our curve might do something like this. So this is when t is equal to b. T is equal to b, so this point right over here is x of b, comma y of b. Let's think about how do we figure out the length of this actual curve, this actual arc length from t equals a to t equals b. Well to think about that, we're gonna zoom in and think about what happens when we have a very small change in t. So a very small change in t. Let's say we're starting at this point right over here and we have a very small change in t. So we go from this point to let's say this point over that very small change in t. It actually would be much smaller than this but if I drew it any smaller, you would have trouble seeing it. So let's say that that is our very small change in our path, in our arc that we are traveling."}, {"video_title": "Parametric curve arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So this is when t is equal to b. T is equal to b, so this point right over here is x of b, comma y of b. Let's think about how do we figure out the length of this actual curve, this actual arc length from t equals a to t equals b. Well to think about that, we're gonna zoom in and think about what happens when we have a very small change in t. So a very small change in t. Let's say we're starting at this point right over here and we have a very small change in t. So we go from this point to let's say this point over that very small change in t. It actually would be much smaller than this but if I drew it any smaller, you would have trouble seeing it. So let's say that that is our very small change in our path, in our arc that we are traveling. And so we want to find this length. Well, we could break it down into how far we've moved in the x direction and how far we've moved in the y direction. So in the x direction, the x direction right over here, we would have moved a very small change in x and what would that be equal to?"}, {"video_title": "Parametric curve arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So let's say that that is our very small change in our path, in our arc that we are traveling. And so we want to find this length. Well, we could break it down into how far we've moved in the x direction and how far we've moved in the y direction. So in the x direction, the x direction right over here, we would have moved a very small change in x and what would that be equal to? Well, that would be the rate of change with which we are changing with respect to t, with which x is changing with respect to t times our very small change in t. And this is a little hand wavy. I'm using differential notation and I'm conceptually using the idea of a differential as an infinitesimally small change in that variable. But this, and so this isn't a formal proof but it's to give us the intuition for how we derive arc length when we're dealing with parametric equations."}, {"video_title": "Parametric curve arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So in the x direction, the x direction right over here, we would have moved a very small change in x and what would that be equal to? Well, that would be the rate of change with which we are changing with respect to t, with which x is changing with respect to t times our very small change in t. And this is a little hand wavy. I'm using differential notation and I'm conceptually using the idea of a differential as an infinitesimally small change in that variable. But this, and so this isn't a formal proof but it's to give us the intuition for how we derive arc length when we're dealing with parametric equations. So this will hopefully make conceptual sense that this is our dx. In fact, we could even write it this way, dx dt. That's the same thing as x prime of t times dt."}, {"video_title": "Parametric curve arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "But this, and so this isn't a formal proof but it's to give us the intuition for how we derive arc length when we're dealing with parametric equations. So this will hopefully make conceptual sense that this is our dx. In fact, we could even write it this way, dx dt. That's the same thing as x prime of t times dt. And then our change in y is going to be the same idea. Our change in y, our infinitesimally small change in y when we have an infinitesimally small change in t. Well, you could view that as your rate of change of y with respect to t times your change in t, your very small change in t, which is going to be equal to, we could write that as y prime of t dt. Now, based on this, what would be the length of our infinitesimally small arc length right over here?"}, {"video_title": "Parametric curve arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "That's the same thing as x prime of t times dt. And then our change in y is going to be the same idea. Our change in y, our infinitesimally small change in y when we have an infinitesimally small change in t. Well, you could view that as your rate of change of y with respect to t times your change in t, your very small change in t, which is going to be equal to, we could write that as y prime of t dt. Now, based on this, what would be the length of our infinitesimally small arc length right over here? Well, that we could just use the Pythagorean theorem. That is going to be the square root of, that's the hypotenuse of this right triangle right over here. So it's gonna be the square root of this squared plus this squared."}, {"video_title": "Parametric curve arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "Now, based on this, what would be the length of our infinitesimally small arc length right over here? Well, that we could just use the Pythagorean theorem. That is going to be the square root of, that's the hypotenuse of this right triangle right over here. So it's gonna be the square root of this squared plus this squared. So it is the square root of, I'm gonna give myself a little bit more space here because I think I'm gonna use a lot of it. So the stuff in blue squared, dx squared, we could rewrite that as x prime of t dt squared plus this squared, which is y prime of t dt squared. And now let's just try to simplify this a little bit."}, {"video_title": "Parametric curve arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So it's gonna be the square root of this squared plus this squared. So it is the square root of, I'm gonna give myself a little bit more space here because I think I'm gonna use a lot of it. So the stuff in blue squared, dx squared, we could rewrite that as x prime of t dt squared plus this squared, which is y prime of t dt squared. And now let's just try to simplify this a little bit. Remember, this is this infinitesimally small arc length right over here. So we can actually factor out a dt squared. It's a term in both of these."}, {"video_title": "Parametric curve arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "And now let's just try to simplify this a little bit. Remember, this is this infinitesimally small arc length right over here. So we can actually factor out a dt squared. It's a term in both of these. And so we can rewrite this as, let me, so I can rewrite this, my big radical sign. So I'm gonna factor out a dt squared here. So we could write this as dt squared times x prime of t squared plus y prime of t squared."}, {"video_title": "Parametric curve arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "It's a term in both of these. And so we can rewrite this as, let me, so I can rewrite this, my big radical sign. So I'm gonna factor out a dt squared here. So we could write this as dt squared times x prime of t squared plus y prime of t squared. And then to be clear, this is being multiplied by all of this stuff right over there. Well, now we can actually, if we have this dt squared under the radical, we can take it out. And so we will have a dt."}, {"video_title": "Parametric curve arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So we could write this as dt squared times x prime of t squared plus y prime of t squared. And then to be clear, this is being multiplied by all of this stuff right over there. Well, now we can actually, if we have this dt squared under the radical, we can take it out. And so we will have a dt. And so this is all going to be equal to the square root of, so the stuff that's still under the radical is going to be x prime of t squared plus y prime of t squared. And now we took out a dt. Now we took out a dt."}, {"video_title": "Parametric curve arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "And so we will have a dt. And so this is all going to be equal to the square root of, so the stuff that's still under the radical is going to be x prime of t squared plus y prime of t squared. And now we took out a dt. Now we took out a dt. I could have written it right over here, but I'm just writing it on the other side. We're just multiplying the two. So this is, once again, just rewriting the expression for this infinitesimally small change in our arc length."}, {"video_title": "Parametric curve arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "Now we took out a dt. I could have written it right over here, but I'm just writing it on the other side. We're just multiplying the two. So this is, once again, just rewriting the expression for this infinitesimally small change in our arc length. Well, what's lucky for us is in calculus, we have the tools for adding up all of these infinitesimally small changes. That's what the definite integral does for us. So what we can do, if we want to add up that plus that plus that plus that, and remember, these are infinitesimally small changes."}, {"video_title": "Parametric curve arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So this is, once again, just rewriting the expression for this infinitesimally small change in our arc length. Well, what's lucky for us is in calculus, we have the tools for adding up all of these infinitesimally small changes. That's what the definite integral does for us. So what we can do, if we want to add up that plus that plus that plus that, and remember, these are infinitesimally small changes. I'm just showing them as not infinitesimal just so that you can kind of think about them. But if you were to add them all up, then we're essentially taking the integral. Now we're integrating with respect to t, and so we're starting at t is equal to a all the way to t is equal to b."}, {"video_title": "Worked example interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "So we have an infinite series here, and the goal of this video is to try to figure out the interval of convergence for this series. And that's another way of saying, for what x values, what range of x values is this series going to converge? And like always, pause this video and see if you can figure it out. So when you look at this series, it doesn't fit cleanly into something like a geometric series or an alternating series. And when I see something like this, I think about the ratio test, because it tends to be pretty general. And to apply the ratio test, we wanna think about the limit, the limit as n approaches infinity of the n plus one-th term divided by the n-th term, and the absolute value of that. And if this thing is less than one, so when this thing is less than one, then we are going to converge."}, {"video_title": "Worked example interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "So when you look at this series, it doesn't fit cleanly into something like a geometric series or an alternating series. And when I see something like this, I think about the ratio test, because it tends to be pretty general. And to apply the ratio test, we wanna think about the limit, the limit as n approaches infinity of the n plus one-th term divided by the n-th term, and the absolute value of that. And if this thing is less than one, so when this thing is less than one, then we are going to converge. And the x values that make this thing greater than one, we are going to diverge. And the x values that make this equal to one, well, then we're gonna be inconclusive, and so we're gonna have to use other techniques to think about whether we're going to converge or diverge. So let's just think about this."}, {"video_title": "Worked example interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "And if this thing is less than one, so when this thing is less than one, then we are going to converge. And the x values that make this thing greater than one, we are going to diverge. And the x values that make this equal to one, well, then we're gonna be inconclusive, and so we're gonna have to use other techniques to think about whether we're going to converge or diverge. So let's just think about this. Let's just evaluate this. So let's do that. Limit as n approaches infinity of the absolute value of a sub n plus one, well, that's gonna be x to the n plus one."}, {"video_title": "Worked example interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's just think about this. Let's just evaluate this. So let's do that. Limit as n approaches infinity of the absolute value of a sub n plus one, well, that's gonna be x to the n plus one. Let me color code this just so we know what we're doing. So this thing right over here is going to be x to the n plus one over n plus one times five to the n plus one. And we're gonna divide that by the nth term."}, {"video_title": "Worked example interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "Limit as n approaches infinity of the absolute value of a sub n plus one, well, that's gonna be x to the n plus one. Let me color code this just so we know what we're doing. So this thing right over here is going to be x to the n plus one over n plus one times five to the n plus one. And we're gonna divide that by the nth term. We're gonna divide that by the nth term, and that's just going to be x to the n over n times five to the n. And we're gonna take the absolute value of this whole thing. Now, let me simplify this. I'll simplify it down here."}, {"video_title": "Worked example interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "And we're gonna divide that by the nth term. We're gonna divide that by the nth term, and that's just going to be x to the n over n times five to the n. And we're gonna take the absolute value of this whole thing. Now, let me simplify this. I'll simplify it down here. So this is the same thing as x to the n plus one over n plus one times five to the n plus one times the reciprocal of this. So it's gonna be n times five to the n over x to the n. And we could simplify this. This is going to be equal to, let's see, divide numerator and denominator by x to the n. You're left with just an x."}, {"video_title": "Worked example interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "I'll simplify it down here. So this is the same thing as x to the n plus one over n plus one times five to the n plus one times the reciprocal of this. So it's gonna be n times five to the n over x to the n. And we could simplify this. This is going to be equal to, let's see, divide numerator and denominator by x to the n. You're left with just an x. And then divide numerator and denominator by five to the n. That is going to be, that's a one, this is a one. And then this is just going to be five to the n plus one divided by five to the n. That's just going to be a five. And so what do we have?"}, {"video_title": "Worked example interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "This is going to be equal to, let's see, divide numerator and denominator by x to the n. You're left with just an x. And then divide numerator and denominator by five to the n. That is going to be, that's a one, this is a one. And then this is just going to be five to the n plus one divided by five to the n. That's just going to be a five. And so what do we have? We have x times n, x times n, over, distribute the five, five n, five n plus one. Oh, let me be careful there. Let me distribute the five."}, {"video_title": "Worked example interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so what do we have? We have x times n, x times n, over, distribute the five, five n, five n plus one. Oh, let me be careful there. Let me distribute the five. Five n plus five. Five times n, five times one. Five n plus one."}, {"video_title": "Worked example interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let me distribute the five. Five n plus five. Five times n, five times one. Five n plus one. Five n plus five. All right, so let me just rewrite that. This is going to be equal to the limit as n approaches infinity of the absolute value of this thing."}, {"video_title": "Worked example interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "Five n plus one. Five n plus five. All right, so let me just rewrite that. This is going to be equal to the limit as n approaches infinity of the absolute value of this thing. And actually, to help us with this limit, let me rewrite it a little bit. Let me divide the numerator and the denominator both by n. I'm not changing the value. I'm doing the same thing to the numerator and the denominator."}, {"video_title": "Worked example interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "This is going to be equal to the limit as n approaches infinity of the absolute value of this thing. And actually, to help us with this limit, let me rewrite it a little bit. Let me divide the numerator and the denominator both by n. I'm not changing the value. I'm doing the same thing to the numerator and the denominator. I'm dividing it by the same value. So if I divide the numerator and the denominator by n, this is going to be the same thing as x over five plus five to the n. And so when I divided the numerator and the denominator by n, it becomes very clear what happens as n approaches infinity. As n approaches infinity, x doesn't change, five doesn't change, but five over n goes to zero."}, {"video_title": "Worked example interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "I'm doing the same thing to the numerator and the denominator. I'm dividing it by the same value. So if I divide the numerator and the denominator by n, this is going to be the same thing as x over five plus five to the n. And so when I divided the numerator and the denominator by n, it becomes very clear what happens as n approaches infinity. As n approaches infinity, x doesn't change, five doesn't change, but five over n goes to zero. And so this limit is going to be equal to x over five. So that's a pretty neat, clean thing. So now we can use this to think about it."}, {"video_title": "Worked example interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "As n approaches infinity, x doesn't change, five doesn't change, but five over n goes to zero. And so this limit is going to be equal to x over five. So that's a pretty neat, clean thing. So now we can use this to think about it. Actually, let me write this. This is going to be the absolute value of x over five. And so now we can think about under which conditions is the absolute value of x over five going to be less than one, and we're gonna definitely converge?"}, {"video_title": "Worked example interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "So now we can use this to think about it. Actually, let me write this. This is going to be the absolute value of x over five. And so now we can think about under which conditions is the absolute value of x over five going to be less than one, and we're gonna definitely converge? Under what conditions are we gonna be greater than one and definitely diverge? And then under what conditions is it inconclusive? So let's just see when we know we can converge."}, {"video_title": "Worked example interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so now we can think about under which conditions is the absolute value of x over five going to be less than one, and we're gonna definitely converge? Under what conditions are we gonna be greater than one and definitely diverge? And then under what conditions is it inconclusive? So let's just see when we know we can converge. So the absolute value of x over five is less than one. This is our convergence situation. Well, that's the same thing as saying that x negative one is less than x over five, which is less than one."}, {"video_title": "Worked example interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's just see when we know we can converge. So the absolute value of x over five is less than one. This is our convergence situation. Well, that's the same thing as saying that x negative one is less than x over five, which is less than one. And you multiply all the sides by five. This is the same thing as negative five is less than x, which is less than five. So if we know that this is true, this is definitely gonna be part of the interval of convergence."}, {"video_title": "Worked example interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, that's the same thing as saying that x negative one is less than x over five, which is less than one. And you multiply all the sides by five. This is the same thing as negative five is less than x, which is less than five. So if we know that this is true, this is definitely gonna be part of the interval of convergence. We know that if x meets these constraints, then our series is going to converge. But we're not done yet. We have to think about the inconclusive case."}, {"video_title": "Worked example interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "So if we know that this is true, this is definitely gonna be part of the interval of convergence. We know that if x meets these constraints, then our series is going to converge. But we're not done yet. We have to think about the inconclusive case. So let's think about the scenario where the absolute value of x over five, absolute value of x over five is equal to one. Or another way of thinking about this, this means that x over five is equal to one, or x over five is equal to negative one. And this means that x is equal to five, or x is equal to negative five."}, {"video_title": "Worked example interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "We have to think about the inconclusive case. So let's think about the scenario where the absolute value of x over five, absolute value of x over five is equal to one. Or another way of thinking about this, this means that x over five is equal to one, or x over five is equal to negative one. And this means that x is equal to five, or x is equal to negative five. So these are the two inconclusive cases using the ratio test. So let's test them out individually by looking back at the series and just substituting x equals five, or x equals negative five. So in the first scenario, let me find a new color here."}, {"video_title": "Worked example interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "And this means that x is equal to five, or x is equal to negative five. So these are the two inconclusive cases using the ratio test. So let's test them out individually by looking back at the series and just substituting x equals five, or x equals negative five. So in the first scenario, let me find a new color here. Let me use red. So in the first scenario of x equals five, let's go to our series. Then the series is going to be the sum from n equals one to infinity of five to the n over n times five to the n. Well, this is just going to be equal to the sum, n equals one to infinity, of one over n. And this is a harmonic series."}, {"video_title": "Worked example interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "So in the first scenario, let me find a new color here. Let me use red. So in the first scenario of x equals five, let's go to our series. Then the series is going to be the sum from n equals one to infinity of five to the n over n times five to the n. Well, this is just going to be equal to the sum, n equals one to infinity, of one over n. And this is a harmonic series. This is the p-series where p is equal to one. And we know for p-series, if p is equal to one, that's going to diverge. So, and we know the harmonic series."}, {"video_title": "Worked example interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "Then the series is going to be the sum from n equals one to infinity of five to the n over n times five to the n. Well, this is just going to be equal to the sum, n equals one to infinity, of one over n. And this is a harmonic series. This is the p-series where p is equal to one. And we know for p-series, if p is equal to one, that's going to diverge. So, and we know the harmonic series. We've done it in other videos. This definitely diverges. So this diverges."}, {"video_title": "Worked example interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "So, and we know the harmonic series. We've done it in other videos. This definitely diverges. So this diverges. And you can do that by p-series convergence test. If the p for a p-series is one, well, you're gonna diverge. So now let's think about, so five is definitely not part of our interval of convergence."}, {"video_title": "Worked example interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "So this diverges. And you can do that by p-series convergence test. If the p for a p-series is one, well, you're gonna diverge. So now let's think about, so five is definitely not part of our interval of convergence. Now let's think about x equals negative five. When x equals negative five, let me get another color going here. When x is equal to negative five, then this thing is going to be equal to the sum from n equals one to infinity of negative five to the n. And actually, let me just write that as, I'll write it out."}, {"video_title": "Worked example interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "So now let's think about, so five is definitely not part of our interval of convergence. Now let's think about x equals negative five. When x equals negative five, let me get another color going here. When x is equal to negative five, then this thing is going to be equal to the sum from n equals one to infinity of negative five to the n. And actually, let me just write that as, I'll write it out. Negative five to the n over n times five to the n. This is the same thing as the sum from n equals one to infinity of, we could write this as negative one to the n times five to the, how many times five to the n over n times five to the n. And now this thing, this is an alternating harmonic series. And so you could actually use the, you might already know that that converges, or you could use the alternating series test. And the alternating series test, it might be a little bit clearer if I write it like this, that this is an alternating series."}, {"video_title": "Worked example interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "When x is equal to negative five, then this thing is going to be equal to the sum from n equals one to infinity of negative five to the n. And actually, let me just write that as, I'll write it out. Negative five to the n over n times five to the n. This is the same thing as the sum from n equals one to infinity of, we could write this as negative one to the n times five to the, how many times five to the n over n times five to the n. And now this thing, this is an alternating harmonic series. And so you could actually use the, you might already know that that converges, or you could use the alternating series test. And the alternating series test, it might be a little bit clearer if I write it like this, that this is an alternating series. So in an alternating series test, if we see that this thing is monotonically decreasing, and the limit as n approaches infinity is zero, this thing converges. The alternating harmonic series actually converges. So this converges."}, {"video_title": "Worked example interval of convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "And the alternating series test, it might be a little bit clearer if I write it like this, that this is an alternating series. So in an alternating series test, if we see that this thing is monotonically decreasing, and the limit as n approaches infinity is zero, this thing converges. The alternating harmonic series actually converges. So this converges. So given that this converges, you could view this as this boundary here, we would include that in our interval of convergence. So x doesn't just have to be strictly greater than negative five. It could be greater than or equal to negative five, but it has to be less than five."}, {"video_title": "nth term divergence test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And the first one I'm going to go through right now is perhaps the most basic, and we'll hopefully see the most intuitive. And this is the divergence test. And the divergence test won't tell us if a series will converge, but it can tell us if something will definitely diverge. So first let me write it in kind of mathy notation, and then we'll look at an actual concrete example of it. So the divergence test tells us that if, if the limit as n approaches infinity of a sub n does not equal zero, then the infinite series going from n equals one to infinity of a sub n will diverge. Sub n will diverge. Will diverge."}, {"video_title": "nth term divergence test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So first let me write it in kind of mathy notation, and then we'll look at an actual concrete example of it. So the divergence test tells us that if, if the limit as n approaches infinity of a sub n does not equal zero, then the infinite series going from n equals one to infinity of a sub n will diverge. Sub n will diverge. Will diverge. And we've already gone through what it means to diverge, that this sum is either going to go unbounded to positive infinity or unbounded to negative infinity, or it'll just oscillate between values, it'll never really approach a given sum or a given value. So that's what the divergence test tells us, and you're probably thinking, okay, all right, I can kind of get what this says, but where is this actually useful? And to see where it can be useful, let's look at a candidate series and see if we can figure out if it's going to diverge."}, {"video_title": "nth term divergence test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Will diverge. And we've already gone through what it means to diverge, that this sum is either going to go unbounded to positive infinity or unbounded to negative infinity, or it'll just oscillate between values, it'll never really approach a given sum or a given value. So that's what the divergence test tells us, and you're probably thinking, okay, all right, I can kind of get what this says, but where is this actually useful? And to see where it can be useful, let's look at a candidate series and see if we can figure out if it's going to diverge. So let's say I had the series, so the sum from n equals one to infinity, and in general it doesn't always have to be n equals one, it could be n equals five, it could be n equals zero. The key is is that this is an infinite series that we're talking about, so that's why we care about the limit as n approaches infinity. And so let's say we have the sum of four n squared minus n to the third over seven, over seven minus three n to the third power."}, {"video_title": "nth term divergence test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And to see where it can be useful, let's look at a candidate series and see if we can figure out if it's going to diverge. So let's say I had the series, so the sum from n equals one to infinity, and in general it doesn't always have to be n equals one, it could be n equals five, it could be n equals zero. The key is is that this is an infinite series that we're talking about, so that's why we care about the limit as n approaches infinity. And so let's say we have the sum of four n squared minus n to the third over seven, over seven minus three n to the third power. So given what we know about the divergence test, is this series going to converge or diverge? Well, let's just look at this, let's look at what we're taking the sums of. So this is essentially, or this is our a sub n if we're trying to match to the definition or I guess the explanation of that divergence test."}, {"video_title": "nth term divergence test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so let's say we have the sum of four n squared minus n to the third over seven, over seven minus three n to the third power. So given what we know about the divergence test, is this series going to converge or diverge? Well, let's just look at this, let's look at what we're taking the sums of. So this is essentially, or this is our a sub n if we're trying to match to the definition or I guess the explanation of that divergence test. So let's think about what the limit, the limit as n approaches infinity of four n squared minus n to the third over seven minus three n to the third is. And I encourage you to pause the video and think about that. Well, there's a couple of ways to think about it."}, {"video_title": "nth term divergence test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So this is essentially, or this is our a sub n if we're trying to match to the definition or I guess the explanation of that divergence test. So let's think about what the limit, the limit as n approaches infinity of four n squared minus n to the third over seven minus three n to the third is. And I encourage you to pause the video and think about that. Well, there's a couple of ways to think about it. One way is say, look, as n goes to infinity, the highest degree terms in the numerator and the denominator are the ones that matter. And so this is going to approach negative n to the third over negative three n to the third, which will just be, which would approach negative one over negative three, which would be positive 1 3rd. Or we could say if we want to do a little bit more, or do a little bit more systematically, limit as n approaches infinity, we can divide both the numerator and the denominator by n to the third."}, {"video_title": "nth term divergence test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, there's a couple of ways to think about it. One way is say, look, as n goes to infinity, the highest degree terms in the numerator and the denominator are the ones that matter. And so this is going to approach negative n to the third over negative three n to the third, which will just be, which would approach negative one over negative three, which would be positive 1 3rd. Or we could say if we want to do a little bit more, or do a little bit more systematically, limit as n approaches infinity, we can divide both the numerator and the denominator by n to the third. So if we divide the numerator by n to the third, this first term is going to be four over n minus one over seven over n to the third minus three. And here it becomes clear, the limit as n approaches infinity, that's going to go to zero, that's going to go to zero. And so you're going to be left with negative one over negative three, which is equal to 1 3rd."}, {"video_title": "nth term divergence test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Or we could say if we want to do a little bit more, or do a little bit more systematically, limit as n approaches infinity, we can divide both the numerator and the denominator by n to the third. So if we divide the numerator by n to the third, this first term is going to be four over n minus one over seven over n to the third minus three. And here it becomes clear, the limit as n approaches infinity, that's going to go to zero, that's going to go to zero. And so you're going to be left with negative one over negative three, which is equal to 1 3rd. So notice the limit of a sub n as n approaches infinity in this case is not equal to zero. Therefore, this sum will, this infinite series will diverge. Diverge."}, {"video_title": "nth term divergence test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so you're going to be left with negative one over negative three, which is equal to 1 3rd. So notice the limit of a sub n as n approaches infinity in this case is not equal to zero. Therefore, this sum will, this infinite series will diverge. Diverge. Now let's think for a second why this makes a ton of sense. Well, the only way that you're going to converge, that you're going to settle, remember you're taking an infinite sum, you're taking an infinite sum of things. So the only reasonable way that something might be able to converge to a finite value is if every extra term you're adding is getting smaller and smaller and smaller as approaching zero."}, {"video_title": "nth term divergence test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Diverge. Now let's think for a second why this makes a ton of sense. Well, the only way that you're going to converge, that you're going to settle, remember you're taking an infinite sum, you're taking an infinite sum of things. So the only reasonable way that something might be able to converge to a finite value is if every extra term you're adding is getting smaller and smaller and smaller as approaching zero. If as n approaches infinity, you go unbounded or if it's even equal to 1 3rd, this means that for very large n's, you just keep adding things that are getting closer and closer to 1 3rd. Well, if you add an infinite number of 1 3rds together, you're going to go to infinity, you're going to be unbounded, you are going to diverge. So that's all this is telling us."}, {"video_title": "nth term divergence test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So the only reasonable way that something might be able to converge to a finite value is if every extra term you're adding is getting smaller and smaller and smaller as approaching zero. If as n approaches infinity, you go unbounded or if it's even equal to 1 3rd, this means that for very large n's, you just keep adding things that are getting closer and closer to 1 3rd. Well, if you add an infinite number of 1 3rds together, you're going to go to infinity, you're going to be unbounded, you are going to diverge. So that's all this is telling us. Look, in order for something to converge, if you're taking an infinite sum of them, at some point these things are going to have to get really, really, really close to zero. If at some point they're not getting close to zero, there's no way it's going to converge, that thing will diverge. So hopefully that makes a little sense."}, {"video_title": "nth term divergence test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So that's all this is telling us. Look, in order for something to converge, if you're taking an infinite sum of them, at some point these things are going to have to get really, really, really close to zero. If at some point they're not getting close to zero, there's no way it's going to converge, that thing will diverge. So hopefully that makes a little sense. And that also gives you, I guess, another insight on what the divergence test can't do. The divergence test can be used to show that something will diverge, but if something, I guess you could say, passes the divergence test, it doesn't, or I guess fails the divergence test, or if this isn't true, it doesn't mean that the thing is going to converge. And so let me give you an example of that."}, {"video_title": "nth term divergence test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So hopefully that makes a little sense. And that also gives you, I guess, another insight on what the divergence test can't do. The divergence test can be used to show that something will diverge, but if something, I guess you could say, passes the divergence test, it doesn't, or I guess fails the divergence test, or if this isn't true, it doesn't mean that the thing is going to converge. And so let me give you an example of that. So this right over here, from n equals one to infinity of one over n, and this is actually the harmonic series right over here, this, if we look at, if we try to apply the divergence test, we would say, okay, well, what's the limit as n approaches infinity of one over n? Well, hey, that is zero. As n approaches infinity, this is equal to zero."}, {"video_title": "nth term divergence test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so let me give you an example of that. So this right over here, from n equals one to infinity of one over n, and this is actually the harmonic series right over here, this, if we look at, if we try to apply the divergence test, we would say, okay, well, what's the limit as n approaches infinity of one over n? Well, hey, that is zero. As n approaches infinity, this is equal to zero. And so we could say, well, it's kind of failing the divergence test. So we're not, just by using the divergence test, we can't prove that this thing is going to diverge, but that doesn't mean that it doesn't diverge. It actually turns out, and we prove this in several videos, it actually turns out that this thing does diverge."}, {"video_title": "nth term divergence test Series AP Calculus BC Khan Academy.mp3", "Sentence": "As n approaches infinity, this is equal to zero. And so we could say, well, it's kind of failing the divergence test. So we're not, just by using the divergence test, we can't prove that this thing is going to diverge, but that doesn't mean that it doesn't diverge. It actually turns out, and we prove this in several videos, it actually turns out that this thing does diverge. This thing does, does diverge. It's just that the divergence test isn't enough. It's not enough of a tool to let us know for sure that this diverged."}, {"video_title": "nth term divergence test Series AP Calculus BC Khan Academy.mp3", "Sentence": "It actually turns out, and we prove this in several videos, it actually turns out that this thing does diverge. This thing does, does diverge. It's just that the divergence test isn't enough. It's not enough of a tool to let us know for sure that this diverged. We'll see the comparison test and the integral test can either be used to prove that this, in fact, does diverge. And so you can definitely not say that if something, if this does not apply for something, if you're taking the limit as n approaches infinity and it does go to zero, that still might diverge. It doesn't necessarily mean that it converges."}, {"video_title": "nth term divergence test Series AP Calculus BC Khan Academy.mp3", "Sentence": "It's not enough of a tool to let us know for sure that this diverged. We'll see the comparison test and the integral test can either be used to prove that this, in fact, does diverge. And so you can definitely not say that if something, if this does not apply for something, if you're taking the limit as n approaches infinity and it does go to zero, that still might diverge. It doesn't necessarily mean that it converges. Now, there are things that do converge that where they do approach zero. So for example, if I take the sum from n equals one to infinity of one over n squared, if I try to apply the divergence test, I have the limit as n approaches infinity of one over n squared. Well, this does equal zero."}, {"video_title": "nth term divergence test Series AP Calculus BC Khan Academy.mp3", "Sentence": "It doesn't necessarily mean that it converges. Now, there are things that do converge that where they do approach zero. So for example, if I take the sum from n equals one to infinity of one over n squared, if I try to apply the divergence test, I have the limit as n approaches infinity of one over n squared. Well, this does equal zero. This gets really, really large. This thing is going to approach zero. And so once again, this is, I guess, failing the test for divergence."}, {"video_title": "nth term divergence test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, this does equal zero. This gets really, really large. This thing is going to approach zero. And so once again, this is, I guess, failing the test for divergence. Just with that alone, you don't know for sure that this thing is going to converge. It still might diverge. The divergence test just might not have been enough."}, {"video_title": "nth term divergence test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so once again, this is, I guess, failing the test for divergence. Just with that alone, you don't know for sure that this thing is going to converge. It still might diverge. The divergence test just might not have been enough. Now, it does turn out, and once again, we prove this in future videos, that this thing does converge. This thing does converge, but not because it, I guess you could say, fails the divergence test. We have to use a different test to show that this does, in fact, converge, just as we have to use a different test to show that this does, in fact, diverge."}, {"video_title": "nth term divergence test Series AP Calculus BC Khan Academy.mp3", "Sentence": "The divergence test just might not have been enough. Now, it does turn out, and once again, we prove this in future videos, that this thing does converge. This thing does converge, but not because it, I guess you could say, fails the divergence test. We have to use a different test to show that this does, in fact, converge, just as we have to use a different test to show that this does, in fact, diverge. Where the divergence test is useful is for the things that actually pass the divergence test. When you actually find that the limit as n approaches infinity of a sub n does not equal zero, like this case right over here. In this case, the divergence test helps us because it helps us make the conclusion that this series definitely, definitely, that this series definitely diverges."}, {"video_title": "Worked examples interpreting definite integrals in context AP Calculus AB Khan Academy.mp3", "Sentence": "What does three plus the definite integral from one to five of R of T DT equals 19 mean? And we have some choices, so like always, pause the video and see if you can work through it. All right, now let's work through this together. So they tell us that she made $3,000 in the first month, and we also see this three here, so that's interesting. Maybe they represent the same thing, we don't know for sure yet. But let's look at this definite integral. The definite integral from one to five of R of T DT."}, {"video_title": "Worked examples interpreting definite integrals in context AP Calculus AB Khan Academy.mp3", "Sentence": "So they tell us that she made $3,000 in the first month, and we also see this three here, so that's interesting. Maybe they represent the same thing, we don't know for sure yet. But let's look at this definite integral. The definite integral from one to five of R of T DT. This is the area under this rate curve. R of T is the rate at which Julia makes revenue on a monthly basis. So if you take the area under that rate curve, that's going to give you the net change in revenue from month one to month five, how much that increased."}, {"video_title": "Worked examples interpreting definite integrals in context AP Calculus AB Khan Academy.mp3", "Sentence": "The definite integral from one to five of R of T DT. This is the area under this rate curve. R of T is the rate at which Julia makes revenue on a monthly basis. So if you take the area under that rate curve, that's going to give you the net change in revenue from month one to month five, how much that increased. So if you add that to the amount that she made in month one, well that tells you the total she makes from essentially time zero all the way to month five. And they're saying that is equal to 19. So let's see which of these choices are consistent with that."}, {"video_title": "Worked examples interpreting definite integrals in context AP Calculus AB Khan Academy.mp3", "Sentence": "So if you take the area under that rate curve, that's going to give you the net change in revenue from month one to month five, how much that increased. So if you add that to the amount that she made in month one, well that tells you the total she makes from essentially time zero all the way to month five. And they're saying that is equal to 19. So let's see which of these choices are consistent with that. Julia made an additional $19,000 between months one and five. Choice A would be correct if you didn't see this three over here, because just the definite integral is the additional between months one and five, but that's not what this expression says. It says three plus this is equal to 19."}, {"video_title": "Worked examples interpreting definite integrals in context AP Calculus AB Khan Academy.mp3", "Sentence": "So let's see which of these choices are consistent with that. Julia made an additional $19,000 between months one and five. Choice A would be correct if you didn't see this three over here, because just the definite integral is the additional between months one and five, but that's not what this expression says. It says three plus this is equal to 19. If it said Julia made an additional $16,000, well then that would make sense, because you could subtract three from both sides and you'd get that result, but that's not what they're saying. Julia made an average of $19,000 per month. Well once again, that's also not right, because we already just said from the beginning, from time zero all the way until the fifth month, she made a total of $19,000, not the average per month is 19,000."}, {"video_title": "Worked examples interpreting definite integrals in context AP Calculus AB Khan Academy.mp3", "Sentence": "It says three plus this is equal to 19. If it said Julia made an additional $16,000, well then that would make sense, because you could subtract three from both sides and you'd get that result, but that's not what they're saying. Julia made an average of $19,000 per month. Well once again, that's also not right, because we already just said from the beginning, from time zero all the way until the fifth month, she made a total of $19,000, not the average per month is 19,000. Julia made $19,000 in the fifth month. Once again, this is not just saying what happened in the fifth month. This is saying we have the $3,000 from the first month, and then we have the additional from month, between months one and month five."}, {"video_title": "Worked examples interpreting definite integrals in context AP Calculus AB Khan Academy.mp3", "Sentence": "Well once again, that's also not right, because we already just said from the beginning, from time zero all the way until the fifth month, she made a total of $19,000, not the average per month is 19,000. Julia made $19,000 in the fifth month. Once again, this is not just saying what happened in the fifth month. This is saying we have the $3,000 from the first month, and then we have the additional from month, between months one and month five. So that's not that, so this better be our choice. By the end of the fifth month, Julia had made a total of $19,000. Yes, that is correct."}, {"video_title": "Worked examples interpreting definite integrals in context AP Calculus AB Khan Academy.mp3", "Sentence": "This is saying we have the $3,000 from the first month, and then we have the additional from month, between months one and month five. So that's not that, so this better be our choice. By the end of the fifth month, Julia had made a total of $19,000. Yes, that is correct. She made $3,000 in month one, and then as we go between month one to the end of month five, to the end of the fifth month, she has made a total of $19,000. Let's do another one of these. So here we're told the function K of t gives the amount of ketchup in kilograms produced in a sauce factory by time in hours on a given day."}, {"video_title": "Worked examples interpreting definite integrals in context AP Calculus AB Khan Academy.mp3", "Sentence": "Yes, that is correct. She made $3,000 in month one, and then as we go between month one to the end of month five, to the end of the fifth month, she has made a total of $19,000. Let's do another one of these. So here we're told the function K of t gives the amount of ketchup in kilograms produced in a sauce factory by time in hours on a given day. So this is really quantity as a function of time. It isn't rate. What does the definite integral from zero to four of K prime of t dt represent?"}, {"video_title": "Worked examples interpreting definite integrals in context AP Calculus AB Khan Academy.mp3", "Sentence": "So here we're told the function K of t gives the amount of ketchup in kilograms produced in a sauce factory by time in hours on a given day. So this is really quantity as a function of time. It isn't rate. What does the definite integral from zero to four of K prime of t dt represent? Once again, pause the video and see if you can work through it. Well, K of t is the amount of ketchup as a function of time. So K prime of t, that's going to be the rate at which our amount of ketchup is changing as a function of time."}, {"video_title": "Worked examples interpreting definite integrals in context AP Calculus AB Khan Academy.mp3", "Sentence": "What does the definite integral from zero to four of K prime of t dt represent? Once again, pause the video and see if you can work through it. Well, K of t is the amount of ketchup as a function of time. So K prime of t, that's going to be the rate at which our amount of ketchup is changing as a function of time. But once again, when you're taking the area under the rate curve, that tells you the net change in the original quantity, in the amount of ketchup. And it's the net change between time zero and time four. So let's see which of these choices match up to that."}, {"video_title": "Worked examples interpreting definite integrals in context AP Calculus AB Khan Academy.mp3", "Sentence": "So K prime of t, that's going to be the rate at which our amount of ketchup is changing as a function of time. But once again, when you're taking the area under the rate curve, that tells you the net change in the original quantity, in the amount of ketchup. And it's the net change between time zero and time four. So let's see which of these choices match up to that. The average rate of change of the ketchup production over the first four hours, no, that does not tell us the average rate of change. There's other ways to calculate that. The time it takes to produce four kilograms of ketchup."}, {"video_title": "Worked examples interpreting definite integrals in context AP Calculus AB Khan Academy.mp3", "Sentence": "So let's see which of these choices match up to that. The average rate of change of the ketchup production over the first four hours, no, that does not tell us the average rate of change. There's other ways to calculate that. The time it takes to produce four kilograms of ketchup. So does this represent the time it takes to represent four kilograms of ketchup? No, this four is a time right over here. This is gonna tell you how much ketchup gets produced from time zero to time four."}, {"video_title": "Worked examples interpreting definite integrals in context AP Calculus AB Khan Academy.mp3", "Sentence": "The time it takes to produce four kilograms of ketchup. So does this represent the time it takes to represent four kilograms of ketchup? No, this four is a time right over here. This is gonna tell you how much ketchup gets produced from time zero to time four. The instantaneous rate of production at t equals four. No, this would be K prime of four. That's not what this integral represents."}, {"video_title": "Functions continuous at specific x-values Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "which of the following functions are continuous at x equals three? Well, as we said in the previous video, in the previous example, in order to be continuous at a point, you at least have to be defined at that point. We saw our definition of continuity. F is continuous at a if and only if the limit of f as x approaches a is equal to f of a. So over here in this case, we could say that a function is continuous at x equals three so f is continuous at x equals three if and only if, if and only if, the limit as x approaches three of f of x is equal to f of three. Now let's look at this first function right over here, natural log of x minus three. Well, try to evaluate, and it's not f now, it's g, try to evaluate g of three."}, {"video_title": "Functions continuous at specific x-values Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "F is continuous at a if and only if the limit of f as x approaches a is equal to f of a. So over here in this case, we could say that a function is continuous at x equals three so f is continuous at x equals three if and only if, if and only if, the limit as x approaches three of f of x is equal to f of three. Now let's look at this first function right over here, natural log of x minus three. Well, try to evaluate, and it's not f now, it's g, try to evaluate g of three. G of three, let me write it here, g of three is equal to the natural log of zero, three minus three. This is not defined. You can't raise e to any power to get to zero."}, {"video_title": "Functions continuous at specific x-values Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Well, try to evaluate, and it's not f now, it's g, try to evaluate g of three. G of three, let me write it here, g of three is equal to the natural log of zero, three minus three. This is not defined. You can't raise e to any power to get to zero. You try to go to, you know, you could say negative infinity, but that's not, this is not defined. And so if this isn't even defined at x equals three, there's no way that it's going to be continuous at x equals three. So we could rule this one out."}, {"video_title": "Functions continuous at specific x-values Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "You can't raise e to any power to get to zero. You try to go to, you know, you could say negative infinity, but that's not, this is not defined. And so if this isn't even defined at x equals three, there's no way that it's going to be continuous at x equals three. So we could rule this one out. Now f of x is equal to e to the x minus three. Well, this is just a shifted over version of e to the x. This is defined for all real numbers, and as we saw in the previous example, it's reasonable to say it's continuous for all real numbers, and you could even do this little test here."}, {"video_title": "Functions continuous at specific x-values Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So we could rule this one out. Now f of x is equal to e to the x minus three. Well, this is just a shifted over version of e to the x. This is defined for all real numbers, and as we saw in the previous example, it's reasonable to say it's continuous for all real numbers, and you could even do this little test here. The limit of e to the x minus three as x approaches three, well, that is going to be, that is going to be e to the three minus three, or e to the zero, or one. And so f is the only one that is continuous. And once again, it's good to think about what's going on here visually if you like."}, {"video_title": "Functions continuous at specific x-values Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "This is defined for all real numbers, and as we saw in the previous example, it's reasonable to say it's continuous for all real numbers, and you could even do this little test here. The limit of e to the x minus three as x approaches three, well, that is going to be, that is going to be e to the three minus three, or e to the zero, or one. And so f is the only one that is continuous. And once again, it's good to think about what's going on here visually if you like. Both of these are, you could think of them, this is a shifted over version of ln of x, this is a shifted over version of e to the x, and so if we like, we could draw ourselves some axes. So that's our y-axis. This is our x-axis."}, {"video_title": "Functions continuous at specific x-values Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And once again, it's good to think about what's going on here visually if you like. Both of these are, you could think of them, this is a shifted over version of ln of x, this is a shifted over version of e to the x, and so if we like, we could draw ourselves some axes. So that's our y-axis. This is our x-axis. And actually, let me draw some points here. So that's one, that is one, that is, let's see, two, three, two, and three. And let's see, I said these are shifted over versions, so actually, this is maybe not the best way to draw it."}, {"video_title": "Functions continuous at specific x-values Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "This is our x-axis. And actually, let me draw some points here. So that's one, that is one, that is, let's see, two, three, two, and three. And let's see, I said these are shifted over versions, so actually, this is maybe not the best way to draw it. So let me draw it, this is one, two, three, four, five, and six. And on this axis, I want to make them on the same scale, so let's say this is one, two, three. I'm gonna draw one, two, three, I'm gonna draw a dotted line right over here."}, {"video_title": "Functions continuous at specific x-values Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And let's see, I said these are shifted over versions, so actually, this is maybe not the best way to draw it. So let me draw it, this is one, two, three, four, five, and six. And on this axis, I want to make them on the same scale, so let's say this is one, two, three. I'm gonna draw one, two, three, I'm gonna draw a dotted line right over here. So g of x, ln of x minus three is gonna look something like this. If you put three in it, it's not defined. If you put four in it, ln of four, well, that's, sorry, ln of four minus one, so that's gonna be ln of, sorry, ln of four minus three is, actually, let me just draw a table here, I'm confusing you."}, {"video_title": "Functions continuous at specific x-values Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "I'm gonna draw one, two, three, I'm gonna draw a dotted line right over here. So g of x, ln of x minus three is gonna look something like this. If you put three in it, it's not defined. If you put four in it, ln of four, well, that's, sorry, ln of four minus one, so that's gonna be ln of, sorry, ln of four minus three is, actually, let me just draw a table here, I'm confusing you. So if I say x, and I say g of x, so at three, you're undefined. At four, this is ln of one, ln of one, which is equal to zero. So it's right over there."}, {"video_title": "Functions continuous at specific x-values Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "If you put four in it, ln of four, well, that's, sorry, ln of four minus one, so that's gonna be ln of, sorry, ln of four minus three is, actually, let me just draw a table here, I'm confusing you. So if I say x, and I say g of x, so at three, you're undefined. At four, this is ln of one, ln of one, which is equal to zero. So it's right over there. So g of x is gonna look something like, something like that. And so you can see, at three, you have this discontinuity there. It's not even defined to the left of three."}, {"video_title": "Functions continuous at specific x-values Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So it's right over there. So g of x is gonna look something like, something like that. And so you can see, at three, you have this discontinuity there. It's not even defined to the left of three. Now f of x is a little bit more straightforward. F of three is gonna be e to the three minus three, or e to the zero, so it's going to be one. So it's gonna look something like this."}, {"video_title": "Functions continuous at specific x-values Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "It's not even defined to the left of three. Now f of x is a little bit more straightforward. F of three is gonna be e to the three minus three, or e to the zero, so it's going to be one. So it's gonna look something like this. It's gonna look something, something like that. Like that, there's no jumps, there's no gaps. It is going to be continuous at, frankly, all real numbers, so for sure, it's going to be continuous at three."}, {"video_title": "Derivative of a parametric function.mp3", "Sentence": "T is equal to one, figure out what x and y are, and all the other t's, and then you get this pretty cool looking graph. But the goal in this video isn't just to appreciate the coolness of graphs or curves defined by parametric equations, but we actually wanna do some calculus. And in particular, we wanna find the derivative, we wanna find the derivative of y with respect to x, the derivative of y with respect to x, when t, when t is equal to negative 1 3rd. And if you are inspired, I encourage you to pause and try to solve this. And I'm about to do it with you in case you already did or you just want me to. All right. So the key is, is well, how do you find the derivative with respect to x, derivative of y with respect to x when they're both defined in terms of t?"}, {"video_title": "Derivative of a parametric function.mp3", "Sentence": "And if you are inspired, I encourage you to pause and try to solve this. And I'm about to do it with you in case you already did or you just want me to. All right. So the key is, is well, how do you find the derivative with respect to x, derivative of y with respect to x when they're both defined in terms of t? And the key realization is the derivative of y with respect to x, with respect to x, is going to be equal to, is going to be equal to the derivative of y with respect to t over the derivative of x with respect to t. If you were to view these differentials as numbers, well, this would actually work out mathematically. Now, it gets a little bit non-rigorous when you start to do that. But if you thought of it that, it's an easy way of thinking about why this actually might make sense."}, {"video_title": "Derivative of a parametric function.mp3", "Sentence": "So the key is, is well, how do you find the derivative with respect to x, derivative of y with respect to x when they're both defined in terms of t? And the key realization is the derivative of y with respect to x, with respect to x, is going to be equal to, is going to be equal to the derivative of y with respect to t over the derivative of x with respect to t. If you were to view these differentials as numbers, well, this would actually work out mathematically. Now, it gets a little bit non-rigorous when you start to do that. But if you thought of it that, it's an easy way of thinking about why this actually might make sense. The derivative of something versus something else is equal to the derivative of y with respect to t over x with respect to t. All right. So how does that help us? Well, we can figure out the derivative of x with respect to t and derivative of y with respect to t. The derivative of x with respect to t is just going to be equal to, let's see, the derivative of the outside with respect to the inside, it's going to be two sine, whoops, derivative of sine is cosine, two cosine of one plus three t times the derivative of the inside with respect to t. So that's going to be derivative of one is just zero."}, {"video_title": "Derivative of a parametric function.mp3", "Sentence": "But if you thought of it that, it's an easy way of thinking about why this actually might make sense. The derivative of something versus something else is equal to the derivative of y with respect to t over x with respect to t. All right. So how does that help us? Well, we can figure out the derivative of x with respect to t and derivative of y with respect to t. The derivative of x with respect to t is just going to be equal to, let's see, the derivative of the outside with respect to the inside, it's going to be two sine, whoops, derivative of sine is cosine, two cosine of one plus three t times the derivative of the inside with respect to t. So that's going to be derivative of one is just zero. Derivative of three t with respect to t is three. So times three. That's the derivative of x with respect to t. I just used a chain rule here."}, {"video_title": "Derivative of a parametric function.mp3", "Sentence": "Well, we can figure out the derivative of x with respect to t and derivative of y with respect to t. The derivative of x with respect to t is just going to be equal to, let's see, the derivative of the outside with respect to the inside, it's going to be two sine, whoops, derivative of sine is cosine, two cosine of one plus three t times the derivative of the inside with respect to t. So that's going to be derivative of one is just zero. Derivative of three t with respect to t is three. So times three. That's the derivative of x with respect to t. I just used a chain rule here. Derivative of the outside, two sine of something with respect to the inside. So derivative of this outside, two sine of something with respect to one plus three t is that right over there. And the derivative of the inside with respect to t is just our three."}, {"video_title": "Derivative of a parametric function.mp3", "Sentence": "That's the derivative of x with respect to t. I just used a chain rule here. Derivative of the outside, two sine of something with respect to the inside. So derivative of this outside, two sine of something with respect to one plus three t is that right over there. And the derivative of the inside with respect to t is just our three. Now, derivative of y with respect to t is a little bit more straightforward. Derivative of y with respect to t, we just apply the power rule here. Three times two is six t to the three minus one power, six t squared."}, {"video_title": "Derivative of a parametric function.mp3", "Sentence": "And the derivative of the inside with respect to t is just our three. Now, derivative of y with respect to t is a little bit more straightforward. Derivative of y with respect to t, we just apply the power rule here. Three times two is six t to the three minus one power, six t squared. So this is going to be equal to six t squared, six t squared over, well, we have the two times the three, so we have six times cosine of one plus three t. And then our sixes cancel out. And we are left with, we are left with t squared over cosine of one plus three t. And if we care when t is equal to negative 1 3rd, when t is equal to negative 1 3rd, this is going to be equal to, well, this is going to be equal to negative 1 3rd squared, negative 1 3rd squared over, over, over the cosine of one plus three times negative 1 3rd is negative one. So it's one plus negative one, so it's the cosine of zero."}, {"video_title": "Derivative of a parametric function.mp3", "Sentence": "Three times two is six t to the three minus one power, six t squared. So this is going to be equal to six t squared, six t squared over, well, we have the two times the three, so we have six times cosine of one plus three t. And then our sixes cancel out. And we are left with, we are left with t squared over cosine of one plus three t. And if we care when t is equal to negative 1 3rd, when t is equal to negative 1 3rd, this is going to be equal to, well, this is going to be equal to negative 1 3rd squared, negative 1 3rd squared over, over, over the cosine of one plus three times negative 1 3rd is negative one. So it's one plus negative one, so it's the cosine of zero. And the cosine of zero is just going to be one. So this is going to be equal to positive, positive 1 9th. And let's see if we can visualize what's going on here."}, {"video_title": "Derivative of a parametric function.mp3", "Sentence": "So it's one plus negative one, so it's the cosine of zero. And the cosine of zero is just going to be one. So this is going to be equal to positive, positive 1 9th. And let's see if we can visualize what's going on here. So let me draw a little table here. So, so I'm gonna plot, I'm gonna think about t and x and y. So t and x and y."}, {"video_title": "Derivative of a parametric function.mp3", "Sentence": "And let's see if we can visualize what's going on here. So let me draw a little table here. So, so I'm gonna plot, I'm gonna think about t and x and y. So t and x and y. So when t is equal to negative 1 3rd, well, our x is going to be, this is going to be sine of zero. So our x is going to be zero. And our y is going to be what?"}, {"video_title": "Derivative of a parametric function.mp3", "Sentence": "So t and x and y. So when t is equal to negative 1 3rd, well, our x is going to be, this is going to be sine of zero. So our x is going to be zero. And our y is going to be what? Two over, or negative two over 27. So we're talking about, we are talking about the point zero comma negative two over 27. So that is that point right over there."}, {"video_title": "Derivative of a parametric function.mp3", "Sentence": "And our y is going to be what? Two over, or negative two over 27. So we're talking about, we are talking about the point zero comma negative two over 27. So that is that point right over there. So that's the point where we're trying to find the slope of the tangent line. And it's telling us that that slope is 1 9th. That slope is 1 9th."}, {"video_title": "Derivative of a parametric function.mp3", "Sentence": "So that is that point right over there. So that's the point where we're trying to find the slope of the tangent line. And it's telling us that that slope is 1 9th. That slope is 1 9th. So if we move, I guess one way to think about it is if we move four, one, two, three, four and a half, we're gonna move up half. So if I wanted to draw the tangent line right there, it would look something like, something like that. Something, something, something like that."}, {"video_title": "Derivative of a parametric function.mp3", "Sentence": "That slope is 1 9th. So if we move, I guess one way to think about it is if we move four, one, two, three, four and a half, we're gonna move up half. So if I wanted to draw the tangent line right there, it would look something like, something like that. Something, something, something like that. Let's see, if we go one, two, three, four and a half, so yeah, just like, that is pretty close. So that's what we just figured out. We figured out that the slope of the tangent line right at that point is 1 9th."}, {"video_title": "Derivative of log_x (for any positive base a\u00c3\u0082\u00c2\u00ad1) AP Calculus AB Khan Academy.mp3", "Sentence": "What I want to do in this video is use that knowledge that we've seen in other videos to figure out what the derivative with respect to x is of a logarithm of an arbitrary base. So I'm just gonna call that log base a of x. So how do we figure this out? Well, the key thing is, is what you might be familiar with from your algebra or your pre-calculus classes, which is having a change of base. So if I have some, I'll do it over here, log base a of b, and I want to change it to a different base. Let's say I want to change it to base c. This is the same thing as log base c of b divided by log base, log base c of a. Log base c of b divided by log base c of a. And this is a really useful thing."}, {"video_title": "Derivative of log_x (for any positive base a\u00c3\u0082\u00c2\u00ad1) AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the key thing is, is what you might be familiar with from your algebra or your pre-calculus classes, which is having a change of base. So if I have some, I'll do it over here, log base a of b, and I want to change it to a different base. Let's say I want to change it to base c. This is the same thing as log base c of b divided by log base, log base c of a. Log base c of b divided by log base c of a. And this is a really useful thing. If you've never seen it before, you now have just seen it, this change of base, and we prove it in other videos on Khan Academy. But it's really useful because, for example, your calculator has a log button, but the log on your calculator is log base 10. So if you do log, if you press 10 into your, or 100 into your calculator and press log, you will get a two there."}, {"video_title": "Derivative of log_x (for any positive base a\u00c3\u0082\u00c2\u00ad1) AP Calculus AB Khan Academy.mp3", "Sentence": "And this is a really useful thing. If you've never seen it before, you now have just seen it, this change of base, and we prove it in other videos on Khan Academy. But it's really useful because, for example, your calculator has a log button, but the log on your calculator is log base 10. So if you do log, if you press 10 into your, or 100 into your calculator and press log, you will get a two there. So whenever you just see log of 100, it's implicitly base 10. And you also have a button for natural log, which is log base e. Natural log of x is equal to log base e of x. But sometimes you want to find all sorts of different base logarithms, and this is how you do it."}, {"video_title": "Derivative of log_x (for any positive base a\u00c3\u0082\u00c2\u00ad1) AP Calculus AB Khan Academy.mp3", "Sentence": "So if you do log, if you press 10 into your, or 100 into your calculator and press log, you will get a two there. So whenever you just see log of 100, it's implicitly base 10. And you also have a button for natural log, which is log base e. Natural log of x is equal to log base e of x. But sometimes you want to find all sorts of different base logarithms, and this is how you do it. So if you're using your calculator and you wanted to find, if you wanted to find what log base three of eight is, you would say, you would type in your calculator log of eight and log of three. Or, let me write it this way, and log of three, where both of these are implicitly base 10. And you'd get the same value if you did natural log of eight divided by natural log of three, which you might also have on your calculator."}, {"video_title": "Derivative of log_x (for any positive base a\u00c3\u0082\u00c2\u00ad1) AP Calculus AB Khan Academy.mp3", "Sentence": "But sometimes you want to find all sorts of different base logarithms, and this is how you do it. So if you're using your calculator and you wanted to find, if you wanted to find what log base three of eight is, you would say, you would type in your calculator log of eight and log of three. Or, let me write it this way, and log of three, where both of these are implicitly base 10. And you'd get the same value if you did natural log of eight divided by natural log of three, which you might also have on your calculator. And what we're gonna do in this video is leverage the natural log, because we know what the derivative of the natural log is. So this derivative is the same thing as the derivative with respect to x of, well, log base a of x can be rewritten as natural log of x over natural log of a. And now, natural log of a, that's just a number."}, {"video_title": "Derivative of log_x (for any positive base a\u00c3\u0082\u00c2\u00ad1) AP Calculus AB Khan Academy.mp3", "Sentence": "And you'd get the same value if you did natural log of eight divided by natural log of three, which you might also have on your calculator. And what we're gonna do in this video is leverage the natural log, because we know what the derivative of the natural log is. So this derivative is the same thing as the derivative with respect to x of, well, log base a of x can be rewritten as natural log of x over natural log of a. And now, natural log of a, that's just a number. I could rewrite this, I could rewrite this as, write it this way, one over natural log of a times natural log of x. And what's the derivative of that? Well, we could just take the constant out."}, {"video_title": "Derivative of log_x (for any positive base a\u00c3\u0082\u00c2\u00ad1) AP Calculus AB Khan Academy.mp3", "Sentence": "And now, natural log of a, that's just a number. I could rewrite this, I could rewrite this as, write it this way, one over natural log of a times natural log of x. And what's the derivative of that? Well, we could just take the constant out. One over natural log of a, that's just a number. So we're gonna get one over the natural log of a times the derivative, times the derivative with respect to x of natural log of x, of natural log of x, which we already know is one over x. So this thing right over here is one over x."}, {"video_title": "Derivative of log_x (for any positive base a\u00c3\u0082\u00c2\u00ad1) AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we could just take the constant out. One over natural log of a, that's just a number. So we're gonna get one over the natural log of a times the derivative, times the derivative with respect to x of natural log of x, of natural log of x, which we already know is one over x. So this thing right over here is one over x. So what we get is one over natural log of a times one over x, which we could write as one over natural log of a times x, times x, which is a really useful thing to know. So now we could take all sorts of derivatives. So if I were to tell you f of x is equal to log base seven of x, well, now we could say, well, f prime of x is going to be one over the natural log of seven times x."}, {"video_title": "Derivative of log_x (for any positive base a\u00c3\u0082\u00c2\u00ad1) AP Calculus AB Khan Academy.mp3", "Sentence": "So this thing right over here is one over x. So what we get is one over natural log of a times one over x, which we could write as one over natural log of a times x, times x, which is a really useful thing to know. So now we could take all sorts of derivatives. So if I were to tell you f of x is equal to log base seven of x, well, now we could say, well, f prime of x is going to be one over the natural log of seven times x. If we had a constant out front, if we had, for example, g of x, g of x is equal to negative three times log base, I don't know, log base pi. Pi is a number. Log base pi of x, well, g prime of x would be equal to one over the, oh, let me be careful."}, {"video_title": "Proof of the derivative of cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "What I want to do in this video is make a visual argument as to why the derivative with respect to x of cosine of x is equal to negative sine of x. And we're gonna base this argument based on a previous proof we made that the derivative with respect to x of sine of x is equal to cosine of x. So we're gonna assume this over here. I encourage you to watch that video. That's actually a fairly involved proof that proves this. But if we assume this, I'm gonna make a visual argument that this right over here is true, that the derivative with respect to x of cosine of x is negative sine of x. So right over here, we see sine of x in red and we see cosine of x in blue."}, {"video_title": "Proof of the derivative of cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "I encourage you to watch that video. That's actually a fairly involved proof that proves this. But if we assume this, I'm gonna make a visual argument that this right over here is true, that the derivative with respect to x of cosine of x is negative sine of x. So right over here, we see sine of x in red and we see cosine of x in blue. And we're assuming that this blue graph is showing the derivative, the slope of the tangent line for any x value of the red graph. And we've gotten an intuition for that in previous videos. Now what I'm gonna do next is I'm gonna shift both of these graphs to the left by pi over two."}, {"video_title": "Proof of the derivative of cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So right over here, we see sine of x in red and we see cosine of x in blue. And we're assuming that this blue graph is showing the derivative, the slope of the tangent line for any x value of the red graph. And we've gotten an intuition for that in previous videos. Now what I'm gonna do next is I'm gonna shift both of these graphs to the left by pi over two. Shift it to the left by pi over two. And I'm also gonna shift the blue graph to the left by pi over two. And so what am I going to get?"}, {"video_title": "Proof of the derivative of cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Now what I'm gonna do next is I'm gonna shift both of these graphs to the left by pi over two. Shift it to the left by pi over two. And I'm also gonna shift the blue graph to the left by pi over two. And so what am I going to get? Well, the blue graph is gonna look like this one right over here. And if it was cosine of x up here, we can now say that this is equal to, y is equal to cosine of x plus pi over two. This is the blue graph, cosine of x, shifted to the left by pi over two."}, {"video_title": "Proof of the derivative of cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And so what am I going to get? Well, the blue graph is gonna look like this one right over here. And if it was cosine of x up here, we can now say that this is equal to, y is equal to cosine of x plus pi over two. This is the blue graph, cosine of x, shifted to the left by pi over two. And this is y is equal to sine of x plus pi over two. Now the visual argument is, all I did is I shifted both of these graphs to the left by pi over two. So it should still be the case that the derivative of the red graph is the blue graph."}, {"video_title": "Proof of the derivative of cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "This is the blue graph, cosine of x, shifted to the left by pi over two. And this is y is equal to sine of x plus pi over two. Now the visual argument is, all I did is I shifted both of these graphs to the left by pi over two. So it should still be the case that the derivative of the red graph is the blue graph. So we should still be able to say that the derivative with respect to x of the red graph, sine of x plus pi over two, that that is equal to the blue graph. That that is equal to cosine of x plus pi over two. Now what is sine of x plus pi over two?"}, {"video_title": "Proof of the derivative of cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So it should still be the case that the derivative of the red graph is the blue graph. So we should still be able to say that the derivative with respect to x of the red graph, sine of x plus pi over two, that that is equal to the blue graph. That that is equal to cosine of x plus pi over two. Now what is sine of x plus pi over two? Well, that's the same thing as cosine of x. So as you can see, this red graph is the same thing as cosine of x. We know that from our trig identities, and you can also see it intuitively or graphically just by looking at these graphs."}, {"video_title": "Proof of the derivative of cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Now what is sine of x plus pi over two? Well, that's the same thing as cosine of x. So as you can see, this red graph is the same thing as cosine of x. We know that from our trig identities, and you can also see it intuitively or graphically just by looking at these graphs. And what is cosine of x plus pi over two? Well, once again, from our trig identities, we know that that is the exact same thing as negative sine of x. So there you have it, the visual argument."}, {"video_title": "Proof of the derivative of cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "We know that from our trig identities, and you can also see it intuitively or graphically just by looking at these graphs. And what is cosine of x plus pi over two? Well, once again, from our trig identities, we know that that is the exact same thing as negative sine of x. So there you have it, the visual argument. Just start with this knowledge, shift both of these graphs to the left by pi over two. It should still be true that the derivative with respect to x of sine of x plus pi over two is equal to cosine of x plus pi over two. And this is the same thing as saying what we have right over here."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "In this video, I want to familiarize you with the idea of a limit, which is a super important idea. It's really the idea that all of calculus is based upon. But despite being so super important, it's actually a really, really, really, really simple idea. So let me draw a function here. Actually, let me define a function here. A kind of a simple function. So let's define f of x."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "So let me draw a function here. Actually, let me define a function here. A kind of a simple function. So let's define f of x. Let's say that f of x is going to be x minus 1 over x minus 1. And you might say, hey, Sal, look, I have the same thing in the numerator and the denominator. If I have something divided by itself, that would just be equal to 1."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "So let's define f of x. Let's say that f of x is going to be x minus 1 over x minus 1. And you might say, hey, Sal, look, I have the same thing in the numerator and the denominator. If I have something divided by itself, that would just be equal to 1. Can't I just simplify this to f of x equals 1? And I would say, well, you're almost true. The difference between f of x equals 1 and this thing right over here is that this thing is undefined when x is equal to 1."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "If I have something divided by itself, that would just be equal to 1. Can't I just simplify this to f of x equals 1? And I would say, well, you're almost true. The difference between f of x equals 1 and this thing right over here is that this thing is undefined when x is equal to 1. Because if you set, let me define it right, let me write it over here. If you have f of, sorry, not f of 0, if you have f of 1, what happens? In the numerator, you get 1 minus 1, which is, let me just write it down."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "The difference between f of x equals 1 and this thing right over here is that this thing is undefined when x is equal to 1. Because if you set, let me define it right, let me write it over here. If you have f of, sorry, not f of 0, if you have f of 1, what happens? In the numerator, you get 1 minus 1, which is, let me just write it down. In the numerator, you get 0. And in the denominator, you get 1 minus 1, which is also 0. And so anything divided by 0, including 0 divided by 0, this is undefined."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "In the numerator, you get 1 minus 1, which is, let me just write it down. In the numerator, you get 0. And in the denominator, you get 1 minus 1, which is also 0. And so anything divided by 0, including 0 divided by 0, this is undefined. So you can make the simplification. You can say that this is the same thing as f of x is equal to 1, but you would have to add the constraint that x cannot be equal to 1. Now this and this are equivalent."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "And so anything divided by 0, including 0 divided by 0, this is undefined. So you can make the simplification. You can say that this is the same thing as f of x is equal to 1, but you would have to add the constraint that x cannot be equal to 1. Now this and this are equivalent. Both of these are going to be equal to 1 for all other x's other than 1, but at x equals 1, it becomes undefined. This is undefined and this one's undefined. So how would I graph this function?"}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "Now this and this are equivalent. Both of these are going to be equal to 1 for all other x's other than 1, but at x equals 1, it becomes undefined. This is undefined and this one's undefined. So how would I graph this function? So let me graph it. So that is my y is equal to f of x-axis. y is equal to f of x-axis."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "So how would I graph this function? So let me graph it. So that is my y is equal to f of x-axis. y is equal to f of x-axis. And then this over here is my x-axis. x-axis. And then let's say that this is the point x is equal to 1."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "y is equal to f of x-axis. And then this over here is my x-axis. x-axis. And then let's say that this is the point x is equal to 1. This over here would be x is equal to negative 1. This is y is equal to 1. Right up there, I could do negative 1, but that won't matter much relative to this function right over here."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "And then let's say that this is the point x is equal to 1. This over here would be x is equal to negative 1. This is y is equal to 1. Right up there, I could do negative 1, but that won't matter much relative to this function right over here. And let me graph it. So it's essentially, for any x other than 1, f of x is going to be equal to 1. So it's going to look like this."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "Right up there, I could do negative 1, but that won't matter much relative to this function right over here. And let me graph it. So it's essentially, for any x other than 1, f of x is going to be equal to 1. So it's going to look like this. It's going to look like this. Except at 1. At 1, f of x is undefined."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "So it's going to look like this. It's going to look like this. Except at 1. At 1, f of x is undefined. So I'm going to put a little bit of a gap right over here. This circle to signify that this function is not defined. We don't know what this function equals at 1."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "At 1, f of x is undefined. So I'm going to put a little bit of a gap right over here. This circle to signify that this function is not defined. We don't know what this function equals at 1. We never defined it. This definition of the function doesn't tell us what to do with 1. It's literally undefined."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "We don't know what this function equals at 1. We never defined it. This definition of the function doesn't tell us what to do with 1. It's literally undefined. Literally undefined when x is equal to 1. So this is the function right over here. And so, once again, if someone were to ask you, what is f of 1, you'd go, and let's say that even if this was the function definition, you'd go, okay, x is equal to 1."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "It's literally undefined. Literally undefined when x is equal to 1. So this is the function right over here. And so, once again, if someone were to ask you, what is f of 1, you'd go, and let's say that even if this was the function definition, you'd go, okay, x is equal to 1. Oh, wait, there's a gap in my function over here. It is undefined. So let me write it again."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "And so, once again, if someone were to ask you, what is f of 1, you'd go, and let's say that even if this was the function definition, you'd go, okay, x is equal to 1. Oh, wait, there's a gap in my function over here. It is undefined. So let me write it again. It's kind of redundant, but I'll rewrite it. f of 1 is undefined. But what if I were to ask you, what is the function approaching as x equals 1?"}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "So let me write it again. It's kind of redundant, but I'll rewrite it. f of 1 is undefined. But what if I were to ask you, what is the function approaching as x equals 1? Now, this is starting to touch on the idea of a limit. So as x gets closer and closer to 1, so as we get closer and closer x is to 1, as we get closer and closer x is to 1, what is the function approaching? Well, this entire time, the function, what's it getting closer and closer to?"}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "But what if I were to ask you, what is the function approaching as x equals 1? Now, this is starting to touch on the idea of a limit. So as x gets closer and closer to 1, so as we get closer and closer x is to 1, as we get closer and closer x is to 1, what is the function approaching? Well, this entire time, the function, what's it getting closer and closer to? On the left-hand side, no matter how close you get to 1, as long as you're not at 1, you're actually at f of x is equal to 1. Over here, from the right-hand side, you get the same thing. So you could say, and we'll get more and more familiar with this idea as we do more examples, that the limit as x, and lim, short for limit, as x approaches 1 of f of x is equal to, as we get closer, we can get unbelievably, we can get infinitely close to 1, as long as we're not at 1, and our function is going to be equal to 1."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "Well, this entire time, the function, what's it getting closer and closer to? On the left-hand side, no matter how close you get to 1, as long as you're not at 1, you're actually at f of x is equal to 1. Over here, from the right-hand side, you get the same thing. So you could say, and we'll get more and more familiar with this idea as we do more examples, that the limit as x, and lim, short for limit, as x approaches 1 of f of x is equal to, as we get closer, we can get unbelievably, we can get infinitely close to 1, as long as we're not at 1, and our function is going to be equal to 1. It's getting closer and closer to 1. It's actually at 1 the entire time. So in this case, we could say the limit as x approaches 1 of f of x is 1."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "So you could say, and we'll get more and more familiar with this idea as we do more examples, that the limit as x, and lim, short for limit, as x approaches 1 of f of x is equal to, as we get closer, we can get unbelievably, we can get infinitely close to 1, as long as we're not at 1, and our function is going to be equal to 1. It's getting closer and closer to 1. It's actually at 1 the entire time. So in this case, we could say the limit as x approaches 1 of f of x is 1. So once again, it has very fancy notation, but it's just saying, look, what is the function approaching as x gets closer and closer to 1? Let me do another example where we're dealing with a curve, just so that you have the general idea. So let's say that I have the function f of x, and let me just, just for the sake of variety, let me call it g of x."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "So in this case, we could say the limit as x approaches 1 of f of x is 1. So once again, it has very fancy notation, but it's just saying, look, what is the function approaching as x gets closer and closer to 1? Let me do another example where we're dealing with a curve, just so that you have the general idea. So let's say that I have the function f of x, and let me just, just for the sake of variety, let me call it g of x. Let's say that we have g of x is equal to, I could define it this way, we could define it as x squared when x does not equal, I don't know, when x does not equal 2, and let's say that when x equals 2, when x equals 2, it is equal to 1. So once again, a kind of an interesting function that, as you'll see, is not fully continuous. It has a discontinuity."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "So let's say that I have the function f of x, and let me just, just for the sake of variety, let me call it g of x. Let's say that we have g of x is equal to, I could define it this way, we could define it as x squared when x does not equal, I don't know, when x does not equal 2, and let's say that when x equals 2, when x equals 2, it is equal to 1. So once again, a kind of an interesting function that, as you'll see, is not fully continuous. It has a discontinuity. Let me graph it. So this is my y equals f of x axis. This is my x axis right over here."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "It has a discontinuity. Let me graph it. So this is my y equals f of x axis. This is my x axis right over here. Let me draw x equals 2. Let's say this is x equals 1. This is x equals 2."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "This is my x axis right over here. Let me draw x equals 2. Let's say this is x equals 1. This is x equals 2. This is negative 1. This is negative 2. And then let me draw, so everywhere except x equals 2, it's equal to x squared."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "This is x equals 2. This is negative 1. This is negative 2. And then let me draw, so everywhere except x equals 2, it's equal to x squared. So let me draw it like this. So it's going to be a parabola. It'll look something like this."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "And then let me draw, so everywhere except x equals 2, it's equal to x squared. So let me draw it like this. So it's going to be a parabola. It'll look something like this. It's going to look something, let me draw a better version of the parabola. So it'll look something like this. It'll look something like this."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "It'll look something like this. It's going to look something, let me draw a better version of the parabola. So it'll look something like this. It'll look something like this. Not the most beautifully drawn parabola in the history of drawing parabolas, but I think it'll give you the idea. I think you know what a parabola looks like, hopefully. It should be symmetric."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "It'll look something like this. Not the most beautifully drawn parabola in the history of drawing parabolas, but I think it'll give you the idea. I think you know what a parabola looks like, hopefully. It should be symmetric. Let me redraw it, because that's kind of ugly. Okay, so let me, that's looking better. Okay, all right, there you go."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "It should be symmetric. Let me redraw it, because that's kind of ugly. Okay, so let me, that's looking better. Okay, all right, there you go. All right, now, this would be the graph of just x squared, but it's not x squared when x is equal to 2. So once again, when x is equal to 2, we should have a little bit of a discontinuity here. So I'll draw a gap right over there, because when x equals 2, the function is equal to 1."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "Okay, all right, there you go. All right, now, this would be the graph of just x squared, but it's not x squared when x is equal to 2. So once again, when x is equal to 2, we should have a little bit of a discontinuity here. So I'll draw a gap right over there, because when x equals 2, the function is equal to 1. When x is equal to 2, so let's say that, and I'm not doing them on the same scale, but let's say that, so this, on the graph of f of x is equal to x squared, this would be 4, this would be 2, this would be 1, this would be 3. So when x is equal to 2, our function is equal to 1. So when x is equal to 2, our function is equal to 1."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "So I'll draw a gap right over there, because when x equals 2, the function is equal to 1. When x is equal to 2, so let's say that, and I'm not doing them on the same scale, but let's say that, so this, on the graph of f of x is equal to x squared, this would be 4, this would be 2, this would be 1, this would be 3. So when x is equal to 2, our function is equal to 1. So when x is equal to 2, our function is equal to 1. So this is a bit of a bizarre function, but we can define it this way. You can define a function however you like to define it. And so notice, it's just like the graph of x, of f of x is equal to x squared, except when you get to 2, it has this gap, because that's, you don't use the f of x is equal to x squared when x is equal to 2, you use f of x, or I should say g of x, you use g of x is equal to 1."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "So when x is equal to 2, our function is equal to 1. So this is a bit of a bizarre function, but we can define it this way. You can define a function however you like to define it. And so notice, it's just like the graph of x, of f of x is equal to x squared, except when you get to 2, it has this gap, because that's, you don't use the f of x is equal to x squared when x is equal to 2, you use f of x, or I should say g of x, you use g of x is equal to 1. I've been saying f of x, I apologize for that. You use g of x is equal to 1. So then at 2, just at 2, just exactly at 2, it drops down to 1, and then it keeps going along the function g of x is equal to, or I should say, along the function x squared."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "And so notice, it's just like the graph of x, of f of x is equal to x squared, except when you get to 2, it has this gap, because that's, you don't use the f of x is equal to x squared when x is equal to 2, you use f of x, or I should say g of x, you use g of x is equal to 1. I've been saying f of x, I apologize for that. You use g of x is equal to 1. So then at 2, just at 2, just exactly at 2, it drops down to 1, and then it keeps going along the function g of x is equal to, or I should say, along the function x squared. So my question to you, so there's a couple of things. If I were to just evaluate the function g of 2, well, you look at this definition. Okay, when x equals 2, I use this situation right over here, and it tells me it's going to be equal to 1."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "So then at 2, just at 2, just exactly at 2, it drops down to 1, and then it keeps going along the function g of x is equal to, or I should say, along the function x squared. So my question to you, so there's a couple of things. If I were to just evaluate the function g of 2, well, you look at this definition. Okay, when x equals 2, I use this situation right over here, and it tells me it's going to be equal to 1. Let me ask a more interesting question, or perhaps a more interesting question. What is the limit as x approaches 2 of g of x? Once again, fancy notation, but it's asking something pretty, pretty, pretty simple."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "Okay, when x equals 2, I use this situation right over here, and it tells me it's going to be equal to 1. Let me ask a more interesting question, or perhaps a more interesting question. What is the limit as x approaches 2 of g of x? Once again, fancy notation, but it's asking something pretty, pretty, pretty simple. It's saying as x gets closer and closer to 2, as you get closer and closer, and this isn't a rigorous definition, we'll do that in future videos. As x gets closer and closer to 2, what is g of x approaching? So if you get to 1.9 and then 1.999 and then 1.999999 and then 1.999999, what is g of x approaching?"}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "Once again, fancy notation, but it's asking something pretty, pretty, pretty simple. It's saying as x gets closer and closer to 2, as you get closer and closer, and this isn't a rigorous definition, we'll do that in future videos. As x gets closer and closer to 2, what is g of x approaching? So if you get to 1.9 and then 1.999 and then 1.999999 and then 1.999999, what is g of x approaching? Or if you were to go from the positive direction, if you were to say 2.1, what's g of 2.1? What's g of 2.01? What's g of 2.001?"}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "So if you get to 1.9 and then 1.999 and then 1.999999 and then 1.999999, what is g of x approaching? Or if you were to go from the positive direction, if you were to say 2.1, what's g of 2.1? What's g of 2.01? What's g of 2.001? What is that approaching as we get closer and closer to it? And you can see it visually just by drawing the graph. As g gets closer and closer to 2, and if we were to follow it along the graph, we see that we are approaching 4."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "What's g of 2.001? What is that approaching as we get closer and closer to it? And you can see it visually just by drawing the graph. As g gets closer and closer to 2, and if we were to follow it along the graph, we see that we are approaching 4. Even though that's not where the function is, the function drops down to 1, the limit of g of x as x approaches 2 is equal to 4. And you can even do this numerically using a calculator. And let me do that, because I think that will be interesting."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "As g gets closer and closer to 2, and if we were to follow it along the graph, we see that we are approaching 4. Even though that's not where the function is, the function drops down to 1, the limit of g of x as x approaches 2 is equal to 4. And you can even do this numerically using a calculator. And let me do that, because I think that will be interesting. So let me get a calculator out. Let me get my trusty TI-85 out. Let me..."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "And let me do that, because I think that will be interesting. So let me get a calculator out. Let me get my trusty TI-85 out. Let me... So here is my calculator. And you can numerically say, okay, what's it going to approach as you approach x equals 2? So let's try 1.9."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "Let me... So here is my calculator. And you can numerically say, okay, what's it going to approach as you approach x equals 2? So let's try 1.9. For x is equal to 1.9, you would use this top clause right over here. So you'd have 1.9 squared. And so you'd get 3.61."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "So let's try 1.9. For x is equal to 1.9, you would use this top clause right over here. So you'd have 1.9 squared. And so you'd get 3.61. Well, what if you get even closer to 2? So 1.99. And once again, let me square that."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "And so you'd get 3.61. Well, what if you get even closer to 2? So 1.99. And once again, let me square that. Well, now I'm at 3.96. Well, what if I do 1.999? And I square that."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "And once again, let me square that. Well, now I'm at 3.96. Well, what if I do 1.999? And I square that. And I square that. I'm going to get 3.996. Notice I'm getting closer and closer and closer to our point."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "And I square that. And I square that. I'm going to get 3.996. Notice I'm getting closer and closer and closer to our point. And if I got really close, 1.9999999999 squared, what am I going to get to? It's not actually going to be exactly 4. This calculator just rounded things up."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "Notice I'm getting closer and closer and closer to our point. And if I got really close, 1.9999999999 squared, what am I going to get to? It's not actually going to be exactly 4. This calculator just rounded things up. It's going to get to a number really, really, really, really, really, really, really close to 4. And we can do something from the positive direction, too. And it actually has to be the same number when we approach from the below what we're trying to approach and above what we're trying to approach."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "This calculator just rounded things up. It's going to get to a number really, really, really, really, really, really, really close to 4. And we can do something from the positive direction, too. And it actually has to be the same number when we approach from the below what we're trying to approach and above what we're trying to approach. So if we try 2.1 squared, we get 4.4. If we do 2.0... Let's go a couple of steps ahead. 2.01."}, {"video_title": "Introduction to limits Limits Differential Calculus Khan Academy.mp3", "Sentence": "And it actually has to be the same number when we approach from the below what we're trying to approach and above what we're trying to approach. So if we try 2.1 squared, we get 4.4. If we do 2.0... Let's go a couple of steps ahead. 2.01. So this is much closer to 2 now squared. Now we're getting much closer to 4. So the closer we get to 2, the closer it seems like we're getting to 4."}, {"video_title": "Justification with the intermediate value theorem table AP Calculus AB Khan Academy.mp3", "Sentence": "Can we use the intermediate value theorem to say that the equation f of x is equal to zero has a solution where four is less than or equal to x is less than or equal to six? If so, write a justification. So pause this video and see if you can think about this on your own before we do it together. Okay, well let's just visualize what's going on and visually think about the intermediate value theorem. So if that's my y-axis there, and then let's say that this is my x-axis right over here, we've been given some points over here. We know when x is equal to zero, f of x is equal to zero. Let me draw those."}, {"video_title": "Justification with the intermediate value theorem table AP Calculus AB Khan Academy.mp3", "Sentence": "Okay, well let's just visualize what's going on and visually think about the intermediate value theorem. So if that's my y-axis there, and then let's say that this is my x-axis right over here, we've been given some points over here. We know when x is equal to zero, f of x is equal to zero. Let me draw those. So we have that point. When x is equal to two, y, or f of x, y equals f of x is going to be equal to negative two. So we have a negative two right over there."}, {"video_title": "Justification with the intermediate value theorem table AP Calculus AB Khan Academy.mp3", "Sentence": "Let me draw those. So we have that point. When x is equal to two, y, or f of x, y equals f of x is going to be equal to negative two. So we have a negative two right over there. When x is equal to four, so three, four, f of x is equal to three, one, two, three. I'm doing them on a slightly different scale so that I can show everything. And when x is equal to six, so five, six, f of x is equal to seven, so three, four, five, six, seven."}, {"video_title": "Justification with the intermediate value theorem table AP Calculus AB Khan Academy.mp3", "Sentence": "So we have a negative two right over there. When x is equal to four, so three, four, f of x is equal to three, one, two, three. I'm doing them on a slightly different scale so that I can show everything. And when x is equal to six, so five, six, f of x is equal to seven, so three, four, five, six, seven. So right over here. Now they also tell us that our function is continuous. So one intuitive way of thinking about continuity is I can connect all of these dots without lifting my pencil."}, {"video_title": "Justification with the intermediate value theorem table AP Calculus AB Khan Academy.mp3", "Sentence": "And when x is equal to six, so five, six, f of x is equal to seven, so three, four, five, six, seven. So right over here. Now they also tell us that our function is continuous. So one intuitive way of thinking about continuity is I can connect all of these dots without lifting my pencil. So the function might look, I'm just gonna make up some stuff. It might look something, anything, like what I just drew just now. And it could have even wilder fluctuations."}, {"video_title": "Justification with the intermediate value theorem table AP Calculus AB Khan Academy.mp3", "Sentence": "So one intuitive way of thinking about continuity is I can connect all of these dots without lifting my pencil. So the function might look, I'm just gonna make up some stuff. It might look something, anything, like what I just drew just now. And it could have even wilder fluctuations. But that is what my f looks like. Now the intermediate value theorem says, hey, pick a closed interval. And here we're picking the closed interval from four to six."}, {"video_title": "Justification with the intermediate value theorem table AP Calculus AB Khan Academy.mp3", "Sentence": "And it could have even wilder fluctuations. But that is what my f looks like. Now the intermediate value theorem says, hey, pick a closed interval. And here we're picking the closed interval from four to six. So let me look at that. So this is one, two, three, four here. This is six here."}, {"video_title": "Justification with the intermediate value theorem table AP Calculus AB Khan Academy.mp3", "Sentence": "And here we're picking the closed interval from four to six. So let me look at that. So this is one, two, three, four here. This is six here. So we're gonna look at this closed interval. And the intermediate value theorem tells us that look, if we're continuous over that closed interval, our function f is going to take on every value between f of four, f of four, which in this case, so this is f of four is equal to three, and f of six, and f of six, which is equal to seven. F of six, which is equal to seven."}, {"video_title": "Justification with the intermediate value theorem table AP Calculus AB Khan Academy.mp3", "Sentence": "This is six here. So we're gonna look at this closed interval. And the intermediate value theorem tells us that look, if we're continuous over that closed interval, our function f is going to take on every value between f of four, f of four, which in this case, so this is f of four is equal to three, and f of six, and f of six, which is equal to seven. F of six, which is equal to seven. And so if someone said, hey, is there going to be a solution to f of x is equal to, say, five over this interval? Yes, over this interval for some x, you're going to have f of x being equal to five. But they're not asking us for an f of x equaling something between these two values."}, {"video_title": "Justification with the intermediate value theorem table AP Calculus AB Khan Academy.mp3", "Sentence": "F of six, which is equal to seven. And so if someone said, hey, is there going to be a solution to f of x is equal to, say, five over this interval? Yes, over this interval for some x, you're going to have f of x being equal to five. But they're not asking us for an f of x equaling something between these two values. They're asking us for an f of x equaling zero. Zero isn't between f of four and f of six. And so we cannot use the intermediate value theorem here."}, {"video_title": "Justification with the intermediate value theorem table AP Calculus AB Khan Academy.mp3", "Sentence": "But they're not asking us for an f of x equaling something between these two values. They're asking us for an f of x equaling zero. Zero isn't between f of four and f of six. And so we cannot use the intermediate value theorem here. And so if we wanted to write it out, we could say f is continuous, is continuous, but, but zero is not between, between f of four and f of six. So the intermediate value theorem does not apply. All right, let's do the second one."}, {"video_title": "Justification with the intermediate value theorem table AP Calculus AB Khan Academy.mp3", "Sentence": "And so we cannot use the intermediate value theorem here. And so if we wanted to write it out, we could say f is continuous, is continuous, but, but zero is not between, between f of four and f of six. So the intermediate value theorem does not apply. All right, let's do the second one. So here they say, can we use the intermediate value theorem to say that there is a value c such that f of c equals zero and two is less than or equal to c is less than or equal to four? If so, write a justification. So we are given that f is continuous."}, {"video_title": "Justification with the intermediate value theorem table AP Calculus AB Khan Academy.mp3", "Sentence": "All right, let's do the second one. So here they say, can we use the intermediate value theorem to say that there is a value c such that f of c equals zero and two is less than or equal to c is less than or equal to four? If so, write a justification. So we are given that f is continuous. So let me write that down. We are given that f is continuous. And if you want to be over that interval, but they're telling us it's continuous in general."}, {"video_title": "Justification with the intermediate value theorem table AP Calculus AB Khan Academy.mp3", "Sentence": "So we are given that f is continuous. So let me write that down. We are given that f is continuous. And if you want to be over that interval, but they're telling us it's continuous in general. And then we can just look at what is the value of the function at these endpoints. So our interval goes from two to four. So we're talking about this closed interval right over here."}, {"video_title": "Justification with the intermediate value theorem table AP Calculus AB Khan Academy.mp3", "Sentence": "And if you want to be over that interval, but they're telling us it's continuous in general. And then we can just look at what is the value of the function at these endpoints. So our interval goes from two to four. So we're talking about this closed interval right over here. We know that f of two, f of two is going to be equal to negative two, we see it in that table. And what's f of four? F of four is equal to three."}, {"video_title": "Justification with the intermediate value theorem table AP Calculus AB Khan Academy.mp3", "Sentence": "So we're talking about this closed interval right over here. We know that f of two, f of two is going to be equal to negative two, we see it in that table. And what's f of four? F of four is equal to three. F of four is equal to three. So zero is between f of two and f of four. And you can see it visually here."}, {"video_title": "Justification with the intermediate value theorem table AP Calculus AB Khan Academy.mp3", "Sentence": "F of four is equal to three. F of four is equal to three. So zero is between f of two and f of four. And you can see it visually here. There's no way to draw between this point and that point without picking up your pen, without crossing the x-axis, without having a point where your function is equal to zero. And so we can say, so according to the intermediate value theorem, there is a value c such that f of c is equal to zero and two is less than or equal to c is less than or equal to four. So all we're saying is, hey, there must be a value c, and the way I drew it here, that c value is right over here where c is between two and four, where f of c is equal to zero."}, {"video_title": "Worked example separable equation with an implicit solution Khan Academy.mp3", "Sentence": "We're given a differential equation right over here, cosine of y plus two, this whole thing, times the derivative of y with respect to x, is equal to two x, and we're given that for a particular solution, when x is equal to one, y of one is equal to zero. And we're asked, what is x when y is equal to pi? So the first thing I like to look at when I see a differential equation is, is it separable? Can I get all the y's and dy's on one side, and can I get all the x's and dx's on the other side? And this one seems like it is. If I multiply both sides by dx, where you can view dx as the x differential of an infinitely small change in x, well then you get cosine of y plus two times dy is equal to two x times dx. So just like that, I've been able to, all I did is I multiplied both sides of this times dx."}, {"video_title": "Worked example separable equation with an implicit solution Khan Academy.mp3", "Sentence": "Can I get all the y's and dy's on one side, and can I get all the x's and dx's on the other side? And this one seems like it is. If I multiply both sides by dx, where you can view dx as the x differential of an infinitely small change in x, well then you get cosine of y plus two times dy is equal to two x times dx. So just like that, I've been able to, all I did is I multiplied both sides of this times dx. And I was able to separate the y's and the dy's from the x's and the dx's. And now I can integrate both sides. So if I integrate both sides, what am I going to get?"}, {"video_title": "Worked example separable equation with an implicit solution Khan Academy.mp3", "Sentence": "So just like that, I've been able to, all I did is I multiplied both sides of this times dx. And I was able to separate the y's and the dy's from the x's and the dx's. And now I can integrate both sides. So if I integrate both sides, what am I going to get? So the antiderivative of cosine of y with respect to y, with respect to y is sine of y. And then the antiderivative of two with respect to y is two y. And that is going to be equal to, well the antiderivative of two x with respect to x is x squared."}, {"video_title": "Worked example separable equation with an implicit solution Khan Academy.mp3", "Sentence": "So if I integrate both sides, what am I going to get? So the antiderivative of cosine of y with respect to y, with respect to y is sine of y. And then the antiderivative of two with respect to y is two y. And that is going to be equal to, well the antiderivative of two x with respect to x is x squared. And we can't forget that we could say a different constant on either side, but it serves our purpose just to say plus c on one side. And so this is a general solution to the separable differential equation. And then we can find the particular one by substituting in, when x is equal to one, y is equal to zero."}, {"video_title": "Worked example separable equation with an implicit solution Khan Academy.mp3", "Sentence": "And that is going to be equal to, well the antiderivative of two x with respect to x is x squared. And we can't forget that we could say a different constant on either side, but it serves our purpose just to say plus c on one side. And so this is a general solution to the separable differential equation. And then we can find the particular one by substituting in, when x is equal to one, y is equal to zero. So let's do that to solve for c. So we get, or when y is equal to zero, x is equal to one. So sine of zero plus two times zero, all I did is I substituted in the zero for y, is equal to x squared, well now x is one, is equal to one squared plus c. Well sine of zero zero, two times zero zero, all of that's just going to be zero. So we get zero is equal to one plus c, or c is equal to negative one."}, {"video_title": "Worked example separable equation with an implicit solution Khan Academy.mp3", "Sentence": "And then we can find the particular one by substituting in, when x is equal to one, y is equal to zero. So let's do that to solve for c. So we get, or when y is equal to zero, x is equal to one. So sine of zero plus two times zero, all I did is I substituted in the zero for y, is equal to x squared, well now x is one, is equal to one squared plus c. Well sine of zero zero, two times zero zero, all of that's just going to be zero. So we get zero is equal to one plus c, or c is equal to negative one. So now we can write down the particular solution to this differential equation that meets these conditions. So we get, let me write it over here. Sine of y plus two y is equal to x squared, and our constant is negative one, so minus one."}, {"video_title": "Worked example separable equation with an implicit solution Khan Academy.mp3", "Sentence": "So we get zero is equal to one plus c, or c is equal to negative one. So now we can write down the particular solution to this differential equation that meets these conditions. So we get, let me write it over here. Sine of y plus two y is equal to x squared, and our constant is negative one, so minus one. And now what is x when y is equal to pi? So sine of pi plus two times pi is equal to x squared minus one. See sine of pi is equal to zero, and so we get, see we can add one to both sides, and we get two pi plus one is equal to x squared, or we could say that x is equal to the plus or minus square root of two pi plus one."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084 AP Calculus AB Khan Academy.mp3", "Sentence": "Let g of x be equal to the definite integral from zero to x of f of t dt. What is an appropriate calculus-based justification for the fact that g is concave up on the interval, the open interval from five to 10? So concave up. So before I even think about what it means to be concave up, let's just make sure we understand this relationship between g and f. One way to understand it is if we took the derivative of both sides of this equation, we would get that g prime of x is equal to f of x. The derivative of this with respect to x would just be f of x. In fact, the whole reason why we introduce this variable t here is this thing right over here is actually a function of x, because x is this upper bound. And it would've been weird if we had x as an upper bound, or at least confusing, and we're also integrating with respect to x."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084 AP Calculus AB Khan Academy.mp3", "Sentence": "So before I even think about what it means to be concave up, let's just make sure we understand this relationship between g and f. One way to understand it is if we took the derivative of both sides of this equation, we would get that g prime of x is equal to f of x. The derivative of this with respect to x would just be f of x. In fact, the whole reason why we introduce this variable t here is this thing right over here is actually a function of x, because x is this upper bound. And it would've been weird if we had x as an upper bound, or at least confusing, and we're also integrating with respect to x. So we just have to pick kind of another placeholder variable that doesn't have to be t. It could be alpha, it could be gamma, it could be a, b, or c, whatever we choose. But this is still right over here. This is a function of x."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084 AP Calculus AB Khan Academy.mp3", "Sentence": "And it would've been weird if we had x as an upper bound, or at least confusing, and we're also integrating with respect to x. So we just have to pick kind of another placeholder variable that doesn't have to be t. It could be alpha, it could be gamma, it could be a, b, or c, whatever we choose. But this is still right over here. This is a function of x. But when you take the derivative of both sides, you realize that the function f, which is graphed here, and if this were the x-axis, then this would be f of x. If this is the t-axis, then this is y is equal to f of t. But generally, this is the graph of our function f, which you could also view as the graph of g prime. If this is x, this would be g prime of x."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084 AP Calculus AB Khan Academy.mp3", "Sentence": "This is a function of x. But when you take the derivative of both sides, you realize that the function f, which is graphed here, and if this were the x-axis, then this would be f of x. If this is the t-axis, then this is y is equal to f of t. But generally, this is the graph of our function f, which you could also view as the graph of g prime. If this is x, this would be g prime of x. And so we're thinking about the interval from, the open interval from five to 10, and we have g's derivative graphed here, and we wanna know a calculus-based justification from this graph that lets us know that g is concave up. So what does it mean to be concave up? Well, that means that your slope of tangent line, of tangent, tangent, slope of tangent is increasing, or another way of thinking about it, your derivative, derivative is increasing."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084 AP Calculus AB Khan Academy.mp3", "Sentence": "If this is x, this would be g prime of x. And so we're thinking about the interval from, the open interval from five to 10, and we have g's derivative graphed here, and we wanna know a calculus-based justification from this graph that lets us know that g is concave up. So what does it mean to be concave up? Well, that means that your slope of tangent line, of tangent, tangent, slope of tangent is increasing, or another way of thinking about it, your derivative, derivative is increasing. Or another way to think about it, if your derivative is increasing over an interval, then you're concave up on that interval. And so we here, we have a graph of the derivative, and it is indeed increasing over that interval. So our calculus-based justification that we'd wanna use is that, look, f, which is g prime, is increasing on that interval."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084 AP Calculus AB Khan Academy.mp3", "Sentence": "Well, that means that your slope of tangent line, of tangent, tangent, slope of tangent is increasing, or another way of thinking about it, your derivative, derivative is increasing. Or another way to think about it, if your derivative is increasing over an interval, then you're concave up on that interval. And so we here, we have a graph of the derivative, and it is indeed increasing over that interval. So our calculus-based justification that we'd wanna use is that, look, f, which is g prime, is increasing on that interval. The derivative is increasing on that interval, which means that the original function is concave up. F is positive on that interval, that's not a sufficient calculus-based justification. Because if your derivative is positive, that just means your original function is increasing."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084 AP Calculus AB Khan Academy.mp3", "Sentence": "So our calculus-based justification that we'd wanna use is that, look, f, which is g prime, is increasing on that interval. The derivative is increasing on that interval, which means that the original function is concave up. F is positive on that interval, that's not a sufficient calculus-based justification. Because if your derivative is positive, that just means your original function is increasing. It doesn't tell you that your original function was concave up. F is concave up on the interval. Well, just because your derivative is concave up doesn't mean that your original function is concave up."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084 AP Calculus AB Khan Academy.mp3", "Sentence": "Because if your derivative is positive, that just means your original function is increasing. It doesn't tell you that your original function was concave up. F is concave up on the interval. Well, just because your derivative is concave up doesn't mean that your original function is concave up. In fact, you could have a situation like this where you're concave up over that interval, but for much of that interval right over here, if this was our graph of f or g prime, we are decreasing, and if we're decreasing over much of that interval, then actually on this part, our original function would be concave down. The graph of g has a cup u shape on the interval. Well, if we had the graph of g, this would be a justification, but it wouldn't be a calculus-based justification."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084 AP Calculus AB Khan Academy.mp3", "Sentence": "Well, just because your derivative is concave up doesn't mean that your original function is concave up. In fact, you could have a situation like this where you're concave up over that interval, but for much of that interval right over here, if this was our graph of f or g prime, we are decreasing, and if we're decreasing over much of that interval, then actually on this part, our original function would be concave down. The graph of g has a cup u shape on the interval. Well, if we had the graph of g, this would be a justification, but it wouldn't be a calculus-based justification. Let's do more of these. So this next one says, so we have the exact same setup, which actually all of these examples will have, g of x is equal to this thing here. What is an appropriate calculus-based justification for the fact that g has a relative minimum at x equals eight?"}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084 AP Calculus AB Khan Academy.mp3", "Sentence": "Well, if we had the graph of g, this would be a justification, but it wouldn't be a calculus-based justification. Let's do more of these. So this next one says, so we have the exact same setup, which actually all of these examples will have, g of x is equal to this thing here. What is an appropriate calculus-based justification for the fact that g has a relative minimum at x equals eight? So once again, they've graphed f here, which is the same thing as the derivative of g, and so if we have the graph of the derivative, how can we say, how do we know that we have a relative minimum at x equals eight? Well, if the fact that we cross, that we're at the x-axis, that x equals, that y is equal to zero, that the function, that the derivative is equal to zero at x equals eight, that tells us that the slope of the tangent line of g at that point is zero, but that alone does not tell us we have a relative minimum point. In order to have a relative minimum point, our derivative has to cross from being negative to positive."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084 AP Calculus AB Khan Academy.mp3", "Sentence": "What is an appropriate calculus-based justification for the fact that g has a relative minimum at x equals eight? So once again, they've graphed f here, which is the same thing as the derivative of g, and so if we have the graph of the derivative, how can we say, how do we know that we have a relative minimum at x equals eight? Well, if the fact that we cross, that we're at the x-axis, that x equals, that y is equal to zero, that the function, that the derivative is equal to zero at x equals eight, that tells us that the slope of the tangent line of g at that point is zero, but that alone does not tell us we have a relative minimum point. In order to have a relative minimum point, our derivative has to cross from being negative to positive. Why is that valuable? Because think about, if your derivative goes from being negative to positive, that means your original function goes from decreasing to increasing, goes from decreasing to increasing, and so you would have a relative minimum point. And the choice that describes that, this is starting to get there, but this alone isn't enough for a relative minimum point."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084 AP Calculus AB Khan Academy.mp3", "Sentence": "In order to have a relative minimum point, our derivative has to cross from being negative to positive. Why is that valuable? Because think about, if your derivative goes from being negative to positive, that means your original function goes from decreasing to increasing, goes from decreasing to increasing, and so you would have a relative minimum point. And the choice that describes that, this is starting to get there, but this alone isn't enough for a relative minimum point. F is negative before x equals eight and positive after x equals eight. That's exactly what we just described. Let's see about these."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084 AP Calculus AB Khan Academy.mp3", "Sentence": "And the choice that describes that, this is starting to get there, but this alone isn't enough for a relative minimum point. F is negative before x equals eight and positive after x equals eight. That's exactly what we just described. Let's see about these. F is concave up on the interval around x equals six. Well, x equals six is a little bit unrelated to that. There's an interval in the graph of g around x equals eight where g of eight is the smallest value."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084 AP Calculus AB Khan Academy.mp3", "Sentence": "Let's see about these. F is concave up on the interval around x equals six. Well, x equals six is a little bit unrelated to that. There's an interval in the graph of g around x equals eight where g of eight is the smallest value. Well, this would be a justification for a relative minimum, but it is not calculus-based. So let's get, I'll rule that one out as well. Let's do one more of these."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084 AP Calculus AB Khan Academy.mp3", "Sentence": "There's an interval in the graph of g around x equals eight where g of eight is the smallest value. Well, this would be a justification for a relative minimum, but it is not calculus-based. So let's get, I'll rule that one out as well. Let's do one more of these. So same setup, although we have a different f and g here, and we see it every time with the graph. What is an appropriate calculus-based justification for the fact that g is positive on the interval from the closed interval from seven to 12? So the positive on the closed interval from seven to 12."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084 AP Calculus AB Khan Academy.mp3", "Sentence": "Let's do one more of these. So same setup, although we have a different f and g here, and we see it every time with the graph. What is an appropriate calculus-based justification for the fact that g is positive on the interval from the closed interval from seven to 12? So the positive on the closed interval from seven to 12. So this is interesting. Let's just remind ourselves. Here we're gonna think a little bit deeper about what it means to be this definite integral from zero to x."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084 AP Calculus AB Khan Academy.mp3", "Sentence": "So the positive on the closed interval from seven to 12. So this is interesting. Let's just remind ourselves. Here we're gonna think a little bit deeper about what it means to be this definite integral from zero to x. So if we think about what happens when x is equal to seven, x is equal to seven, when x is equal to seven, or another way of thinking about it, g of seven is going to be the integral from zero to seven of f of t dt. And so the integral from zero to seven, if this was a t-axis, and once again, t is just kind of a placeholder variable to help us keep this x up here, but we're really talking about this area right over here. And because from zero to seven, this function is above the x-axis, this is going to be a positive area."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084 AP Calculus AB Khan Academy.mp3", "Sentence": "Here we're gonna think a little bit deeper about what it means to be this definite integral from zero to x. So if we think about what happens when x is equal to seven, x is equal to seven, when x is equal to seven, or another way of thinking about it, g of seven is going to be the integral from zero to seven of f of t dt. And so the integral from zero to seven, if this was a t-axis, and once again, t is just kind of a placeholder variable to help us keep this x up here, but we're really talking about this area right over here. And because from zero to seven, this function is above the x-axis, this is going to be a positive area. This is a positive area. And as we go from seven to 12, we're not adding any more area, but we're also not taking any away. So actually, g of seven all the way to g of 12 is going to be the same positive value because we're not adding any more value."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084 AP Calculus AB Khan Academy.mp3", "Sentence": "And because from zero to seven, this function is above the x-axis, this is going to be a positive area. This is a positive area. And as we go from seven to 12, we're not adding any more area, but we're also not taking any away. So actually, g of seven all the way to g of 12 is going to be the same positive value because we're not adding any more value. When I say g of 12, g of 12 is going to be actually equal to g of seven because, once again, no added area right here, positive or negative. So let's see which of these choices match. For any x value in the interval from seven to 12, the value of f of x is zero."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084 AP Calculus AB Khan Academy.mp3", "Sentence": "So actually, g of seven all the way to g of 12 is going to be the same positive value because we're not adding any more value. When I say g of 12, g of 12 is going to be actually equal to g of seven because, once again, no added area right here, positive or negative. So let's see which of these choices match. For any x value in the interval from seven to 12, the value of f of x is zero. That is true, but that doesn't mean that we were positive. For example, before that interval, if our function did something like this, then we would have had negative area up to that point, and so these would be negative values. So I would rule that out."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084 AP Calculus AB Khan Academy.mp3", "Sentence": "For any x value in the interval from seven to 12, the value of f of x is zero. That is true, but that doesn't mean that we were positive. For example, before that interval, if our function did something like this, then we would have had negative area up to that point, and so these would be negative values. So I would rule that out. For any x value in the interval from seven to 12, the closed interval, the value of g of x is positive. For any x value in the interval from seven to 12, the value of g of x is positive. That is true, so I like this one."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084 AP Calculus AB Khan Academy.mp3", "Sentence": "So I would rule that out. For any x value in the interval from seven to 12, the closed interval, the value of g of x is positive. For any x value in the interval from seven to 12, the value of g of x is positive. That is true, so I like this one. Let me see these other ones. F is positive over the closed interval from zero to seven, and it is non-negative over seven to 12. I like this one as well."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084 AP Calculus AB Khan Academy.mp3", "Sentence": "That is true, so I like this one. Let me see these other ones. F is positive over the closed interval from zero to seven, and it is non-negative over seven to 12. I like this one as well. And actually, the reason why I would rule out this first one, this first one has nothing to do with the derivative, and so it's not a calculus-based justification, so I would rule that one out. This one is good. This is the exact rationale that I was talking about."}, {"video_title": "Interpreting behavior of _ from graph of _'=\u00c3\u0083\u00c2\u0084 AP Calculus AB Khan Academy.mp3", "Sentence": "I like this one as well. And actually, the reason why I would rule out this first one, this first one has nothing to do with the derivative, and so it's not a calculus-based justification, so I would rule that one out. This one is good. This is the exact rationale that I was talking about. F is positive from zero to seven, so it develops all this positive area, and it's non-negative over the interval, and so we are going to stay positive this entire time for g, which is the area under f and above the x-axis from zero to our whatever x we wanna pick. So I like this choice here. F is neither concave up nor concave down over the interval from, the closed interval from seven to 12."}, {"video_title": "Worked example Logistic model word problem Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "The population P of T of bacteria in a Petri dish satisfies the logistic differential equation. The rate of change of population with respect to time is equal to two times the population times the difference between six and the population divided by 8,000, where T is measured in hours and the initial population is 700 bacteria. What is the carrying capacity of the population and what is the population's size when it's growing the fastest? All right, so in order to even attempt to answer these questions, and at any point, if you're inspired, definitely pause the video and try to answer it by yourself, let's just remind ourselves what we're talking about or what they're talking about with the logistic differential equation and the carrying capacity. So in general, a logistic differential equation is one where we're saying the rate of change of, and it's often referring to population, so let's just stick with population. So the rate of change of our population with respect to time is proportional to the product of the population and the difference between what's known as the carrying capacity and the population. Now, why is this a model that you will see a lot, and especially why is it useful for studying things like populations?"}, {"video_title": "Worked example Logistic model word problem Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "All right, so in order to even attempt to answer these questions, and at any point, if you're inspired, definitely pause the video and try to answer it by yourself, let's just remind ourselves what we're talking about or what they're talking about with the logistic differential equation and the carrying capacity. So in general, a logistic differential equation is one where we're saying the rate of change of, and it's often referring to population, so let's just stick with population. So the rate of change of our population with respect to time is proportional to the product of the population and the difference between what's known as the carrying capacity and the population. Now, why is this a model that you will see a lot, and especially why is it useful for studying things like populations? Well, when a population is small, the environment really isn't limiting it, and so assuming it starts from some non-zero value, this thing is, this thing grows, this thing is not going to get much smaller, and so our population is going to, our rate of change is going to increase, and so let me just draw a little graph here to show the typical solution to a logistic differential equation. So this is our population, this is time. So when our population is low, let's say it's gonna start from some non-zero value, if it was zero, what would happen?"}, {"video_title": "Worked example Logistic model word problem Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "Now, why is this a model that you will see a lot, and especially why is it useful for studying things like populations? Well, when a population is small, the environment really isn't limiting it, and so assuming it starts from some non-zero value, this thing is, this thing grows, this thing is not going to get much smaller, and so our population is going to, our rate of change is going to increase, and so let me just draw a little graph here to show the typical solution to a logistic differential equation. So this is our population, this is time. So when our population is low, let's say it's gonna start from some non-zero value, if it was zero, what would happen? Well, then our rate of change would just be zero, and our population would never grow, and that makes sense. If you have no bunnies on your island, then there never will be any bunnies on your island, but if you have a few bunnies, well, initially, their rate of change, the rate of population is gonna keep accelerating as this thing grows, it's gonna keep accelerating, but then at some point, your environment is going to limit how many bunnies, for example, or bacteria can grow in your environment, because once the population gets close to A, this thing right over here is going to approach zero, and it's going to make our rate of change smaller and smaller and smaller, and so you can imagine, in the limiting case, as P gets very, very, very, very, very close to A, as P gets very, very, very close to A, you can imagine as T approaches infinity, our rate of change is going to approach zero. So one way to think about it is our population would asymptote towards the carrying capacity."}, {"video_title": "Worked example Logistic model word problem Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So when our population is low, let's say it's gonna start from some non-zero value, if it was zero, what would happen? Well, then our rate of change would just be zero, and our population would never grow, and that makes sense. If you have no bunnies on your island, then there never will be any bunnies on your island, but if you have a few bunnies, well, initially, their rate of change, the rate of population is gonna keep accelerating as this thing grows, it's gonna keep accelerating, but then at some point, your environment is going to limit how many bunnies, for example, or bacteria can grow in your environment, because once the population gets close to A, this thing right over here is going to approach zero, and it's going to make our rate of change smaller and smaller and smaller, and so you can imagine, in the limiting case, as P gets very, very, very, very, very close to A, as P gets very, very, very close to A, you can imagine as T approaches infinity, our rate of change is going to approach zero. So one way to think about it is our population would asymptote towards the carrying capacity. That is A right over here, that is our carrying capacity. So there's a couple of ways of answering this first question. One way is we can actually put our logistic differential equation in this form, and then we can recognize what the carrying capacity is."}, {"video_title": "Worked example Logistic model word problem Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So one way to think about it is our population would asymptote towards the carrying capacity. That is A right over here, that is our carrying capacity. So there's a couple of ways of answering this first question. One way is we can actually put our logistic differential equation in this form, and then we can recognize what the carrying capacity is. The other way is to think about, well, what happens as T approaches infinity? As T approaches infinity, this thing approaches zero, and so we can think from this logistic differential equation what P values would make this thing be zero based on this differential equation, or when this thing approaches zero, what P values would this approach? So let's do it both ways."}, {"video_title": "Worked example Logistic model word problem Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "One way is we can actually put our logistic differential equation in this form, and then we can recognize what the carrying capacity is. The other way is to think about, well, what happens as T approaches infinity? As T approaches infinity, this thing approaches zero, and so we can think from this logistic differential equation what P values would make this thing be zero based on this differential equation, or when this thing approaches zero, what P values would this approach? So let's do it both ways. So one way, let me write it in this form right over here. So it's close, this is six minus P over 8,000. Can we make it in the form where we just have some number minus P?"}, {"video_title": "Worked example Logistic model word problem Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So let's do it both ways. So one way, let me write it in this form right over here. So it's close, this is six minus P over 8,000. Can we make it in the form where we just have some number minus P? Well, if we multiplied this times 8,000, and then we divided this by 8,000, we wouldn't be changing the value of that expression. So let's do that. If we divide this by 8,000, you have dP dt is equal to two P over 8,000 is P over 4,000 times, and now let's multiply this expression by 8,000."}, {"video_title": "Worked example Logistic model word problem Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "Can we make it in the form where we just have some number minus P? Well, if we multiplied this times 8,000, and then we divided this by 8,000, we wouldn't be changing the value of that expression. So let's do that. If we divide this by 8,000, you have dP dt is equal to two P over 8,000 is P over 4,000 times, and now let's multiply this expression by 8,000. So six times 8,000 is 48,000, minus P over 8,000 times 8,000 is minus P. And there you have it. We've written it now in this somewhat classic form, and we can see that the carrying capacity is 48,000, I guess we say 48,000 bacteria. So this thing right over here is 48,000."}, {"video_title": "Worked example Logistic model word problem Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "If we divide this by 8,000, you have dP dt is equal to two P over 8,000 is P over 4,000 times, and now let's multiply this expression by 8,000. So six times 8,000 is 48,000, minus P over 8,000 times 8,000 is minus P. And there you have it. We've written it now in this somewhat classic form, and we can see that the carrying capacity is 48,000, I guess we say 48,000 bacteria. So this thing right over here is 48,000. Another way we could think about it is, well, the carrying capacity is what happens as T approaches infinity. So as T approaches infinity, this thing right over here approaches zero. You can see the rate of change approaches zero."}, {"video_title": "Worked example Logistic model word problem Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So this thing right over here is 48,000. Another way we could think about it is, well, the carrying capacity is what happens as T approaches infinity. So as T approaches infinity, this thing right over here approaches zero. You can see the rate of change approaches zero. So when that approaches zero, what does P approach? So we can just solve for this, six minus P over 8,000. Well, there's two situations, two P's, for which our derivative is equal to zero."}, {"video_title": "Worked example Logistic model word problem Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "You can see the rate of change approaches zero. So when that approaches zero, what does P approach? So we can just solve for this, six minus P over 8,000. Well, there's two situations, two P's, for which our derivative is equal to zero. There's one case when this is equal to zero, in which case our population is zero, or the other case is when this is equal to zero. So let's do that. Six minus P over 8,000 is equal to zero, or we could say P over 8,000 is equal to six, and so we can say multiply both sides by 8,000."}, {"video_title": "Worked example Logistic model word problem Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "Well, there's two situations, two P's, for which our derivative is equal to zero. There's one case when this is equal to zero, in which case our population is zero, or the other case is when this is equal to zero. So let's do that. Six minus P over 8,000 is equal to zero, or we could say P over 8,000 is equal to six, and so we can say multiply both sides by 8,000. P is equal to 48,000, which is exactly what we had right over there. So now let's answer the second part. What is the population's size when it's growing the fastest?"}, {"video_title": "Worked example Logistic model word problem Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "Six minus P over 8,000 is equal to zero, or we could say P over 8,000 is equal to six, and so we can say multiply both sides by 8,000. P is equal to 48,000, which is exactly what we had right over there. So now let's answer the second part. What is the population's size when it's growing the fastest? So intuitively, you can see when that is right over here. The population, the rate of change is gonna grow, grow, grow, grow, grow, but then as we approach the carrying capacity, the rate of change is gonna start slowing down. So your maximum rate of change, it's growing the fastest, is right about, right about there, but how do we figure out it exactly?"}, {"video_title": "Worked example Logistic model word problem Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "What is the population's size when it's growing the fastest? So intuitively, you can see when that is right over here. The population, the rate of change is gonna grow, grow, grow, grow, grow, but then as we approach the carrying capacity, the rate of change is gonna start slowing down. So your maximum rate of change, it's growing the fastest, is right about, right about there, but how do we figure out it exactly? Well, you could go back to the logistic differential equation. You can see that it's really, our rate of change is a function, you could view it as a function of our population right over here, and this is actually a quadratic expression. This is a concave downward quadratic expression."}, {"video_title": "Worked example Logistic model word problem Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So your maximum rate of change, it's growing the fastest, is right about, right about there, but how do we figure out it exactly? Well, you could go back to the logistic differential equation. You can see that it's really, our rate of change is a function, you could view it as a function of our population right over here, and this is actually a quadratic expression. This is a concave downward quadratic expression. It would look like this. If you were graphing, if you were graphing rate of change, let me do that. If you were to graph rate of change, dP, dT, as a function of population, well, when the population is small, so when the population is, say, 700, or that's where we're starting, well, let's just, I'll speak in generalities."}, {"video_title": "Worked example Logistic model word problem Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "This is a concave downward quadratic expression. It would look like this. If you were graphing, if you were graphing rate of change, let me do that. If you were to graph rate of change, dP, dT, as a function of population, well, when the population is small, so when the population is, say, 700, or that's where we're starting, well, let's just, I'll speak in generalities. When your population is small, dP, dT, your rate is small, but then it increases, and then some point around there, it will start, the rate of change starts to decrease, and it approaches zero as our population is approaching the carrying capacity. So this right over here, for example, would actually be our carrying capacity. So one way to think about, well, what is this maximum point right over here?"}, {"video_title": "Worked example Logistic model word problem Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "If you were to graph rate of change, dP, dT, as a function of population, well, when the population is small, so when the population is, say, 700, or that's where we're starting, well, let's just, I'll speak in generalities. When your population is small, dP, dT, your rate is small, but then it increases, and then some point around there, it will start, the rate of change starts to decrease, and it approaches zero as our population is approaching the carrying capacity. So this right over here, for example, would actually be our carrying capacity. So one way to think about, well, what is this maximum point right over here? And there's a couple of ways that you can approach it with calculus and even algebra. We have many tools for identifying this maximum point, which is really just the vertex of this downward opening, this concave downward parabola. So let's just do that, and let's find this vertex."}, {"video_title": "Worked example Logistic model word problem Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So one way to think about, well, what is this maximum point right over here? And there's a couple of ways that you can approach it with calculus and even algebra. We have many tools for identifying this maximum point, which is really just the vertex of this downward opening, this concave downward parabola. So let's just do that, and let's find this vertex. And the vertex is just halfway between the zeros of this quadratic. So let's find the P values that make this equal to zero, which we actually just figured out. Well, the maximum point, the vertex, is just going to be halfway between that."}, {"video_title": "Worked example Logistic model word problem Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So let's just do that, and let's find this vertex. And the vertex is just halfway between the zeros of this quadratic. So let's find the P values that make this equal to zero, which we actually just figured out. Well, the maximum point, the vertex, is just going to be halfway between that. So when the population is zero, our rate of change is zero. When the population is A, which we know is 48,000, our rate of change is zero. And so our carrying capacity, sorry, our maximum rate of change is gonna happen halfway between those two points, which is 24,000."}, {"video_title": "Worked example Logistic model word problem Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "Well, the maximum point, the vertex, is just going to be halfway between that. So when the population is zero, our rate of change is zero. When the population is A, which we know is 48,000, our rate of change is zero. And so our carrying capacity, sorry, our maximum rate of change is gonna happen halfway between those two points, which is 24,000. So it is a population of 24,000. When we got that, by really just saying, well, at what point does this quadratic, quadratic is a function of P, when does that hit a maximum point? Well, that's going to be halfway between the zeros."}, {"video_title": "Worked example Logistic model word problem Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "And so our carrying capacity, sorry, our maximum rate of change is gonna happen halfway between those two points, which is 24,000. So it is a population of 24,000. When we got that, by really just saying, well, at what point does this quadratic, quadratic is a function of P, when does that hit a maximum point? Well, that's going to be halfway between the zeros. The zeros happen when P equals zero, and P is equal to 48,000. So that's gonna happen when the population is 24,000. So this type of problem seems very intimidating at first."}, {"video_title": "Worked example Logistic model word problem Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "Well, that's going to be halfway between the zeros. The zeros happen when P equals zero, and P is equal to 48,000. So that's gonna happen when the population is 24,000. So this type of problem seems very intimidating at first. Logistic differential equation, how do I actually solve this and then analyze it? But the key is to, one, recognize the logistic differential equation, see what it's talking about, and then maybe think of term, think of this in terms of our rate of change as a function of our population. And remember that the carrying capacity is what happens as T approaches infinity, and as T approaches infinity, our rate of change approaches zero."}, {"video_title": "Derivatives of inverse functions from table AP Calculus AB Khan Academy.mp3", "Sentence": "So let's just remind ourselves what it means for them to be inverse functions. That means that if I have two sets of numbers, let's say one set right over there, that's another set right over there, and if we view that first set as the domain of g, so if you start with some x right over here, g is going to map from that x to another value which we would call g of x, g of x. That's what the function g does. Now if h is the inverse of g, and frankly vice versa, then h could go from that point g of x back to x. So h would do this. H would get us back to our original value. So that's what the function h would do."}, {"video_title": "Derivatives of inverse functions from table AP Calculus AB Khan Academy.mp3", "Sentence": "Now if h is the inverse of g, and frankly vice versa, then h could go from that point g of x back to x. So h would do this. H would get us back to our original value. So that's what the function h would do. And so we could view this point right over here, we could view it as x, so that is x, but we could also view it as h of g of x. So we could also view this as h of g of x. And I did all of that so that we can really feel good about this idea."}, {"video_title": "Derivatives of inverse functions from table AP Calculus AB Khan Academy.mp3", "Sentence": "So that's what the function h would do. And so we could view this point right over here, we could view it as x, so that is x, but we could also view it as h of g of x. So we could also view this as h of g of x. And I did all of that so that we can really feel good about this idea. If someone tells you that g and h are inverse functions, that means that h of g of x is x. So g of x is x, h of g of x, h of g of x is equal to x, or you could have gone the other way around. You could have started with a, well you could have done it multiple different ways, but also g of h of x. I could have just swapped these letters here, the letters h and g are somewhat arbitrary."}, {"video_title": "Derivatives of inverse functions from table AP Calculus AB Khan Academy.mp3", "Sentence": "And I did all of that so that we can really feel good about this idea. If someone tells you that g and h are inverse functions, that means that h of g of x is x. So g of x is x, h of g of x, h of g of x is equal to x, or you could have gone the other way around. You could have started with a, well you could have done it multiple different ways, but also g of h of x. I could have just swapped these letters here, the letters h and g are somewhat arbitrary. So you could have also said that g of h of x is equal to x. So g of h of x is equal to x. And then they give us some information."}, {"video_title": "Derivatives of inverse functions from table AP Calculus AB Khan Academy.mp3", "Sentence": "You could have started with a, well you could have done it multiple different ways, but also g of h of x. I could have just swapped these letters here, the letters h and g are somewhat arbitrary. So you could have also said that g of h of x is equal to x. So g of h of x is equal to x. And then they give us some information. The following table lists a few values of g, h, and g prime. Alright, so they want us to evaluate h prime of three. They don't even give us h prime of three, how do we figure it out?"}, {"video_title": "Derivatives of inverse functions from table AP Calculus AB Khan Academy.mp3", "Sentence": "And then they give us some information. The following table lists a few values of g, h, and g prime. Alright, so they want us to evaluate h prime of three. They don't even give us h prime of three, how do we figure it out? They gave us g prime and h and g. How do we figure this out? Well here we're going to actually derive something based on the chain rule. And this isn't the type of problem that you'll see a lot of, but it is interesting."}, {"video_title": "Derivatives of inverse functions from table AP Calculus AB Khan Academy.mp3", "Sentence": "They don't even give us h prime of three, how do we figure it out? They gave us g prime and h and g. How do we figure this out? Well here we're going to actually derive something based on the chain rule. And this isn't the type of problem that you'll see a lot of, but it is interesting. So we're going to work through it, and there's a chance that you might see it in your calculus class. So let's start with either one of these expressions up here. So let's start with the expression, well let's start with the, let's do this one over here."}, {"video_title": "Derivatives of inverse functions from table AP Calculus AB Khan Academy.mp3", "Sentence": "And this isn't the type of problem that you'll see a lot of, but it is interesting. So we're going to work through it, and there's a chance that you might see it in your calculus class. So let's start with either one of these expressions up here. So let's start with the expression, well let's start with the, let's do this one over here. So if we have g of h of x is equal to x, h of x is equal to x, so we put that h of x back there, which is by definition true if g and h's are inverses. Well now let's take the derivative of both sides of this. So let's take the derivative with respect to x of both sides, derivative with respect to x."}, {"video_title": "Derivatives of inverse functions from table AP Calculus AB Khan Academy.mp3", "Sentence": "So let's start with the expression, well let's start with the, let's do this one over here. So if we have g of h of x is equal to x, h of x is equal to x, so we put that h of x back there, which is by definition true if g and h's are inverses. Well now let's take the derivative of both sides of this. So let's take the derivative with respect to x of both sides, derivative with respect to x. And on the left hand side, well we just apply the chain rule. So this would be g prime of h of x, g prime of h of x, times h prime of x, that's just the chain rule right over there, and that would be equal to, what's the derivative with respect to x of x? Well that's just going to be equal to one."}, {"video_title": "Derivatives of inverse functions from table AP Calculus AB Khan Academy.mp3", "Sentence": "So let's take the derivative with respect to x of both sides, derivative with respect to x. And on the left hand side, well we just apply the chain rule. So this would be g prime of h of x, g prime of h of x, times h prime of x, that's just the chain rule right over there, and that would be equal to, what's the derivative with respect to x of x? Well that's just going to be equal to one. So now it's interesting. We need to figure out what h prime of three is. So we can figure out what h of three is, and then we can use that to figure out what g prime of whatever h, g prime of h of three is."}, {"video_title": "Derivatives of inverse functions from table AP Calculus AB Khan Academy.mp3", "Sentence": "Well that's just going to be equal to one. So now it's interesting. We need to figure out what h prime of three is. So we can figure out what h of three is, and then we can use that to figure out what g prime of whatever h, g prime of h of three is. And so we should be able to figure out h prime of x. Or we could just rewrite it this way. We could rewrite that h prime of x is equal to, is equal to one over g prime of h of x."}, {"video_title": "Derivatives of inverse functions from table AP Calculus AB Khan Academy.mp3", "Sentence": "So we can figure out what h of three is, and then we can use that to figure out what g prime of whatever h, g prime of h of three is. And so we should be able to figure out h prime of x. Or we could just rewrite it this way. We could rewrite that h prime of x is equal to, is equal to one over g prime of h of x. Now in some circles, they might encourage you to memorize this, and maybe for the sake of doing this exercise on Khan Academy, you might want to memorize it. But I'll tell you, 20 years after I took calculus, or almost 25 years after I took calculus, this is not something that I retain in my long term memory, but I did retain that you can derive this from just what the definition of inverse functions actually are. But we can use this now if we want to figure out what h prime of three is."}, {"video_title": "Derivatives of inverse functions from table AP Calculus AB Khan Academy.mp3", "Sentence": "We could rewrite that h prime of x is equal to, is equal to one over g prime of h of x. Now in some circles, they might encourage you to memorize this, and maybe for the sake of doing this exercise on Khan Academy, you might want to memorize it. But I'll tell you, 20 years after I took calculus, or almost 25 years after I took calculus, this is not something that I retain in my long term memory, but I did retain that you can derive this from just what the definition of inverse functions actually are. But we can use this now if we want to figure out what h prime of three is. H prime of three is going to be equal to one over, one over, one over g prime of h of three, which I'm guessing that they have given us. So h of three, when x is three, h is four. So that is h of three there."}, {"video_title": "Derivatives of inverse functions from table AP Calculus AB Khan Academy.mp3", "Sentence": "But we can use this now if we want to figure out what h prime of three is. H prime of three is going to be equal to one over, one over, one over g prime of h of three, which I'm guessing that they have given us. So h of three, when x is three, h is four. So that is h of three there. So h of three is four. So now we just have to figure out g prime of four. Well lucky for us, they have given us, when x is equal to four, g prime, g prime is equal to 1 1 2."}, {"video_title": "Derivatives of inverse functions from table AP Calculus AB Khan Academy.mp3", "Sentence": "So that is h of three there. So h of three is four. So now we just have to figure out g prime of four. Well lucky for us, they have given us, when x is equal to four, g prime, g prime is equal to 1 1 2. So g prime of four is equal to 1 1 2. So h prime of three is equal to one over 1 1 2. So one over 1 1 2."}, {"video_title": "Analyzing problems involving definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "The population of a town grows at a rate of r of t is equal to 300 times e to the 0.3 t, people per year, where t is time in years. At time t is equal to two, the town's population is 1,200 people. What is the town's population at t is equal to seven? Which expression can we use to solve the problem? So they don't want us to actually answer the question, they just want us to set up the expression using some symbols from calculus. So why don't you pause this video and try to think about it. So let's just remind us what they've given us."}, {"video_title": "Analyzing problems involving definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Which expression can we use to solve the problem? So they don't want us to actually answer the question, they just want us to set up the expression using some symbols from calculus. So why don't you pause this video and try to think about it. So let's just remind us what they've given us. They've given us the rate function right over here. And so if you want to find a change in population from one time to another time, what you could do is you could take the integral of the rate function from the starting time, t is equal to two years, to t is equal to seven years. So we're gonna take the integral of the rate function, and what this is going to tell us, this is going to tell us the change in population from time two to time seven."}, {"video_title": "Analyzing problems involving definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So let's just remind us what they've given us. They've given us the rate function right over here. And so if you want to find a change in population from one time to another time, what you could do is you could take the integral of the rate function from the starting time, t is equal to two years, to t is equal to seven years. So we're gonna take the integral of the rate function, and what this is going to tell us, this is going to tell us the change in population from time two to time seven. So this is, you could say, let me just write this, this is the change, I'll use delta for change, or let me just write it out. Change in population, change in population, population, but they don't want us to, they're not asking us for the change in population. They want us to know what is the town's population at t is equal to seven."}, {"video_title": "Analyzing problems involving definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So we're gonna take the integral of the rate function, and what this is going to tell us, this is going to tell us the change in population from time two to time seven. So this is, you could say, let me just write this, this is the change, I'll use delta for change, or let me just write it out. Change in population, change in population, population, but they don't want us to, they're not asking us for the change in population. They want us to know what is the town's population at t is equal to seven. So what you would want is, you would want what your population is at t equals two, plus the change in population from two to seven, to get you your population at seven. So they tell us the population at time t equals two, the town's population is 1,200 people. So if you want the population at t is equal to seven, it's going to be 1,200 plus how whatever the change in population, if you take the integral of the rate function, you are, and this is the rate of population, this integral is gonna give you the change in population from time t equals two to t equals seven."}, {"video_title": "Analyzing problems involving definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "They want us to know what is the town's population at t is equal to seven. So what you would want is, you would want what your population is at t equals two, plus the change in population from two to seven, to get you your population at seven. So they tell us the population at time t equals two, the town's population is 1,200 people. So if you want the population at t is equal to seven, it's going to be 1,200 plus how whatever the change in population, if you take the integral of the rate function, you are, and this is the rate of population, this integral is gonna give you the change in population from time t equals two to t equals seven. So we can see clearly that is choice D right over here. These other choices, we can look at them really quick, choice B is just the change in population. So that's assuming that this is, and this is actually increasing."}, {"video_title": "Analyzing problems involving definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So if you want the population at t is equal to seven, it's going to be 1,200 plus how whatever the change in population, if you take the integral of the rate function, you are, and this is the rate of population, this integral is gonna give you the change in population from time t equals two to t equals seven. So we can see clearly that is choice D right over here. These other choices, we can look at them really quick, choice B is just the change in population. So that's assuming that this is, and this is actually increasing. So this would tell us how much of the population increase from t equals two to t equals seven. So that's not what we want. We want what the actual population is."}, {"video_title": "Analyzing problems involving definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So that's assuming that this is, and this is actually increasing. So this would tell us how much of the population increase from t equals two to t equals seven. So that's not what we want. We want what the actual population is. This is how much the population increases from time zero to time seven. Now you might say, well, wouldn't that be the town's population? Well, that would be the town's population if they had no people at time zero, but you can't assume that."}, {"video_title": "Analyzing problems involving definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "We want what the actual population is. This is how much the population increases from time zero to time seven. Now you might say, well, wouldn't that be the town's population? Well, that would be the town's population if they had no people at time zero, but you can't assume that. Maybe the town got settled by 10 people or by 1,000 people or who knows whatever else. So right over there. And this is taking the derivative of the rate function, which is, it's actually a little hard to think about what is this."}, {"video_title": "Analyzing problems involving definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Well, that would be the town's population if they had no people at time zero, but you can't assume that. Maybe the town got settled by 10 people or by 1,000 people or who knows whatever else. So right over there. And this is taking the derivative of the rate function, which is, it's actually a little hard to think about what is this. This is the rate of change of the rate at time seven minus the rate of change of the rate at time two. So I would rule that one out as well. Let's do one more of these."}, {"video_title": "Analyzing problems involving definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "And this is taking the derivative of the rate function, which is, it's actually a little hard to think about what is this. This is the rate of change of the rate at time seven minus the rate of change of the rate at time two. So I would rule that one out as well. Let's do one more of these. So here we have, here we have the depth of water in a tank is changing at a rate of, r of t is equal to 0.3 t, centimeters per minute, where t is the time in minutes. At time t equals zero, the depth of the water is 35 centimeters. What is the change in the water's depth during the fourth minute?"}, {"video_title": "Analyzing problems involving definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Let's do one more of these. So here we have, here we have the depth of water in a tank is changing at a rate of, r of t is equal to 0.3 t, centimeters per minute, where t is the time in minutes. At time t equals zero, the depth of the water is 35 centimeters. What is the change in the water's depth during the fourth minute? So let's pause the video again and see if you can figure this out again and figure out what expression can we use to solve the problem. The problem being, what is the change in the water's depth during the fourth minute? Alright, so we've just talked about if you're trying to find the change in a value, you could take the integral of the rate function over the appropriate time."}, {"video_title": "Analyzing problems involving definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "What is the change in the water's depth during the fourth minute? So let's pause the video again and see if you can figure this out again and figure out what expression can we use to solve the problem. The problem being, what is the change in the water's depth during the fourth minute? Alright, so we've just talked about if you're trying to find the change in a value, you could take the integral of the rate function over the appropriate time. So we're talking about during the fourth minute. So we definitely wanna take the integral of the rate function. And we just have to think about the bounds and all of the choices here taking the integral of the rate function."}, {"video_title": "Analyzing problems involving definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Alright, so we've just talked about if you're trying to find the change in a value, you could take the integral of the rate function over the appropriate time. So we're talking about during the fourth minute. So we definitely wanna take the integral of the rate function. And we just have to think about the bounds and all of the choices here taking the integral of the rate function. So really the interesting part is during the fourth minute. So let me just draw a little number line here and we can think about what the fourth minute looks like. Or actually, let me just draw the whole thing."}, {"video_title": "Analyzing problems involving definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "And we just have to think about the bounds and all of the choices here taking the integral of the rate function. So really the interesting part is during the fourth minute. So let me just draw a little number line here and we can think about what the fourth minute looks like. Or actually, let me just draw the whole thing. So let's say this is r of t right over there. We could say y is equal to r of t. And this is t. And let's see, the first minute goes from zero to one, second minute goes from one to two, third minute goes from two to three, fourth minute goes from three to four. The rate function, this actually just looks like a straight up linear rate function, looks something like this."}, {"video_title": "Analyzing problems involving definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Or actually, let me just draw the whole thing. So let's say this is r of t right over there. We could say y is equal to r of t. And this is t. And let's see, the first minute goes from zero to one, second minute goes from one to two, third minute goes from two to three, fourth minute goes from three to four. The rate function, this actually just looks like a straight up linear rate function, looks something like this. And so what is the fourth minute? Well the fourth minute is, the first minute is this one, second, third, fourth. The fourth minute is going from minute three to minute four."}, {"video_title": "Analyzing problems involving definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "The rate function, this actually just looks like a straight up linear rate function, looks something like this. And so what is the fourth minute? Well the fourth minute is, the first minute is this one, second, third, fourth. The fourth minute is going from minute three to minute four. So what we wanna do is the expression that gives us this area right over here under the rate curve. Well this is, our lower bound's going to be three and our upper bound is going to be four. And so there you have it."}, {"video_title": "Analyzing problems involving definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "The fourth minute is going from minute three to minute four. So what we wanna do is the expression that gives us this area right over here under the rate curve. Well this is, our lower bound's going to be three and our upper bound is going to be four. And so there you have it. It is this first choice. You might have been tempted here if you got a little bit confused, hey maybe the fourth minute is after we've crossed that t is equal to four, but no, that would be the fifth minute. This would tell us, this would tell us our change over the first four minutes, not just during the fourth minute."}, {"video_title": "Analyzing problems involving definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "And so there you have it. It is this first choice. You might have been tempted here if you got a little bit confused, hey maybe the fourth minute is after we've crossed that t is equal to four, but no, that would be the fifth minute. This would tell us, this would tell us our change over the first four minutes, not just during the fourth minute. And then this, well this is just gonna be zero. If you're taking, what is the change in the value from three to three? Well it didn't change at all in that, because it's essentially at an instant."}, {"video_title": "Connecting limits and graphical behavior Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So we have the graph of y is equal to g of x right over here, and I want to think about what is the limit as x approaches five of g of x? Well, we've done this multiple times. Let's think about what g of x approaches as x approaches five from the left. G of x is approaching negative six. As x approaches five from the right, g of x looks like it's approaching negative six. So a reasonable estimate based on looking at this graph is that as x approaches five, g of x is approaching negative six. And it's worth noting that that's not what g of five is."}, {"video_title": "Connecting limits and graphical behavior Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "G of x is approaching negative six. As x approaches five from the right, g of x looks like it's approaching negative six. So a reasonable estimate based on looking at this graph is that as x approaches five, g of x is approaching negative six. And it's worth noting that that's not what g of five is. G of five is a different value. But the whole point of this video is to appreciate all that a limit does. A limit only describes the behavior of a function as it approaches a point."}, {"video_title": "Connecting limits and graphical behavior Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And it's worth noting that that's not what g of five is. G of five is a different value. But the whole point of this video is to appreciate all that a limit does. A limit only describes the behavior of a function as it approaches a point. It doesn't tell us exactly what's happening at that point, what g of five is, and it doesn't tell us much about the rest of the function, about the rest of the graph. For example, I could construct many different functions for which the limit as x approaches five is equal to negative six, and they would look very different from g of x. For example, I could say the limit of f of x as x approaches five is equal to negative six, and I can construct an f of x that does this that looks very different than g of x."}, {"video_title": "Connecting limits and graphical behavior Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "A limit only describes the behavior of a function as it approaches a point. It doesn't tell us exactly what's happening at that point, what g of five is, and it doesn't tell us much about the rest of the function, about the rest of the graph. For example, I could construct many different functions for which the limit as x approaches five is equal to negative six, and they would look very different from g of x. For example, I could say the limit of f of x as x approaches five is equal to negative six, and I can construct an f of x that does this that looks very different than g of x. And in fact, if you're up for it, pause this video and see if you could do the same. If you have some graph paper, even just sketch it. Well, the key thing is that the behavior of the function as x approaches five from both sides, from the left and the right, it has to be approaching negative six."}, {"video_title": "Connecting limits and graphical behavior Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "For example, I could say the limit of f of x as x approaches five is equal to negative six, and I can construct an f of x that does this that looks very different than g of x. And in fact, if you're up for it, pause this video and see if you could do the same. If you have some graph paper, even just sketch it. Well, the key thing is that the behavior of the function as x approaches five from both sides, from the left and the right, it has to be approaching negative six. So for example, a function that looks like this, so let me draw f of x, an f of x that looks like this and is even defined right over there and then does something like this, that would work. As we approach from the left, we're approaching negative six as we approach from the right, we are approaching negative six You could have a function like this. Let's say the limit, let's call it h of x as x approaches five is equal to negative six."}, {"video_title": "Connecting limits and graphical behavior Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the key thing is that the behavior of the function as x approaches five from both sides, from the left and the right, it has to be approaching negative six. So for example, a function that looks like this, so let me draw f of x, an f of x that looks like this and is even defined right over there and then does something like this, that would work. As we approach from the left, we're approaching negative six as we approach from the right, we are approaching negative six You could have a function like this. Let's say the limit, let's call it h of x as x approaches five is equal to negative six. You could have a function like this. Maybe it's defined up to there, then it's, you have a circle there, then it keeps going. Maybe it's not defined at all for any of these values."}, {"video_title": "Connecting limits and graphical behavior Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say the limit, let's call it h of x as x approaches five is equal to negative six. You could have a function like this. Maybe it's defined up to there, then it's, you have a circle there, then it keeps going. Maybe it's not defined at all for any of these values. And then maybe down here, it is defined for all x values greater than or equal to four and it just goes right through negative six. So notice, all of these functions as x approaches five, they all have the limit defined and it's equal to negative six, but these functions all look very, very, very different. Now another thing to appreciate is, for a given function, let me delete these, oftentimes we're asked to find the limits as x approaches some type of an interesting value."}, {"video_title": "Connecting limits and graphical behavior Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Maybe it's not defined at all for any of these values. And then maybe down here, it is defined for all x values greater than or equal to four and it just goes right through negative six. So notice, all of these functions as x approaches five, they all have the limit defined and it's equal to negative six, but these functions all look very, very, very different. Now another thing to appreciate is, for a given function, let me delete these, oftentimes we're asked to find the limits as x approaches some type of an interesting value. So for example, x approaches five, five is interesting right over here because we have this point discontinuity. But you could take the limit on an infinite number of points for this function right over here. You could say the limit of g of x as x approaches, not x equals, as x approaches one."}, {"video_title": "Connecting limits and graphical behavior Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Now another thing to appreciate is, for a given function, let me delete these, oftentimes we're asked to find the limits as x approaches some type of an interesting value. So for example, x approaches five, five is interesting right over here because we have this point discontinuity. But you could take the limit on an infinite number of points for this function right over here. You could say the limit of g of x as x approaches, not x equals, as x approaches one. What would that be? Pause the video and try to figure it out. Let's see."}, {"video_title": "Connecting limits and graphical behavior Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "You could say the limit of g of x as x approaches, not x equals, as x approaches one. What would that be? Pause the video and try to figure it out. Let's see. As x approaches one from the left-hand side, it looks like we are approaching this value here. And as x approaches one from the right-hand side, it looks like we are approaching that value there. So that would be equal to g of one."}, {"video_title": "Connecting limits and graphical behavior Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Let's see. As x approaches one from the left-hand side, it looks like we are approaching this value here. And as x approaches one from the right-hand side, it looks like we are approaching that value there. So that would be equal to g of one. That is equal to g of one based on, that would be a reasonable, that's a reasonable conclusion to make looking at this graph. And if we were to estimate that, g of one looks like it's approximately negative 5.1 or 5.2, negative 5.1. We could find the limit of g of x as x approaches pi."}, {"video_title": "Connecting limits and graphical behavior Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So that would be equal to g of one. That is equal to g of one based on, that would be a reasonable, that's a reasonable conclusion to make looking at this graph. And if we were to estimate that, g of one looks like it's approximately negative 5.1 or 5.2, negative 5.1. We could find the limit of g of x as x approaches pi. So pi is right around there. As x approaches pi from the left, we're approaching that value, which looks actually pretty close to the one we just thought about. As we approach from the right, we're approaching that value."}, {"video_title": "Connecting limits and graphical behavior Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "We could find the limit of g of x as x approaches pi. So pi is right around there. As x approaches pi from the left, we're approaching that value, which looks actually pretty close to the one we just thought about. As we approach from the right, we're approaching that value. And once again, in this case, this is gonna be equal to g of pi. We don't have any interesting discontinuities there or anything like that. So there's two big takeaways here."}, {"video_title": "Connecting limits and graphical behavior Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "As we approach from the right, we're approaching that value. And once again, in this case, this is gonna be equal to g of pi. We don't have any interesting discontinuities there or anything like that. So there's two big takeaways here. You can construct many different functions that would have the same limit at a point. And for a given function, you can take the limit at many different points. In fact, an infinite number of different points."}, {"video_title": "Functions defined by integrals switched interval AP Calculus AB Khan Academy.mp3", "Sentence": "Let g of x be equal to the definite integral from zero to x of f of t dt. Now at first when you see this, you're like, wow, this is strange. I have a function that is being defined by an integral, a definite integral, but one of its bounds are x. And you should just say, well, this is okay. A function can be defined any which way. And as we'll see, it's actually quite straightforward to evaluate this. So g of negative two, g of negative two, and I'll do the negative two in a different color, g of negative two, well, what we do is we take this expression right over here, this definite integral, and everywhere we see an x, we replace it with a negative two."}, {"video_title": "Functions defined by integrals switched interval AP Calculus AB Khan Academy.mp3", "Sentence": "And you should just say, well, this is okay. A function can be defined any which way. And as we'll see, it's actually quite straightforward to evaluate this. So g of negative two, g of negative two, and I'll do the negative two in a different color, g of negative two, well, what we do is we take this expression right over here, this definite integral, and everywhere we see an x, we replace it with a negative two. So this is going to be equal to the integral from zero to x of, and I'll write x in a second, f of t dt. Well, x is now negative two. This is now negative two."}, {"video_title": "Functions defined by integrals switched interval AP Calculus AB Khan Academy.mp3", "Sentence": "So g of negative two, g of negative two, and I'll do the negative two in a different color, g of negative two, well, what we do is we take this expression right over here, this definite integral, and everywhere we see an x, we replace it with a negative two. So this is going to be equal to the integral from zero to x of, and I'll write x in a second, f of t dt. Well, x is now negative two. This is now negative two. And so how do we figure out what this is? Now before we even look at this graph, you might say, okay, this is the region under, the area of the region under the graph y equals f of t between negative two and zero, but you have to be careful. Notice our upper bound here is actually a lower number than our lower bound right over here."}, {"video_title": "Functions defined by integrals switched interval AP Calculus AB Khan Academy.mp3", "Sentence": "This is now negative two. And so how do we figure out what this is? Now before we even look at this graph, you might say, okay, this is the region under, the area of the region under the graph y equals f of t between negative two and zero, but you have to be careful. Notice our upper bound here is actually a lower number than our lower bound right over here. So it will be nice to swap those bounds so we can truly view it as the area of the region under f of t above the t-axis between those two bounds. And so when you swap the bounds, this is going to be equal to negative definite integral from negative two, negative two to zero of f of t dt. And now what we have right over here, what I'm squaring off in magenta, this is the area under the curve f between negative two and zero."}, {"video_title": "Functions defined by integrals switched interval AP Calculus AB Khan Academy.mp3", "Sentence": "Notice our upper bound here is actually a lower number than our lower bound right over here. So it will be nice to swap those bounds so we can truly view it as the area of the region under f of t above the t-axis between those two bounds. And so when you swap the bounds, this is going to be equal to negative definite integral from negative two, negative two to zero of f of t dt. And now what we have right over here, what I'm squaring off in magenta, this is the area under the curve f between negative two and zero. So between negative two and zero. So that is this area right over here that we care about. Now what is that going to be?"}, {"video_title": "Functions defined by integrals switched interval AP Calculus AB Khan Academy.mp3", "Sentence": "And now what we have right over here, what I'm squaring off in magenta, this is the area under the curve f between negative two and zero. So between negative two and zero. So that is this area right over here that we care about. Now what is that going to be? Well, you could, there's a bunch of different ways that you could do this. You could split it off into a square and a triangle. The area of this square right over here is four."}, {"video_title": "Functions defined by integrals switched interval AP Calculus AB Khan Academy.mp3", "Sentence": "Now what is that going to be? Well, you could, there's a bunch of different ways that you could do this. You could split it off into a square and a triangle. The area of this square right over here is four. It's two by two. And just make sure to, make sure you look at the units. Sometimes each square doesn't represent one square unit, but in this case it does."}, {"video_title": "Functions defined by integrals switched interval AP Calculus AB Khan Academy.mp3", "Sentence": "The area of this square right over here is four. It's two by two. And just make sure to, make sure you look at the units. Sometimes each square doesn't represent one square unit, but in this case it does. So that's four. And then up here, this is half of four, right? If it was all of this, that would be four."}, {"video_title": "Functions defined by integrals switched interval AP Calculus AB Khan Academy.mp3", "Sentence": "Sometimes each square doesn't represent one square unit, but in this case it does. So that's four. And then up here, this is half of four, right? If it was all of this, that would be four. This triangle is half of four. So this is two right over there. Or you could view this as base times height times 1 1 2, which is gonna be two times two times 1 1 2."}, {"video_title": "Functions defined by integrals switched interval AP Calculus AB Khan Academy.mp3", "Sentence": "If it was all of this, that would be four. This triangle is half of four. So this is two right over there. Or you could view this as base times height times 1 1 2, which is gonna be two times two times 1 1 2. And so this area right over here is six. So this part is six, but we can't forget that negative sign. So this is going to be equal to negative, negative six."}, {"video_title": "Interval of convergence for derivative and integral Series AP Calculus BC Khan Academy.mp3", "Sentence": "And in general, we can do this term by term. What do I mean by that? Well, that means that the derivative of f, f prime of x, is just going to be the derivative of each of these terms. So that's going to be the sum from n equals one to infinity. And let's see, the derivative of x to the n is n times x to the n minus one. So I could write this as n times x to the n minus one all of that over n, and these n's will cancel out. So this is just going to be, this is just going to be x to the n minus one."}, {"video_title": "Interval of convergence for derivative and integral Series AP Calculus BC Khan Academy.mp3", "Sentence": "So that's going to be the sum from n equals one to infinity. And let's see, the derivative of x to the n is n times x to the n minus one. So I could write this as n times x to the n minus one all of that over n, and these n's will cancel out. So this is just going to be, this is just going to be x to the n minus one. So this is taking the derivative with respect to x. Similarly, we could integrate. We could integrate and we could evaluate, we could evaluate the integral of f of x dx."}, {"video_title": "Interval of convergence for derivative and integral Series AP Calculus BC Khan Academy.mp3", "Sentence": "So this is just going to be, this is just going to be x to the n minus one. So this is taking the derivative with respect to x. Similarly, we could integrate. We could integrate and we could evaluate, we could evaluate the integral of f of x dx. And this is going to be equal to some constant plus, if we integrate this term by term. And so this is going to be equal to the sum from n equals one to infinity. And let's see, we increment the exponent, so x to the n plus one, and then we divide by that."}, {"video_title": "Interval of convergence for derivative and integral Series AP Calculus BC Khan Academy.mp3", "Sentence": "We could integrate and we could evaluate, we could evaluate the integral of f of x dx. And this is going to be equal to some constant plus, if we integrate this term by term. And so this is going to be equal to the sum from n equals one to infinity. And let's see, we increment the exponent, so x to the n plus one, and then we divide by that. So times n plus one times this n right over here. So this is a common technique that you will see when dealing with power series. And we're gonna go a little bit more into the details, because you can only do this for x values within the interval of convergence for the power series."}, {"video_title": "Interval of convergence for derivative and integral Series AP Calculus BC Khan Academy.mp3", "Sentence": "And let's see, we increment the exponent, so x to the n plus one, and then we divide by that. So times n plus one times this n right over here. So this is a common technique that you will see when dealing with power series. And we're gonna go a little bit more into the details, because you can only do this for x values within the interval of convergence for the power series. And as we will see, the interval of convergence for these different series is slightly different. The intervals are very similar, but what happens at the end point is different. So I encourage you, pause this video, and see if you can figure out the interval of convergence for each of these series."}, {"video_title": "Interval of convergence for derivative and integral Series AP Calculus BC Khan Academy.mp3", "Sentence": "And we're gonna go a little bit more into the details, because you can only do this for x values within the interval of convergence for the power series. And as we will see, the interval of convergence for these different series is slightly different. The intervals are very similar, but what happens at the end point is different. So I encourage you, pause this video, and see if you can figure out the interval of convergence for each of these series. This is the integral of our original series, and this is the derivative of our original series. So let's start with our original series. Let's figure out the interval of convergence."}, {"video_title": "Interval of convergence for derivative and integral Series AP Calculus BC Khan Academy.mp3", "Sentence": "So I encourage you, pause this video, and see if you can figure out the interval of convergence for each of these series. This is the integral of our original series, and this is the derivative of our original series. So let's start with our original series. Let's figure out the interval of convergence. And we could do that using the ratio test. So the ratio test, we would wanna do the limit, the limit as n approaches infinity, of a sub n plus one, so that's going to be x to the n plus one over n plus one, divided by a sub n, so that's x to the n over n. So we wanna take the absolute value of that. That's going to be the limit as n approaches infinity."}, {"video_title": "Interval of convergence for derivative and integral Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let's figure out the interval of convergence. And we could do that using the ratio test. So the ratio test, we would wanna do the limit, the limit as n approaches infinity, of a sub n plus one, so that's going to be x to the n plus one over n plus one, divided by a sub n, so that's x to the n over n. So we wanna take the absolute value of that. That's going to be the limit as n approaches infinity. Let's see, this is, if you divide this and this by x to the n, that's gonna be a one, and this is just going to be an x, and then this n is going to end up up top, so this is going to be xn over n plus one. And this is equal to the limit as n approaches infinity of, let's see, if we divide the numerator and the denominators here by one over, if we divide both of the numerator and denominator by n, we're gonna get x over one plus one over n. And what is this going to be? Well, this term's gonna go to zero, so this is just going to be equal to the absolute value of x."}, {"video_title": "Interval of convergence for derivative and integral Series AP Calculus BC Khan Academy.mp3", "Sentence": "That's going to be the limit as n approaches infinity. Let's see, this is, if you divide this and this by x to the n, that's gonna be a one, and this is just going to be an x, and then this n is going to end up up top, so this is going to be xn over n plus one. And this is equal to the limit as n approaches infinity of, let's see, if we divide the numerator and the denominators here by one over, if we divide both of the numerator and denominator by n, we're gonna get x over one plus one over n. And what is this going to be? Well, this term's gonna go to zero, so this is just going to be equal to the absolute value of x. And the ratio test tells us that this series is convergent if this right over here is less than one, it's divergent if this is greater than one, and it's inconclusive if this equals one. So we know, let's write that down. We know we are convergent, convergent, convergent for the absolute value of x less than one, when this thing, when it is less than one."}, {"video_title": "Interval of convergence for derivative and integral Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, this term's gonna go to zero, so this is just going to be equal to the absolute value of x. And the ratio test tells us that this series is convergent if this right over here is less than one, it's divergent if this is greater than one, and it's inconclusive if this equals one. So we know, let's write that down. We know we are convergent, convergent, convergent for the absolute value of x less than one, when this thing, when it is less than one. We know that we are divergent, divergent when this thing is greater than one, when the absolute value of x is greater than one, but what about when the absolute value of x is equal to one? That's where the ratio test breaks down, and we have to test that separately. So let's look at the scenario where x is equal to one."}, {"video_title": "Interval of convergence for derivative and integral Series AP Calculus BC Khan Academy.mp3", "Sentence": "We know we are convergent, convergent, convergent for the absolute value of x less than one, when this thing, when it is less than one. We know that we are divergent, divergent when this thing is greater than one, when the absolute value of x is greater than one, but what about when the absolute value of x is equal to one? That's where the ratio test breaks down, and we have to test that separately. So let's look at the scenario where x is equal to one. When x equals to one, this series is the sum from n equals one to infinity of one to the n over n. Well, that's just gonna be one over n. This is the harmonic series, or the p-series, where our p is one, and we've seen in multiple videos that this diverges. So when x equals one, we diverge, but what about when x equals negative one? When x equals negative one, this thing becomes the sum from n equals one to infinity of negative one to the n over n, and this is often known as the alternating harmonic series, and this one, by the alternating series test, this one actually converges, and we've seen that in multiple videos."}, {"video_title": "Interval of convergence for derivative and integral Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's look at the scenario where x is equal to one. When x equals to one, this series is the sum from n equals one to infinity of one to the n over n. Well, that's just gonna be one over n. This is the harmonic series, or the p-series, where our p is one, and we've seen in multiple videos that this diverges. So when x equals one, we diverge, but what about when x equals negative one? When x equals negative one, this thing becomes the sum from n equals one to infinity of negative one to the n over n, and this is often known as the alternating harmonic series, and this one, by the alternating series test, this one actually converges, and we've seen that in multiple videos. So it turns out the interval of convergence for our original thing right over here, our interval of convergence, interval of convergence, convergence here, is we can, x can be, so it could be, x can be greater than or equal to negative one, or I could say negative one is less than or equal to x, because if x is negative one, we still converge, but then x has to be less than one, because right at one, we diverge, so we can't say less than or equal to. So this is the interval of convergence for our original function. What about the interval of convergence for this one right over here, when we take the derivative?"}, {"video_title": "Interval of convergence for derivative and integral Series AP Calculus BC Khan Academy.mp3", "Sentence": "When x equals negative one, this thing becomes the sum from n equals one to infinity of negative one to the n over n, and this is often known as the alternating harmonic series, and this one, by the alternating series test, this one actually converges, and we've seen that in multiple videos. So it turns out the interval of convergence for our original thing right over here, our interval of convergence, interval of convergence, convergence here, is we can, x can be, so it could be, x can be greater than or equal to negative one, or I could say negative one is less than or equal to x, because if x is negative one, we still converge, but then x has to be less than one, because right at one, we diverge, so we can't say less than or equal to. So this is the interval of convergence for our original function. What about the interval of convergence for this one right over here, when we take the derivative? Well, when we take the derivative, this is, this is the same thing as x to the zero plus x to the first plus x to the second, x to the second, and we go on and on and on. Now you might recognize this. This is a geometric series with common ratio of x. Geometric series, series where our common ratio, often denoted by r, is equal to x."}, {"video_title": "Interval of convergence for derivative and integral Series AP Calculus BC Khan Academy.mp3", "Sentence": "What about the interval of convergence for this one right over here, when we take the derivative? Well, when we take the derivative, this is, this is the same thing as x to the zero plus x to the first plus x to the second, x to the second, and we go on and on and on. Now you might recognize this. This is a geometric series with common ratio of x. Geometric series, series where our common ratio, often denoted by r, is equal to x. And we know that a geometric series converges only in the situation where the, where our common ratio, where the absolute value of our common ratio, so converges, converges, only in the situation where the absolute value of our common ratio is less than one. So in this situation, when we took the derivative for f prime of x, our interval of convergence is almost the same. So here our interval, interval of convergence is going to be, x has to be between negative one and one."}, {"video_title": "Interval of convergence for derivative and integral Series AP Calculus BC Khan Academy.mp3", "Sentence": "This is a geometric series with common ratio of x. Geometric series, series where our common ratio, often denoted by r, is equal to x. And we know that a geometric series converges only in the situation where the, where our common ratio, where the absolute value of our common ratio, so converges, converges, only in the situation where the absolute value of our common ratio is less than one. So in this situation, when we took the derivative for f prime of x, our interval of convergence is almost the same. So here our interval, interval of convergence is going to be, x has to be between negative one and one. But it can't be equal to negative one. At negative one, we would actually diverge, and at one, we would diverge. So notice, these are almost the same."}, {"video_title": "Interval of convergence for derivative and integral Series AP Calculus BC Khan Academy.mp3", "Sentence": "So here our interval, interval of convergence is going to be, x has to be between negative one and one. But it can't be equal to negative one. At negative one, we would actually diverge, and at one, we would diverge. So notice, these are almost the same. If we view these as series centered at zero, the radius of convergence is the same. We can go one above, one below, one above, one below. And that's in general true as we take derivatives integrals."}, {"video_title": "Interval of convergence for derivative and integral Series AP Calculus BC Khan Academy.mp3", "Sentence": "So notice, these are almost the same. If we view these as series centered at zero, the radius of convergence is the same. We can go one above, one below, one above, one below. And that's in general true as we take derivatives integrals. But the endpoints of our interval of convergence can be different. And to continue to see this, I encourage you to use the ratio test to figure out one, what is the, well, use a ratio test plus using the boundary conditions, figure out what the interval of convergence is for the anti-derivative, for the integral here. And what you will see is, the radius of convergence is the same."}, {"video_title": "Interval of convergence for derivative and integral Series AP Calculus BC Khan Academy.mp3", "Sentence": "And that's in general true as we take derivatives integrals. But the endpoints of our interval of convergence can be different. And to continue to see this, I encourage you to use the ratio test to figure out one, what is the, well, use a ratio test plus using the boundary conditions, figure out what the interval of convergence is for the anti-derivative, for the integral here. And what you will see is, the radius of convergence is the same. We can go one above zero and one below zero. We have to be in that interval. But as you will see, this one converges for x equals negative one or x equals one."}, {"video_title": "Interval of convergence for derivative and integral Series AP Calculus BC Khan Academy.mp3", "Sentence": "And what you will see is, the radius of convergence is the same. We can go one above zero and one below zero. We have to be in that interval. But as you will see, this one converges for x equals negative one or x equals one. I'll just cut to the chase here. So interval of, let me write that in yellow. The interval of convergence for this top one converges, converges for negative one is less than x is less than or equal to one."}, {"video_title": "Interval of convergence for derivative and integral Series AP Calculus BC Khan Academy.mp3", "Sentence": "But as you will see, this one converges for x equals negative one or x equals one. I'll just cut to the chase here. So interval of, let me write that in yellow. The interval of convergence for this top one converges, converges for negative one is less than x is less than or equal to one. So notice, they all have the same radius of convergence, but the interval of convergence, it differs at the endpoint. And if you wanna prove this one for yourself, I encourage you to use a very similar technique that we used for our original function. Use the ratio test."}, {"video_title": "Interval of convergence for derivative and integral Series AP Calculus BC Khan Academy.mp3", "Sentence": "The interval of convergence for this top one converges, converges for negative one is less than x is less than or equal to one. So notice, they all have the same radius of convergence, but the interval of convergence, it differs at the endpoint. And if you wanna prove this one for yourself, I encourage you to use a very similar technique that we used for our original function. Use the ratio test. You're gonna come to this conclusion right over here, and then test the cases when x is equal to one and x is equal to negative one. And you will see, when x is equal to negative one, you have an alternating p-series, so that's gonna converge. And then when x equals one, you're gonna have a p-series where the denominator has a degree larger than one or something similar to a p-series."}, {"video_title": "Worked example finding a Riemann sum using a table AP Calculus AB Khan Academy.mp3", "Sentence": "To do that, we are given a table of values for f. So I encourage you to pause the video and see if you can come up with an approximation for the area between the x-axis and the graph from x equals one to x equals 10 using a right Riemann sum with three equal subdivisions. So I'm assuming you've had a go at it. So now let's try to do that together. And this is interesting because we don't have a graph of the entire function, but we just have the value of the function at certain points. But as we'll see, this is all we need in order to get an approximation for the area. We don't know how close it is to the actual area with just these points, but it'll give us at least a right Riemann sum for the approximation or an approximation using a right Riemann sum. So let me just draw some axes here because whenever I do Riemann sums, you can do them without graphs, but it helps to think about what's going on if you can visualize it graphically."}, {"video_title": "Worked example finding a Riemann sum using a table AP Calculus AB Khan Academy.mp3", "Sentence": "And this is interesting because we don't have a graph of the entire function, but we just have the value of the function at certain points. But as we'll see, this is all we need in order to get an approximation for the area. We don't know how close it is to the actual area with just these points, but it'll give us at least a right Riemann sum for the approximation or an approximation using a right Riemann sum. So let me just draw some axes here because whenever I do Riemann sums, you can do them without graphs, but it helps to think about what's going on if you can visualize it graphically. So let's see. We are going from x equals one to x equals 10. So this is one, two, three, four, five, six, seven, eight, nine, 10."}, {"video_title": "Worked example finding a Riemann sum using a table AP Calculus AB Khan Academy.mp3", "Sentence": "So let me just draw some axes here because whenever I do Riemann sums, you can do them without graphs, but it helps to think about what's going on if you can visualize it graphically. So let's see. We are going from x equals one to x equals 10. So this is one, two, three, four, five, six, seven, eight, nine, 10. And so they give us the value of f of x when x equals one, when x equals two, three, four, when x equals seven, five, six, seven, eight, nine, 10, and x equals 10. And they tell us that when x is one, we're at six, and we go to eight, to three, five. So let me mark these off."}, {"video_title": "Worked example finding a Riemann sum using a table AP Calculus AB Khan Academy.mp3", "Sentence": "So this is one, two, three, four, five, six, seven, eight, nine, 10. And so they give us the value of f of x when x equals one, when x equals two, three, four, when x equals seven, five, six, seven, eight, nine, 10, and x equals 10. And they tell us that when x is one, we're at six, and we go to eight, to three, five. So let me mark these off. So we're gonna go up to eight. So one, two, three, four, five, six, seven, eight. And so what we know, when x is equal to one, f of one is six."}, {"video_title": "Worked example finding a Riemann sum using a table AP Calculus AB Khan Academy.mp3", "Sentence": "So let me mark these off. So we're gonna go up to eight. So one, two, three, four, five, six, seven, eight. And so what we know, when x is equal to one, f of one is six. So this is one, two, three, four, five, six, seven, eight. So this point right over here is f of one. This is the point one comma six."}, {"video_title": "Worked example finding a Riemann sum using a table AP Calculus AB Khan Academy.mp3", "Sentence": "And so what we know, when x is equal to one, f of one is six. So this is one, two, three, four, five, six, seven, eight. So this point right over here is f of one. This is the point one comma six. And then we have the point four comma eight. Four comma eight will put us right about there. And then we have seven comma three is on our graph, y equals f of x."}, {"video_title": "Worked example finding a Riemann sum using a table AP Calculus AB Khan Academy.mp3", "Sentence": "This is the point one comma six. And then we have the point four comma eight. Four comma eight will put us right about there. And then we have seven comma three is on our graph, y equals f of x. So seven comma three would put us right over there. And then we have 10 comma five. So 10 comma five would put us right over there."}, {"video_title": "Worked example finding a Riemann sum using a table AP Calculus AB Khan Academy.mp3", "Sentence": "And then we have seven comma three is on our graph, y equals f of x. So seven comma three would put us right over there. And then we have 10 comma five. So 10 comma five would put us right over there. That's all we know about the function. We don't know exactly what it looks like. Our function might look like this."}, {"video_title": "Worked example finding a Riemann sum using a table AP Calculus AB Khan Academy.mp3", "Sentence": "So 10 comma five would put us right over there. That's all we know about the function. We don't know exactly what it looks like. Our function might look like this. It might do something like this. Whoops, I drew a part that didn't look like a function. It might do something like this and oscillate really quickly."}, {"video_title": "Worked example finding a Riemann sum using a table AP Calculus AB Khan Academy.mp3", "Sentence": "Our function might look like this. It might do something like this. Whoops, I drew a part that didn't look like a function. It might do something like this and oscillate really quickly. It might be nice and smooth and just kind of go and do something just like that, kind of connect the dots. We don't know. But we can still do the approximation using a right Riemann sum with three equal subdivisions."}, {"video_title": "Worked example finding a Riemann sum using a table AP Calculus AB Khan Academy.mp3", "Sentence": "It might do something like this and oscillate really quickly. It might be nice and smooth and just kind of go and do something just like that, kind of connect the dots. We don't know. But we can still do the approximation using a right Riemann sum with three equal subdivisions. How do we do that? Well, we're thinking about the area from x equals one to x equals 10. So let me make those boundaries clear."}, {"video_title": "Worked example finding a Riemann sum using a table AP Calculus AB Khan Academy.mp3", "Sentence": "But we can still do the approximation using a right Riemann sum with three equal subdivisions. How do we do that? Well, we're thinking about the area from x equals one to x equals 10. So let me make those boundaries clear. So this is from x equals one to x equals 10. And what we want to do is have three equal subdivisions. And there's three very natural subdivisions here."}, {"video_title": "Worked example finding a Riemann sum using a table AP Calculus AB Khan Academy.mp3", "Sentence": "So let me make those boundaries clear. So this is from x equals one to x equals 10. And what we want to do is have three equal subdivisions. And there's three very natural subdivisions here. If we make each of our subdivisions three wide, so this could be a subdivision. And then this is another subdivision. And when you do Riemann sums, you don't have to have equal subdivisions, although that's what you'll often see."}, {"video_title": "Worked example finding a Riemann sum using a table AP Calculus AB Khan Academy.mp3", "Sentence": "And there's three very natural subdivisions here. If we make each of our subdivisions three wide, so this could be a subdivision. And then this is another subdivision. And when you do Riemann sums, you don't have to have equal subdivisions, although that's what you'll often see. So we've just divided going from one to 10 into three equal sections that are three wide. So that's three, this is three, and this is three. And so the question is, how do we define the height of these subdivisions which are going to end up being rectangles?"}, {"video_title": "Worked example finding a Riemann sum using a table AP Calculus AB Khan Academy.mp3", "Sentence": "And when you do Riemann sums, you don't have to have equal subdivisions, although that's what you'll often see. So we've just divided going from one to 10 into three equal sections that are three wide. So that's three, this is three, and this is three. And so the question is, how do we define the height of these subdivisions which are going to end up being rectangles? And that's where the right Riemann sum applies. If we were doing a left Riemann sum, we would use the left boundary of each of the subdivisions and the value of the function there to define the height of the rectangle. So this would be doing a left Riemann sum."}, {"video_title": "Worked example finding a Riemann sum using a table AP Calculus AB Khan Academy.mp3", "Sentence": "And so the question is, how do we define the height of these subdivisions which are going to end up being rectangles? And that's where the right Riemann sum applies. If we were doing a left Riemann sum, we would use the left boundary of each of the subdivisions and the value of the function there to define the height of the rectangle. So this would be doing a left Riemann sum. But we're doing a right Riemann sum. So we use the right boundary of each of these subdivisions to define the height. So our right boundary is when x equals four for this first section."}, {"video_title": "Worked example finding a Riemann sum using a table AP Calculus AB Khan Academy.mp3", "Sentence": "So this would be doing a left Riemann sum. But we're doing a right Riemann sum. So we use the right boundary of each of these subdivisions to define the height. So our right boundary is when x equals four for this first section. What is f of four? It's eight. So we're gonna use that as the height of this first rectangle that's approximating the area for this part of the curve."}, {"video_title": "Worked example finding a Riemann sum using a table AP Calculus AB Khan Academy.mp3", "Sentence": "So our right boundary is when x equals four for this first section. What is f of four? It's eight. So we're gonna use that as the height of this first rectangle that's approximating the area for this part of the curve. Similarly, for this second one, since we're using a right Riemann sum, we use the value of the function at the right boundary, the right boundary seven. So the value of the function is three. So this would be our second rectangle, our second division, I guess, used to approximate the area."}, {"video_title": "Worked example finding a Riemann sum using a table AP Calculus AB Khan Academy.mp3", "Sentence": "So we're gonna use that as the height of this first rectangle that's approximating the area for this part of the curve. Similarly, for this second one, since we're using a right Riemann sum, we use the value of the function at the right boundary, the right boundary seven. So the value of the function is three. So this would be our second rectangle, our second division, I guess, used to approximate the area. And then last but not least, we would use the right boundary of this third subdivision when x equals 10, f of 10 is five. So just like that. And so then our right Riemann approximation using our right Riemann sum with three equal subdivisions to approximate the area, we would just add the area of these rectangles."}, {"video_title": "Worked example finding a Riemann sum using a table AP Calculus AB Khan Academy.mp3", "Sentence": "So this would be our second rectangle, our second division, I guess, used to approximate the area. And then last but not least, we would use the right boundary of this third subdivision when x equals 10, f of 10 is five. So just like that. And so then our right Riemann approximation using our right Riemann sum with three equal subdivisions to approximate the area, we would just add the area of these rectangles. So this first rectangle, let's see, it is three wide. And how high is it? Well, the height here is f of four, which is eight."}, {"video_title": "Worked example finding a Riemann sum using a table AP Calculus AB Khan Academy.mp3", "Sentence": "And so then our right Riemann approximation using our right Riemann sum with three equal subdivisions to approximate the area, we would just add the area of these rectangles. So this first rectangle, let's see, it is three wide. And how high is it? Well, the height here is f of four, which is eight. So this is going to be 24 square units, whatever the units happen to be. This is going to be three times, the height here is three, f of seven is three. So that is nine square units."}, {"video_title": "Worked example finding a Riemann sum using a table AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the height here is f of four, which is eight. So this is going to be 24 square units, whatever the units happen to be. This is going to be three times, the height here is three, f of seven is three. So that is nine square units. And then here, this is three, the width is three times the height, f of 10 is five. So three times five, which gets us 15. And so our approximation of the area would be summing these three values up."}, {"video_title": "Worked example finding a Riemann sum using a table AP Calculus AB Khan Academy.mp3", "Sentence": "So that is nine square units. And then here, this is three, the width is three times the height, f of 10 is five. So three times five, which gets us 15. And so our approximation of the area would be summing these three values up. So this would be 24 plus, let's see, nine plus 15 would give us another 24. So it's 24 plus 24, it gets us to 48. There you go."}, {"video_title": "Worked example finding a Riemann sum using a table AP Calculus AB Khan Academy.mp3", "Sentence": "And so our approximation of the area would be summing these three values up. So this would be 24 plus, let's see, nine plus 15 would give us another 24. So it's 24 plus 24, it gets us to 48. There you go. We, just using that table of values, we've been able to find an approximation. Now once again, we don't know how good of an approximation is, it depends on what the function is doing. There's a world where it could be a very good approximation."}, {"video_title": "Worked example finding a Riemann sum using a table AP Calculus AB Khan Academy.mp3", "Sentence": "There you go. We, just using that table of values, we've been able to find an approximation. Now once again, we don't know how good of an approximation is, it depends on what the function is doing. There's a world where it could be a very good approximation. Maybe the function does something like this. Maybe the function just happens, let me make it a little bit, maybe the function does something like this, where in this case, what we just did would be a very good approximation. Or maybe the function does something like this, where in this situation, this might be a very bad approximation."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "And what we want to do in this video is find the derivative of f. And the key here is to recognize that f can actually be viewed as a composition of two functions. And we can diagram that out. What's going on here? Well, if you input an x into our function f, what's the first thing that you do? Well, you take the square root of it. So if we start off with some x, you input it, the first thing that you do is you take the square root of it. So you are going to take the square root of the input to produce the square root of x."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "Well, if you input an x into our function f, what's the first thing that you do? Well, you take the square root of it. So if we start off with some x, you input it, the first thing that you do is you take the square root of it. So you are going to take the square root of the input to produce the square root of x. And then what do you do? Well, you take the square root, and then you take the natural log of that. And so then you take the natural log of that."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "So you are going to take the square root of the input to produce the square root of x. And then what do you do? Well, you take the square root, and then you take the natural log of that. And so then you take the natural log of that. So you could view that as inputting it into another function that takes the natural log of whatever is inputted in. I'm making these little squares to show what you do with the input. And then what do you produce?"}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "And so then you take the natural log of that. So you could view that as inputting it into another function that takes the natural log of whatever is inputted in. I'm making these little squares to show what you do with the input. And then what do you produce? Well, you produce the natural log of the square root of x. Natural log of the square root of x, which is equal to f of x. So you could view f of x as this entire, as this entire set, or this entire, I guess you could say, this combination of functions right over there."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "And then what do you produce? Well, you produce the natural log of the square root of x. Natural log of the square root of x, which is equal to f of x. So you could view f of x as this entire, as this entire set, or this entire, I guess you could say, this combination of functions right over there. And that is f of x, which is essentially a composition of two functions. You're inputting into one function, then taking that output and inputting it into another. So you could have a function u here, which takes the square root of whatever it's input in, is."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "So you could view f of x as this entire, as this entire set, or this entire, I guess you could say, this combination of functions right over there. And that is f of x, which is essentially a composition of two functions. You're inputting into one function, then taking that output and inputting it into another. So you could have a function u here, which takes the square root of whatever it's input in, is. So u of x is equal to the square root of x. And then you take another, and then you take that output and input it into another function that we could call v. And what does v do? Well, it takes the natural log of whatever the input is."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "So you could have a function u here, which takes the square root of whatever it's input in, is. So u of x is equal to the square root of x. And then you take another, and then you take that output and input it into another function that we could call v. And what does v do? Well, it takes the natural log of whatever the input is. In this case, in the case of f, or in the case of how I've just diagrammed it, v is taking the natural log, the input happens to be square root of x. So it outputs the natural log of the square root of x. If we wanted to write v with x as an input, we would just say, well, that's just the natural log."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "Well, it takes the natural log of whatever the input is. In this case, in the case of f, or in the case of how I've just diagrammed it, v is taking the natural log, the input happens to be square root of x. So it outputs the natural log of the square root of x. If we wanted to write v with x as an input, we would just say, well, that's just the natural log. That is just the natural log of x. And as you can see here, f of x, and I color-coded it ahead of time, is equal to, f of x, f of x is equal to the natural log of the square root of x. So that is v of the square root of x, or v of u of x."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "If we wanted to write v with x as an input, we would just say, well, that's just the natural log. That is just the natural log of x. And as you can see here, f of x, and I color-coded it ahead of time, is equal to, f of x, f of x is equal to the natural log of the square root of x. So that is v of the square root of x, or v of u of x. So it is a composition, which tells you that, okay, if I'm trying to find the derivative here, the chain rule is going to be very, very, very, very useful. And the chain rule tells us that f prime of x is going to be equal to the derivative of, you can view it as the outside function, with respect to this inside function. So it's going to be v prime of u of x, v prime of u of x, times the derivative of this inside function with respect to x."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "So that is v of the square root of x, or v of u of x. So it is a composition, which tells you that, okay, if I'm trying to find the derivative here, the chain rule is going to be very, very, very, very useful. And the chain rule tells us that f prime of x is going to be equal to the derivative of, you can view it as the outside function, with respect to this inside function. So it's going to be v prime of u of x, v prime of u of x, times the derivative of this inside function with respect to x. So that's just u prime, u prime of x. So how do we evaluate these things? Well, we know how to take the derivative of u of x and v of x. u prime of x here is going to be equal to, well, remember, square root of x is just the same thing as x to the 1 1\u20442 power, so we can use the power rule, bring the 1 1\u20442 out front, so it becomes 1 1\u20442 x to the, and then take off one out of that exponent, so that's 1 1\u20442 minus one is negative 1 1\u20442 power."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "So it's going to be v prime of u of x, v prime of u of x, times the derivative of this inside function with respect to x. So that's just u prime, u prime of x. So how do we evaluate these things? Well, we know how to take the derivative of u of x and v of x. u prime of x here is going to be equal to, well, remember, square root of x is just the same thing as x to the 1 1\u20442 power, so we can use the power rule, bring the 1 1\u20442 out front, so it becomes 1 1\u20442 x to the, and then take off one out of that exponent, so that's 1 1\u20442 minus one is negative 1 1\u20442 power. And what is v of x, sorry, what is v prime of x? Well, the derivative of the natural log of x is one over x. We show that in other videos."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we know how to take the derivative of u of x and v of x. u prime of x here is going to be equal to, well, remember, square root of x is just the same thing as x to the 1 1\u20442 power, so we can use the power rule, bring the 1 1\u20442 out front, so it becomes 1 1\u20442 x to the, and then take off one out of that exponent, so that's 1 1\u20442 minus one is negative 1 1\u20442 power. And what is v of x, sorry, what is v prime of x? Well, the derivative of the natural log of x is one over x. We show that in other videos. And so we now know what u prime of x is. What is, we know what v prime of x is, but what is v prime of u of x? Well, v prime of u of x, wherever we see the x, we replace it, let me write that a little bit neater, we replace that with a u of x."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "We show that in other videos. And so we now know what u prime of x is. What is, we know what v prime of x is, but what is v prime of u of x? Well, v prime of u of x, wherever we see the x, we replace it, let me write that a little bit neater, we replace that with a u of x. So v prime of u of x is going to be equal to, is going to be equal to one over u of x, one over u of x, u of x, which is equal to, which is equal to one over, u of x is just the square root of x, one over the square root of x. So this thing right over here, we have figured out is one over the square root of x, and this thing, u prime of x, we figured out is 1 1\u20442 times x to the negative 1 1\u20442, and x to the negative 1 1\u20442, I could rewrite that as 1 1\u20442 times one over x to the 1 1\u20442, which is the same thing as 1 1\u20442 times one over the square root of x, or I could write that as one over two square roots of x. So what is this thing going to be?"}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "Well, v prime of u of x, wherever we see the x, we replace it, let me write that a little bit neater, we replace that with a u of x. So v prime of u of x is going to be equal to, is going to be equal to one over u of x, one over u of x, u of x, which is equal to, which is equal to one over, u of x is just the square root of x, one over the square root of x. So this thing right over here, we have figured out is one over the square root of x, and this thing, u prime of x, we figured out is 1 1\u20442 times x to the negative 1 1\u20442, and x to the negative 1 1\u20442, I could rewrite that as 1 1\u20442 times one over x to the 1 1\u20442, which is the same thing as 1 1\u20442 times one over the square root of x, or I could write that as one over two square roots of x. So what is this thing going to be? Well, this is going to be equal to, in green, v prime of u of x is one over the square root of x times, times u prime of x is one over two times the square root of x. Now what is this going to be equal to? Well, this is going to be equal to, this is just algebra at this point, one over, we have our two, and square root of x times square root of x is just x."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "So what is this thing going to be? Well, this is going to be equal to, in green, v prime of u of x is one over the square root of x times, times u prime of x is one over two times the square root of x. Now what is this going to be equal to? Well, this is going to be equal to, this is just algebra at this point, one over, we have our two, and square root of x times square root of x is just x. So it just simplifies to one over two x. So hopefully this made sense, and I intentionally diagrammed it out so you start to get that muscle in your brain going of recognizing the composite functions, and then making a little bit more sense of some of these expressions of the chain rule that you might see in your calculus class or in your calculus textbook. But as you get more practice, you'll be able to do it essentially without having to write out all of this."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "Well, this is going to be equal to, this is just algebra at this point, one over, we have our two, and square root of x times square root of x is just x. So it just simplifies to one over two x. So hopefully this made sense, and I intentionally diagrammed it out so you start to get that muscle in your brain going of recognizing the composite functions, and then making a little bit more sense of some of these expressions of the chain rule that you might see in your calculus class or in your calculus textbook. But as you get more practice, you'll be able to do it essentially without having to write out all of this. You'll say, okay, look, I have a composition. This is the natural log of the square root of x. This is v of u of x."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "But as you get more practice, you'll be able to do it essentially without having to write out all of this. You'll say, okay, look, I have a composition. This is the natural log of the square root of x. This is v of u of x. So what I want to do is I want to take the derivative of this outside function with respect to this inside function. So the derivative of natural log of something with respect to that something is one over that something. The derivative of natural log of something with respect to that something is one over that something."}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "This is v of u of x. So what I want to do is I want to take the derivative of this outside function with respect to this inside function. So the derivative of natural log of something with respect to that something is one over that something. The derivative of natural log of something with respect to that something is one over that something. So that's what we just did here. One way to think about it. What would natural log of x be?"}, {"video_title": "Worked example Derivative of ln(\u00c3\u0083\u00c2\u0083x) using the chain rule AP Calculus AB Khan Academy.mp3", "Sentence": "The derivative of natural log of something with respect to that something is one over that something. So that's what we just did here. One way to think about it. What would natural log of x be? Well, that'd be one over x, but it's not natural log of x. It's one over square root of x. So it's going to be one over the square root of x."}, {"video_title": "Expression for combined area of triangle and square Differential Calculus Khan Academy.mp3", "Sentence": "Let's say that I have a 100 meter long wire. So that is my wire right over there. It is 100 meters. And I'm going to make a cut someplace on this wire. And so let's say I make the cut right over there. With the left section of wire, I'm going to obviously cut it in two. With the left section, I am going to construct an equilateral triangle."}, {"video_title": "Expression for combined area of triangle and square Differential Calculus Khan Academy.mp3", "Sentence": "And I'm going to make a cut someplace on this wire. And so let's say I make the cut right over there. With the left section of wire, I'm going to obviously cut it in two. With the left section, I am going to construct an equilateral triangle. And with the right section, I am going to construct a square. I am going to construct a square. And my question for you and for me is, where do we make this cut in order to minimize the combined areas of this triangle and this square?"}, {"video_title": "Expression for combined area of triangle and square Differential Calculus Khan Academy.mp3", "Sentence": "With the left section, I am going to construct an equilateral triangle. And with the right section, I am going to construct a square. I am going to construct a square. And my question for you and for me is, where do we make this cut in order to minimize the combined areas of this triangle and this square? Well, let's figure out, let's define a variable that we're trying to minimize or that we're trying to optimize with respect to. So let's say that the variable x is the number of meters that we decide to cut from the left. So if we did that, then this length for the triangle would be x meters."}, {"video_title": "Expression for combined area of triangle and square Differential Calculus Khan Academy.mp3", "Sentence": "And my question for you and for me is, where do we make this cut in order to minimize the combined areas of this triangle and this square? Well, let's figure out, let's define a variable that we're trying to minimize or that we're trying to optimize with respect to. So let's say that the variable x is the number of meters that we decide to cut from the left. So if we did that, then this length for the triangle would be x meters. And the length for the square would be, well, if we use x up for the left-hand side, we're going to have 100 minus x for the right-hand side. And so what would the dimensions of the triangle and the square be? Well, the triangle sides are going to be x over 3, x over 3, and x over 3 is an equilateral triangle."}, {"video_title": "Expression for combined area of triangle and square Differential Calculus Khan Academy.mp3", "Sentence": "So if we did that, then this length for the triangle would be x meters. And the length for the square would be, well, if we use x up for the left-hand side, we're going to have 100 minus x for the right-hand side. And so what would the dimensions of the triangle and the square be? Well, the triangle sides are going to be x over 3, x over 3, and x over 3 is an equilateral triangle. And the square is going to be 100 minus x over 4 by 100 minus x over 4. Now, it's easy to figure out an expression for the area of this square in terms of x. But let's think about what the area of an equilateral triangle might be as a function of the length of its sides."}, {"video_title": "Expression for combined area of triangle and square Differential Calculus Khan Academy.mp3", "Sentence": "Well, the triangle sides are going to be x over 3, x over 3, and x over 3 is an equilateral triangle. And the square is going to be 100 minus x over 4 by 100 minus x over 4. Now, it's easy to figure out an expression for the area of this square in terms of x. But let's think about what the area of an equilateral triangle might be as a function of the length of its sides. So let me do a little bit of an aside right over here. So let's say we have an equilateral triangle just like that. And its sides are length s, s, and s. Now, we know that the area of a triangle is 1 half times the base times the height."}, {"video_title": "Expression for combined area of triangle and square Differential Calculus Khan Academy.mp3", "Sentence": "But let's think about what the area of an equilateral triangle might be as a function of the length of its sides. So let me do a little bit of an aside right over here. So let's say we have an equilateral triangle just like that. And its sides are length s, s, and s. Now, we know that the area of a triangle is 1 half times the base times the height. So in this case, the height we could consider to be an altitude. If we were to drop an altitude just like this, this length right over here, this is the height. And this would be perpendicular just like that."}, {"video_title": "Expression for combined area of triangle and square Differential Calculus Khan Academy.mp3", "Sentence": "And its sides are length s, s, and s. Now, we know that the area of a triangle is 1 half times the base times the height. So in this case, the height we could consider to be an altitude. If we were to drop an altitude just like this, this length right over here, this is the height. And this would be perpendicular just like that. So our area is going to be equal to 1 half times our base is s, 1 half times s times whatever our height is. Now, how can we express h as a function of s? Well, to do that, we just have to remind ourselves that what we've drawn over here is a right triangle."}, {"video_title": "Expression for combined area of triangle and square Differential Calculus Khan Academy.mp3", "Sentence": "And this would be perpendicular just like that. So our area is going to be equal to 1 half times our base is s, 1 half times s times whatever our height is. Now, how can we express h as a function of s? Well, to do that, we just have to remind ourselves that what we've drawn over here is a right triangle. It's the left half of this equilateral triangle. And we know what this bottom side of this right triangle is. This altitude splits this side exactly into two."}, {"video_title": "Expression for combined area of triangle and square Differential Calculus Khan Academy.mp3", "Sentence": "Well, to do that, we just have to remind ourselves that what we've drawn over here is a right triangle. It's the left half of this equilateral triangle. And we know what this bottom side of this right triangle is. This altitude splits this side exactly into two. So this right over here has length s over 2. So to figure out what h is, we can just use the Pythagorean theorem. We would have h squared plus s over 2 squared is going to be equal to the hypotenuse squared."}, {"video_title": "Expression for combined area of triangle and square Differential Calculus Khan Academy.mp3", "Sentence": "This altitude splits this side exactly into two. So this right over here has length s over 2. So to figure out what h is, we can just use the Pythagorean theorem. We would have h squared plus s over 2 squared is going to be equal to the hypotenuse squared. It is going to be equal to s squared. So you would get h squared plus s squared over 4 is equal to s squared. Subtract s squared over 4 from both sides, and you get h squared is equal to s squared minus s squared over 4."}, {"video_title": "Expression for combined area of triangle and square Differential Calculus Khan Academy.mp3", "Sentence": "We would have h squared plus s over 2 squared is going to be equal to the hypotenuse squared. It is going to be equal to s squared. So you would get h squared plus s squared over 4 is equal to s squared. Subtract s squared over 4 from both sides, and you get h squared is equal to s squared minus s squared over 4. Now, to do this, I could call s squared. I could call this 4s squared over 4, just to be able to have a common denominator. And 4s squared minus s squared over 4 is going to be equal to 3s squared over 4."}, {"video_title": "Expression for combined area of triangle and square Differential Calculus Khan Academy.mp3", "Sentence": "Subtract s squared over 4 from both sides, and you get h squared is equal to s squared minus s squared over 4. Now, to do this, I could call s squared. I could call this 4s squared over 4, just to be able to have a common denominator. And 4s squared minus s squared over 4 is going to be equal to 3s squared over 4. So we get h squared is equal to 3s squared over 4. Now we can take the principal root of both sides, and we get h is equal to the square root of 3 times s over 2. So now we can just substitute back right over here, and we get our area is equal to 1 half s times h. Well, h is this business, so it's s times this."}, {"video_title": "Expression for combined area of triangle and square Differential Calculus Khan Academy.mp3", "Sentence": "And 4s squared minus s squared over 4 is going to be equal to 3s squared over 4. So we get h squared is equal to 3s squared over 4. Now we can take the principal root of both sides, and we get h is equal to the square root of 3 times s over 2. So now we can just substitute back right over here, and we get our area is equal to 1 half s times h. Well, h is this business, so it's s times this. So it's 1 half times s times the square root of 3s over 2, which is going to be equal to s times s is s squared. So it's going to be square root of 3s squared over 2 times 2 over 4. So this is the area of an equilateral triangle as the function of the length of its sides."}, {"video_title": "Expression for combined area of triangle and square Differential Calculus Khan Academy.mp3", "Sentence": "So now we can just substitute back right over here, and we get our area is equal to 1 half s times h. Well, h is this business, so it's s times this. So it's 1 half times s times the square root of 3s over 2, which is going to be equal to s times s is s squared. So it's going to be square root of 3s squared over 2 times 2 over 4. So this is the area of an equilateral triangle as the function of the length of its sides. So what's the area of this business going to be? So the area of our little equilateral triangle, let me write combined area. Let me do that in a neutral color."}, {"video_title": "Expression for combined area of triangle and square Differential Calculus Khan Academy.mp3", "Sentence": "So this is the area of an equilateral triangle as the function of the length of its sides. So what's the area of this business going to be? So the area of our little equilateral triangle, let me write combined area. Let me do that in a neutral color. So let me do that in white. So the combined area, I'll write it a sub c, is going to be equal to the area of my triangle, a sub t, plus the area of my square. Well, the area of my triangle, we know what it's going to be."}, {"video_title": "Expression for combined area of triangle and square Differential Calculus Khan Academy.mp3", "Sentence": "Let me do that in a neutral color. So let me do that in white. So the combined area, I'll write it a sub c, is going to be equal to the area of my triangle, a sub t, plus the area of my square. Well, the area of my triangle, we know what it's going to be. It's going to be square root of 3 times the length of a side squared divided by 4. So it's going to be square root of 3. Let me do that in that same yellow color."}, {"video_title": "Expression for combined area of triangle and square Differential Calculus Khan Academy.mp3", "Sentence": "Well, the area of my triangle, we know what it's going to be. It's going to be square root of 3 times the length of a side squared divided by 4. So it's going to be square root of 3. Let me do that in that same yellow color. It's going to be, make sure I switch colors, it's going to be square root of 3 over 4 times the side squared, times x over 3 squared. All I did is the length of a side is x over 3. We already know what the area is."}, {"video_title": "Expression for combined area of triangle and square Differential Calculus Khan Academy.mp3", "Sentence": "Let me do that in that same yellow color. It's going to be, make sure I switch colors, it's going to be square root of 3 over 4 times the side squared, times x over 3 squared. All I did is the length of a side is x over 3. We already know what the area is. It's square root of 3 over 4 times the length of a side squared. And then the area of this square right over here, the area of the square is just going to be 100 minus x over 4 squared. So our area, our combined area, maybe I could write like this."}, {"video_title": "Expression for combined area of triangle and square Differential Calculus Khan Academy.mp3", "Sentence": "We already know what the area is. It's square root of 3 over 4 times the length of a side squared. And then the area of this square right over here, the area of the square is just going to be 100 minus x over 4 squared. So our area, our combined area, maybe I could write like this. Our combined area as a function of where we make the cut is all of this business right over here. And this is what we need to minimize. So we need to minimize that right over there."}, {"video_title": "Horizontal area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "In fact, we can write this top expression as being a function of y, and this second one, just to make it different, we could view this as g of y. Once again, it is a function of y. And what we're concerning ourselves with in this video is how do we find this area in this light blue color between these two curves? And I encourage you to pause the video and try to work through it. All right, so a huge hint here is we're going to want to integrate with respect to y, a definite integral where our bounds are in terms of y. So for example, this is this lower point of intersection right over here. This would be our lower bound in terms of y."}, {"video_title": "Horizontal area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "And I encourage you to pause the video and try to work through it. All right, so a huge hint here is we're going to want to integrate with respect to y, a definite integral where our bounds are in terms of y. So for example, this is this lower point of intersection right over here. This would be our lower bound in terms of y. Let's call that y one. And then this up here, this would be our upper y bound. So if we think about where do these two curves intersect, and we look at the y coordinates of those intersections, well, that gives us two nice bounds for our integral."}, {"video_title": "Horizontal area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "This would be our lower bound in terms of y. Let's call that y one. And then this up here, this would be our upper y bound. So if we think about where do these two curves intersect, and we look at the y coordinates of those intersections, well, that gives us two nice bounds for our integral. So we're gonna take our integral from y one to y two, from y one to y two, y two. And we're going to integrate with respect to y, dy. And so what are we going to sum up?"}, {"video_title": "Horizontal area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So if we think about where do these two curves intersect, and we look at the y coordinates of those intersections, well, that gives us two nice bounds for our integral. So we're gonna take our integral from y one to y two, from y one to y two, y two. And we're going to integrate with respect to y, dy. And so what are we going to sum up? Well, when we integrate, we can think about taking the sum of infinitely thin rectangles. And in this case, it would be infinitely flat rectangles since we're thinking about dy. So dy would be the height of each of these rectangles."}, {"video_title": "Horizontal area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "And so what are we going to sum up? Well, when we integrate, we can think about taking the sum of infinitely thin rectangles. And in this case, it would be infinitely flat rectangles since we're thinking about dy. So dy would be the height of each of these rectangles. And what would be, in this case, the width or the length of this rectangle right over here? Well, over this interval from y one to y two, our blue function, f of y, takes on larger x values than g of y. So this length right over here, this would be f of y, f of y, this x value, minus this x value, minus g of y."}, {"video_title": "Horizontal area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So dy would be the height of each of these rectangles. And what would be, in this case, the width or the length of this rectangle right over here? Well, over this interval from y one to y two, our blue function, f of y, takes on larger x values than g of y. So this length right over here, this would be f of y, f of y, this x value, minus this x value, minus g of y. So this is going to be f of y minus g of y, g of y. Well, we know what f of y and g of y are. Really, the trickiest part is is figuring out these points of intersection."}, {"video_title": "Horizontal area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So this length right over here, this would be f of y, f of y, this x value, minus this x value, minus g of y. So this is going to be f of y minus g of y, g of y. Well, we know what f of y and g of y are. Really, the trickiest part is is figuring out these points of intersection. So let's think about where these two curves intersect. They are both equal to x, so we can set these two y expressions equal to each other. So we know that negative, let me do it in that other color."}, {"video_title": "Horizontal area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Really, the trickiest part is is figuring out these points of intersection. So let's think about where these two curves intersect. They are both equal to x, so we can set these two y expressions equal to each other. So we know that negative, let me do it in that other color. So we know that negative y squared plus three y plus 11 is going to be equal to this, is going to be equal to y squared plus y minus one. So let's just subtract all of this from both sides so that on the right side we have a zero and on the left side we just have a quadratic. So let's subtract y squared, let's subtract y, and then subtract negative one, which is just adding one, is, and over here we're going to do the same thing, minus y plus one."}, {"video_title": "Horizontal area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So we know that negative, let me do it in that other color. So we know that negative y squared plus three y plus 11 is going to be equal to this, is going to be equal to y squared plus y minus one. So let's just subtract all of this from both sides so that on the right side we have a zero and on the left side we just have a quadratic. So let's subtract y squared, let's subtract y, and then subtract negative one, which is just adding one, is, and over here we're going to do the same thing, minus y plus one. And what we are left with is going to be hopefully a straightforward quadratic. So let's see, this is going to be negative two y squared plus two y, am I doing that right? Yeah, plus two y, plus 12 is equal to zero."}, {"video_title": "Horizontal area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So let's subtract y squared, let's subtract y, and then subtract negative one, which is just adding one, is, and over here we're going to do the same thing, minus y plus one. And what we are left with is going to be hopefully a straightforward quadratic. So let's see, this is going to be negative two y squared plus two y, am I doing that right? Yeah, plus two y, plus 12 is equal to zero. And then this over here I can factor out a negative two, and I get negative two times y squared minus y minus six is equal to zero. This we can factor from inspection. What two numbers, when we add, equal negative one?"}, {"video_title": "Horizontal area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "Yeah, plus two y, plus 12 is equal to zero. And then this over here I can factor out a negative two, and I get negative two times y squared minus y minus six is equal to zero. This we can factor from inspection. What two numbers, when we add, equal negative one? When we take their product we get negative six. Well that would be negative three and two. So this is going to be negative two times y minus three times y plus two, that's just straightforward factoring a polynomial, a quadratic."}, {"video_title": "Horizontal area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "What two numbers, when we add, equal negative one? When we take their product we get negative six. Well that would be negative three and two. So this is going to be negative two times y minus three times y plus two, that's just straightforward factoring a polynomial, a quadratic. Did I do that right? Yep, that looks right, is equal to zero. So what are the points of intersection?"}, {"video_title": "Horizontal area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be negative two times y minus three times y plus two, that's just straightforward factoring a polynomial, a quadratic. Did I do that right? Yep, that looks right, is equal to zero. So what are the points of intersection? The points of intersection are going to be y is equal to three and y is equal to negative two. So this right over here is y is equal to negative two, and then the upper bound is y is equal to three. So now we just have to evaluate this from negative two all the way until three."}, {"video_title": "Horizontal area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So what are the points of intersection? The points of intersection are going to be y is equal to three and y is equal to negative two. So this right over here is y is equal to negative two, and then the upper bound is y is equal to three. So now we just have to evaluate this from negative two all the way until three. So let's do that, I'm going to clear this out so I get a little bit of real estate. So this is equal to the integral from negative two to three of negative y squared plus three y plus 11 minus all of this stuff. So if we just distribute a negative sign here, it's minus y squared minus y plus one and then we have a dy, dy."}, {"video_title": "Horizontal area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So now we just have to evaluate this from negative two all the way until three. So let's do that, I'm going to clear this out so I get a little bit of real estate. So this is equal to the integral from negative two to three of negative y squared plus three y plus 11 minus all of this stuff. So if we just distribute a negative sign here, it's minus y squared minus y plus one and then we have a dy, dy. This is equal to the definite integral from negative two to positive three of, let's see, negative y squared minus y squared, negative two y squared and then three y minus y is going to be plus two y and then 11 plus one plus 12. We saw this just now when we were trying to solve for y, dy. And so what is that going to be equal to?"}, {"video_title": "Horizontal area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So if we just distribute a negative sign here, it's minus y squared minus y plus one and then we have a dy, dy. This is equal to the definite integral from negative two to positive three of, let's see, negative y squared minus y squared, negative two y squared and then three y minus y is going to be plus two y and then 11 plus one plus 12. We saw this just now when we were trying to solve for y, dy. And so what is that going to be equal to? Well, we just take the antiderivative here. This is going to be, let's see, negative two, let's increment the exponent, y to the third, divide by that exponent, reverse power rule, plus two y squared divided by two, which is just y squared, just the reverse power rule, and then plus 12 y and we're going to evaluate that at three and at negative two. So if we evaluate that at three, we are going to get, let's see, negative two times 27 over three plus nine plus 36 and then we are going to want to subtract minus, all of this evaluated at negative two."}, {"video_title": "Horizontal area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "And so what is that going to be equal to? Well, we just take the antiderivative here. This is going to be, let's see, negative two, let's increment the exponent, y to the third, divide by that exponent, reverse power rule, plus two y squared divided by two, which is just y squared, just the reverse power rule, and then plus 12 y and we're going to evaluate that at three and at negative two. So if we evaluate that at three, we are going to get, let's see, negative two times 27 over three plus nine plus 36 and then we are going to want to subtract minus, all of this evaluated at negative two. So it's going to be negative two times negative eight over three plus four minus 24. So we just have a little bit of mathematics ahead of us. So let's see, this is going to be 27 divided by three is nine so this is negative 18, negative 18 plus nine is going to be negative nine plus 36."}, {"video_title": "Horizontal area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So if we evaluate that at three, we are going to get, let's see, negative two times 27 over three plus nine plus 36 and then we are going to want to subtract minus, all of this evaluated at negative two. So it's going to be negative two times negative eight over three plus four minus 24. So we just have a little bit of mathematics ahead of us. So let's see, this is going to be 27 divided by three is nine so this is negative 18, negative 18 plus nine is going to be negative nine plus 36. All of that is going to be equal to, so the stuff in blue is equal to 27, right? Did I do that right? You got negative 18 plus nine, yep, 27."}, {"video_title": "Horizontal area between curves Applications of definite integrals AP Calculus AB Khan Academy.mp3", "Sentence": "So let's see, this is going to be 27 divided by three is nine so this is negative 18, negative 18 plus nine is going to be negative nine plus 36. All of that is going to be equal to, so the stuff in blue is equal to 27, right? Did I do that right? You got negative 18 plus nine, yep, 27. And then all the stuff in red over here, we have, this is going to be negative times a negative, so it's 16 over three plus four minus 24. So that is going to be 16 over three and then minus 20, but then we have this negative out here so if we distribute that, we'll get plus 20 minus, we could say instead of 16 over three, we could rewrite that as five and 1 3rd, minus five and 1 3rd and so what is that going to get us? Let me scroll down a little bit or let me go to the right so I have a little bit more real estate here."}, {"video_title": "Exponential functions differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "What is the derivative of y, derivative of y, with respect to x? And like always, pause this video and see if you can figure it out. Well, based on how this has been color-coded ahead of time, you might immediately recognize that this is a composite function, or it could be viewed as a composite function. If you had a v of x, which if you had a function v of x, which is equal to seven to the x power, and you had another function u of x, u of x, which is equal to x squared minus x, then what we have right over here, y, y is equal to seven to something. So it's equal to v of, and it's not just v of x, it's v of u of x. Instead of an x here, you have the whole function u of x, x squared minus x. So it's v of u of x, and the chain rule tells us that the derivative of y with respect to x, and you'll see different notations here, sometimes you'll see it written as the derivative of v with respect to u, so v prime of u of x, times the derivative of u with respect to x, so that's one way you could do it, or you could say that this is equal to, this is equal to the derivative, the derivative of v with respect to x, derivative, or sorry, derivative of v with respect to u, dv du, times the derivative of u with respect to x, derivative of u with respect to x."}, {"video_title": "Exponential functions differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "If you had a v of x, which if you had a function v of x, which is equal to seven to the x power, and you had another function u of x, u of x, which is equal to x squared minus x, then what we have right over here, y, y is equal to seven to something. So it's equal to v of, and it's not just v of x, it's v of u of x. Instead of an x here, you have the whole function u of x, x squared minus x. So it's v of u of x, and the chain rule tells us that the derivative of y with respect to x, and you'll see different notations here, sometimes you'll see it written as the derivative of v with respect to u, so v prime of u of x, times the derivative of u with respect to x, so that's one way you could do it, or you could say that this is equal to, this is equal to the derivative, the derivative of v with respect to x, derivative, or sorry, derivative of v with respect to u, dv du, times the derivative of u with respect to x, derivative of u with respect to x. And so either way, we can apply that right over here. So what's the derivative of v with respect to u? What is v prime of u of x?"}, {"video_title": "Exponential functions differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So it's v of u of x, and the chain rule tells us that the derivative of y with respect to x, and you'll see different notations here, sometimes you'll see it written as the derivative of v with respect to u, so v prime of u of x, times the derivative of u with respect to x, so that's one way you could do it, or you could say that this is equal to, this is equal to the derivative, the derivative of v with respect to x, derivative, or sorry, derivative of v with respect to u, dv du, times the derivative of u with respect to x, derivative of u with respect to x. And so either way, we can apply that right over here. So what's the derivative of v with respect to u? What is v prime of u of x? Well, we know, we know, let me actually write it right over here. If v of x is equal to seven to the x power, v prime of x would be equal to, and we proved this in other videos where we take derivatives, exponentials of bases other than e, this is going to be the natural log of seven, times seven to the x power. So if we are taking v prime of u of x, then notice, instead of an x everywhere, we're going to have a u of x everywhere."}, {"video_title": "Exponential functions differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "What is v prime of u of x? Well, we know, we know, let me actually write it right over here. If v of x is equal to seven to the x power, v prime of x would be equal to, and we proved this in other videos where we take derivatives, exponentials of bases other than e, this is going to be the natural log of seven, times seven to the x power. So if we are taking v prime of u of x, then notice, instead of an x everywhere, we're going to have a u of x everywhere. So this right over here, this is going to be natural log of seven times seven to the, instead of saying seven to the x power, remember, we're taking v prime of u of x. So it's going to be seven to the x squared minus x power. X squared, x squared minus x power."}, {"video_title": "Exponential functions differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So if we are taking v prime of u of x, then notice, instead of an x everywhere, we're going to have a u of x everywhere. So this right over here, this is going to be natural log of seven times seven to the, instead of saying seven to the x power, remember, we're taking v prime of u of x. So it's going to be seven to the x squared minus x power. X squared, x squared minus x power. And then we want to multiply that times the derivative of u with respect to x. So u prime of x, well that's going to be two x to the first, which is just two x, minus one. So we're going to multiply this times two x, two x minus one."}, {"video_title": "Exponential functions differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "X squared, x squared minus x power. And then we want to multiply that times the derivative of u with respect to x. So u prime of x, well that's going to be two x to the first, which is just two x, minus one. So we're going to multiply this times two x, two x minus one. So there you have it. That is the derivative of y with respect to x. You could, we could try to simplify this or I guess re-express it in different ways, but the main thing to realize is, look, we're just going to take the derivative of the seven to the, this to the u of x power with respect to u of x."}, {"video_title": "Exponential functions differentiation Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So we're going to multiply this times two x, two x minus one. So there you have it. That is the derivative of y with respect to x. You could, we could try to simplify this or I guess re-express it in different ways, but the main thing to realize is, look, we're just going to take the derivative of the seven to the, this to the u of x power with respect to u of x. So we treat the u of x the way that we would have treated an x right over here. So it's going to be natural log of seven times seven to the u of x power. We take that and multiply that times u prime of x."}, {"video_title": "Secant lines & average rate of change Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So let me write that down. We want to know the average rate of change of y with respect to x over the interval from x going from one to three, and it's a closed interval where x could be one and x could be equal to three. Well we could do this even without looking at the graph. If I were to just make a table here where if this is x and this is y is equal to x squared, when x is equal to one, y is equal to one squared, which is just one. You see that right over there. And when x is equal to three, y is equal to three squared, which is equal to nine. And so you can see when x is equal to three, y is equal to nine."}, {"video_title": "Secant lines & average rate of change Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "If I were to just make a table here where if this is x and this is y is equal to x squared, when x is equal to one, y is equal to one squared, which is just one. You see that right over there. And when x is equal to three, y is equal to three squared, which is equal to nine. And so you can see when x is equal to three, y is equal to nine. And to figure out the average rate of change of y with respect to x, you say okay, well what's my change in x? Well we can see very clearly that our change in x over this interval is equal to positive two. Well what's our change in y over the same interval?"}, {"video_title": "Secant lines & average rate of change Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And so you can see when x is equal to three, y is equal to nine. And to figure out the average rate of change of y with respect to x, you say okay, well what's my change in x? Well we can see very clearly that our change in x over this interval is equal to positive two. Well what's our change in y over the same interval? Our change in y is equal to, when x went increased by two from one to three, y increases by eight. So it's gonna be a positive eight. So what is our average rate of change?"}, {"video_title": "Secant lines & average rate of change Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Well what's our change in y over the same interval? Our change in y is equal to, when x went increased by two from one to three, y increases by eight. So it's gonna be a positive eight. So what is our average rate of change? Well it's gonna be our change in y over our change in x, which is equal to eight over two, which is equal to four. So that would be our average rate of change. Over that interval, on average, every time x increases by one, y is increasing by four."}, {"video_title": "Secant lines & average rate of change Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So what is our average rate of change? Well it's gonna be our change in y over our change in x, which is equal to eight over two, which is equal to four. So that would be our average rate of change. Over that interval, on average, every time x increases by one, y is increasing by four. And how did we calculate that? We looked at our change in x. Let me draw that here."}, {"video_title": "Secant lines & average rate of change Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Over that interval, on average, every time x increases by one, y is increasing by four. And how did we calculate that? We looked at our change in x. Let me draw that here. We looked at our change in x. And we looked at our change in y, which would be this right over here. And we calculated change in y over change of x for average rate of change."}, {"video_title": "Secant lines & average rate of change Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Let me draw that here. We looked at our change in x. And we looked at our change in y, which would be this right over here. And we calculated change in y over change of x for average rate of change. Now this might be looking fairly familiar to you, because you're used to thinking about change in y over change in x as the slope of a line connecting two points. And that's indeed what we did calculate. If you were to draw a secant line between these two points, we essentially just calculated the slope of that secant line."}, {"video_title": "Secant lines & average rate of change Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And we calculated change in y over change of x for average rate of change. Now this might be looking fairly familiar to you, because you're used to thinking about change in y over change in x as the slope of a line connecting two points. And that's indeed what we did calculate. If you were to draw a secant line between these two points, we essentially just calculated the slope of that secant line. And so the average rate of change between two points, that is the same thing as the slope of the secant line. And by looking at the secant line in comparison to the curve over that interval, it hopefully gives you a visual intuition for what even average rate of change means. Because in the beginning part of the interval, you see that the secant line is actually increasing at a faster rate."}, {"video_title": "Secant lines & average rate of change Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "If you were to draw a secant line between these two points, we essentially just calculated the slope of that secant line. And so the average rate of change between two points, that is the same thing as the slope of the secant line. And by looking at the secant line in comparison to the curve over that interval, it hopefully gives you a visual intuition for what even average rate of change means. Because in the beginning part of the interval, you see that the secant line is actually increasing at a faster rate. But then as we get closer to three, it looks like our yellow curve is increasing at a faster rate than the secant line, and then they eventually catch up. And so that's why the slope of the secant line is the average rate of change. Is it the exact rate of change at every point?"}, {"video_title": "Secant lines & average rate of change Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Because in the beginning part of the interval, you see that the secant line is actually increasing at a faster rate. But then as we get closer to three, it looks like our yellow curve is increasing at a faster rate than the secant line, and then they eventually catch up. And so that's why the slope of the secant line is the average rate of change. Is it the exact rate of change at every point? Absolutely not. The curve's rate of change is constantly changing. It's at a slower rate of change in the beginning part of this interval, and then it's actually increasing at a higher rate as we get closer and closer to three."}, {"video_title": "Secant lines & average rate of change Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Is it the exact rate of change at every point? Absolutely not. The curve's rate of change is constantly changing. It's at a slower rate of change in the beginning part of this interval, and then it's actually increasing at a higher rate as we get closer and closer to three. So over the interval, their change in y over the change in x is exactly the same. Now one question you might be wondering is, why are you learning this in a calculus class? Couldn't you have learned this in an algebra class?"}, {"video_title": "Secant lines & average rate of change Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "It's at a slower rate of change in the beginning part of this interval, and then it's actually increasing at a higher rate as we get closer and closer to three. So over the interval, their change in y over the change in x is exactly the same. Now one question you might be wondering is, why are you learning this in a calculus class? Couldn't you have learned this in an algebra class? The answer is yes. But what's going to be interesting, and it's really one of the foundational ideas of calculus, is, well, what happens as these points get closer and closer together? We found the average rate of change between one and three, or the slope of the secant line from one comma one to three comma nine."}, {"video_title": "Secant lines & average rate of change Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Couldn't you have learned this in an algebra class? The answer is yes. But what's going to be interesting, and it's really one of the foundational ideas of calculus, is, well, what happens as these points get closer and closer together? We found the average rate of change between one and three, or the slope of the secant line from one comma one to three comma nine. But what instead if you found the slope of the secant line between two comma four and three comma nine? So what if you found this slope? But what if you wanted to get even closer?"}, {"video_title": "Secant lines & average rate of change Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "We found the average rate of change between one and three, or the slope of the secant line from one comma one to three comma nine. But what instead if you found the slope of the secant line between two comma four and three comma nine? So what if you found this slope? But what if you wanted to get even closer? Let's say you wanted to find the slope of the secant line between the point 2.5, 6.25, and three comma nine. And what if you just kept getting closer and closer and closer? Well then, the slopes of these secant lines are gonna get closer and closer to the slope of the tangent line at x equals three."}, {"video_title": "Secant lines & average rate of change Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "But what if you wanted to get even closer? Let's say you wanted to find the slope of the secant line between the point 2.5, 6.25, and three comma nine. And what if you just kept getting closer and closer and closer? Well then, the slopes of these secant lines are gonna get closer and closer to the slope of the tangent line at x equals three. And if we can figure out the slope of the tangent line, well then we're in business. Because then we're not talking about average rate of change, we're gonna be talking about instantaneous rate of change, which is one of the central ideas. That is the derivative, and we're going to get there soon."}, {"video_title": "Related rates Falling ladder Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So I've got a 10-foot ladder that's leaning against a wall, but it's on very slick ground, and it starts to slide outward. And right at the moment that we're looking at this ladder, the base of the ladder is 8 feet away from the base of the wall, and it's sliding outward at 4 feet per second. And we'll assume that the top of the ladder kind of glides along the side of the wall. It stays kind of in contact with the wall and moves straight down. And we see right over here, the arrow is moving straight down. And our question is, how fast is it moving straight down at that moment? So let's think about this a little bit."}, {"video_title": "Related rates Falling ladder Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "It stays kind of in contact with the wall and moves straight down. And we see right over here, the arrow is moving straight down. And our question is, how fast is it moving straight down at that moment? So let's think about this a little bit. What do we know and what do we not know? So if we call the distance between the base of the wall and the base of the ladder, let's call that x. We know right now x is equal to 8 feet."}, {"video_title": "Related rates Falling ladder Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let's think about this a little bit. What do we know and what do we not know? So if we call the distance between the base of the wall and the base of the ladder, let's call that x. We know right now x is equal to 8 feet. We also know the rate at which x is changing with respect to time. The rate at which x is changing with respect to time is 4 feet per second. So we could call this dx dt."}, {"video_title": "Related rates Falling ladder Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We know right now x is equal to 8 feet. We also know the rate at which x is changing with respect to time. The rate at which x is changing with respect to time is 4 feet per second. So we could call this dx dt. Now let's call the distance between the top of the ladder and the base of the ladder h. Let's call that h. So what we're really trying to figure out is what dh dt is, given that we know all of this other information. So let's see if we can come up with a relationship between x and h, and then take the derivative with respect to time, maybe using the chain rule, and see if we can solve for dh dt knowing all of this information. Well, we know the relationship between x and h at any time because of the Pythagorean theorem."}, {"video_title": "Related rates Falling ladder Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So we could call this dx dt. Now let's call the distance between the top of the ladder and the base of the ladder h. Let's call that h. So what we're really trying to figure out is what dh dt is, given that we know all of this other information. So let's see if we can come up with a relationship between x and h, and then take the derivative with respect to time, maybe using the chain rule, and see if we can solve for dh dt knowing all of this information. Well, we know the relationship between x and h at any time because of the Pythagorean theorem. We can assume this is a right angle. So we know that x squared plus h squared is going to be equal to the length of the ladder squared, is going to be equal to 100. And what we care about is the rate at which these things change with respect to time."}, {"video_title": "Related rates Falling ladder Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we know the relationship between x and h at any time because of the Pythagorean theorem. We can assume this is a right angle. So we know that x squared plus h squared is going to be equal to the length of the ladder squared, is going to be equal to 100. And what we care about is the rate at which these things change with respect to time. So let's take the derivative with respect to time of both sides of this. So we're doing a little bit of implicit differentiation. So what's the derivative with respect to time of x squared?"}, {"video_title": "Related rates Falling ladder Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And what we care about is the rate at which these things change with respect to time. So let's take the derivative with respect to time of both sides of this. So we're doing a little bit of implicit differentiation. So what's the derivative with respect to time of x squared? Well, the derivative of x squared with respect to x is 2x. And we're going to have to multiply that times the derivative of x with respect to t, dx dt. Just to be clear, this is a chain rule."}, {"video_title": "Related rates Falling ladder Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So what's the derivative with respect to time of x squared? Well, the derivative of x squared with respect to x is 2x. And we're going to have to multiply that times the derivative of x with respect to t, dx dt. Just to be clear, this is a chain rule. This is the derivative of x squared with respect to x, which is 2x, times dx dt to get the derivative of x squared with respect to time, just the chain rule. Now similarly, what's the derivative of h squared with respect to time? Well, that's just going to be 2h."}, {"video_title": "Related rates Falling ladder Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Just to be clear, this is a chain rule. This is the derivative of x squared with respect to x, which is 2x, times dx dt to get the derivative of x squared with respect to time, just the chain rule. Now similarly, what's the derivative of h squared with respect to time? Well, that's just going to be 2h. The derivative of h squared with respect to h is 2h times the derivative of h with respect to time. Once again, this right over here is the derivative of h squared with respect to h times the derivative of h with respect to time, which gives us the derivative of h squared with respect to time. And what do we get on the right-hand side of our equation?"}, {"video_title": "Related rates Falling ladder Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, that's just going to be 2h. The derivative of h squared with respect to h is 2h times the derivative of h with respect to time. Once again, this right over here is the derivative of h squared with respect to h times the derivative of h with respect to time, which gives us the derivative of h squared with respect to time. And what do we get on the right-hand side of our equation? Well, the length of our ladder isn't changing. This 100 isn't going to change with respect to time. The derivative of a constant is just equal 0."}, {"video_title": "Related rates Falling ladder Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And what do we get on the right-hand side of our equation? Well, the length of our ladder isn't changing. This 100 isn't going to change with respect to time. The derivative of a constant is just equal 0. So now we have it, a relationship between the rate of change of h with respect to time, the rate of change of x with respect to time, and then at a given point in time when the length of x is x and h is h. But do we know what h is when x is equal to 8 feet? Well, we can figure it out. When x is equal to 8 feet, we can use the Pythagorean theorem again."}, {"video_title": "Related rates Falling ladder Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "The derivative of a constant is just equal 0. So now we have it, a relationship between the rate of change of h with respect to time, the rate of change of x with respect to time, and then at a given point in time when the length of x is x and h is h. But do we know what h is when x is equal to 8 feet? Well, we can figure it out. When x is equal to 8 feet, we can use the Pythagorean theorem again. We get 8 feet squared plus h squared is going to be equal to 100. So 8 squared is 64. Subtract it from both sides."}, {"video_title": "Related rates Falling ladder Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "When x is equal to 8 feet, we can use the Pythagorean theorem again. We get 8 feet squared plus h squared is going to be equal to 100. So 8 squared is 64. Subtract it from both sides. You get 8 squared is equal to 36. Take the positive square root. A negative square root doesn't make sense because then the ladder would be below the ground."}, {"video_title": "Related rates Falling ladder Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Subtract it from both sides. You get 8 squared is equal to 36. Take the positive square root. A negative square root doesn't make sense because then the ladder would be below the ground. It would be somehow underground. So we get h is equal to 6. So this is something that was essentially given by the problem."}, {"video_title": "Related rates Falling ladder Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "A negative square root doesn't make sense because then the ladder would be below the ground. It would be somehow underground. So we get h is equal to 6. So this is something that was essentially given by the problem. So now we know. We can look at this original thing right over here. We know what x is."}, {"video_title": "Related rates Falling ladder Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So this is something that was essentially given by the problem. So now we know. We can look at this original thing right over here. We know what x is. That was given. Right now, x is 8 feet. We know the rate of change of x with respect to time."}, {"video_title": "Related rates Falling ladder Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We know what x is. That was given. Right now, x is 8 feet. We know the rate of change of x with respect to time. It's 4 feet per second. We know what h is right now. It is 6."}, {"video_title": "Related rates Falling ladder Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We know the rate of change of x with respect to time. It's 4 feet per second. We know what h is right now. It is 6. So then we can solve for the rate of h with respect to time. So let's do that. So we get 2 times 8 feet times 4 feet per second plus 2h."}, {"video_title": "Related rates Falling ladder Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "It is 6. So then we can solve for the rate of h with respect to time. So let's do that. So we get 2 times 8 feet times 4 feet per second plus 2h. So plus 2h is going to be plus 2 times our height right now is 6 times the rate at which our height is changing with respect to t is equal to 0. And so we get 2 times 8 times 4 is 64 plus 12 dh dt is equal to 0. We can subtract 64 from both sides."}, {"video_title": "Related rates Falling ladder Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So we get 2 times 8 feet times 4 feet per second plus 2h. So plus 2h is going to be plus 2 times our height right now is 6 times the rate at which our height is changing with respect to t is equal to 0. And so we get 2 times 8 times 4 is 64 plus 12 dh dt is equal to 0. We can subtract 64 from both sides. We get 12 times the derivative of h with respect to time is equal to negative 64. And then we just have to divide both sides by 12. And so now we get a little bit of a drum roll."}, {"video_title": "Related rates Falling ladder Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We can subtract 64 from both sides. We get 12 times the derivative of h with respect to time is equal to negative 64. And then we just have to divide both sides by 12. And so now we get a little bit of a drum roll. The derivative, the rate of change of h with respect to time is equal to negative 64 divided by 12 is equal to negative 64 over 12, which is the same thing as negative 16 over 3. Yep, that's right, which is equal to negative 5 and 1 third feet per second. So we're done."}, {"video_title": "Related rates Falling ladder Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And so now we get a little bit of a drum roll. The derivative, the rate of change of h with respect to time is equal to negative 64 divided by 12 is equal to negative 64 over 12, which is the same thing as negative 16 over 3. Yep, that's right, which is equal to negative 5 and 1 third feet per second. So we're done. But let's just do a reality check. Does that make sense that we got a negative value right over here? Well, our height is decreasing."}, {"video_title": "Switching bounds of definite integral AP Calculus AB Khan Academy.mp3", "Sentence": "We've already seen one definition of the definite integral, and many of them are closely related to this definition that we've already seen, is, well, look, the definite integral from a to b of f of x, d of x is this area shaded in blue, and we can approximate it by splitting it into n rectangles, so let's say that's the first rectangle, one, that's the second rectangle, two, and you're gonna go all the way to the nth rectangle, so this would be the n minus oneth rectangle, and for the sake of this argument I'm gonna make in this video, I'm gonna assume that they're all the same width, so this is the nth rectangle, and they all have the same width, and we see there are definitions of integration where you don't have to have the same width here, but let's say that each of those widths are delta x, and the way that we calculate delta x is we take b minus a, we take b minus a, and we divide it by n, we divide it by n, which is common sense, or this is what you learned in division, we're just taking this length and dividing it by n to get n equal spacings, which is n equal spacings of delta x, and so if you do this, you say, okay, we see this multiple times, you can approximate it, you can approximate this area using these rectangles as the sum from i equals one to n, so you're summing, you're summing n of these rectangles areas, where the height of each of these rectangles are gonna be f of x sub i, where x sub i is the point at which you're taking the function value to find out its height, so that could be x sub one, x sub two, x sub three, so on and so forth, and you're multiplying that times your delta x, times your delta x, so you take x sub two, f of x sub two is that height right there, f of x sub two is that height right there, you multiply it times delta x, you get the area, and we saw that when we looked at Riemann sums and using that to approximate, and we said, hey, the one definition of the definite integral is that since this is the area, this is going to be the limit as n approaches infinity of this, where delta x is defined as that, so let me just copy and paste that, so copy and paste, where that, so that's one way to think about it. Now, given this definition, what do you think, what do you think this, or maybe another way to think about it, how do you think this expression that I'm writing right over here, based on this definition, should relate to this expression? So notice, all I've done is I've, instead of going from a to b, I'm now going from b to a, I'm now going from b to a. How do you think these two things should relate? And I encourage you to look at all of this to come to that conclusion, and pause the video to do so. Well, let's just think about what's going to happen. This is going to be, if I were to literally just take this, if I were to literally just take this and copy and paste it, which is exactly what I'm going to do, if I just took this, by definition, since I swapped these two bounds, I am going to want to swap these two."}, {"video_title": "Switching bounds of definite integral AP Calculus AB Khan Academy.mp3", "Sentence": "How do you think these two things should relate? And I encourage you to look at all of this to come to that conclusion, and pause the video to do so. Well, let's just think about what's going to happen. This is going to be, if I were to literally just take this, if I were to literally just take this and copy and paste it, which is exactly what I'm going to do, if I just took this, by definition, since I swapped these two bounds, I am going to want to swap these two. Instead of b minus a, it's going to be a minus b now. It's going to be, it's going to be a minus b. So each of these are going, this value right over here, let me make these color-coded maybe, so this orange delta x, this orange delta x is going to be the negative of this green, of this green delta x."}, {"video_title": "Switching bounds of definite integral AP Calculus AB Khan Academy.mp3", "Sentence": "This is going to be, if I were to literally just take this, if I were to literally just take this and copy and paste it, which is exactly what I'm going to do, if I just took this, by definition, since I swapped these two bounds, I am going to want to swap these two. Instead of b minus a, it's going to be a minus b now. It's going to be, it's going to be a minus b. So each of these are going, this value right over here, let me make these color-coded maybe, so this orange delta x, this orange delta x is going to be the negative of this green, of this green delta x. This is the negative of that right over there. And everything else is the same. So what am I going to end up doing?"}, {"video_title": "Switching bounds of definite integral AP Calculus AB Khan Academy.mp3", "Sentence": "So each of these are going, this value right over here, let me make these color-coded maybe, so this orange delta x, this orange delta x is going to be the negative of this green, of this green delta x. This is the negative of that right over there. And everything else is the same. So what am I going to end up doing? Well, I'm essentially going to end up having the negative value of this. So this is going to be equal to the negative of the integral from a to b of f of x dx. And so this is the result we get, which is another really important integration property that if you swap, if you swap the bounds of integration, and it really just comes from this idea."}, {"video_title": "Switching bounds of definite integral AP Calculus AB Khan Academy.mp3", "Sentence": "So what am I going to end up doing? Well, I'm essentially going to end up having the negative value of this. So this is going to be equal to the negative of the integral from a to b of f of x dx. And so this is the result we get, which is another really important integration property that if you swap, if you swap the bounds of integration, and it really just comes from this idea. Instead of delta x being b minus a, if you swap the bounds of integration, it's going to be a minus b. You're going to get the negative delta x, or the negative of your original delta x, which is going to give you the negative of this original value right over here. And once again, this is a really, really useful integration property where you're trying to make sense of some integrals and even sometimes solve some of them."}, {"video_title": "Derivative of sin(ln(x_)) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So one way you could think of it, if you set f of x as being equal to sine of x, and g of x being the natural log of x, and let's see, fg, let's say h of x, h of x equaling x squared, then this thing right over here is the exact same thing as trying to take the derivative with respect to x of f of g of h of x. And what I want to do is kind of think about it how I would do it in my head, without having to write all the chain rule notation. So the way I would think about this, if I were doing this in my head, is the derivative of this outer function of f with respect to the level of composition directly below it. So the derivative of sine of x is cosine of x. But instead of it being cosine of x, it's going to be cosine of whatever was inside of here. So it's going to be cosine of natural log. Let me write that in that same color."}, {"video_title": "Derivative of sin(ln(x_)) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So the derivative of sine of x is cosine of x. But instead of it being cosine of x, it's going to be cosine of whatever was inside of here. So it's going to be cosine of natural log. Let me write that in that same color. Cosine of natural log of x squared. I'm going to do x in that same yellow color. Cosine of x squared."}, {"video_title": "Derivative of sin(ln(x_)) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Let me write that in that same color. Cosine of natural log of x squared. I'm going to do x in that same yellow color. Cosine of x squared. And so you could really view this part, what I just write over here, as f prime. This is f prime of g of h of x. This is f prime of g of h of x."}, {"video_title": "Derivative of sin(ln(x_)) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Cosine of x squared. And so you could really view this part, what I just write over here, as f prime. This is f prime of g of h of x. This is f prime of g of h of x. If you want to keep track of things. So I just took the derivative of the outer with respect to whatever was inside of it. And now I have to take the derivative of the inside with respect to x."}, {"video_title": "Derivative of sin(ln(x_)) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "This is f prime of g of h of x. If you want to keep track of things. So I just took the derivative of the outer with respect to whatever was inside of it. And now I have to take the derivative of the inside with respect to x. But now we have another composite function. So we're going to multiply this times. We're going to do the chain rule again."}, {"video_title": "Derivative of sin(ln(x_)) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And now I have to take the derivative of the inside with respect to x. But now we have another composite function. So we're going to multiply this times. We're going to do the chain rule again. The derivative of, we're going to take the derivative of ln with respect to x squared. So the derivative of ln of x is 1 over x. But now we're going to have 1 over not x, but 1 over x squared."}, {"video_title": "Derivative of sin(ln(x_)) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We're going to do the chain rule again. The derivative of, we're going to take the derivative of ln with respect to x squared. So the derivative of ln of x is 1 over x. But now we're going to have 1 over not x, but 1 over x squared. So to be clear, this part right over here is g prime of not x. If it was g prime of x, this would be 1 over x. But instead of an x, we have our h of x there."}, {"video_title": "Derivative of sin(ln(x_)) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "But now we're going to have 1 over not x, but 1 over x squared. So to be clear, this part right over here is g prime of not x. If it was g prime of x, this would be 1 over x. But instead of an x, we have our h of x there. We have our x squared. So it's g prime of x squared. And then finally, we can take the derivative of our inner function."}, {"video_title": "Derivative of sin(ln(x_)) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "But instead of an x, we have our h of x there. We have our x squared. So it's g prime of x squared. And then finally, we can take the derivative of our inner function. Let me write it. So we could write this as g prime of h of x. And finally, we just have to take the derivative of our innermost function with respect to x."}, {"video_title": "Derivative of sin(ln(x_)) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And then finally, we can take the derivative of our inner function. Let me write it. So we could write this as g prime of h of x. And finally, we just have to take the derivative of our innermost function with respect to x. So the derivative of x squared with respect to x is 2x. So times h prime of x. Let me make everything clear."}, {"video_title": "Derivative of sin(ln(x_)) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And finally, we just have to take the derivative of our innermost function with respect to x. So the derivative of x squared with respect to x is 2x. So times h prime of x. Let me make everything clear. So what we have right over here in purple, this, this, and this are the same things. One expressed concretely, one expressed abstractly. This, this, and this are the same thing, expressed concretely and abstractly."}, {"video_title": "Derivative of sin(ln(x_)) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Let me make everything clear. So what we have right over here in purple, this, this, and this are the same things. One expressed concretely, one expressed abstractly. This, this, and this are the same thing, expressed concretely and abstractly. And then finally, this and this are the same thing, expressed concretely and abstractly. But then we're done. All we have to do to be done is to just simplify this."}, {"video_title": "Derivative of sin(ln(x_)) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "This, this, and this are the same thing, expressed concretely and abstractly. And then finally, this and this are the same thing, expressed concretely and abstractly. But then we're done. All we have to do to be done is to just simplify this. So if we just change the order in which we're multiplying, we have 2x over x squared. So I can cancel some out. So this 2x over x squared is the same thing as 2 over x."}, {"video_title": "Derivative of sin(ln(x_)) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "All we have to do to be done is to just simplify this. So if we just change the order in which we're multiplying, we have 2x over x squared. So I can cancel some out. So this 2x over x squared is the same thing as 2 over x. And we're multiplying it times all of this business. So we're left with 2 over x. This goes away."}, {"video_title": "Derivative of sin(ln(x_)) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So this 2x over x squared is the same thing as 2 over x. And we're multiplying it times all of this business. So we're left with 2 over x. This goes away. 2 over x times the cosine of the natural log of x squared. So it seemed like a very daunting derivative. But we just say, OK, what's the derivative of sine of something with respect to that something?"}, {"video_title": "Derivative of sin(ln(x_)) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "This goes away. 2 over x times the cosine of the natural log of x squared. So it seemed like a very daunting derivative. But we just say, OK, what's the derivative of sine of something with respect to that something? Well, that's cosine of that something. And then we go in one layer. What's the derivative of that something?"}, {"video_title": "Derivative of sin(ln(x_)) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "But we just say, OK, what's the derivative of sine of something with respect to that something? Well, that's cosine of that something. And then we go in one layer. What's the derivative of that something? Well, in that something, we have another composition. So the derivative of ln of x or ln of something with respect to another something, well, that's going to be 1 over the something. So we had gotten a 1 over x squared here."}, {"video_title": "Derivative of sin(ln(x_)) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "What's the derivative of that something? Well, in that something, we have another composition. So the derivative of ln of x or ln of something with respect to another something, well, that's going to be 1 over the something. So we had gotten a 1 over x squared here. That squared got canceled out. And then finally, the derivative of this innermost function. It's like peeling an onion."}, {"video_title": "Derivative of sin(ln(x_)) Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So we had gotten a 1 over x squared here. That squared got canceled out. And then finally, the derivative of this innermost function. It's like peeling an onion. The derivative of this inner function with respect to x, which was just 2x, which we got right over here. This was 1 over x squared. This was 2x before we did any canceling out."}, {"video_title": "Formal definition for limit of a sequence Series AP Calculus BC Khan Academy.mp3", "Sentence": "And what we'll see is it's actually very similar to the definition of any function as the limit approaches infinity. And this is because these sequences really can be just viewed as a function of their indices. So let's say, let me draw an arbitrary sequence right over here. So actually, let me draw it like this, just to make it clear what the limit is approaching. So let me draw a sequence that is jumping around a little bit. So let's say when n is equal to 1, a sub 1 is there. When n is equal to 2, a sub 2 is there."}, {"video_title": "Formal definition for limit of a sequence Series AP Calculus BC Khan Academy.mp3", "Sentence": "So actually, let me draw it like this, just to make it clear what the limit is approaching. So let me draw a sequence that is jumping around a little bit. So let's say when n is equal to 1, a sub 1 is there. When n is equal to 2, a sub 2 is there. When n is equal to 3, a sub 3 is over there. When n is equal to 4, a sub 4 is over here. When n is equal to 5, a sub 5 is over here."}, {"video_title": "Formal definition for limit of a sequence Series AP Calculus BC Khan Academy.mp3", "Sentence": "When n is equal to 2, a sub 2 is there. When n is equal to 3, a sub 3 is over there. When n is equal to 4, a sub 4 is over here. When n is equal to 5, a sub 5 is over here. And it looks like as n is, so this is 1, 2, 3, 4, 5. And so it looks like as n gets bigger and bigger and bigger, a sub n seems to be approaching some value. It seems to be getting closer and closer."}, {"video_title": "Formal definition for limit of a sequence Series AP Calculus BC Khan Academy.mp3", "Sentence": "When n is equal to 5, a sub 5 is over here. And it looks like as n is, so this is 1, 2, 3, 4, 5. And so it looks like as n gets bigger and bigger and bigger, a sub n seems to be approaching some value. It seems to be getting closer and closer. It seems to be converging to some value l right over here. But what we need to do is come up with a definition of what does it really mean to converge to l. So let's say for any, so we're going to say that you converge to l. If for any epsilon greater than 0, for any positive epsilon, you can come up, you can get, or you can, there is, or maybe I'll write this way. For any positive epsilon, there is a positive M, capital M, such that if lowercase n is greater than capital M, then the distance between a sub n and our limit, this l right over here, the distance between those two points is less than epsilon."}, {"video_title": "Formal definition for limit of a sequence Series AP Calculus BC Khan Academy.mp3", "Sentence": "It seems to be getting closer and closer. It seems to be converging to some value l right over here. But what we need to do is come up with a definition of what does it really mean to converge to l. So let's say for any, so we're going to say that you converge to l. If for any epsilon greater than 0, for any positive epsilon, you can come up, you can get, or you can, there is, or maybe I'll write this way. For any positive epsilon, there is a positive M, capital M, such that if lowercase n is greater than capital M, then the distance between a sub n and our limit, this l right over here, the distance between those two points is less than epsilon. If you can do this for any epsilon, for any epsilon greater than 0, there's a positive M such that if n is greater than M, then the distance between a sub n and our limit is less than epsilon. Then we can say that the limit of a sub n as n approaches infinity is equal to l, and we can say that a sub n converges. So let's parse this."}, {"video_title": "Formal definition for limit of a sequence Series AP Calculus BC Khan Academy.mp3", "Sentence": "For any positive epsilon, there is a positive M, capital M, such that if lowercase n is greater than capital M, then the distance between a sub n and our limit, this l right over here, the distance between those two points is less than epsilon. If you can do this for any epsilon, for any epsilon greater than 0, there's a positive M such that if n is greater than M, then the distance between a sub n and our limit is less than epsilon. Then we can say that the limit of a sub n as n approaches infinity is equal to l, and we can say that a sub n converges. So let's parse this. So here I was making the claim that a sub n is approaching this l right over here. I tried to draw it as a horizontal line. This definition of what it means to converge, for a sequence to converge, says look, for any epsilon greater than 0."}, {"video_title": "Formal definition for limit of a sequence Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's parse this. So here I was making the claim that a sub n is approaching this l right over here. I tried to draw it as a horizontal line. This definition of what it means to converge, for a sequence to converge, says look, for any epsilon greater than 0. So let me pick an epsilon greater than 0. So I'm going to go to l plus epsilon. Actually, let me do it right over here."}, {"video_title": "Formal definition for limit of a sequence Series AP Calculus BC Khan Academy.mp3", "Sentence": "This definition of what it means to converge, for a sequence to converge, says look, for any epsilon greater than 0. So let me pick an epsilon greater than 0. So I'm going to go to l plus epsilon. Actually, let me do it right over here. Let's say this is l plus epsilon, and let's say this is right here. This is l minus epsilon. So let me draw those two bounds right over here."}, {"video_title": "Formal definition for limit of a sequence Series AP Calculus BC Khan Academy.mp3", "Sentence": "Actually, let me do it right over here. Let's say this is l plus epsilon, and let's say this is right here. This is l minus epsilon. So let me draw those two bounds right over here. And so I picked an epsilon here. So for any arbitrary positive epsilon I pick, we can find a positive M. We can find a positive M. So let's say that that is our M right over there. So that as long as our n is greater than our M, then our a sub n is within epsilon of l. So being within epsilon of l is essentially being in this range."}, {"video_title": "Formal definition for limit of a sequence Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let me draw those two bounds right over here. And so I picked an epsilon here. So for any arbitrary positive epsilon I pick, we can find a positive M. We can find a positive M. So let's say that that is our M right over there. So that as long as our n is greater than our M, then our a sub n is within epsilon of l. So being within epsilon of l is essentially being in this range. This right over here is just saying, look, the distance between a sub n and l is less than epsilon. So that would be any of these. Anything that's in this between l minus epsilon and l plus epsilon, the distance between that and our limit is going to be less than epsilon."}, {"video_title": "Formal definition for limit of a sequence Series AP Calculus BC Khan Academy.mp3", "Sentence": "So that as long as our n is greater than our M, then our a sub n is within epsilon of l. So being within epsilon of l is essentially being in this range. This right over here is just saying, look, the distance between a sub n and l is less than epsilon. So that would be any of these. Anything that's in this between l minus epsilon and l plus epsilon, the distance between that and our limit is going to be less than epsilon. And we see right over here, at least visually, if we pick M there, if you take an n that's larger than that M, if n is equal to 3, a sub n seems to be close enough. If n is 4, a sub n is even getting closer. It's within our epsilon."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So the first thing I want to think about is why this little special case for n not equaling 0? What happens if n equals 0? So let's just think of the situation. Let's try to take the derivative with respect to x of x to the 0 power. Well, what is x to the 0 power going to be? Well, we can assume that x for this case right over here is not equal to 0. 0 to the 0, weird things happen at that point."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Let's try to take the derivative with respect to x of x to the 0 power. Well, what is x to the 0 power going to be? Well, we can assume that x for this case right over here is not equal to 0. 0 to the 0, weird things happen at that point. But if x does not equal 0, what is x to the 0 power going to be? Well, this is the same thing as the derivative with respect to x of 1. x to the 0 power is just going to be 1. And so what is the derivative with respect to x of 1?"}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "0 to the 0, weird things happen at that point. But if x does not equal 0, what is x to the 0 power going to be? Well, this is the same thing as the derivative with respect to x of 1. x to the 0 power is just going to be 1. And so what is the derivative with respect to x of 1? And to answer that question, I'll just graph it. I'll just graph f of x equals 1 to make it a little bit clearer. So that's my y-axis."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And so what is the derivative with respect to x of 1? And to answer that question, I'll just graph it. I'll just graph f of x equals 1 to make it a little bit clearer. So that's my y-axis. This is my x-axis. And let me graph y equals 1, or f of x equals 1. So that's 1 right over there."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So that's my y-axis. This is my x-axis. And let me graph y equals 1, or f of x equals 1. So that's 1 right over there. f of x equals 1 is just a horizontal line. So that right over there is the graph y is equal to f of x, which is equal to 1. Now, remember, the derivative, one way to conceptualize it, is just the slope of the tangent line at any point."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So that's 1 right over there. f of x equals 1 is just a horizontal line. So that right over there is the graph y is equal to f of x, which is equal to 1. Now, remember, the derivative, one way to conceptualize it, is just the slope of the tangent line at any point. So what is the slope of the tangent line at this point? And actually, what's the slope at every point? Well, this is a line, so the slope doesn't change."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Now, remember, the derivative, one way to conceptualize it, is just the slope of the tangent line at any point. So what is the slope of the tangent line at this point? And actually, what's the slope at every point? Well, this is a line, so the slope doesn't change. It has a constant slope, and it's a completely horizontal line. It has a slope of 0. So the slope at every point over here, slope is going to be equal to 0."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, this is a line, so the slope doesn't change. It has a constant slope, and it's a completely horizontal line. It has a slope of 0. So the slope at every point over here, slope is going to be equal to 0. So the slope of this line at any point is just going to be equal to 0. And that's actually going to be true for any constant. The derivative, if I had a function, let's say I had f of x is equal to 3."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So the slope at every point over here, slope is going to be equal to 0. So the slope of this line at any point is just going to be equal to 0. And that's actually going to be true for any constant. The derivative, if I had a function, let's say I had f of x is equal to 3. Let's say that that's y is equal to 3. What's the derivative of y with respect to x going to be equal to? And I'm intentionally showing you all the different ways of the notation for derivatives."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "The derivative, if I had a function, let's say I had f of x is equal to 3. Let's say that that's y is equal to 3. What's the derivative of y with respect to x going to be equal to? And I'm intentionally showing you all the different ways of the notation for derivatives. So what's the derivative of y with respect to x? It can also be written as y prime. What's that going to be equal to?"}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And I'm intentionally showing you all the different ways of the notation for derivatives. So what's the derivative of y with respect to x? It can also be written as y prime. What's that going to be equal to? Well, it's the slope at any given point. And you see that no matter what x you're looking at, the slope here is going to be 0. So it's going to be 0."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "What's that going to be equal to? Well, it's the slope at any given point. And you see that no matter what x you're looking at, the slope here is going to be 0. So it's going to be 0. So it's not just x to the 0. If you take the derivative of any constant, you're going to get 0. So let me write that."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So it's going to be 0. So it's not just x to the 0. If you take the derivative of any constant, you're going to get 0. So let me write that. Derivative with respect to x of any constant. So let's say of a, where this is just a constant, that's going to be equal to 0. So pretty straightforward idea."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So let me write that. Derivative with respect to x of any constant. So let's say of a, where this is just a constant, that's going to be equal to 0. So pretty straightforward idea. Now let's explore a few more properties. Let's say I want to take the derivative with respect to x of, well, let's use the same a. Let's say I have some constant times some function."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So pretty straightforward idea. Now let's explore a few more properties. Let's say I want to take the derivative with respect to x of, well, let's use the same a. Let's say I have some constant times some function. Well, derivatives work out quite well. You can actually take this little scalar multiplier, this little constant, and take it out of the derivative. This is going to be equal to a."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say I have some constant times some function. Well, derivatives work out quite well. You can actually take this little scalar multiplier, this little constant, and take it out of the derivative. This is going to be equal to a. It's going to be equal to a. I didn't want to do that magenta color. It's going to be equal to a times the derivative of f of x. a times the derivative of f of x. Let me do that in blue color."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "This is going to be equal to a. It's going to be equal to a. I didn't want to do that magenta color. It's going to be equal to a times the derivative of f of x. a times the derivative of f of x. Let me do that in blue color. Of f of x. And the other way to denote the derivative of f of x is to just say that this is the same thing. This is equal to a times, this thing right over here is the exact same thing as f prime of x."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Let me do that in blue color. Of f of x. And the other way to denote the derivative of f of x is to just say that this is the same thing. This is equal to a times, this thing right over here is the exact same thing as f prime of x. Now, this might all look like really fancy notation, but I think if I gave you an example, it might make some sense. So what about if I were to ask you the derivative with respect to x of 2 times x to the fifth power? Well, this property that I just articulated says, well, this is going to be the same thing as 2 times the derivative."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "This is equal to a times, this thing right over here is the exact same thing as f prime of x. Now, this might all look like really fancy notation, but I think if I gave you an example, it might make some sense. So what about if I were to ask you the derivative with respect to x of 2 times x to the fifth power? Well, this property that I just articulated says, well, this is going to be the same thing as 2 times the derivative. This is going to be the same thing as 2 times the derivative of x to the fifth. 2 times the derivative with respect to x of x to the fifth. Essentially, I could just take this scalar multiplier and put it in front of the derivative."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, this property that I just articulated says, well, this is going to be the same thing as 2 times the derivative. This is going to be the same thing as 2 times the derivative of x to the fifth. 2 times the derivative with respect to x of x to the fifth. Essentially, I could just take this scalar multiplier and put it in front of the derivative. So this right here, this is the derivative with respect to x of x to the fifth. And we know how to do that using the power rule. This is going to be equal to 2 times, let me write that, I want to keep it consistent with the colors."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Essentially, I could just take this scalar multiplier and put it in front of the derivative. So this right here, this is the derivative with respect to x of x to the fifth. And we know how to do that using the power rule. This is going to be equal to 2 times, let me write that, I want to keep it consistent with the colors. This is going to be 2 times derivative of x to the fifth. Well, the power rule tells us n is 5. It's going to be 5x to the 5 minus 1, or 5x to the fourth power."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "This is going to be equal to 2 times, let me write that, I want to keep it consistent with the colors. This is going to be 2 times derivative of x to the fifth. Well, the power rule tells us n is 5. It's going to be 5x to the 5 minus 1, or 5x to the fourth power. So it's going to be 5x to the fourth power, which is going to be equal to 2 times 5 is 10x to the fourth. So 2x to the fifth, you can literally just say, OK, the power rule tells me the derivative of that is 5x to the fourth. 5 times 2 is 10."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "It's going to be 5x to the 5 minus 1, or 5x to the fourth power. So it's going to be 5x to the fourth power, which is going to be equal to 2 times 5 is 10x to the fourth. So 2x to the fifth, you can literally just say, OK, the power rule tells me the derivative of that is 5x to the fourth. 5 times 2 is 10. So that simplifies our life a good bit. We can now, using the power rule and this one property, take the derivative of anything that takes the form ax to the n power. Now let's think about another very useful derivative property."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "5 times 2 is 10. So that simplifies our life a good bit. We can now, using the power rule and this one property, take the derivative of anything that takes the form ax to the n power. Now let's think about another very useful derivative property. And these don't just apply to the power rule. They apply to any derivative, but they are especially useful for the power rule because it allows us to construct polynomials and take the derivatives of them. If I were to take the derivative of the sum of two functions, so the derivative of, let's say, one function is f of x, and then the other function is g of x, it's lucky for us that this ends up being the same thing as the derivative of f of x plus the derivative of g of x."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's think about another very useful derivative property. And these don't just apply to the power rule. They apply to any derivative, but they are especially useful for the power rule because it allows us to construct polynomials and take the derivatives of them. If I were to take the derivative of the sum of two functions, so the derivative of, let's say, one function is f of x, and then the other function is g of x, it's lucky for us that this ends up being the same thing as the derivative of f of x plus the derivative of g of x. So this is the same thing as f. Actually, let me use that derivative operator just to make it clear. It's the same thing as the derivative with respect to x of f of x plus the derivative with respect to x of g of x. So we put f of x right over here and put g of x right over there."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "If I were to take the derivative of the sum of two functions, so the derivative of, let's say, one function is f of x, and then the other function is g of x, it's lucky for us that this ends up being the same thing as the derivative of f of x plus the derivative of g of x. So this is the same thing as f. Actually, let me use that derivative operator just to make it clear. It's the same thing as the derivative with respect to x of f of x plus the derivative with respect to x of g of x. So we put f of x right over here and put g of x right over there. And so with the other notation, we can say this is going to be the same thing. Derivative with respect to x of f of x, we can write as f prime of x. And the derivative with respect to x of g of x, we can write as g prime of x."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So we put f of x right over here and put g of x right over there. And so with the other notation, we can say this is going to be the same thing. Derivative with respect to x of f of x, we can write as f prime of x. And the derivative with respect to x of g of x, we can write as g prime of x. Now, once again, this might look like kind of fancy notation to you, but when you see an example, it'll make it pretty clear. If I want to take the derivative with respect to x of, let's say, x to the third power plus x to the negative 4 power, this just tells us that the derivative of the sum is just the sum of the derivatives. So we can take the derivative of this term using the power rule."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And the derivative with respect to x of g of x, we can write as g prime of x. Now, once again, this might look like kind of fancy notation to you, but when you see an example, it'll make it pretty clear. If I want to take the derivative with respect to x of, let's say, x to the third power plus x to the negative 4 power, this just tells us that the derivative of the sum is just the sum of the derivatives. So we can take the derivative of this term using the power rule. So it's going to be 3x squared. And to that, we can add the derivative of this thing right over here. So it's going to be plus, that's a different shade of blue, plus, and over here is negative 4, so it's plus negative 4 times x to the negative 4 minus 1, or x to the negative 5 power."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So we can take the derivative of this term using the power rule. So it's going to be 3x squared. And to that, we can add the derivative of this thing right over here. So it's going to be plus, that's a different shade of blue, plus, and over here is negative 4, so it's plus negative 4 times x to the negative 4 minus 1, or x to the negative 5 power. So we have, and I can just simplify a little bit, this is going to be equal to 3x squared minus 4x to the negative 5. And so now we have all the tools we need in our toolkit to essentially take the derivative of any polynomial. So let's give ourselves a little practice there."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So it's going to be plus, that's a different shade of blue, plus, and over here is negative 4, so it's plus negative 4 times x to the negative 4 minus 1, or x to the negative 5 power. So we have, and I can just simplify a little bit, this is going to be equal to 3x squared minus 4x to the negative 5. And so now we have all the tools we need in our toolkit to essentially take the derivative of any polynomial. So let's give ourselves a little practice there. So let's say that I have, and I'll do it in white, let's say that f of x is equal to 2x to the third power minus 7x squared plus 3x minus 100. What is f prime of x? What is the derivative of f with respect to x going to be?"}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So let's give ourselves a little practice there. So let's say that I have, and I'll do it in white, let's say that f of x is equal to 2x to the third power minus 7x squared plus 3x minus 100. What is f prime of x? What is the derivative of f with respect to x going to be? Well, we can use the properties that we just said. The derivative of this is just going to be 2 times the derivative of x to the third. Derivative of x to the third is going to be 3x squared, so it's going to be 2 times 3x squared."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "What is the derivative of f with respect to x going to be? Well, we can use the properties that we just said. The derivative of this is just going to be 2 times the derivative of x to the third. Derivative of x to the third is going to be 3x squared, so it's going to be 2 times 3x squared. What's the derivative of negative 7x squared going to be? Well, it's just going to be negative 7 times the derivative of x squared, which is 2x. What is the derivative of 3x going to be?"}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Derivative of x to the third is going to be 3x squared, so it's going to be 2 times 3x squared. What's the derivative of negative 7x squared going to be? Well, it's just going to be negative 7 times the derivative of x squared, which is 2x. What is the derivative of 3x going to be? Well, it's just going to be 3 times the derivative of x or 3 times the derivative of x to the first. The derivative of x to the first is just 1, so this is just going to be plus 3 times. could say 1x to the 0, but that's just 1."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "What is the derivative of 3x going to be? Well, it's just going to be 3 times the derivative of x or 3 times the derivative of x to the first. The derivative of x to the first is just 1, so this is just going to be plus 3 times. could say 1x to the 0, but that's just 1. And then finally, what's the derivative of a constant going to be? Let me do that in a different color. What's the derivative of a constant going to be?"}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "could say 1x to the 0, but that's just 1. And then finally, what's the derivative of a constant going to be? Let me do that in a different color. What's the derivative of a constant going to be? Well, we covered that at the beginning of this video. The derivative of any constant is just going to be 0. So plus 0."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "What's the derivative of a constant going to be? Well, we covered that at the beginning of this video. The derivative of any constant is just going to be 0. So plus 0. And so now we are ready to simplify. The derivative of f is going to be 2 times 3x squared is just 6x squared. Negative 7 times 2x is negative 14x plus 3."}, {"video_title": "Differentiating polynomials Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So plus 0. And so now we are ready to simplify. The derivative of f is going to be 2 times 3x squared is just 6x squared. Negative 7 times 2x is negative 14x plus 3. And we don't have to write the 0 there. And we're done. We now have all the properties in our tool belt to find the derivative of any polynomial, and actually things that even go beyond polynomials."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on lower bound Khan Academy.mp3", "Sentence": "And you might say, well, it looks like the fundamental theorem of calculus might apply. But I'm used to seeing the x or the function x as the upper bound, not as the lower bound. How do I deal with this? And the key realization is to realize what happens when you switch bounds for a definite integral. And I'll do a little bit of an aside to review that. So if I'm taking the definite integral from a to b of f of t dt, we know that this is capital F, the antiderivative of F evaluated at b, minus the antiderivative of F evaluated at a. This is essentially corollary to the fundamental theorem, or it's the fundamental theorem part two, the second fundamental theorem of calculus."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on lower bound Khan Academy.mp3", "Sentence": "And the key realization is to realize what happens when you switch bounds for a definite integral. And I'll do a little bit of an aside to review that. So if I'm taking the definite integral from a to b of f of t dt, we know that this is capital F, the antiderivative of F evaluated at b, minus the antiderivative of F evaluated at a. This is essentially corollary to the fundamental theorem, or it's the fundamental theorem part two, the second fundamental theorem of calculus. This is how we evaluate definite integrals. Now, let's think about what the negative of this is. So the negative of that, of a to b of f of t dt, is just going to be equal to the negative of this, which is equal to the negative of F of b minus F of a, which is equal to capital F of a minus capital F of b."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on lower bound Khan Academy.mp3", "Sentence": "This is essentially corollary to the fundamental theorem, or it's the fundamental theorem part two, the second fundamental theorem of calculus. This is how we evaluate definite integrals. Now, let's think about what the negative of this is. So the negative of that, of a to b of f of t dt, is just going to be equal to the negative of this, which is equal to the negative of F of b minus F of a, which is equal to capital F of a minus capital F of b. All I did is distribute the negative sign and then switch the two terms. But this right over here is equal to the definite integral from, instead of a to b, but from b to a of F of t dt. So notice, when you put a negative, that's just like switching the signs or switching the boundaries."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on lower bound Khan Academy.mp3", "Sentence": "So the negative of that, of a to b of f of t dt, is just going to be equal to the negative of this, which is equal to the negative of F of b minus F of a, which is equal to capital F of a minus capital F of b. All I did is distribute the negative sign and then switch the two terms. But this right over here is equal to the definite integral from, instead of a to b, but from b to a of F of t dt. So notice, when you put a negative, that's just like switching the signs or switching the boundaries. Or if you switch the boundaries, they're the negatives of each other. So we can go back to our original problem. We can rewrite this as being equal to the derivative with respect to x of, instead of this, it'll be the negative of the same definite integral but with the boundaries switched."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on lower bound Khan Academy.mp3", "Sentence": "So notice, when you put a negative, that's just like switching the signs or switching the boundaries. Or if you switch the boundaries, they're the negatives of each other. So we can go back to our original problem. We can rewrite this as being equal to the derivative with respect to x of, instead of this, it'll be the negative of the same definite integral but with the boundaries switched. The negative of x, with the upper boundary is x, the lower boundary is 3, of the square root of the absolute value of cosine t dt, which is equal to, we can take the negative out front, negative times negative times the derivative with respect to x of all of this business. I should just copy and paste that. So I'll just copy and paste."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on lower bound Khan Academy.mp3", "Sentence": "We can rewrite this as being equal to the derivative with respect to x of, instead of this, it'll be the negative of the same definite integral but with the boundaries switched. The negative of x, with the upper boundary is x, the lower boundary is 3, of the square root of the absolute value of cosine t dt, which is equal to, we can take the negative out front, negative times negative times the derivative with respect to x of all of this business. I should just copy and paste that. So I'll just copy and paste. Let me paste it. So times the derivative with respect to x of all of that. And now the fundamental theorem of calculus directly applies."}, {"video_title": "Finding derivative with fundamental theorem of calculus x is on lower bound Khan Academy.mp3", "Sentence": "So I'll just copy and paste. Let me paste it. So times the derivative with respect to x of all of that. And now the fundamental theorem of calculus directly applies. This is going to be equal to, we deserve a drum roll now, this is going to be equal to the negative. Can't forget the negative. And the fundamental theorem of calculus tells us that that's just going to be this function as a function of x."}, {"video_title": "Worked example problem involving definite integral (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "We are told the population of a town grows at a rate of e to the 1.2t power minus 2t people per year, where t is the number of years. At t equals two years, the town has 1,500 people. So first they ask us, approximately by how many people does the population grow between t equals two and t equals five, and then what is the town's population at t equals five years? And if we actually figure out this first question, the second question is actually pretty straightforward. We figure out the amount that it grows and then add it to what we were at at t equals two, add it to 1,500. So pause this video and see if you can figure it out. So the key here is to appreciate that this right over here is expressing the rate of how fast the population is growing and we've been seen in multiple videos now, let me just draw, do a quick review of this notion of a rate curve."}, {"video_title": "Worked example problem involving definite integral (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "And if we actually figure out this first question, the second question is actually pretty straightforward. We figure out the amount that it grows and then add it to what we were at at t equals two, add it to 1,500. So pause this video and see if you can figure it out. So the key here is to appreciate that this right over here is expressing the rate of how fast the population is growing and we've been seen in multiple videos now, let me just draw, do a quick review of this notion of a rate curve. So those are my axes and this is my t axis, my time axis. And so this is showing me how my rate of change changes as a function of time. So let's say it's something like this."}, {"video_title": "Worked example problem involving definite integral (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "So the key here is to appreciate that this right over here is expressing the rate of how fast the population is growing and we've been seen in multiple videos now, let me just draw, do a quick review of this notion of a rate curve. So those are my axes and this is my t axis, my time axis. And so this is showing me how my rate of change changes as a function of time. So let's say it's something like this. So once again, if I said at this time right over here, this is my rate, this doesn't tell me, for example, what my population is, this tells me what is my rate of change of a population. And we have seen in previous videos that if you wanna figure out the change in the thing that the rate is, that the rate is the rate of change of, say, the change in population, you would find the area under the rate curve between those two appropriate times. And why does that make sense?"}, {"video_title": "Worked example problem involving definite integral (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say it's something like this. So once again, if I said at this time right over here, this is my rate, this doesn't tell me, for example, what my population is, this tells me what is my rate of change of a population. And we have seen in previous videos that if you wanna figure out the change in the thing that the rate is, that the rate is the rate of change of, say, the change in population, you would find the area under the rate curve between those two appropriate times. And why does that make sense? Well, imagine a very small change in time right over here. If you have a very small change in time and if you assume that your rate is approximately constant over that very small change in time, well then your change in, let's say we're measuring the rate of change of population here, your accumulation, you could say, is going to be your rate times your change in time, which would be the area of this rectangle. And so that would be the, roughly, that would be the area under the curve over that very, very small change in time."}, {"video_title": "Worked example problem involving definite integral (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "And why does that make sense? Well, imagine a very small change in time right over here. If you have a very small change in time and if you assume that your rate is approximately constant over that very small change in time, well then your change in, let's say we're measuring the rate of change of population here, your accumulation, you could say, is going to be your rate times your change in time, which would be the area of this rectangle. And so that would be the, roughly, that would be the area under the curve over that very, very small change in time. So what we really wanna do is find the area under this curve from t equals two to t equals five. And we have seen multiple times in calculus how to express that. So the definite integral from t is equal to two to t is equal to five of this expression of e to the 1.2t minus two t dt, dt."}, {"video_title": "Worked example problem involving definite integral (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "And so that would be the, roughly, that would be the area under the curve over that very, very small change in time. So what we really wanna do is find the area under this curve from t equals two to t equals five. And we have seen multiple times in calculus how to express that. So the definite integral from t is equal to two to t is equal to five of this expression of e to the 1.2t minus two t dt, dt. So if we just evaluate that, that will be the answer to this first question. So what is this going to be? Well, let's actually work it out."}, {"video_title": "Worked example problem involving definite integral (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "So the definite integral from t is equal to two to t is equal to five of this expression of e to the 1.2t minus two t dt, dt. So if we just evaluate that, that will be the answer to this first question. So what is this going to be? Well, let's actually work it out. So what is the antiderivative of e to the 1.2t? Well, let me just try to do it over here. So if I am trying to calculate, let me write it as e to the five, or actually 6 5ths t, 12 tenths is the same thing as 6 5ths, 6 5ths t dt."}, {"video_title": "Worked example problem involving definite integral (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "Well, let's actually work it out. So what is the antiderivative of e to the 1.2t? Well, let me just try to do it over here. So if I am trying to calculate, let me write it as e to the five, or actually 6 5ths t, 12 tenths is the same thing as 6 5ths, 6 5ths t dt. So this is an indefinite integral. I'm just trying to figure out the antiderivative here. Well, if I had a 6 5ths right over here, then u substitution, or sometimes you would say the inverse chain rule, would be very appropriate."}, {"video_title": "Worked example problem involving definite integral (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "So if I am trying to calculate, let me write it as e to the five, or actually 6 5ths t, 12 tenths is the same thing as 6 5ths, 6 5ths t dt. So this is an indefinite integral. I'm just trying to figure out the antiderivative here. Well, if I had a 6 5ths right over here, then u substitution, or sometimes you would say the inverse chain rule, would be very appropriate. Well, we could put a 6 5ths there if we write a 5 6ths right over here, 5 6ths times 6 5ths, and we can take constants in and out of the integral like this, scaling constants, I should say. Well, now, so this is going to be equal to 5 6ths, this 5 6ths right over here. This antiderivative is pretty straightforward."}, {"video_title": "Worked example problem involving definite integral (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "Well, if I had a 6 5ths right over here, then u substitution, or sometimes you would say the inverse chain rule, would be very appropriate. Well, we could put a 6 5ths there if we write a 5 6ths right over here, 5 6ths times 6 5ths, and we can take constants in and out of the integral like this, scaling constants, I should say. Well, now, so this is going to be equal to 5 6ths, this 5 6ths right over here. This antiderivative is pretty straightforward. Since I have the derivative of 6 5ths t right over here, I can find the antiderivative with respect to 6 5ths t, which is essentially I'm doing u substitution. If you had to do u substitution, you would make that right over there u, and then that, and that would be your du. But needless to say, this would be 5 6ths times e to the 6 5ths t, and if you're thinking about the indefinite integral, you would then have a plus c here, of course, and you can verify that the derivative of this is indeed e to the 1.2 t. So this is going to be equal to, so this part right over here the antiderivative is 5 6th e to the 6 5ths t, and then this part right over here, the antiderivative of 2t is t squared, so minus t squared, and we are going to evaluate that at five and two and find the difference."}, {"video_title": "Worked example problem involving definite integral (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "This antiderivative is pretty straightforward. Since I have the derivative of 6 5ths t right over here, I can find the antiderivative with respect to 6 5ths t, which is essentially I'm doing u substitution. If you had to do u substitution, you would make that right over there u, and then that, and that would be your du. But needless to say, this would be 5 6ths times e to the 6 5ths t, and if you're thinking about the indefinite integral, you would then have a plus c here, of course, and you can verify that the derivative of this is indeed e to the 1.2 t. So this is going to be equal to, so this part right over here the antiderivative is 5 6th e to the 6 5ths t, and then this part right over here, the antiderivative of 2t is t squared, so minus t squared, and we are going to evaluate that at five and two and find the difference. So let's evaluate this at when t is equal to five. Well, you are going to, let me color code this a little bit. When t is equal to five, you get 5 6ths times e to the 6 5ths times five is e to the 6th minus five squared, so minus 25, and so from that, I want to subtract."}, {"video_title": "Worked example problem involving definite integral (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "But needless to say, this would be 5 6ths times e to the 6 5ths t, and if you're thinking about the indefinite integral, you would then have a plus c here, of course, and you can verify that the derivative of this is indeed e to the 1.2 t. So this is going to be equal to, so this part right over here the antiderivative is 5 6th e to the 6 5ths t, and then this part right over here, the antiderivative of 2t is t squared, so minus t squared, and we are going to evaluate that at five and two and find the difference. So let's evaluate this at when t is equal to five. Well, you are going to, let me color code this a little bit. When t is equal to five, you get 5 6ths times e to the 6 5ths times five is e to the 6th minus five squared, so minus 25, and so from that, I want to subtract. When we evaluate it at two, we get 5 6ths e to the 6 5ths times two is the same thing as 12 5ths, or we could say that's 2.4, 2.4 minus four. Two squared is four, and so what do we get? Well, there's a couple of ways that we could do this."}, {"video_title": "Worked example problem involving definite integral (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "When t is equal to five, you get 5 6ths times e to the 6 5ths times five is e to the 6th minus five squared, so minus 25, and so from that, I want to subtract. When we evaluate it at two, we get 5 6ths e to the 6 5ths times two is the same thing as 12 5ths, or we could say that's 2.4, 2.4 minus four. Two squared is four, and so what do we get? Well, there's a couple of ways that we could do this. So we could write this as, let me write it this way. We could write this so we have a 5 6ths and a 5 6ths, so we could write this as 5 6ths times e to the 6th minus e to the 2.4 minus, because we distribute that negative sign, e to the 2.4, e to the 2.4 power, and then we have minus 25. Let me just send another color to keep track of it."}, {"video_title": "Worked example problem involving definite integral (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "Well, there's a couple of ways that we could do this. So we could write this as, let me write it this way. We could write this so we have a 5 6ths and a 5 6ths, so we could write this as 5 6ths times e to the 6th minus e to the 2.4 minus, because we distribute that negative sign, e to the 2.4, e to the 2.4 power, and then we have minus 25. Let me just send another color to keep track of it. We have minus 25, and then you have minus negative four, so that would be plus four, so that would be minus 21, and I would need a calculator to figure this out, so let me do that. Let me get my calculator on this computer, and there we go, and so let's see. If we want to find e to the 6th power, that's 430, okay, so now let me figure out, so minus, I would say 22.4, that looks like a 24."}, {"video_title": "Worked example problem involving definite integral (algebraic) AP Calculus AB Khan Academy.mp3", "Sentence": "Let me just send another color to keep track of it. We have minus 25, and then you have minus negative four, so that would be plus four, so that would be minus 21, and I would need a calculator to figure this out, so let me do that. Let me get my calculator on this computer, and there we go, and so let's see. If we want to find e to the 6th power, that's 430, okay, so now let me figure out, so minus, I would say 22.4, that looks like a 24. I'll correct it as soon as I get back to that screen. E to that power, e to the 2.4 power, and I get equals, so what's in parentheses is this number right here, so times 5 6ths, times five, divided by six, is equal to that minus 21, minus 21, is equal to this, so if I round to the nearest hundredth, it's going to be approximately 306.00, so this is approximately 306.00, so approximately by how many people does a population grow between t equals two and t equals five? Well, by approximately 306 people."}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "What we have here is the graph of r is equal to sine of two theta in polar coordinates. And if polar coordinates look unfamiliar to you or if you need to brush up on them, I encourage you to do a search for polar coordinates in Khan Academy or look at our pre-calculus section. But I'll give you a little bit of a primer here. Let's just familiarize ourself why this graph looks the way it does. When, so what we're doing for any point here, we could obviously specify these points in terms of x and y coordinates, but we could also specify them in terms of an angle and a radius. So for example, this would have some x coordinate and some y coordinate, or we could draw a line from the origin to that point, right over here, and specify it with some angle theta and some r, which is the distance from the origin to that point. And just to familiarize ourselves with this curve, let's just see why it's intuitive."}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "Let's just familiarize ourself why this graph looks the way it does. When, so what we're doing for any point here, we could obviously specify these points in terms of x and y coordinates, but we could also specify them in terms of an angle and a radius. So for example, this would have some x coordinate and some y coordinate, or we could draw a line from the origin to that point, right over here, and specify it with some angle theta and some r, which is the distance from the origin to that point. And just to familiarize ourselves with this curve, let's just see why it's intuitive. So when theta is zero, r is going to be zero. Sine of two times zero is just zero. So our r, we're just gonna be at the origin."}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "And just to familiarize ourselves with this curve, let's just see why it's intuitive. So when theta is zero, r is going to be zero. Sine of two times zero is just zero. So our r, we're just gonna be at the origin. And then as theta gets larger, our r gets larger. And so we start tracing out this petal of this flower or clover looking thing. So it starts looking like that."}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "So our r, we're just gonna be at the origin. And then as theta gets larger, our r gets larger. And so we start tracing out this petal of this flower or clover looking thing. So it starts looking like that. And we could keep going all the way. What happens when theta is equal to pi over four? When theta is equal to pi over four right over there, well sine of two times pi over four is sine of pi over two."}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "So it starts looking like that. And we could keep going all the way. What happens when theta is equal to pi over four? When theta is equal to pi over four right over there, well sine of two times pi over four is sine of pi over two. r is equal to one. So we reach a kind of a maximum r there. And then as theta increases, our r once again starts to get smaller and smaller and smaller."}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "When theta is equal to pi over four right over there, well sine of two times pi over four is sine of pi over two. r is equal to one. So we reach a kind of a maximum r there. And then as theta increases, our r once again starts to get smaller and smaller and smaller. Now we're going to do this in a calculus context. So the first question might be, well how do we express the rate of change of r with respect to theta? Pause this video and see if you can figure it out."}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "And then as theta increases, our r once again starts to get smaller and smaller and smaller. Now we're going to do this in a calculus context. So the first question might be, well how do we express the rate of change of r with respect to theta? Pause this video and see if you can figure it out. What is r prime of theta? Well there's really nothing new here because you just have one variable as a function of another. You just use the chain rule."}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "Pause this video and see if you can figure it out. What is r prime of theta? Well there's really nothing new here because you just have one variable as a function of another. You just use the chain rule. Take the derivative with respect to theta right over here. So the derivative of sine of two theta with respect to two theta is going to be cosine of two theta. And then you multiply that times the derivative of two theta with respect to theta, which is two."}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "You just use the chain rule. Take the derivative with respect to theta right over here. So the derivative of sine of two theta with respect to two theta is going to be cosine of two theta. And then you multiply that times the derivative of two theta with respect to theta, which is two. So we could just say times two here or we could write a two out front. All right, that was interesting. But let's see if we can express this curve in terms of x's and y's."}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "And then you multiply that times the derivative of two theta with respect to theta, which is two. So we could just say times two here or we could write a two out front. All right, that was interesting. But let's see if we can express this curve in terms of x's and y's. And then think about those derivatives. So one primer, a review from pre-calculus, is that when you want to go between the polar world and the, I guess you could say rectangular world, you have to remember the transformation that y is equal to r sine of theta and that x is equal to r cosine of theta. Now just as a really quick primer, why does that make sense?"}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "But let's see if we can express this curve in terms of x's and y's. And then think about those derivatives. So one primer, a review from pre-calculus, is that when you want to go between the polar world and the, I guess you could say rectangular world, you have to remember the transformation that y is equal to r sine of theta and that x is equal to r cosine of theta. Now just as a really quick primer, why does that make sense? Well, let's just take one of these angle r combinations right over here. So let's say this is theta and that is our r. Well, this is our, the height of that side is going to be our y and then the length of this side is going to be our x. Well, we know from trigonometry, from our unit circle definition, the Sohcahtoa definition of our trig functions, sine of theta is opposite over hypotenuse."}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "Now just as a really quick primer, why does that make sense? Well, let's just take one of these angle r combinations right over here. So let's say this is theta and that is our r. Well, this is our, the height of that side is going to be our y and then the length of this side is going to be our x. Well, we know from trigonometry, from our unit circle definition, the Sohcahtoa definition of our trig functions, sine of theta is opposite over hypotenuse. Sine of theta is equal to y over our hypotenuse, which is r, and cosine of theta is equal to the adjacent or x over r. And you just have to multiply both sides of these equations by r to get to what we have right over there. And once again, if this is going too fast, this is a review of just polar coordinates from pre-calculus. But now we can use these to express in terms, purely in terms of theta."}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "Well, we know from trigonometry, from our unit circle definition, the Sohcahtoa definition of our trig functions, sine of theta is opposite over hypotenuse. Sine of theta is equal to y over our hypotenuse, which is r, and cosine of theta is equal to the adjacent or x over r. And you just have to multiply both sides of these equations by r to get to what we have right over there. And once again, if this is going too fast, this is a review of just polar coordinates from pre-calculus. But now we can use these to express in terms, purely in terms of theta. How do we do that? Well, we know that r is equal to sine of two theta, so you just have to replace these r's with sine of two theta. So y would be equal to sine of two theta, sine of two theta times sine of theta, times sine of theta, and x is going to be equal to sine of two theta, sine of two theta times cosine of theta, cosine of theta times cosine of theta, just like that."}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "But now we can use these to express in terms, purely in terms of theta. How do we do that? Well, we know that r is equal to sine of two theta, so you just have to replace these r's with sine of two theta. So y would be equal to sine of two theta, sine of two theta times sine of theta, times sine of theta, and x is going to be equal to sine of two theta, sine of two theta times cosine of theta, cosine of theta times cosine of theta, just like that. But now we can use these expressions to find the rate of change of y with respect to theta, find a general expression for it. Pause the video and see if you can do that. All right, let's work through it together."}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "So y would be equal to sine of two theta, sine of two theta times sine of theta, times sine of theta, and x is going to be equal to sine of two theta, sine of two theta times cosine of theta, cosine of theta times cosine of theta, just like that. But now we can use these expressions to find the rate of change of y with respect to theta, find a general expression for it. Pause the video and see if you can do that. All right, let's work through it together. Well, this is, once again, we're just gonna use our derivative techniques. So I could write y prime of theta, the derivative of y with respect to theta. Just gonna use the product rule right over here."}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "All right, let's work through it together. Well, this is, once again, we're just gonna use our derivative techniques. So I could write y prime of theta, the derivative of y with respect to theta. Just gonna use the product rule right over here. The derivative of this first expression is two cosine of two theta, cosine of two theta. We've already seen that. That's just coming out of the chain rule."}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "Just gonna use the product rule right over here. The derivative of this first expression is two cosine of two theta, cosine of two theta. We've already seen that. That's just coming out of the chain rule. And then times the second expression, sine of theta, and then plus the first expression, sine of two theta, sine of two theta, times the derivative of the second expression. Derivative of the sine of theta is cosine of theta. Fair enough."}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "That's just coming out of the chain rule. And then times the second expression, sine of theta, and then plus the first expression, sine of two theta, sine of two theta, times the derivative of the second expression. Derivative of the sine of theta is cosine of theta. Fair enough. And we can do the same thing for x. X prime of theta, derivative of the first expression, it is going to be two times cosine of two theta, two times cosine of two theta, times the second expression, cosine of theta, and then you're gonna have the first expression, sine of two theta, times the derivative of the second expression, which is negative sine of theta, negative sine of theta. So we can use this. We can actually evaluate these at points."}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "Fair enough. And we can do the same thing for x. X prime of theta, derivative of the first expression, it is going to be two times cosine of two theta, two times cosine of two theta, times the second expression, cosine of theta, and then you're gonna have the first expression, sine of two theta, times the derivative of the second expression, which is negative sine of theta, negative sine of theta. So we can use this. We can actually evaluate these at points. For example, we could say, well, what's happening when theta is equal to pi over four? So when theta's pi over four, I'll do that in black right over here, we are going to be at this point right over there. Well, let's evaluate it."}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "We can actually evaluate these at points. For example, we could say, well, what's happening when theta is equal to pi over four? So when theta's pi over four, I'll do that in black right over here, we are going to be at this point right over there. Well, let's evaluate it. So if I were to say y prime of pi over four is equal to, let's see, this is going to be equal to two cosine of pi over two, two times pi over four, times sine of pi over four, plus sine of, two times pi over four is sine of pi over two, times cosine of pi over four. Cosine of pi over four. What is this going to be equal to?"}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "Well, let's evaluate it. So if I were to say y prime of pi over four is equal to, let's see, this is going to be equal to two cosine of pi over two, two times pi over four, times sine of pi over four, plus sine of, two times pi over four is sine of pi over two, times cosine of pi over four. Cosine of pi over four. What is this going to be equal to? Well, cosine of pi over two is zero. So if that's zero, all of this stuff's gonna be zero. And here's sine of pi over two, this is one."}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "What is this going to be equal to? Well, cosine of pi over two is zero. So if that's zero, all of this stuff's gonna be zero. And here's sine of pi over two, this is one. Cosine of pi over four is square root of two over two. Square root of two over two. So this is going to be equal to square root of two over two."}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "And here's sine of pi over two, this is one. Cosine of pi over four is square root of two over two. Square root of two over two. So this is going to be equal to square root of two over two. And actually, just for the sake of saving some space, I'll just write it right over here. It's going to be equal to square root of two over two. Well, we could do the same exercise with x."}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "So this is going to be equal to square root of two over two. And actually, just for the sake of saving some space, I'll just write it right over here. It's going to be equal to square root of two over two. Well, we could do the same exercise with x. We could say x prime of pi over four. Let's see, we're still gonna have two times cosine of two times pi over four. So that's going to be two times cosine of pi over two."}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "Well, we could do the same exercise with x. We could say x prime of pi over four. Let's see, we're still gonna have two times cosine of two times pi over four. So that's going to be two times cosine of pi over two. This first part right over here is gonna look the same, so that first term's gonna be zero. Then we're gonna have minus, so this is all gonna be zero. So then we're gonna have minus sine of pi over, two times pi over four is sine of pi over, sine of pi over two, and then times sine of pi over four."}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "So that's going to be two times cosine of pi over two. This first part right over here is gonna look the same, so that first term's gonna be zero. Then we're gonna have minus, so this is all gonna be zero. So then we're gonna have minus sine of pi over, two times pi over four is sine of pi over, sine of pi over two, and then times sine of pi over four. Sine of pi over four. Now, this is just going to be one, and so this is going to be equal to, and this is square root of two over two as well, so this is negative square root of two over two. Now let's see why that makes sense."}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "So then we're gonna have minus sine of pi over, two times pi over four is sine of pi over, sine of pi over two, and then times sine of pi over four. Sine of pi over four. Now, this is just going to be one, and so this is going to be equal to, and this is square root of two over two as well, so this is negative square root of two over two. Now let's see why that makes sense. So let's think about what happens as theta increases here. If you increase theta a little bit from pi over four, if you increase it just a little bit, your y coordinate continues to increase, so it makes sense you have a positive slope here. But what happens to your x coordinate as theta increases a little bit, as theta goes from there to there?"}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "Now let's see why that makes sense. So let's think about what happens as theta increases here. If you increase theta a little bit from pi over four, if you increase it just a little bit, your y coordinate continues to increase, so it makes sense you have a positive slope here. But what happens to your x coordinate as theta increases a little bit, as theta goes from there to there? Well, then your x coordinate starts to decrease when theta increases, so that's why it makes sense that you have a negative rate of change right over here. Now, the next question that you might say is, well, I wanna find the rate of change of y with respect to x, because I wanna figure out the slope of the tangent line right over there. And it looks like it has a slope of negative one, but how would we actually calculate it?"}, {"video_title": "Polar functions derivatives Advanced derivatives AP Calculus BC Khan Academy.mp3", "Sentence": "But what happens to your x coordinate as theta increases a little bit, as theta goes from there to there? Well, then your x coordinate starts to decrease when theta increases, so that's why it makes sense that you have a negative rate of change right over here. Now, the next question that you might say is, well, I wanna find the rate of change of y with respect to x, because I wanna figure out the slope of the tangent line right over there. And it looks like it has a slope of negative one, but how would we actually calculate it? Well, one way to think about it is, the derivative of y, the derivative of y with respect to x is going to be equal to the derivative of y with respect to theta over the derivative of x with respect to theta. And so at that value, so at theta is equal to pi over four, theta is equal to pi over four, this is going to be equal to positive square root of two over two over negative square root of two over two, negative square root of two over two. And this all simplifies to being equal to negative one, which makes sense, this does look indeed like a tangent line that has a slope of negative one."}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "I looked up the phonetic spelling before making this video. I'm assuming I'm still mispronouncing it. And I apologize to all the French speakers out there ahead of time. But he was a famous French philosopher mathematician who lived in medieval France. He lived in the 1300s. And he's famous for his proof that the harmonic series actually diverges. And just as a little bit of review, this is a harmonic series, one plus one half plus one third plus one fourth plus one fifth."}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "But he was a famous French philosopher mathematician who lived in medieval France. He lived in the 1300s. And he's famous for his proof that the harmonic series actually diverges. And just as a little bit of review, this is a harmonic series, one plus one half plus one third plus one fourth plus one fifth. And it's always been in my brain, the first time that I saw the harmonic series, it wasn't obvious to me whether it converged or diverged. It looks like, well gee, all these terms are positive, but they're going towards zero. So I could imagine that this thing could converge."}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "And just as a little bit of review, this is a harmonic series, one plus one half plus one third plus one fourth plus one fifth. And it's always been in my brain, the first time that I saw the harmonic series, it wasn't obvious to me whether it converged or diverged. It looks like, well gee, all these terms are positive, but they're going towards zero. So I could imagine that this thing could converge. But he proved otherwise. He proved, one of the most famous and most elegant proofs in mathematics, that it does indeed diverge. And the way that he did this is he replaced every term in the harmonic series with a term that is less than or equal to that term."}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "So I could imagine that this thing could converge. But he proved otherwise. He proved, one of the most famous and most elegant proofs in mathematics, that it does indeed diverge. And the way that he did this is he replaced every term in the harmonic series with a term that is less than or equal to that term. And then if he proves, and then by proving that his new series diverges, and it's less than or equal to this series, or each of the terms are less than or equal to each of the corresponding terms of here, then he says, therefore, by the comparison test, this must diverge. So how did he construct that? Well, one way to think about it is he replaced each of the terms in the harmonic series with the largest power of 1 1\u20442 that is less than or equal to that term."}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "And the way that he did this is he replaced every term in the harmonic series with a term that is less than or equal to that term. And then if he proves, and then by proving that his new series diverges, and it's less than or equal to this series, or each of the terms are less than or equal to each of the corresponding terms of here, then he says, therefore, by the comparison test, this must diverge. So how did he construct that? Well, one way to think about it is he replaced each of the terms in the harmonic series with the largest power of 1 1\u20442 that is less than or equal to that term. So what's the largest power of 1 1\u20442 that is less than or equal to one? Well, one is a power of 1 1\u20442. So that is, 1 1\u20442 to the zero power is one."}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, one way to think about it is he replaced each of the terms in the harmonic series with the largest power of 1 1\u20442 that is less than or equal to that term. So what's the largest power of 1 1\u20442 that is less than or equal to one? Well, one is a power of 1 1\u20442. So that is, 1 1\u20442 to the zero power is one. So one is the largest power of 1 1\u20442 that is less than or equal to one. So I'll just write the one there. And now, what's the largest power of 1 1\u20442 that is less than or equal to 1 1\u20442?"}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "So that is, 1 1\u20442 to the zero power is one. So one is the largest power of 1 1\u20442 that is less than or equal to one. So I'll just write the one there. And now, what's the largest power of 1 1\u20442 that is less than or equal to 1 1\u20442? Well, that's just going to be 1 1\u20442. That's just 1 1\u20442 to the first power. Now, what's the largest power of, what's the largest power of 1 1\u20442 that is less than or equal to 1 3rd?"}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "And now, what's the largest power of 1 1\u20442 that is less than or equal to 1 1\u20442? Well, that's just going to be 1 1\u20442. That's just 1 1\u20442 to the first power. Now, what's the largest power of, what's the largest power of 1 1\u20442 that is less than or equal to 1 3rd? Well, 1 1\u20442 is larger than 1 3rd. It's not less than 1 3rd. We want it to be less than 1 3rd."}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "Now, what's the largest power of, what's the largest power of 1 1\u20442 that is less than or equal to 1 3rd? Well, 1 1\u20442 is larger than 1 3rd. It's not less than 1 3rd. We want it to be less than 1 3rd. So the next power of 1 1\u20442 is, or I should say the power of 1 1\u20442, the largest power of 1 1\u20442 that is less than or equal to 1 3rd is 1 4th. So replace 1 3rd with 1 4th, and of course, replace 1 4th with 1 4th. And then we get to 1 5th."}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "We want it to be less than 1 3rd. So the next power of 1 1\u20442 is, or I should say the power of 1 1\u20442, the largest power of 1 1\u20442 that is less than or equal to 1 3rd is 1 4th. So replace 1 3rd with 1 4th, and of course, replace 1 4th with 1 4th. And then we get to 1 5th. What's the largest power of 1 1\u20442 that is less than or equal to 1 5th? Once again, 1 4th is greater than 1 5th. The largest power of 1 1\u20442 that is less than or equal to 1 5th is 1 8th."}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "And then we get to 1 5th. What's the largest power of 1 1\u20442 that is less than or equal to 1 5th? Once again, 1 4th is greater than 1 5th. The largest power of 1 1\u20442 that is less than or equal to 1 5th is 1 8th. So he replaced that with 1 8th. He of course would replace it for the same reason, that with 1 8th. He would replace that one with 1 8th."}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "The largest power of 1 1\u20442 that is less than or equal to 1 5th is 1 8th. So he replaced that with 1 8th. He of course would replace it for the same reason, that with 1 8th. He would replace that one with 1 8th. And of course, 1 8th, the largest power of 1 1\u20442 that is less than or equal to 1 8th is 1 8th. And then what would he replace 1 9th with? Well, he would replace 1 9th with 1 16th by the exact same argument."}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "He would replace that one with 1 8th. And of course, 1 8th, the largest power of 1 1\u20442 that is less than or equal to 1 8th is 1 8th. And then what would he replace 1 9th with? Well, he would replace 1 9th with 1 16th by the exact same argument. And you'd keep going all the way until you get to 1 16th. So you would essentially have eight 1 16ths in a row. Well, what's interesting here?"}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, he would replace 1 9th with 1 16th by the exact same argument. And you'd keep going all the way until you get to 1 16th. So you would essentially have eight 1 16ths in a row. Well, what's interesting here? Well, let's first verify that we can use the comparison test here. So in this first series, each of the terms are non-negative. In the second series, each of the terms are non-negative."}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, what's interesting here? Well, let's first verify that we can use the comparison test here. So in this first series, each of the terms are non-negative. In the second series, each of the terms are non-negative. And we also see that each of the corresponding terms in a harmonic series is larger than each of the, or greater than or equal to the corresponding terms in this series. We constructed it this way. These are equal."}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "In the second series, each of the terms are non-negative. And we also see that each of the corresponding terms in a harmonic series is larger than each of the, or greater than or equal to the corresponding terms in this series. We constructed it this way. These are equal. This one is equal. This is greater than this. 1 3rd is greater than 1 4th."}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "These are equal. This one is equal. This is greater than this. 1 3rd is greater than 1 4th. 1 4th is equal to 1 4th. 1 5th is greater than 1 8th. 1 6th is greater than 1 8th."}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "1 3rd is greater than 1 4th. 1 4th is equal to 1 4th. 1 5th is greater than 1 8th. 1 6th is greater than 1 8th. 1 7th is greater than 1 8th. 1 8th is equal to 1 8th. And so one way to think about it is this is the, each of the corresponding terms in this new constructed series are smaller."}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "1 6th is greater than 1 8th. 1 7th is greater than 1 8th. 1 8th is equal to 1 8th. And so one way to think about it is this is the, each of the corresponding terms in this new constructed series are smaller. And I'm gonna just call it S in this infinite sum. And of course we keep going on and on and on. Maybe I should do that in magenta."}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so one way to think about it is this is the, each of the corresponding terms in this new constructed series are smaller. And I'm gonna just call it S in this infinite sum. And of course we keep going on and on and on. Maybe I should do that in magenta. So we see that each of the corresponding terms here are smaller than the corresponding terms up here. And they're all positive. So if we can prove, if we can prove that S, that this sum right over here diverges, then by the comparison test, the larger series, the harmonic series here, the one where the corresponding terms are larger, that must also diverge."}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "Maybe I should do that in magenta. So we see that each of the corresponding terms here are smaller than the corresponding terms up here. And they're all positive. So if we can prove, if we can prove that S, that this sum right over here diverges, then by the comparison test, the larger series, the harmonic series here, the one where the corresponding terms are larger, that must also diverge. And how do we do that? Well let's just actually just take these sums. This is going to be, so let me write it."}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "So if we can prove, if we can prove that S, that this sum right over here diverges, then by the comparison test, the larger series, the harmonic series here, the one where the corresponding terms are larger, that must also diverge. And how do we do that? Well let's just actually just take these sums. This is going to be, so let me write it. So S is going to be equal to one plus 1 1. 1 4th plus 1 4th, what's that? Well it's 2 4ths or 1 1."}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "This is going to be, so let me write it. So S is going to be equal to one plus 1 1. 1 4th plus 1 4th, what's that? Well it's 2 4ths or 1 1. I think you see what's going on here. This is exciting. What's 1 8th plus 1 8th plus 1 8th plus 1 8th?"}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well it's 2 4ths or 1 1. I think you see what's going on here. This is exciting. What's 1 8th plus 1 8th plus 1 8th plus 1 8th? Well that's 4 8ths or 1 1. What's 1 16th plus 1 16th? And we're gonna go all the way until we get to one, although we're gonna have eight of these."}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "What's 1 8th plus 1 8th plus 1 8th plus 1 8th? Well that's 4 8ths or 1 1. What's 1 16th plus 1 16th? And we're gonna go all the way until we get to one, although we're gonna have eight of these. Well that's gonna be 8 16ths or 1 1. 1 1. And then you're gonna have 16 1 32nds or 1 1."}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "And we're gonna go all the way until we get to one, although we're gonna have eight of these. Well that's gonna be 8 16ths or 1 1. 1 1. And then you're gonna have 16 1 32nds or 1 1. And so we're essentially just going to be adding 1 1. And we start with one, we just keep adding one plus 1 1 plus 1 1 plus 1 1 plus 1 1. Well this is clearly going to be equal to, or this is going to, this is unbounded."}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "And then you're gonna have 16 1 32nds or 1 1. And so we're essentially just going to be adding 1 1. And we start with one, we just keep adding one plus 1 1 plus 1 1 plus 1 1 plus 1 1. Well this is clearly going to be equal to, or this is going to, this is unbounded. This is, you could say that this is equal to infinity. This is equal to infinity. Or another way to think about it is s clearly diverges."}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well this is clearly going to be equal to, or this is going to, this is unbounded. This is, you could say that this is equal to infinity. This is equal to infinity. Or another way to think about it is s clearly diverges. S clearly diverges. And since s is, I guess we could say, each of its corresponding terms, or each of its terms are smaller than the corresponding terms in the harmonic series, we can then say that the harmonic series diverges. The harmonic series diverges."}, {"video_title": "Proof harmonic series diverges Series AP Calculus BC Khan Academy.mp3", "Sentence": "Or another way to think about it is s clearly diverges. S clearly diverges. And since s is, I guess we could say, each of its corresponding terms, or each of its terms are smaller than the corresponding terms in the harmonic series, we can then say that the harmonic series diverges. The harmonic series diverges. Diverges. There's no way that this thing over here can converge. If this thing is, each of its corresponding terms are smaller, and you could even think of this kind of, the sum as being smaller, but this sum goes to infinity, so this one must also go to infinity."}, {"video_title": "Derivatives of sin(x) and cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And I've graphed y is equal to cosine of x in blue and y is equal to sine of x in red. We're not going to prove what the derivatives are, but we're gonna know what they are and get an intuitive sense, and in future videos we'll actually do a proof. So let's start with sine of x. So the derivative can be viewed as a slope of the tangent line. So for example, at this point right over here, it looks like the slope of our tangent line should be zero. So our derivative function should be zero at that x value. Similarly, over here, it looks like the derivative is zero."}, {"video_title": "Derivatives of sin(x) and cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So the derivative can be viewed as a slope of the tangent line. So for example, at this point right over here, it looks like the slope of our tangent line should be zero. So our derivative function should be zero at that x value. Similarly, over here, it looks like the derivative is zero. The slope of the tangent line would be zero. So whatever our derivative function is at that x value, it should be equal to zero. If we look right over here on sine of x, it looks like the slope of the tangent line would be pretty close to one."}, {"video_title": "Derivatives of sin(x) and cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Similarly, over here, it looks like the derivative is zero. The slope of the tangent line would be zero. So whatever our derivative function is at that x value, it should be equal to zero. If we look right over here on sine of x, it looks like the slope of the tangent line would be pretty close to one. If that is the case, then in our derivative function, when x is equal to zero, that derivative function should be equal to one. Similarly, over here, it looks like the slope of the tangent line is negative one, which tells us that the derivative function should be hitting the value of negative one at that x value. So you're probably seeing something interesting emerge."}, {"video_title": "Derivatives of sin(x) and cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "If we look right over here on sine of x, it looks like the slope of the tangent line would be pretty close to one. If that is the case, then in our derivative function, when x is equal to zero, that derivative function should be equal to one. Similarly, over here, it looks like the slope of the tangent line is negative one, which tells us that the derivative function should be hitting the value of negative one at that x value. So you're probably seeing something interesting emerge. Everywhere, while we're trying to plot the slope of the tangent line, it seems to coincide with y is equal to cosine of x. And it is indeed the case that the derivative of sine of x is equal to cosine of x. And you can see that it makes sense not just at the points we tried, but even in the trends."}, {"video_title": "Derivatives of sin(x) and cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So you're probably seeing something interesting emerge. Everywhere, while we're trying to plot the slope of the tangent line, it seems to coincide with y is equal to cosine of x. And it is indeed the case that the derivative of sine of x is equal to cosine of x. And you can see that it makes sense not just at the points we tried, but even in the trends. If you look at sine of x here, the slope is one, but then it becomes less and less and less positive all the way until it becomes zero. Cosine of x, the value of the function, is one, and it becomes less and less positive all the way until it equals zero. And you could keep doing that type of analysis to feel good about it."}, {"video_title": "Derivatives of sin(x) and cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And you can see that it makes sense not just at the points we tried, but even in the trends. If you look at sine of x here, the slope is one, but then it becomes less and less and less positive all the way until it becomes zero. Cosine of x, the value of the function, is one, and it becomes less and less positive all the way until it equals zero. And you could keep doing that type of analysis to feel good about it. In another video, we're going to prove this more rigorously. So now let's think about cosine of x. So cosine of x, right over here, the slope of the tangent line looks like it is zero."}, {"video_title": "Derivatives of sin(x) and cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And you could keep doing that type of analysis to feel good about it. In another video, we're going to prove this more rigorously. So now let's think about cosine of x. So cosine of x, right over here, the slope of the tangent line looks like it is zero. And so its derivative function needs to be zero at that point. So hey, maybe it's sine of x. Let's keep trying this."}, {"video_title": "Derivatives of sin(x) and cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So cosine of x, right over here, the slope of the tangent line looks like it is zero. And so its derivative function needs to be zero at that point. So hey, maybe it's sine of x. Let's keep trying this. So over here, cosine of x, it looks like the slope of the tangent line is negative one. And so we would want the derivative to go through that point right over there. All right, this is starting to seem, it doesn't seem like the derivative of cosine of x could be sine of x."}, {"video_title": "Derivatives of sin(x) and cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Let's keep trying this. So over here, cosine of x, it looks like the slope of the tangent line is negative one. And so we would want the derivative to go through that point right over there. All right, this is starting to seem, it doesn't seem like the derivative of cosine of x could be sine of x. In fact, this is the opposite of what sine of x is doing. Sine of x is at one, not negative one at that point. But that's an interesting theory."}, {"video_title": "Derivatives of sin(x) and cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "All right, this is starting to seem, it doesn't seem like the derivative of cosine of x could be sine of x. In fact, this is the opposite of what sine of x is doing. Sine of x is at one, not negative one at that point. But that's an interesting theory. Maybe the derivative of cosine of x is negative sine of x. So let's plot that. So this does seem to coincide."}, {"video_title": "Derivatives of sin(x) and cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "But that's an interesting theory. Maybe the derivative of cosine of x is negative sine of x. So let's plot that. So this does seem to coincide. The derivative of cosine of x here looks like negative one, the slope of the tangent line. And negative sine of this x value is negative one. Over here, the derivative of cosine of x looks like it is zero."}, {"video_title": "Derivatives of sin(x) and cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So this does seem to coincide. The derivative of cosine of x here looks like negative one, the slope of the tangent line. And negative sine of this x value is negative one. Over here, the derivative of cosine of x looks like it is zero. And negative sine of x is indeed zero. So it actually turns out that it is the case, that the derivative of cosine of x is negative sine of x. So these are really good to know."}, {"video_title": "Derivatives of sin(x) and cos(x) Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Over here, the derivative of cosine of x looks like it is zero. And negative sine of x is indeed zero. So it actually turns out that it is the case, that the derivative of cosine of x is negative sine of x. So these are really good to know. These are kind of fundamental trigonometric derivatives to know, we'll be able to derive other things for them. And hopefully this video gives you a good intuitive sense of why this is true. And in future videos, we will prove it rigorously."}, {"video_title": "Conditional & absolute convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "We use the alternating series test in that video to prove that it converges. So this thing converges, so this converges by alternating series test. By alternating alternating series. That looks a little bit messy. Alternating series test. If you want to review that, go watch the video on the alternating series test. Now let's take a little bit about what happens if we were to take the absolute value of each of these terms."}, {"video_title": "Conditional & absolute convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "That looks a little bit messy. Alternating series test. If you want to review that, go watch the video on the alternating series test. Now let's take a little bit about what happens if we were to take the absolute value of each of these terms. So if you were to take the absolute value of each of these terms, so if you were to take the sum from n equals one to infinity of the absolute value of negative one to the n plus one over n, well what is this going to be equal to? Well, this numerator is either going to be one or negative one. The absolute value of that is always going to be one, so it's going to be that over and n is always positive."}, {"video_title": "Conditional & absolute convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "Now let's take a little bit about what happens if we were to take the absolute value of each of these terms. So if you were to take the absolute value of each of these terms, so if you were to take the sum from n equals one to infinity of the absolute value of negative one to the n plus one over n, well what is this going to be equal to? Well, this numerator is either going to be one or negative one. The absolute value of that is always going to be one, so it's going to be that over and n is always positive. We're going from one to infinity, so it's just going to be equal to the sum. It's going to be equal to the sum from n equals one to infinity of one over n. And this is just the famous harmonic series. And there's this video that we have, and you should look it up on Khan Academy if you don't believe me, on the famous proof that the harmonic series diverges."}, {"video_title": "Conditional & absolute convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "The absolute value of that is always going to be one, so it's going to be that over and n is always positive. We're going from one to infinity, so it's just going to be equal to the sum. It's going to be equal to the sum from n equals one to infinity of one over n. And this is just the famous harmonic series. And there's this video that we have, and you should look it up on Khan Academy if you don't believe me, on the famous proof that the harmonic series diverges. So the harmonic series is one plus one half plus one third. This thing right over here diverges. And so when you see a series that converges, but if you were to take the absolute value of each of its terms and then that diverges, we say that this series converges conditionally."}, {"video_title": "Conditional & absolute convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "And there's this video that we have, and you should look it up on Khan Academy if you don't believe me, on the famous proof that the harmonic series diverges. So the harmonic series is one plus one half plus one third. This thing right over here diverges. And so when you see a series that converges, but if you were to take the absolute value of each of its terms and then that diverges, we say that this series converges conditionally. You can say it converges, but you could also say it converges conditionally. And the condition is, I guess you could say that we're not taking the absolute value of each of the terms. And if something converges when you take the absolute value as well, then you say it converges absolutely."}, {"video_title": "Conditional & absolute convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so when you see a series that converges, but if you were to take the absolute value of each of its terms and then that diverges, we say that this series converges conditionally. You can say it converges, but you could also say it converges conditionally. And the condition is, I guess you could say that we're not taking the absolute value of each of the terms. And if something converges when you take the absolute value as well, then you say it converges absolutely. So let's look at an example of that. If I were to take this series, let's do a geometric series, that might be fun. Actually, I'm using these colors too much, let me use another color."}, {"video_title": "Conditional & absolute convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "And if something converges when you take the absolute value as well, then you say it converges absolutely. So let's look at an example of that. If I were to take this series, let's do a geometric series, that might be fun. Actually, I'm using these colors too much, let me use another color. Let's take the sum from n equals one to infinity of negative one half to the n plus one power. We know this is a geometric series where the absolute value of our common ratio is less than one. We know that this converges."}, {"video_title": "Conditional & absolute convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "Actually, I'm using these colors too much, let me use another color. Let's take the sum from n equals one to infinity of negative one half to the n plus one power. We know this is a geometric series where the absolute value of our common ratio is less than one. We know that this converges. And if we were to take the absolute value of each of these terms, so if you were to take the sum, let me do that in a different color just to mix things up a little bit. If you were to take the absolute value of each of these terms, so the absolute value of negative one half, absolute value of negative one half to the n plus one power, this is going to be the same thing as the sum from n equals one to infinity of one half to the n plus one. And here, once again, the common ratio, the absolute value of the common ratio is less than one, and we've studied this when we looked at geometric series."}, {"video_title": "Conditional & absolute convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "We know that this converges. And if we were to take the absolute value of each of these terms, so if you were to take the sum, let me do that in a different color just to mix things up a little bit. If you were to take the absolute value of each of these terms, so the absolute value of negative one half, absolute value of negative one half to the n plus one power, this is going to be the same thing as the sum from n equals one to infinity of one half to the n plus one. And here, once again, the common ratio, the absolute value of the common ratio is less than one, and we've studied this when we looked at geometric series. This also converges. This also converges. So when we took the absolute value of the terms, it still converged."}, {"video_title": "Conditional & absolute convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "And here, once again, the common ratio, the absolute value of the common ratio is less than one, and we've studied this when we looked at geometric series. This also converges. This also converges. So when we took the absolute value of the terms, it still converged. So for this one, we can say that this converges absolutely. So we've talked a lot already about convergence or divergence, and that's all been good. And what we're doing in this video is we're introducing a nuance or a flavors of convergence."}, {"video_title": "Conditional & absolute convergence Series AP Calculus BC Khan Academy.mp3", "Sentence": "So when we took the absolute value of the terms, it still converged. So for this one, we can say that this converges absolutely. So we've talked a lot already about convergence or divergence, and that's all been good. And what we're doing in this video is we're introducing a nuance or a flavors of convergence. So you can converge, but it might be interesting to say, well, would it still converge if we took the absolute value of the terms? If it won't, if you converge but it doesn't converge when you take the absolute value of the terms, then you say it converges conditionally. If it converges and it still converges when you take the absolute value of the terms, then we say it converges absolutely."}, {"video_title": "Proof d dx(x^n) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "I just did several videos on the binomial theorem, and now that they're done, I think now is a good time to do the proof of the derivative of the general form. So let's take the derivative of x to the n. And now that we know the binomial theorem, we have the tools to do it. So how do we take the derivative? Well, what's the classic definition of the derivative? It is the limit as delta x approaches 0 of f of x plus delta x, right? So f of x plus delta x in this situation is x plus delta x to the nth power, right? Minus f of x."}, {"video_title": "Proof d dx(x^n) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Well, what's the classic definition of the derivative? It is the limit as delta x approaches 0 of f of x plus delta x, right? So f of x plus delta x in this situation is x plus delta x to the nth power, right? Minus f of x. Well, f of x here is just x to the n. All of that over delta x. And now that we know the binomial theorem, we can figure out what the expansion of x plus delta x is to the nth power. And if you don't know the binomial theorem, go to my pre-calculus playlist and watch the videos on the binomial theorem."}, {"video_title": "Proof d dx(x^n) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Minus f of x. Well, f of x here is just x to the n. All of that over delta x. And now that we know the binomial theorem, we can figure out what the expansion of x plus delta x is to the nth power. And if you don't know the binomial theorem, go to my pre-calculus playlist and watch the videos on the binomial theorem. So the binomial theorem tells us that this is equal to, and let me, I'm going to need some space for this one, the limit as delta x approaches 0. And what's the binomial theorem tell us? This is going to be equal to, I'm just going to do the numerator, x to the n plus n choose 1."}, {"video_title": "Proof d dx(x^n) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And if you don't know the binomial theorem, go to my pre-calculus playlist and watch the videos on the binomial theorem. So the binomial theorem tells us that this is equal to, and let me, I'm going to need some space for this one, the limit as delta x approaches 0. And what's the binomial theorem tell us? This is going to be equal to, I'm just going to do the numerator, x to the n plus n choose 1. And once again, review the binomial theorem if this looks like Latin to you and you don't know Latin. n choose 1 of x to the n minus 1 delta x plus n choose 2 x to the n minus 2 delta x squared. And then plus, and we have a bunch of digits."}, {"video_title": "Proof d dx(x^n) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "This is going to be equal to, I'm just going to do the numerator, x to the n plus n choose 1. And once again, review the binomial theorem if this looks like Latin to you and you don't know Latin. n choose 1 of x to the n minus 1 delta x plus n choose 2 x to the n minus 2 delta x squared. And then plus, and we have a bunch of digits. And in this proof, we don't have to go through all the digits, but the binomial theorem tells us what they are. And of course, the last digit, we just keep adding, is going to be 1, well, it would be n choose n, which is 1. Let me just write that down."}, {"video_title": "Proof d dx(x^n) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And then plus, and we have a bunch of digits. And in this proof, we don't have to go through all the digits, but the binomial theorem tells us what they are. And of course, the last digit, we just keep adding, is going to be 1, well, it would be n choose n, which is 1. Let me just write that down. n choose n is going to be x to the 0 times delta x to the n. So that's the binomial expansion. And let me switch back to minus. So all of that, that green, that's x plus delta x to the n, so minus x to the n. Minus x to the n power, that's x to the n, I know I squashed it there."}, {"video_title": "Proof d dx(x^n) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Let me just write that down. n choose n is going to be x to the 0 times delta x to the n. So that's the binomial expansion. And let me switch back to minus. So all of that, that green, that's x plus delta x to the n, so minus x to the n. Minus x to the n power, that's x to the n, I know I squashed it there. All of that over delta x. Let's see if we can simplify. So first of all, we have an x to the n here, and at the very end, we subtract out an x to the n, so these two cancel out."}, {"video_title": "Proof d dx(x^n) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So all of that, that green, that's x plus delta x to the n, so minus x to the n. Minus x to the n power, that's x to the n, I know I squashed it there. All of that over delta x. Let's see if we can simplify. So first of all, we have an x to the n here, and at the very end, we subtract out an x to the n, so these two cancel out. And then if we look at every term here, every term in the numerator has a delta x, right? So we can divide the numerator and the denominator, essentially, by delta x. This is the same thing as 1 over delta x times this whole thing."}, {"video_title": "Proof d dx(x^n) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So first of all, we have an x to the n here, and at the very end, we subtract out an x to the n, so these two cancel out. And then if we look at every term here, every term in the numerator has a delta x, right? So we can divide the numerator and the denominator, essentially, by delta x. This is the same thing as 1 over delta x times this whole thing. So that is equal to, let me, the limit as delta x approaches 0 of, so we divide the top and the bottom by delta x, or we multiply the numerator times 1 over delta x, we get n choose 1 x to the n minus 1. What's delta x divided by delta x? Well, that's just 1, right?"}, {"video_title": "Proof d dx(x^n) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "This is the same thing as 1 over delta x times this whole thing. So that is equal to, let me, the limit as delta x approaches 0 of, so we divide the top and the bottom by delta x, or we multiply the numerator times 1 over delta x, we get n choose 1 x to the n minus 1. What's delta x divided by delta x? Well, that's just 1, right? Plus n choose 2 x to the n minus 2. This is delta x squared, but when we divide by delta x, we just get a delta x here. And then we keep having a bunch of terms."}, {"video_title": "Proof d dx(x^n) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Well, that's just 1, right? Plus n choose 2 x to the n minus 2. This is delta x squared, but when we divide by delta x, we just get a delta x here. And then we keep having a bunch of terms. We're going to divide all of them by delta x, right? And then the last term is delta x to the n. But now we're going to divide that by delta x. So the last term becomes n choose n. x to the 0 is 1, so we can ignore that."}, {"video_title": "Proof d dx(x^n) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And then we keep having a bunch of terms. We're going to divide all of them by delta x, right? And then the last term is delta x to the n. But now we're going to divide that by delta x. So the last term becomes n choose n. x to the 0 is 1, so we can ignore that. Delta x to the n divided by delta x. Well, that's delta x to the n minus 1, right? And then what are we doing now?"}, {"video_title": "Proof d dx(x^n) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So the last term becomes n choose n. x to the 0 is 1, so we can ignore that. Delta x to the n divided by delta x. Well, that's delta x to the n minus 1, right? And then what are we doing now? Well, remember, we're taking the limit as delta x approaches 0. So as delta x approaches 0, pretty much every term that has a delta x in it, it becomes 0, right? We just, when you multiply by 0, you get 0."}, {"video_title": "Proof d dx(x^n) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "And then what are we doing now? Well, remember, we're taking the limit as delta x approaches 0. So as delta x approaches 0, pretty much every term that has a delta x in it, it becomes 0, right? We just, when you multiply by 0, you get 0. So every term, this first term has no delta x in it, but every other term does. Every other term, even after we divide it by delta x, has a delta x in it. So that's a 0."}, {"video_title": "Proof d dx(x^n) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "We just, when you multiply by 0, you get 0. So every term, this first term has no delta x in it, but every other term does. Every other term, even after we divide it by delta x, has a delta x in it. So that's a 0. Every term is 0. All of the other, I don't know, what is it, n minus 1 terms, they're all 0. So all we're left with is that this is equal to n choose 1 of x to the n minus 1."}, {"video_title": "Proof d dx(x^n) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So that's a 0. Every term is 0. All of the other, I don't know, what is it, n minus 1 terms, they're all 0. So all we're left with is that this is equal to n choose 1 of x to the n minus 1. And what's n choose 1? That equals n factorial over 1 factorial divided by n minus 1 factorial times x to the n minus 1. 1 factorial is 1."}, {"video_title": "Proof d dx(x^n) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So all we're left with is that this is equal to n choose 1 of x to the n minus 1. And what's n choose 1? That equals n factorial over 1 factorial divided by n minus 1 factorial times x to the n minus 1. 1 factorial is 1. And if I have 7 factorial divided by 6 factorial, that's just 7. Or if I have 3 factorial divided by 2 factorial, that's just 3. You can work it out."}, {"video_title": "Proof d dx(x^n) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "1 factorial is 1. And if I have 7 factorial divided by 6 factorial, that's just 7. Or if I have 3 factorial divided by 2 factorial, that's just 3. You can work it out. 10 factorial divided by 9 factorial, that's 10. So n factorial divided by n minus 1 factorial, that's just equal to n. So this is equal to n times x to the n minus 1. So that's the derivative of x to the n. n times x to the n minus 1."}, {"video_title": "Proof d dx(x^n) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "You can work it out. 10 factorial divided by 9 factorial, that's 10. So n factorial divided by n minus 1 factorial, that's just equal to n. So this is equal to n times x to the n minus 1. So that's the derivative of x to the n. n times x to the n minus 1. So we just proved the derivative for any positive integer, x to the power n, where n is any positive integer. And we see later it actually works for all, actually, real numbers in the exponent. So I will see you in a future video."}, {"video_title": "Proof d dx(x^n) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So that's the derivative of x to the n. n times x to the n minus 1. So we just proved the derivative for any positive integer, x to the power n, where n is any positive integer. And we see later it actually works for all, actually, real numbers in the exponent. So I will see you in a future video. Oh, and another thing I wanted to point out is, I said that we had to know the binomial theorem. But if you think about it, we really didn't even have to know the binomial theorem. Because we knew in any binomial expansion, I mean, you'd have to know a little bit."}, {"video_title": "Proof d dx(x^n) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "So I will see you in a future video. Oh, and another thing I wanted to point out is, I said that we had to know the binomial theorem. But if you think about it, we really didn't even have to know the binomial theorem. Because we knew in any binomial expansion, I mean, you'd have to know a little bit. But if you did a little experimentation, you would realize that whenever you expand a plus b to the nth power, the first term is going to be a to the n. And then the second term is going to be plus n a to the n minus 1 b. And then you could keep having a bunch of terms. But these are the only terms that are relevant in this proof, because all the other terms get canceled out when delta x approaches 0."}, {"video_title": "Proof d dx(x^n) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "Because we knew in any binomial expansion, I mean, you'd have to know a little bit. But if you did a little experimentation, you would realize that whenever you expand a plus b to the nth power, the first term is going to be a to the n. And then the second term is going to be plus n a to the n minus 1 b. And then you could keep having a bunch of terms. But these are the only terms that are relevant in this proof, because all the other terms get canceled out when delta x approaches 0. So if you just knew that, you could have done this. But it's much better to do it with the binomial theorem. Ignore what I just said if it confused you."}, {"video_title": "Proof d dx(x^n) Taking derivatives Differential Calculus Khan Academy.mp3", "Sentence": "But these are the only terms that are relevant in this proof, because all the other terms get canceled out when delta x approaches 0. So if you just knew that, you could have done this. But it's much better to do it with the binomial theorem. Ignore what I just said if it confused you. I'm just saying that we could have just said, oh, the rest of these terms just all go to 0. Anyway, hopefully you found that fulfilling. I will see you in future videos."}, {"video_title": "Worked example area between curves AP Calculus AB Khan Academy.mp3", "Sentence": "So the key here is you might recognize, hey, this is an area between curves. A definite integral might be useful. So I'll just set up the definite integral sign. And so first we need to think about what are our left and right boundaries of our region? Well, it looks like the left boundary is where the two graphs intersect right over here, and the right boundary is where they intersect right over there. Well, what is this point of intersection? It looks like it is negative one comma negative two."}, {"video_title": "Worked example area between curves AP Calculus AB Khan Academy.mp3", "Sentence": "And so first we need to think about what are our left and right boundaries of our region? Well, it looks like the left boundary is where the two graphs intersect right over here, and the right boundary is where they intersect right over there. Well, what is this point of intersection? It looks like it is negative one comma negative two. We can verify that. In this red curve, if x is negative one, let's see, you square that, you'll get one minus three. You do indeed get y is equal to negative two."}, {"video_title": "Worked example area between curves AP Calculus AB Khan Academy.mp3", "Sentence": "It looks like it is negative one comma negative two. We can verify that. In this red curve, if x is negative one, let's see, you square that, you'll get one minus three. You do indeed get y is equal to negative two. And in this blue function, when x is equal to negative one, you get one minus four plus one. Once again, you do indeed get y is equal to negative two. And the same thing is true when x is equal to one."}, {"video_title": "Worked example area between curves AP Calculus AB Khan Academy.mp3", "Sentence": "You do indeed get y is equal to negative two. And in this blue function, when x is equal to negative one, you get one minus four plus one. Once again, you do indeed get y is equal to negative two. And the same thing is true when x is equal to one. One minus three, negative two. One minus four plus one, negative two. So our bounds are indeed, we're going from x equals negative one to x equals positive one."}, {"video_title": "Worked example area between curves AP Calculus AB Khan Academy.mp3", "Sentence": "And the same thing is true when x is equal to one. One minus three, negative two. One minus four plus one, negative two. So our bounds are indeed, we're going from x equals negative one to x equals positive one. And now let's think about our upper and lower bounds. Over that interval, this blue graph is our upper bound. And so we would subtract the lower bound from the upper bound."}, {"video_title": "Worked example area between curves AP Calculus AB Khan Academy.mp3", "Sentence": "So our bounds are indeed, we're going from x equals negative one to x equals positive one. And now let's think about our upper and lower bounds. Over that interval, this blue graph is our upper bound. And so we would subtract the lower bound from the upper bound. So we would have x to the fourth minus four x squared plus one. And from that, we will subtract x squared minus three dx. And in many other videos, we have talked about why you do this, why this makes sense to just subtract the lower graph from the upper graph when you're finding the area between them."}, {"video_title": "Worked example area between curves AP Calculus AB Khan Academy.mp3", "Sentence": "And so we would subtract the lower bound from the upper bound. So we would have x to the fourth minus four x squared plus one. And from that, we will subtract x squared minus three dx. And in many other videos, we have talked about why you do this, why this makes sense to just subtract the lower graph from the upper graph when you're finding the area between them. But now we just have to evaluate this definite integral. So let's just get down to business. All right."}, {"video_title": "Worked example area between curves AP Calculus AB Khan Academy.mp3", "Sentence": "And in many other videos, we have talked about why you do this, why this makes sense to just subtract the lower graph from the upper graph when you're finding the area between them. But now we just have to evaluate this definite integral. So let's just get down to business. All right. So we have the integral from negative one to one. And so we have x to the fourth, x to the fourth. And now we have minus four x squared."}, {"video_title": "Worked example area between curves AP Calculus AB Khan Academy.mp3", "Sentence": "All right. So we have the integral from negative one to one. And so we have x to the fourth, x to the fourth. And now we have minus four x squared. And then when you distribute this negative sign, you're gonna subtract another x squared. So you're gonna have minus five x squared. And then you have plus one."}, {"video_title": "Worked example area between curves AP Calculus AB Khan Academy.mp3", "Sentence": "And now we have minus four x squared. And then when you distribute this negative sign, you're gonna subtract another x squared. So you're gonna have minus five x squared. And then you have plus one. And then you're gonna subtract a negative three. So it's gonna be one plus three. So it's gonna be plus four dx."}, {"video_title": "Worked example area between curves AP Calculus AB Khan Academy.mp3", "Sentence": "And then you have plus one. And then you're gonna subtract a negative three. So it's gonna be one plus three. So it's gonna be plus four dx. And just to be clear, I should put parentheses right over there because it's really the dx is being multiplied by this entire expression. And so let's see. Let's find the antiderivative of this."}, {"video_title": "Worked example area between curves AP Calculus AB Khan Academy.mp3", "Sentence": "So it's gonna be plus four dx. And just to be clear, I should put parentheses right over there because it's really the dx is being multiplied by this entire expression. And so let's see. Let's find the antiderivative of this. This should be pretty straightforward. We're just gonna use the reverse power rule multiple times. So this is going to be, the antiderivative of x to the fourth is x to the fifth over five."}, {"video_title": "Worked example area between curves AP Calculus AB Khan Academy.mp3", "Sentence": "Let's find the antiderivative of this. This should be pretty straightforward. We're just gonna use the reverse power rule multiple times. So this is going to be, the antiderivative of x to the fourth is x to the fifth over five. We just incremented the exponent and divided by that incremented exponent minus, same idea here, five x to the third over three plus four x. And then we are going to evaluate it at one and then subtract from that it evaluated at negative one. So let's first evaluate it at one."}, {"video_title": "Worked example area between curves AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be, the antiderivative of x to the fourth is x to the fifth over five. We just incremented the exponent and divided by that incremented exponent minus, same idea here, five x to the third over three plus four x. And then we are going to evaluate it at one and then subtract from that it evaluated at negative one. So let's first evaluate it at one. We're gonna get 1 5th minus 5 3rds plus four. And now let us evaluate it at negative one. So minus, let's see, if this is negative one, we're gonna negative 1 5th, negative 1 5th, and this is gonna be plus 5 3rds, plus 5 3rds, and then this is going to be minus four, minus four."}, {"video_title": "Worked example area between curves AP Calculus AB Khan Academy.mp3", "Sentence": "So let's first evaluate it at one. We're gonna get 1 5th minus 5 3rds plus four. And now let us evaluate it at negative one. So minus, let's see, if this is negative one, we're gonna negative 1 5th, negative 1 5th, and this is gonna be plus 5 3rds, plus 5 3rds, and then this is going to be minus four, minus four. But then when you distribute the negative sign, this, so we're gonna distribute this over all of these terms. And so this is going to be, if we make this positive, this will be positive. This one will be negative, and then this one will be positive."}, {"video_title": "Worked example area between curves AP Calculus AB Khan Academy.mp3", "Sentence": "So minus, let's see, if this is negative one, we're gonna negative 1 5th, negative 1 5th, and this is gonna be plus 5 3rds, plus 5 3rds, and then this is going to be minus four, minus four. But then when you distribute the negative sign, this, so we're gonna distribute this over all of these terms. And so this is going to be, if we make this positive, this will be positive. This one will be negative, and then this one will be positive. So you have 1 5th plus 1 5th, which is going to be 2 5ths, that and that, and then minus 5 3rds, minus 5 3rds. So minus 10 over three, and then four plus four. So plus eight."}, {"video_title": "Worked example area between curves AP Calculus AB Khan Academy.mp3", "Sentence": "This one will be negative, and then this one will be positive. So you have 1 5th plus 1 5th, which is going to be 2 5ths, that and that, and then minus 5 3rds, minus 5 3rds. So minus 10 over three, and then four plus four. So plus eight. And also we just need to simplify this. This is going to be, let's see, it's gonna be eight, and then if I write, so plus, I'm gonna write these two with a denominator of 15, because that's the common denominator of three and five. Let's see, 2 5ths is 6 15ths."}, {"video_title": "Worked example area between curves AP Calculus AB Khan Academy.mp3", "Sentence": "So plus eight. And also we just need to simplify this. This is going to be, let's see, it's gonna be eight, and then if I write, so plus, I'm gonna write these two with a denominator of 15, because that's the common denominator of three and five. Let's see, 2 5ths is 6 15ths. Yeah, that's right, five times three is 15, two times three is six. And then 10 3rds, let's see, if we multiply the denominator times five, we have to multiply the numerator times five, so this is gonna be 50 15ths. And so what's 6 15ths minus 50 15ths?"}, {"video_title": "Worked example area between curves AP Calculus AB Khan Academy.mp3", "Sentence": "Let's see, 2 5ths is 6 15ths. Yeah, that's right, five times three is 15, two times three is six. And then 10 3rds, let's see, if we multiply the denominator times five, we have to multiply the numerator times five, so this is gonna be 50 15ths. And so what's 6 15ths minus 50 15ths? So this is going to be equal to eight minus six minus 50 is minus 44. Minus 44 over 15. And so what is 44 over 15?"}, {"video_title": "Worked example area between curves AP Calculus AB Khan Academy.mp3", "Sentence": "And so what's 6 15ths minus 50 15ths? So this is going to be equal to eight minus six minus 50 is minus 44. Minus 44 over 15. And so what is 44 over 15? 44 over 15 is equal to two and 14 15ths. So that's really what we're subtracting. We're gonna subtract two and 14 15ths."}, {"video_title": "Worked example area between curves AP Calculus AB Khan Academy.mp3", "Sentence": "And so what is 44 over 15? 44 over 15 is equal to two and 14 15ths. So that's really what we're subtracting. We're gonna subtract two and 14 15ths. So if you subtract two from this, you would get six minus 14 over 15, because we still have to subtract the 14 15ths. And then six minus 14 15ths is going to be equal to five and 1 15th. So just like that, we were indeed able to figure out this area."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "50 meters. And on this cliff, I have a car. And this car is not just sitting on the cliff, it's driving off of it. Very dramatic problem. So let's see, I have this car here, and it's driving off of this cliff at 5 meters per second. 5 meters per second. And I want to know, what is the path of this car as it falls off of the cliff?"}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "Very dramatic problem. So let's see, I have this car here, and it's driving off of this cliff at 5 meters per second. 5 meters per second. And I want to know, what is the path of this car as it falls off of the cliff? So let's set up a little coordinate axis here. Let's say that I get my y-axis. Let's say that this is the y-axis."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "And I want to know, what is the path of this car as it falls off of the cliff? So let's set up a little coordinate axis here. Let's say that I get my y-axis. Let's say that this is the y-axis. Right there. And then this will be my x-axis. x-axis."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "Let's say that this is the y-axis. Right there. And then this will be my x-axis. x-axis. This is y, this is x. Let's say that this is the point, well we know, this is a 50 meter high cliff, maybe y equals 0 is sea level. So this would be 50 right here."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "x-axis. This is y, this is x. Let's say that this is the point, well we know, this is a 50 meter high cliff, maybe y equals 0 is sea level. So this would be 50 right here. And let's say that this point right here on the cliff, that's at x is equal to 10. So this point right here is the point 10, 50. And let's say that the car is right at this point, right about to drive off the cliff, at time is equal to 0."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "So this would be 50 right here. And let's say that this point right here on the cliff, that's at x is equal to 10. So this point right here is the point 10, 50. And let's say that the car is right at this point, right about to drive off the cliff, at time is equal to 0. So this is at time equal to 0. So t for time. Time is equal to 0."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "And let's say that the car is right at this point, right about to drive off the cliff, at time is equal to 0. So this is at time equal to 0. So t for time. Time is equal to 0. So my question is, what happens to this car as it drives off the cliff? So this is a bit of a physics problem, and I won't go deep into the physics, and I won't prove some of the equations, and I encourage you to watch the kinematics videos, the projectile motion videos, if you want to know where the equations come from. But the point here is just to get the equations of what the graph looks like."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "Time is equal to 0. So my question is, what happens to this car as it drives off the cliff? So this is a bit of a physics problem, and I won't go deep into the physics, and I won't prove some of the equations, and I encourage you to watch the kinematics videos, the projectile motion videos, if you want to know where the equations come from. But the point here is just to get the equations of what the graph looks like. So if I wanted to know x as a function of time, x is a function of time, a suitably vibrant color. So x as a function of time is going to be what? Well, we're going to assume that we're on a planet that has no air, we're in a vacuum."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "But the point here is just to get the equations of what the graph looks like. So if I wanted to know x as a function of time, x is a function of time, a suitably vibrant color. So x as a function of time is going to be what? Well, we're going to assume that we're on a planet that has no air, we're in a vacuum. So if we start off in the x direction at 5 meters per second to the right, we won't be decelerated by air or friction or anything else. Newton's laws of motion, stays in motion unless it's affected by a net force, and there won't be any net force in the x direction. So it's just going to keep moving to the right at 5 meters per second, and position or distance is just equal to velocity times time."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "Well, we're going to assume that we're on a planet that has no air, we're in a vacuum. So if we start off in the x direction at 5 meters per second to the right, we won't be decelerated by air or friction or anything else. Newton's laws of motion, stays in motion unless it's affected by a net force, and there won't be any net force in the x direction. So it's just going to keep moving to the right at 5 meters per second, and position or distance is just equal to velocity times time. Our velocity is 5 times time. And of course, it didn't start at x is equal to 0, right? This is at time equals 0."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "So it's just going to keep moving to the right at 5 meters per second, and position or distance is just equal to velocity times time. Our velocity is 5 times time. And of course, it didn't start at x is equal to 0, right? This is at time equals 0. So it started at x is equal to 10. So you want to know it's kind of x of 0, where it started off, so plus 10. And this should be a little intuitive for you, right?"}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "This is at time equals 0. So it started at x is equal to 10. So you want to know it's kind of x of 0, where it started off, so plus 10. And this should be a little intuitive for you, right? At time is equal to 0, this term cancels out where x is equal to 10. That makes sense. At time is equal to 1, we should be a little bit, we'll be 5 meters further out, so on and so forth."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "And this should be a little intuitive for you, right? At time is equal to 0, this term cancels out where x is equal to 10. That makes sense. At time is equal to 1, we should be a little bit, we'll be 5 meters further out, so on and so forth. Fair enough. That's x as a function of the parameter time. You probably realize that this is a video on parametric equations, not physics, so it's nice to early on say the word parameter."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "At time is equal to 1, we should be a little bit, we'll be 5 meters further out, so on and so forth. Fair enough. That's x as a function of the parameter time. You probably realize that this is a video on parametric equations, not physics, so it's nice to early on say the word parameter. And time tends to be the parameter when people talk about parametric equations, although it could be anything. It could be radius or angle or who knows what else. So let's figure out what y is as a function of time."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "You probably realize that this is a video on parametric equations, not physics, so it's nice to early on say the word parameter. And time tends to be the parameter when people talk about parametric equations, although it could be anything. It could be radius or angle or who knows what else. So let's figure out what y is as a function of time. So y as a function of time, it's going to be equal to the initial y position, or y of 0, which is 50. We're 50 meters up in the air, plus our initial velocity in the y direction, and we don't actually have any initial velocity in the y direction. The car isn't jumping or isn't diving."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "So let's figure out what y is as a function of time. So y as a function of time, it's going to be equal to the initial y position, or y of 0, which is 50. We're 50 meters up in the air, plus our initial velocity in the y direction, and we don't actually have any initial velocity in the y direction. The car isn't jumping or isn't diving. It's just moving horizontally to the right, and the cliff is supporting it. So it has no y velocity, but if you were curious, it would be the y velocity times time. But since there's no y velocity times time, at least initially, I'll put nothing there."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "The car isn't jumping or isn't diving. It's just moving horizontally to the right, and the cliff is supporting it. So it has no y velocity, but if you were curious, it would be the y velocity times time. But since there's no y velocity times time, at least initially, I'll put nothing there. Plus the acceleration of gravity, so times time squared over 2. We want to figure out the sign. And just so you know, it's nice to touch on the physics a little bit, just so you know where these formulas come from and you know the motivation behind why you would even use a parametric equation."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "But since there's no y velocity times time, at least initially, I'll put nothing there. Plus the acceleration of gravity, so times time squared over 2. We want to figure out the sign. And just so you know, it's nice to touch on the physics a little bit, just so you know where these formulas come from and you know the motivation behind why you would even use a parametric equation. Gravity goes downwards in this example, and downwards in this example is in the minus y direction, where y is decreasing. And the real, and it's not exact, but gravity is normally 9.8 meters per second squared in most textbooks. But for the sake of simplifying this, we'll say that it's approximately 10 meters per second squared."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "And just so you know, it's nice to touch on the physics a little bit, just so you know where these formulas come from and you know the motivation behind why you would even use a parametric equation. Gravity goes downwards in this example, and downwards in this example is in the minus y direction, where y is decreasing. And the real, and it's not exact, but gravity is normally 9.8 meters per second squared in most textbooks. But for the sake of simplifying this, we'll say that it's approximately 10 meters per second squared. That's how fast everything will be accelerated downward on this planet. Since it has no air, let's assume it's a planet with a little bit more mass than Earth. And since it's going downwards, its direction is negative."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "But for the sake of simplifying this, we'll say that it's approximately 10 meters per second squared. That's how fast everything will be accelerated downward on this planet. Since it has no air, let's assume it's a planet with a little bit more mass than Earth. And since it's going downwards, its direction is negative. So in our formula up here, it's our initial position. We add no velocity times time, so I won't put that there, minus 10 meters per second squared times t squared over 2. And you can watch the projectile motion videos to figure out how I got these formulas right there, but that's not the point of this."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "And since it's going downwards, its direction is negative. So in our formula up here, it's our initial position. We add no velocity times time, so I won't put that there, minus 10 meters per second squared times t squared over 2. And you can watch the projectile motion videos to figure out how I got these formulas right there, but that's not the point of this. The point of this is to graph what happens to the cars and learn a little bit about parametric equations. So what is the path of this car as it falls off the cliff? Let's make a table here."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "And you can watch the projectile motion videos to figure out how I got these formulas right there, but that's not the point of this. The point of this is to graph what happens to the cars and learn a little bit about parametric equations. So what is the path of this car as it falls off the cliff? Let's make a table here. I'll make a table. And I'll do the... So x and y are a function of this third parameter, t. So we're going to set t at different values, and we're going to figure out what x and y are equal to."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "Let's make a table here. I'll make a table. And I'll do the... So x and y are a function of this third parameter, t. So we're going to set t at different values, and we're going to figure out what x and y are equal to. And I'm just going to arbitrarily pick some t's. t is equal to 0, t is equal to 1, 2, and 3. At time is equal to 0, what is x?"}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "So x and y are a function of this third parameter, t. So we're going to set t at different values, and we're going to figure out what x and y are equal to. And I'm just going to arbitrarily pick some t's. t is equal to 0, t is equal to 1, 2, and 3. At time is equal to 0, what is x? x of 0, this is 0, x is equal to 10 meters. At time is equal to 1, what is x? This is x of 1, right?"}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "At time is equal to 0, what is x? x of 0, this is 0, x is equal to 10 meters. At time is equal to 1, what is x? This is x of 1, right? If I wanted to write that notation. So 5 times 1 is 5, plus 10 is 15. x of 2, 5 times 2 plus 10, that's 20. And it makes sense."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "This is x of 1, right? If I wanted to write that notation. So 5 times 1 is 5, plus 10 is 15. x of 2, 5 times 2 plus 10, that's 20. And it makes sense. Every second we're getting 5 meters more to the right, or x is increasing by 5 meters. So when t is equal to 3, 15 plus 10 is 25. Easy enough."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "And it makes sense. Every second we're getting 5 meters more to the right, or x is increasing by 5 meters. So when t is equal to 3, 15 plus 10 is 25. Easy enough. The y is a little bit more complicated. And just to simplify this, this is the same thing as 5, right? 10 divided by 2."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "Easy enough. The y is a little bit more complicated. And just to simplify this, this is the same thing as 5, right? 10 divided by 2. So 50 minus 5t squared. So time is equal to 0. This term cancels out."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "10 divided by 2. So 50 minus 5t squared. So time is equal to 0. This term cancels out. We just have 50 meters up in the air. At time is equal to 1, 1 squared is 1, times 5 is 5. 50 minus 5 is 40 or 45 meters in the air."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "This term cancels out. We just have 50 meters up in the air. At time is equal to 1, 1 squared is 1, times 5 is 5. 50 minus 5 is 40 or 45 meters in the air. Is that right? Yeah, time, 1, 50, right? And then at time is equal to 2, 2 squared is 4, 4 times 5 is 20."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "50 minus 5 is 40 or 45 meters in the air. Is that right? Yeah, time, 1, 50, right? And then at time is equal to 2, 2 squared is 4, 4 times 5 is 20. 50 minus 20 is 30. And then finally, at time is equal to 3, and I just say finally because that's the last number we picked. Time is equal to 3, 3 squared is 9, 9 times 5 is 45."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "And then at time is equal to 2, 2 squared is 4, 4 times 5 is 20. 50 minus 20 is 30. And then finally, at time is equal to 3, and I just say finally because that's the last number we picked. Time is equal to 3, 3 squared is 9, 9 times 5 is 45. 50 minus 45 is 5. So let's plot these points. So time is equal to 0, that's what we got right there."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "Time is equal to 3, 3 squared is 9, 9 times 5 is 45. 50 minus 45 is 5. So let's plot these points. So time is equal to 0, that's what we got right there. At time is equal to 1, we're at x is equal to 15. That's roughly, see this is 5, 10, 15. Let me do all of them."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "So time is equal to 0, that's what we got right there. At time is equal to 1, we're at x is equal to 15. That's roughly, see this is 5, 10, 15. Let me do all of them. 15, 20, 25. And then the y-axis, let me label that while we're at it. So this is roughly 10, 20, 30, 40, 50."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "Let me do all of them. 15, 20, 25. And then the y-axis, let me label that while we're at it. So this is roughly 10, 20, 30, 40, 50. So at time is equal to 0, we're at 10, 50. That's that point right there. At time is equal to 1, we're at 15, 45."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "So this is roughly 10, 20, 30, 40, 50. So at time is equal to 0, we're at 10, 50. That's that point right there. At time is equal to 1, we're at 15, 45. So x is 15, y is 45, which is right about there. So this is t is equal to 1. At time is equal to 2, we're at the coordinate 20, 30."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "At time is equal to 1, we're at 15, 45. So x is 15, y is 45, which is right about there. So this is t is equal to 1. At time is equal to 2, we're at the coordinate 20, 30. So 20 and 30, it's right about there. So this is at time is equal to 2. And then at time is equal to 3, we're at 25, 5."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "At time is equal to 2, we're at the coordinate 20, 30. So 20 and 30, it's right about there. So this is at time is equal to 2. And then at time is equal to 3, we're at 25, 5. So we're right there, 25, 5. Time is equal to 3. And if we kept going on, at some point we're going to hit the ground."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "And then at time is equal to 3, we're at 25, 5. So we're right there, 25, 5. Time is equal to 3. And if we kept going on, at some point we're going to hit the ground. You can figure out, actually, set this equal to 0 and you figure out the exact time you hit the ground. Actually, let's do that. If this is equal to 0, 50, you get t is equal to the square root of 10, which is a little over 3 seconds, which makes sense."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "And if we kept going on, at some point we're going to hit the ground. You can figure out, actually, set this equal to 0 and you figure out the exact time you hit the ground. Actually, let's do that. If this is equal to 0, 50, you get t is equal to the square root of 10, which is a little over 3 seconds, which makes sense. A little over 3 seconds, we're going to be hitting the ground. But anyway, what's the path of this car? Well, it's going to look something like this."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "If this is equal to 0, 50, you get t is equal to the square root of 10, which is a little over 3 seconds, which makes sense. A little over 3 seconds, we're going to be hitting the ground. But anyway, what's the path of this car? Well, it's going to look something like this. It's going to look like this. It starts getting accelerated downwards, and then plunk, it hits the ground at 3 point something seconds. Now, what was interesting here is that by setting the parameter, not only did we get the curve, right?"}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "Well, it's going to look something like this. It's going to look like this. It starts getting accelerated downwards, and then plunk, it hits the ground at 3 point something seconds. Now, what was interesting here is that by setting the parameter, not only did we get the curve, right? We got this curve, which is kind of half of a parabola, half of a downward-shaping parabola, and we can actually eliminate the t and just get the equation for that parabola, and we'll do that in future videos. But what was interesting, by making it a parametric equation, we know the direction of the car. If you just saw this graph without the car and everything else I drew, you would know which way the car is falling."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "Now, what was interesting here is that by setting the parameter, not only did we get the curve, right? We got this curve, which is kind of half of a parabola, half of a downward-shaping parabola, and we can actually eliminate the t and just get the equation for that parabola, and we'll do that in future videos. But what was interesting, by making it a parametric equation, we know the direction of the car. If you just saw this graph without the car and everything else I drew, you would know which way the car is falling. But now we know that as t is increasing, we're going in that direction, so we can draw some arrows here. So because it's a parametric equation, we can draw some arrows. And then the most important thing is we know exactly where the car is at any time t. You can substitute t is equal to 1.25 seconds, and you know exactly where the car is."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "If you just saw this graph without the car and everything else I drew, you would know which way the car is falling. But now we know that as t is increasing, we're going in that direction, so we can draw some arrows here. So because it's a parametric equation, we can draw some arrows. And then the most important thing is we know exactly where the car is at any time t. You can substitute t is equal to 1.25 seconds, and you know exactly where the car is. So you can plot these points, and you can kind of get a sense that as time goes on, we're getting accelerated downwards, and that's why for every second further, especially the y distance gets further and further apart. Anyway, I just wanted to give you this example. Although this was a good physics problem, the intention wasn't to teach you physics."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "And then the most important thing is we know exactly where the car is at any time t. You can substitute t is equal to 1.25 seconds, and you know exactly where the car is. So you can plot these points, and you can kind of get a sense that as time goes on, we're getting accelerated downwards, and that's why for every second further, especially the y distance gets further and further apart. Anyway, I just wanted to give you this example. Although this was a good physics problem, the intention wasn't to teach you physics. The intention is to give you the motivation behind why parametric equations even exist. Parametric equations. These two things are parametric equations."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "Although this was a good physics problem, the intention wasn't to teach you physics. The intention is to give you the motivation behind why parametric equations even exist. Parametric equations. These two things are parametric equations. We defined x and y as a function of a third parameter t instead of defining y in terms of x or x in terms of y like we've done every other time since then. And this is super useful. I mean, you can imagine when you have really hard physics problems where you want to figure out the three-dimensional position of something."}, {"video_title": "Parametric equations 1 Parametric equations and polar coordinates Precalculus Khan Academy.mp3", "Sentence": "These two things are parametric equations. We defined x and y as a function of a third parameter t instead of defining y in terms of x or x in terms of y like we've done every other time since then. And this is super useful. I mean, you can imagine when you have really hard physics problems where you want to figure out the three-dimensional position of something. Then you'll have x as a function of t, y as a function of t, z as a function of t. All sorts of interesting problems come out of using parametric equations, not just in physics. But anyway, I thought a good place to start is the motivation because the first time I learned parametric equations, I was like, why mess up my nice and simple world of x's and y's while introducing a third parameter t? This is why, because you can figure out the path of things."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "Let's now attempt to find a solution for the logistic differential equation. And we already found some constant solutions. We can think through that a little bit, just as a little bit of a review from the last few videos. So this is the t-axis and this is the n-axis. We already saw that if n of zero, if at time equals zero, our population is zero, there's no one to reproduce, and this differential equation is consistent with that because if n is zero, this thing is going to be zero, and so our rate of change is going to be zero with respect to time, and so our population just won't change. It'll just stay at zero, which is nice because that's what would actually happen in a real population. So that gives us the constant solution that n of t is equal to zero."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "So this is the t-axis and this is the n-axis. We already saw that if n of zero, if at time equals zero, our population is zero, there's no one to reproduce, and this differential equation is consistent with that because if n is zero, this thing is going to be zero, and so our rate of change is going to be zero with respect to time, and so our population just won't change. It'll just stay at zero, which is nice because that's what would actually happen in a real population. So that gives us the constant solution that n of t is equal to zero. That's one solution to this differential equation. Not that interesting. A zero population will never grow or change."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "So that gives us the constant solution that n of t is equal to zero. That's one solution to this differential equation. Not that interesting. A zero population will never grow or change. The other constant solution is, well, what if our population started at the maximum of what the environment could sustain? And in that situation, this term is going to be k over k, which is one. One minus one is zero, and so in there, the population would also not change."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "A zero population will never grow or change. The other constant solution is, well, what if our population started at the maximum of what the environment could sustain? And in that situation, this term is going to be k over k, which is one. One minus one is zero, and so in there, the population would also not change. It would just stay at k, and so the rate of change would just stay at zero. So that's another constant solution that we start at the maximum population and then this differential equation tells us a scenario that never changes. But we said, hey, look, well, there could be something interesting that happens if our initial condition, which we called n sub naught, that's our n of zero, time equals zero if we start someplace in between, maybe closer, much below our eventual ceiling, I guess, because if we're a lot below our eventual ceiling, this thing is going to be a small fraction."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "One minus one is zero, and so in there, the population would also not change. It would just stay at k, and so the rate of change would just stay at zero. So that's another constant solution that we start at the maximum population and then this differential equation tells us a scenario that never changes. But we said, hey, look, well, there could be something interesting that happens if our initial condition, which we called n sub naught, that's our n of zero, time equals zero if we start someplace in between, maybe closer, much below our eventual ceiling, I guess, because if we're a lot below our eventual ceiling, this thing is going to be a small fraction. This thing is going to be close to one, and so the rate of change is going to be pretty close to being proportional to n, which is going to, and when we're thinking that, we can kind of, it might look something like that. As n increases, our rate of change increases, but then as n approaches k, this thing is going to approach one, this thing is going to approach zero, and so it's going to overwhelm this and our rate of change is going to approach zero. And so we could imagine a scenario where we asymptote towards k. And let's see if we can solve this to find this, this n of t, this n of t, to find the actual analytic expression for this n of t right over here, because this is interesting."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "But we said, hey, look, well, there could be something interesting that happens if our initial condition, which we called n sub naught, that's our n of zero, time equals zero if we start someplace in between, maybe closer, much below our eventual ceiling, I guess, because if we're a lot below our eventual ceiling, this thing is going to be a small fraction. This thing is going to be close to one, and so the rate of change is going to be pretty close to being proportional to n, which is going to, and when we're thinking that, we can kind of, it might look something like that. As n increases, our rate of change increases, but then as n approaches k, this thing is going to approach one, this thing is going to approach zero, and so it's going to overwhelm this and our rate of change is going to approach zero. And so we could imagine a scenario where we asymptote towards k. And let's see if we can solve this to find this, this n of t, this n of t, to find the actual analytic expression for this n of t right over here, because this is interesting. This is what could be used to model populations that would be more consistent with a Malthusian mindset. So let's see if we can do that. And to do that, we just have to realize this is a completely different, this is a separable differential equation, and we're assuming as a function of t, we're going to solve for an n of t that satisfies this, and so we just have to separate it from the explicit t's, but there are no explicit t's here, so it's quite easy to do."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "And so we could imagine a scenario where we asymptote towards k. And let's see if we can solve this to find this, this n of t, this n of t, to find the actual analytic expression for this n of t right over here, because this is interesting. This is what could be used to model populations that would be more consistent with a Malthusian mindset. So let's see if we can do that. And to do that, we just have to realize this is a completely different, this is a separable differential equation, and we're assuming as a function of t, we're going to solve for an n of t that satisfies this, and so we just have to separate it from the explicit t's, but there are no explicit t's here, so it's quite easy to do. So what I'm going to do is I'm going to take this part right over here, I'm going to divide both sides by that. I'm going to leave the r on the right-hand side. That'll make things, I think, a little bit easier as we try to solve for n of t, so let me just do that."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "And to do that, we just have to realize this is a completely different, this is a separable differential equation, and we're assuming as a function of t, we're going to solve for an n of t that satisfies this, and so we just have to separate it from the explicit t's, but there are no explicit t's here, so it's quite easy to do. So what I'm going to do is I'm going to take this part right over here, I'm going to divide both sides by that. I'm going to leave the r on the right-hand side. That'll make things, I think, a little bit easier as we try to solve for n of t, so let me just do that. So this is going to be equal to one over n times one minus n over k. One minus n over k times dn dt, times dn dt is equal to r, is equal to r, and another way we could think about it, oh, actually, let me just continue to tackle it this way. So we get that, and now what I want to do is take the antiderivative of both sides with respect to t. Well, this is pretty straightforward. That's just going to be rt times some constant, but what's this going to be?"}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "That'll make things, I think, a little bit easier as we try to solve for n of t, so let me just do that. So this is going to be equal to one over n times one minus n over k. One minus n over k times dn dt, times dn dt is equal to r, is equal to r, and another way we could think about it, oh, actually, let me just continue to tackle it this way. So we get that, and now what I want to do is take the antiderivative of both sides with respect to t. Well, this is pretty straightforward. That's just going to be rt times some constant, but what's this going to be? This is kind of this messy thing, and maybe if we could break this out, if we can expand this into two fractions, do a little bit of partial fraction expansion, maybe we can come up with an expression that's a little bit easier to find the antiderivative of, and so I'm hoping that I can find an a and a b where a over n plus b over one minus n over k is equal to this business, is equal to one over n times one minus n over k. Let's see if we can find an a and a b. This is just partial fraction expansion. If this looks unfamiliar to you, I encourage you to review that part on Khan Academy."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "That's just going to be rt times some constant, but what's this going to be? This is kind of this messy thing, and maybe if we could break this out, if we can expand this into two fractions, do a little bit of partial fraction expansion, maybe we can come up with an expression that's a little bit easier to find the antiderivative of, and so I'm hoping that I can find an a and a b where a over n plus b over one minus n over k is equal to this business, is equal to one over n times one minus n over k. Let's see if we can find an a and a b. This is just partial fraction expansion. If this looks unfamiliar to you, I encourage you to review that part on Khan Academy. Do a search for partial fraction expansion on Khan Academy. So how do we do it? Well, let's just add these two right over here."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "If this looks unfamiliar to you, I encourage you to review that part on Khan Academy. Do a search for partial fraction expansion on Khan Academy. So how do we do it? Well, let's just add these two right over here. So this, if we take the sum right over here, this is going to be a times this, which is a minus a over kn plus b times this, plus bn, over the product of these two. So it's just going to be n times one minus n over k. One way to think about it, I just multiply the numerator and the denominator of this one by one minus n over k. So a times one minus n over k is that. N times one minus n over k is that."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "Well, let's just add these two right over here. So this, if we take the sum right over here, this is going to be a times this, which is a minus a over kn plus b times this, plus bn, over the product of these two. So it's just going to be n times one minus n over k. One way to think about it, I just multiply the numerator and the denominator of this one by one minus n over k. So a times one minus n over k is that. N times one minus n over k is that. And I multiply the numerator and denominator of this both by n, bn, and n times that. And then I added the two now that I had the same denominator. So that's just adding fractions with unlike denominators."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "N times one minus n over k is that. And I multiply the numerator and denominator of this both by n, bn, and n times that. And then I added the two now that I had the same denominator. So that's just adding fractions with unlike denominators. It's going to be equal to this, one over n times one minus n over k. And now we can try to think, well, what is a and b going to be equal to? Or what can they be equal to? Well, I have a constant term here."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "So that's just adding fractions with unlike denominators. It's going to be equal to this, one over n times one minus n over k. And now we can try to think, well, what is a and b going to be equal to? Or what can they be equal to? Well, I have a constant term here. I don't have any n term here. I could say that maybe I have a zero times an n. And that actually helps things a little bit. Because maybe we could say that this thing, that this plus this, which are the coefficients on n, are going to sum up to zero."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "Well, I have a constant term here. I don't have any n term here. I could say that maybe I have a zero times an n. And that actually helps things a little bit. Because maybe we could say that this thing, that this plus this, which are the coefficients on n, are going to sum up to zero. And that this, which is our a, is going to be equal to one. And that's pretty nice. So we could say that a is equal to one."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "Because maybe we could say that this thing, that this plus this, which are the coefficients on n, are going to sum up to zero. And that this, which is our a, is going to be equal to one. And that's pretty nice. So we could say that a is equal to one. a is equal to one. And then if a is equal to one, we have negative one over k, negative one over k plus b, plus b is equal to zero. Well, what's b going to be?"}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "So we could say that a is equal to one. a is equal to one. And then if a is equal to one, we have negative one over k, negative one over k plus b, plus b is equal to zero. Well, what's b going to be? Well, b would be equal to one, one over k. So we can rewrite this as, we can rewrite it as one over n, one over n, one over n plus, one over n plus one over k, plus one over k over, over, all over, let me just do this, over one minus n over k. One minus n over k, over k, and then, and then times, times dn dt, dn dt is equal to, is equal to r. So I just did a little bit of partial fraction expansion. I don't wanna give myself a little bit of real estate. Let me clear this out."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "Well, what's b going to be? Well, b would be equal to one, one over k. So we can rewrite this as, we can rewrite it as one over n, one over n, one over n plus, one over n plus one over k, plus one over k over, over, all over, let me just do this, over one minus n over k. One minus n over k, over k, and then, and then times, times dn dt, dn dt is equal to, is equal to r. So I just did a little bit of partial fraction expansion. I don't wanna give myself a little bit of real estate. Let me clear this out. You can rewind the video and review that if you find it necessary. So let's do that. And so how does this help us?"}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "Let me clear this out. You can rewind the video and review that if you find it necessary. So let's do that. And so how does this help us? Well, hey, you know, I kinda, you probably might be recognizing the antiderivative of one over n, and you might even see this. So let's just think through this a little bit. We know the antiderivative of one over n is the natural log, I'm just in a new color, is the natural log of the absolute value of n. And we can see the derivative of that with respect to n is equal to one over n. And if we were to find the derivative of this with respect to t, derivative with respect to t of the natural log of the absolute value of n is equal to, this is going to be the derivative of this with respect to n times the derivative of n with respect to t, times dn dt."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "And so how does this help us? Well, hey, you know, I kinda, you probably might be recognizing the antiderivative of one over n, and you might even see this. So let's just think through this a little bit. We know the antiderivative of one over n is the natural log, I'm just in a new color, is the natural log of the absolute value of n. And we can see the derivative of that with respect to n is equal to one over n. And if we were to find the derivative of this with respect to t, derivative with respect to t of the natural log of the absolute value of n is equal to, this is going to be the derivative of this with respect to n times the derivative of n with respect to t, times dn dt. And we could do the same thing over here. Notice, I have, I have an expression. What would be the derivative of this expression right over here?"}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "We know the antiderivative of one over n is the natural log, I'm just in a new color, is the natural log of the absolute value of n. And we can see the derivative of that with respect to n is equal to one over n. And if we were to find the derivative of this with respect to t, derivative with respect to t of the natural log of the absolute value of n is equal to, this is going to be the derivative of this with respect to n times the derivative of n with respect to t, times dn dt. And we could do the same thing over here. Notice, I have, I have an expression. What would be the derivative of this expression right over here? It would be negative one over k. If I'm talking about the derivative with respect to n. It would be negative one over k. I have a positive one over k here, and I can even make it negative. I can turn this, I can clear this out of the way. Instead of having a positive there, I could have a, I guess I could say a double negative."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "What would be the derivative of this expression right over here? It would be negative one over k. If I'm talking about the derivative with respect to n. It would be negative one over k. I have a positive one over k here, and I can even make it negative. I can turn this, I can clear this out of the way. Instead of having a positive there, I could have a, I guess I could say a double negative. So a double negative, I haven't changed the value of it. And notice, now this is the derivative of this, which makes it very nice for you substitution, or you might be already used to doing this thing in your head. And so we know that the derivative with respect to n, let me write this in a new color."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "Instead of having a positive there, I could have a, I guess I could say a double negative. So a double negative, I haven't changed the value of it. And notice, now this is the derivative of this, which makes it very nice for you substitution, or you might be already used to doing this thing in your head. And so we know that the derivative with respect to n, let me write this in a new color. We know that the derivative with respect to n of the natural log of one minus n over k, once again, this comes straight out of the chain rule, it's going to be the derivative of this with respect to n, which is negative one over k, times the derivative of this whole thing with respect to this thing, which is times one over one minus n over k, which is exactly what we have over here. But if I were to take the derivative of this with respect to t, so the derivative with respect to t of the natural log of one minus n over k, it's going to be the derivative of this with respect to n, which is that, which is what we just found. It's going to be this."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "And so we know that the derivative with respect to n, let me write this in a new color. We know that the derivative with respect to n of the natural log of one minus n over k, once again, this comes straight out of the chain rule, it's going to be the derivative of this with respect to n, which is negative one over k, times the derivative of this whole thing with respect to this thing, which is times one over one minus n over k, which is exactly what we have over here. But if I were to take the derivative of this with respect to t, so the derivative with respect to t of the natural log of one minus n over k, it's going to be the derivative of this with respect to n, which is that, which is what we just found. It's going to be this. So copy and paste. It's going to be that. It's going to be that, times the derivative of n with respect to t, just straight out of the chain rule, dn dt."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "It's going to be this. So copy and paste. It's going to be that. It's going to be that, times the derivative of n with respect to t, just straight out of the chain rule, dn dt. Well, notice, we have a dn dt. We can multiply it times each of these things. And actually, why not?"}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "It's going to be that, times the derivative of n with respect to t, just straight out of the chain rule, dn dt. Well, notice, we have a dn dt. We can multiply it times each of these things. And actually, why not? Let's just do that just to make it clear, because this is really, you know, this isn't so much differential equations, but sometimes some of the calculus that we learned not too long ago, and even frankly, some of the algebra, it's nice to not skip steps. So if I distribute this, edit, paste, whoops. I wanted to copy and paste."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "And actually, why not? Let's just do that just to make it clear, because this is really, you know, this isn't so much differential equations, but sometimes some of the calculus that we learned not too long ago, and even frankly, some of the algebra, it's nice to not skip steps. So if I distribute this, edit, paste, whoops. I wanted to copy and paste. So copy and paste. So I have that. And then I have, and then I have, I can copy and paste."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "I wanted to copy and paste. So copy and paste. So I have that. And then I have, and then I have, I can copy and paste. So that, I'm just distributing, I'm just distributing this business right over here. So I get that. Now let me clean it up a little bit."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "And then I have, and then I have, I can copy and paste. So that, I'm just distributing, I'm just distributing this business right over here. So I get that. Now let me clean it up a little bit. So I have that right over here. And of course, we have this being equal to r. If I take the antiderivative of this with respect to t, well, I'm just gonna get, I'm just gonna get this. I am just going to get this."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "Now let me clean it up a little bit. So I have that right over here. And of course, we have this being equal to r. If I take the antiderivative of this with respect to t, well, I'm just gonna get, I'm just gonna get this. I am just going to get this. Notice the anti, I'm just gonna get this, and plus, or this minus, this minus this right over that. So let's do that. Let's take the antiderivative with respect to t. So the left-hand side, I'm gonna get the natural log, this, the antiderivative of that with respect to t is the natural log of the absolute value of n. And then minus the antiderivative of this with respect to t is the natural log, the natural log of the absolute value of one minus n over k is equal to, oh, and let's not forget, let's say it's, and we're gonna have some constant here."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "I am just going to get this. Notice the anti, I'm just gonna get this, and plus, or this minus, this minus this right over that. So let's do that. Let's take the antiderivative with respect to t. So the left-hand side, I'm gonna get the natural log, this, the antiderivative of that with respect to t is the natural log of the absolute value of n. And then minus the antiderivative of this with respect to t is the natural log, the natural log of the absolute value of one minus n over k is equal to, oh, and let's not forget, let's say it's, and we're gonna have some constant here. I wanna find a pretty general solution. Let's call it c one, is equal to the antiderivative of this with respect to t is r times t, maybe plus some other constant, plus some other constant, just like that. And now I'm going to assume that my n of t meets this assumption right over here."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "Let's take the antiderivative with respect to t. So the left-hand side, I'm gonna get the natural log, this, the antiderivative of that with respect to t is the natural log of the absolute value of n. And then minus the antiderivative of this with respect to t is the natural log, the natural log of the absolute value of one minus n over k is equal to, oh, and let's not forget, let's say it's, and we're gonna have some constant here. I wanna find a pretty general solution. Let's call it c one, is equal to the antiderivative of this with respect to t is r times t, maybe plus some other constant, plus some other constant, just like that. And now I'm going to assume that my n of t meets this assumption right over here. And so I'm going to assume, I'm going to, so assuming, assuming, assume, assume n of t is going to be less than k and greater than zero. That means that this thing, this n is always going to be positive. And if n is always between zero and k, that means that this thing is always going to be positive."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "And now I'm going to assume that my n of t meets this assumption right over here. And so I'm going to assume, I'm going to, so assuming, assuming, assume, assume n of t is going to be less than k and greater than zero. That means that this thing, this n is always going to be positive. And if n is always between zero and k, that means that this thing is always going to be positive. And so that helps me clear things up a little bit. Actually, let me do it in this color. That helps me get rid of that, helps me get rid of, helps me get rid of that."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "And if n is always between zero and k, that means that this thing is always going to be positive. And so that helps me clear things up a little bit. Actually, let me do it in this color. That helps me get rid of that, helps me get rid of, helps me get rid of that. I can instead throw some parentheses here. And why don't I subtract the c one from both sides? So let me, that would get rid of it here."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "That helps me get rid of that, helps me get rid of, helps me get rid of that. I can instead throw some parentheses here. And why don't I subtract the c one from both sides? So let me, that would get rid of it here. So edit, cut, and paste. So I have the c one there, and I know I'm overwriting my own work, which is not making it look as clean as possible. I'm just trying to clean it up a little bit."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "So let me, that would get rid of it here. So edit, cut, and paste. So I have the c one there, and I know I'm overwriting my own work, which is not making it look as clean as possible. I'm just trying to clean it up a little bit. And now, hey, I have an arbitrary constant that I haven't really solved for yet, minus another constant. Well, let me just call this some other constant. Let me just call it a general c. So let me just call this, let me just call this c. So clear, and I'm just going to call that c. I am just going to call that c. And I could even simplify this a little bit."}, {"video_title": "Solving the logistic differential equation part 1 Khan Academy.mp3", "Sentence": "I'm just trying to clean it up a little bit. And now, hey, I have an arbitrary constant that I haven't really solved for yet, minus another constant. Well, let me just call this some other constant. Let me just call it a general c. So let me just call this, let me just call this c. So clear, and I'm just going to call that c. I am just going to call that c. And I could even simplify this a little bit. But actually, I just realized I'm 13 minutes into this video, which is longer than I like to do. So let's continue this in the next video. I was getting excited, because I'm so close, I'm so close to solving for an n of t that satisfies our logistic differential equation."}, {"video_title": "Worked example Maclaurin polynomial Series AP Calculus BC Khan Academy.mp3", "Sentence": "And what we wanna figure out is what is the second degree Maclaurin polynomial of f? And like always, pause this video and see if you could have a go at it. So let's remind ourselves what a Maclaurin polynomial is. A Maclaurin polynomial is just a Taylor polynomial centered at zero. So the form of this second degree Maclaurin polynomial, and we just have to find this Maclaurin expansion until our second degree term, it's going to look like this. So p of x, I'm using p for polynomial. It's going to be our f of zero plus, we could view that as f of zero times x to the zero power."}, {"video_title": "Worked example Maclaurin polynomial Series AP Calculus BC Khan Academy.mp3", "Sentence": "A Maclaurin polynomial is just a Taylor polynomial centered at zero. So the form of this second degree Maclaurin polynomial, and we just have to find this Maclaurin expansion until our second degree term, it's going to look like this. So p of x, I'm using p for polynomial. It's going to be our f of zero plus, we could view that as f of zero times x to the zero power. Well, that's just f of zero. F of zero plus f prime of zero x plus f prime prime of zero divided by, we could think of it as two factorial, but it's really just two. We could think of this as dividing by one factorial, which is just one, this dividing by zero factorial, but that's just one again."}, {"video_title": "Worked example Maclaurin polynomial Series AP Calculus BC Khan Academy.mp3", "Sentence": "It's going to be our f of zero plus, we could view that as f of zero times x to the zero power. Well, that's just f of zero. F of zero plus f prime of zero x plus f prime prime of zero divided by, we could think of it as two factorial, but it's really just two. We could think of this as dividing by one factorial, which is just one, this dividing by zero factorial, but that's just one again. So we have f prime prime of zero, the second derivative evaluated at zero, divided by two x squared. Now, if we wanted a higher degree, we could keep on going, but remember, they're just asking us for the second degree. So this is the form that we're gonna need."}, {"video_title": "Worked example Maclaurin polynomial Series AP Calculus BC Khan Academy.mp3", "Sentence": "We could think of this as dividing by one factorial, which is just one, this dividing by zero factorial, but that's just one again. So we have f prime prime of zero, the second derivative evaluated at zero, divided by two x squared. Now, if we wanted a higher degree, we could keep on going, but remember, they're just asking us for the second degree. So this is the form that we're gonna need. We're gonna have these three terms. So let's see if we can evaluate these, let's see if we can evaluate the function and its derivatives at zero. So f of zero, f of zero is equal to one over the square root of zero plus one."}, {"video_title": "Worked example Maclaurin polynomial Series AP Calculus BC Khan Academy.mp3", "Sentence": "So this is the form that we're gonna need. We're gonna have these three terms. So let's see if we can evaluate these, let's see if we can evaluate the function and its derivatives at zero. So f of zero, f of zero is equal to one over the square root of zero plus one. Well, that's one over the square root of one, the principal root of one, which is positive one, so that's just going to be equal to one. So that right over there is equal to one. Now, let's evaluate f prime of x, and then I'll evaluate f prime of zero."}, {"video_title": "Worked example Maclaurin polynomial Series AP Calculus BC Khan Academy.mp3", "Sentence": "So f of zero, f of zero is equal to one over the square root of zero plus one. Well, that's one over the square root of one, the principal root of one, which is positive one, so that's just going to be equal to one. So that right over there is equal to one. Now, let's evaluate f prime of x, and then I'll evaluate f prime of zero. F prime of x is equal to, well, one over the square root of x plus one, this is the same thing as x plus one, let me write it this way. This is the same thing as, let me actually write it down this way. F of x, another way of writing f of x is, this is the same thing as x plus one to the negative 1 1\u20442, and so if I'm thinking the first derivative of f, well, I could use the chain rule here, the derivative of x plus one with respect to x, well, that's just going to be one, and then I'll take the derivative of this whole thing with respect to x plus one, and I'll just use the power rule there, so it's going to be negative 1 1\u20442 times x plus one to the, and I decrement the exponent, negative 3 1\u20442, and so the first derivative evaluated at zero is just negative 1 1\u20442 times, if this is zero, zero plus one is just one, one to the negative 3 1\u20442, one to the negative 3 1\u20442 power, well, that's just going to be one."}, {"video_title": "Worked example Maclaurin polynomial Series AP Calculus BC Khan Academy.mp3", "Sentence": "Now, let's evaluate f prime of x, and then I'll evaluate f prime of zero. F prime of x is equal to, well, one over the square root of x plus one, this is the same thing as x plus one, let me write it this way. This is the same thing as, let me actually write it down this way. F of x, another way of writing f of x is, this is the same thing as x plus one to the negative 1 1\u20442, and so if I'm thinking the first derivative of f, well, I could use the chain rule here, the derivative of x plus one with respect to x, well, that's just going to be one, and then I'll take the derivative of this whole thing with respect to x plus one, and I'll just use the power rule there, so it's going to be negative 1 1\u20442 times x plus one to the, and I decrement the exponent, negative 3 1\u20442, and so the first derivative evaluated at zero is just negative 1 1\u20442 times, if this is zero, zero plus one is just one, one to the negative 3 1\u20442, one to the negative 3 1\u20442 power, well, that's just going to be one. So this whole thing, f prime of zero is just negative 1 1\u20442, so that is, this right over here is negative 1 1\u20442, and now let's figure out the second derivative. All right, I'll do this, let me do this in this green color. So the second derivative with respect to x, well, I do the same thing again."}, {"video_title": "Worked example Maclaurin polynomial Series AP Calculus BC Khan Academy.mp3", "Sentence": "F of x, another way of writing f of x is, this is the same thing as x plus one to the negative 1 1\u20442, and so if I'm thinking the first derivative of f, well, I could use the chain rule here, the derivative of x plus one with respect to x, well, that's just going to be one, and then I'll take the derivative of this whole thing with respect to x plus one, and I'll just use the power rule there, so it's going to be negative 1 1\u20442 times x plus one to the, and I decrement the exponent, negative 3 1\u20442, and so the first derivative evaluated at zero is just negative 1 1\u20442 times, if this is zero, zero plus one is just one, one to the negative 3 1\u20442, one to the negative 3 1\u20442 power, well, that's just going to be one. So this whole thing, f prime of zero is just negative 1 1\u20442, so that is, this right over here is negative 1 1\u20442, and now let's figure out the second derivative. All right, I'll do this, let me do this in this green color. So the second derivative with respect to x, well, I do the same thing again. The derivative of x plus one with respect to x, that's just one, so I just have to take the derivative of the whole thing with respect to x plus one. So I take my exponent, bring it out front, negative 3 1\u20442 times 1 1\u20442, times negative 1 1\u20442 is going to be positive 3\u20444 times x plus one, and then I decrement the exponent here by one, or by 2 1\u20442, so this is going to be negative 5 1\u20442, and so the second derivative evaluated at zero, well, if this is equal to zero, you're going to have one to the negative 5 1\u20442, which is just one, times 3\u20444, so this is going to be 3\u20444. So this part right over here is 3\u20444, and so you're going to have 3\u20444 divided by two."}, {"video_title": "Worked example Maclaurin polynomial Series AP Calculus BC Khan Academy.mp3", "Sentence": "So the second derivative with respect to x, well, I do the same thing again. The derivative of x plus one with respect to x, that's just one, so I just have to take the derivative of the whole thing with respect to x plus one. So I take my exponent, bring it out front, negative 3 1\u20442 times 1 1\u20442, times negative 1 1\u20442 is going to be positive 3\u20444 times x plus one, and then I decrement the exponent here by one, or by 2 1\u20442, so this is going to be negative 5 1\u20442, and so the second derivative evaluated at zero, well, if this is equal to zero, you're going to have one to the negative 5 1\u20442, which is just one, times 3\u20444, so this is going to be 3\u20444. So this part right over here is 3\u20444, and so you're going to have 3\u20444 divided by two. 3\u20444 divided by two is 3\u2078. So our Taylor, or I should say our Maclaurin polynomial, our second degree Maclaurin polynomial, P of x is going to be equal to, and I'll do it in the same colors, it's going to be equal to one plus, or maybe I'll just write it as minus 1\u20442, minus 1\u20442 x, plus 3\u2078 x squared, plus 3\u2078 x squared, and we are done, there you have it. We have our second degree Maclaurin polynomial of f, which could be used to provide an approximation for our function, especially for x's near zero."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "So let's do it so it looks something like this. So that right over there is y is equal to the square root of x. And let's say I also have the graph of y equals x. So let's say y equals x looks something like this. Looks just like that. y equals x. And what I care about now is the solid I get if I were to rotate the area between these two things around the x-axis."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "So let's say y equals x looks something like this. Looks just like that. y equals x. And what I care about now is the solid I get if I were to rotate the area between these two things around the x-axis. So let's try to visualize it. So the outside is going to be kind of a truffle shape. It's going to look like a truffle shape."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "And what I care about now is the solid I get if I were to rotate the area between these two things around the x-axis. So let's try to visualize it. So the outside is going to be kind of a truffle shape. It's going to look like a truffle shape. And then we hollow out a cone inside of it. So my best attempt to draw this shape. So it's going to look something like..."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "It's going to look like a truffle shape. And then we hollow out a cone inside of it. So my best attempt to draw this shape. So it's going to look something like... So the outside is going to look something like this. It's going to look something like that. And we care about the interval."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "So it's going to look something like... So the outside is going to look something like this. It's going to look something like that. And we care about the interval. We care about the interval between the points that they intersect. So between this point and this point here. So the outside is going to look something like this."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "And we care about the interval. We care about the interval between the points that they intersect. So between this point and this point here. So the outside is going to look something like this. So this is the base of the truffle. That is the base of the truffle. It's going to have this kind of truffle shape on the outside."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "So the outside is going to look something like this. So this is the base of the truffle. That is the base of the truffle. It's going to have this kind of truffle shape on the outside. But I guess maybe we are on some type of a diet. We don't want to eat the entire truffle. So we carve out a cone on the inside."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "It's going to have this kind of truffle shape on the outside. But I guess maybe we are on some type of a diet. We don't want to eat the entire truffle. So we carve out a cone on the inside. So the inside of it is essentially hollow except for this kind of shell part. So we carve out a cone in the center. So we rotate it around the x-axis."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "So we carve out a cone on the inside. So the inside of it is essentially hollow except for this kind of shell part. So we carve out a cone in the center. So we rotate it around the x-axis. Truffle on the outside. Carved out a cone on the inside. So what's going to be the volume of that thing?"}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "So we rotate it around the x-axis. Truffle on the outside. Carved out a cone on the inside. So what's going to be the volume of that thing? So it's essentially, if we take a slice of our figure, this is going to be the wall. And we're essentially going to take the volume of this entire wall that we're rotating around the x-axis. So how do we do that?"}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "So what's going to be the volume of that thing? So it's essentially, if we take a slice of our figure, this is going to be the wall. And we're essentially going to take the volume of this entire wall that we're rotating around the x-axis. So how do we do that? Well, it might dawn on you that if we found the volume of the truffle if it was not carved out, and then subtract from that the volume of the cone, we would essentially find out the volume of the space in between the outside of the truffle and the cone part of the truffle. So how would we do that? Well, so to find the volume of the outer shape, so let me draw it over here."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "So how do we do that? Well, it might dawn on you that if we found the volume of the truffle if it was not carved out, and then subtract from that the volume of the cone, we would essentially find out the volume of the space in between the outside of the truffle and the cone part of the truffle. So how would we do that? Well, so to find the volume of the outer shape, so let me draw it over here. So actually, let me just draw it over here. If we think about the volume of the outer shape, once again, we can use the disk method. So at any given point in time, our radius for one of our disks is going to be equal to the function."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "Well, so to find the volume of the outer shape, so let me draw it over here. So actually, let me just draw it over here. If we think about the volume of the outer shape, once again, we can use the disk method. So at any given point in time, our radius for one of our disks is going to be equal to the function. Let's rotate that disk around. Actually, let me do it in a different color. It's hard to see that disk, since it's in the same magenta."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "So at any given point in time, our radius for one of our disks is going to be equal to the function. Let's rotate that disk around. Actually, let me do it in a different color. It's hard to see that disk, since it's in the same magenta. So this is our radius, and let's rotate it around. So I'm rotating the disk around. This is our face of the disk."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "It's hard to see that disk, since it's in the same magenta. So this is our radius, and let's rotate it around. So I'm rotating the disk around. This is our face of the disk. That's our face of the disk. It's going to have a depth of dx. We've seen this multiple times."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "This is our face of the disk. That's our face of the disk. It's going to have a depth of dx. We've seen this multiple times. It's going to have a depth of dx. So the volume of this disk is going to be our depth, dx, times the area of the face. The area of the face is going to be pi times the radius squared."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "We've seen this multiple times. It's going to have a depth of dx. So the volume of this disk is going to be our depth, dx, times the area of the face. The area of the face is going to be pi times the radius squared. The radius is going to be equal to the value of the outer function. In this case, it's square root of x. So it's going to be pi times our radius squared, which is pi times square root of x squared."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "The area of the face is going to be pi times the radius squared. The radius is going to be equal to the value of the outer function. In this case, it's square root of x. So it's going to be pi times our radius squared, which is pi times square root of x squared. And so if we want to find the volume of the entire outer thimble or truffle or whatever we want to call it, before we even carve out the center, we just take a sum of a bunch of these disks that we've created. So that's one disk. We would have another disk over here, another disk over here."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "So it's going to be pi times our radius squared, which is pi times square root of x squared. And so if we want to find the volume of the entire outer thimble or truffle or whatever we want to call it, before we even carve out the center, we just take a sum of a bunch of these disks that we've created. So that's one disk. We would have another disk over here, another disk over here. For each x, we have another disk. And as we go, as x's get larger and larger, the disks have a larger and larger radius. So we're going to sum up all of those disks, and we're going to take the limit as each of those disks get infinitely thin, and we have an infinite number of them."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "We would have another disk over here, another disk over here. For each x, we have another disk. And as we go, as x's get larger and larger, the disks have a larger and larger radius. So we're going to sum up all of those disks, and we're going to take the limit as each of those disks get infinitely thin, and we have an infinite number of them. But we have to figure out our boundaries of integration. So what are our boundaries of integration? What are the two points right over here where they intersect?"}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "So we're going to sum up all of those disks, and we're going to take the limit as each of those disks get infinitely thin, and we have an infinite number of them. But we have to figure out our boundaries of integration. So what are our boundaries of integration? What are the two points right over here where they intersect? Well, we could just set these two things to be equal to each other. If you just said x is equal to square root of x, when does x equal square root of x? I mean, you could square both sides of this."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "What are the two points right over here where they intersect? Well, we could just set these two things to be equal to each other. If you just said x is equal to square root of x, when does x equal square root of x? I mean, you could square both sides of this. You could say, when does x squared equal x? You could, well, we could keep it there. You could kind of solve this."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "I mean, you could square both sides of this. You could say, when does x squared equal x? You could, well, we could keep it there. You could kind of solve this. There's multiple ways you could do it, but you could solve this kind of just thinking about it. If x is equal to 0, x squared is equal to x, and you see that on the graph right over here. x is equal to 0."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "You could kind of solve this. There's multiple ways you could do it, but you could solve this kind of just thinking about it. If x is equal to 0, x squared is equal to x, and you see that on the graph right over here. x is equal to 0. And also, 1 squared is equal to 1. 1 is equal to the square root of 1. You could have done other things."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "x is equal to 0. And also, 1 squared is equal to 1. 1 is equal to the square root of 1. You could have done other things. You could say, okay, x squared minus x is equal to 0. You could factor out an x. You get x times x minus 1 is equal to 0."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "You could have done other things. You could say, okay, x squared minus x is equal to 0. You could factor out an x. You get x times x minus 1 is equal to 0. And so either one of these could be equal to 0. So x is equal to 0, or x minus 1 is equal to 0. And then you get x equals 0, or x is equal to 1. x is equal to 0, or x equals 1, which gives us our boundaries of integration."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "You get x times x minus 1 is equal to 0. And so either one of these could be equal to 0. So x is equal to 0, or x minus 1 is equal to 0. And then you get x equals 0, or x is equal to 1. x is equal to 0, or x equals 1, which gives us our boundaries of integration. So this goes from x equals 0 to x equals 1. And so for the outside of our shape, we can now figure out the volume. But we're not done."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "And then you get x equals 0, or x is equal to 1. x is equal to 0, or x equals 1, which gives us our boundaries of integration. So this goes from x equals 0 to x equals 1. And so for the outside of our shape, we can now figure out the volume. But we're not done. We also need to figure out the volume of the inside of our shape that we're going to take out. So we're going to subtract out that volume. So we're going to subtract out a volume."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "But we're not done. We also need to figure out the volume of the inside of our shape that we're going to take out. So we're going to subtract out that volume. So we're going to subtract out a volume. Our x values, once again, are going between 0 and 1. And so let's think about those disks. So let's construct a disk on the inside right over here."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "So we're going to subtract out a volume. Our x values, once again, are going between 0 and 1. And so let's think about those disks. So let's construct a disk on the inside right over here. So if I construct a disk on the inside, so now I'm carving out the cone part of it. What is the area of the face of one of those disks? Well, it's going to be pi times the radius squared."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "So let's construct a disk on the inside right over here. So if I construct a disk on the inside, so now I'm carving out the cone part of it. What is the area of the face of one of those disks? Well, it's going to be pi times the radius squared. In this case, the radius is going to be equal to the value of this inner function, which is just x. So pi times, and so this is just y is equal to x. And then we're going to multiply it times the depth, times the depth of each of these disks."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "Well, it's going to be pi times the radius squared. In this case, the radius is going to be equal to the value of this inner function, which is just x. So pi times, and so this is just y is equal to x. And then we're going to multiply it times the depth, times the depth of each of these disks. And each of these disks are going to have a depth of dx. If you imagine a quarter that has an infinitely thin depth right over here, so it's going to be dx. And so the volume of our truffle with the cone carved out is going to be this integral minus this integral right over here."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "And then we're going to multiply it times the depth, times the depth of each of these disks. And each of these disks are going to have a depth of dx. If you imagine a quarter that has an infinitely thin depth right over here, so it's going to be dx. And so the volume of our truffle with the cone carved out is going to be this integral minus this integral right over here. And we could evaluate it just like that, or we could even say, OK, we could factor out a pi out of both of them. Actually, there's multiple ways that we could write it, but let's just evaluate it like this, and then I'll generalize it in the next video. So this is going to be equal to the definite integral from 0 to 1."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "And so the volume of our truffle with the cone carved out is going to be this integral minus this integral right over here. And we could evaluate it just like that, or we could even say, OK, we could factor out a pi out of both of them. Actually, there's multiple ways that we could write it, but let's just evaluate it like this, and then I'll generalize it in the next video. So this is going to be equal to the definite integral from 0 to 1. We take the pi outside. The square root of x squared is going to be x dx minus the integral. We can factor the pi out from 0 to 1 of x squared dx."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "So this is going to be equal to the definite integral from 0 to 1. We take the pi outside. The square root of x squared is going to be x dx minus the integral. We can factor the pi out from 0 to 1 of x squared dx. And we could say this is going to be equal to pi times the antiderivative of x, which is just x squared over 2 evaluated from 0 to 1, minus pi times the antiderivative of x squared, which is x to the third over 3 evaluated from 0 to 1. This expression is equal to, and I'm going to arbitrarily switch colors just because the green is getting monotonous, pi times 1 squared over 2 minus 0 squared over 2. I could write it squared."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "We can factor the pi out from 0 to 1 of x squared dx. And we could say this is going to be equal to pi times the antiderivative of x, which is just x squared over 2 evaluated from 0 to 1, minus pi times the antiderivative of x squared, which is x to the third over 3 evaluated from 0 to 1. This expression is equal to, and I'm going to arbitrarily switch colors just because the green is getting monotonous, pi times 1 squared over 2 minus 0 squared over 2. I could write it squared. 1 squared over 2 minus 0 squared over 2 minus pi times 1 to the third over 3 minus 0 to the third over 3. And so we get this is equal to, let me do it in that same blue color, it's equal to, so this is just simplified, this is just 0 right over here. This is 1 squared over 2, which is just 1 half."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "I could write it squared. 1 squared over 2 minus 0 squared over 2 minus pi times 1 to the third over 3 minus 0 to the third over 3. And so we get this is equal to, let me do it in that same blue color, it's equal to, so this is just simplified, this is just 0 right over here. This is 1 squared over 2, which is just 1 half. So it's just pi over 2, 1 half times pi minus, well this is just 0, this is 1 third minus pi over 3. Minus pi over 3 and then to simplify this it's just really subtracting fractions. So we can find a common denominator."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "This is 1 squared over 2, which is just 1 half. So it's just pi over 2, 1 half times pi minus, well this is just 0, this is 1 third minus pi over 3. Minus pi over 3 and then to simplify this it's just really subtracting fractions. So we can find a common denominator. Common denominator is 6. This is going to be 3 pi over 6. This is 3 pi over 6 minus 2 pi over 6."}, {"video_title": "Solid of revolution between two functions (leading up to the washer method) Khan Academy.mp3", "Sentence": "So we can find a common denominator. Common denominator is 6. This is going to be 3 pi over 6. This is 3 pi over 6 minus 2 pi over 6. Pi over 3 is 2 pi over 6, pi over 2 is 3 pi over 6. And we end up with 3 of something minus 2 of something, you end up with 1 of something. We end up with 1 pi over 6."}, {"video_title": "Interpreting direction of motion from velocity-time graph AP Calculus AB Khan Academy.mp3", "Sentence": "An object is moving along a line. The following graph gives the object's velocity over time. For each point on the graph, is the object moving forward, backward, or neither? So pause this video and see if you can figure that out. All right, now let's do this together. And so we can see these different points on this velocity versus time graph. And the important thing to realize is, is if the velocity is positive, we're moving forward."}, {"video_title": "Interpreting direction of motion from velocity-time graph AP Calculus AB Khan Academy.mp3", "Sentence": "So pause this video and see if you can figure that out. All right, now let's do this together. And so we can see these different points on this velocity versus time graph. And the important thing to realize is, is if the velocity is positive, we're moving forward. If the velocity is negative, we're moving backward. And if the velocity is zero, we're not moving either forward nor backwards, or neither forward nor backwards. So right over here, we see that our velocity is positive."}, {"video_title": "Interpreting direction of motion from velocity-time graph AP Calculus AB Khan Academy.mp3", "Sentence": "And the important thing to realize is, is if the velocity is positive, we're moving forward. If the velocity is negative, we're moving backward. And if the velocity is zero, we're not moving either forward nor backwards, or neither forward nor backwards. So right over here, we see that our velocity is positive. It's a positive two meters per second. So that means that we are moving forward. Now over here, our velocity is zero meters per second."}, {"video_title": "Interpreting direction of motion from velocity-time graph AP Calculus AB Khan Academy.mp3", "Sentence": "So right over here, we see that our velocity is positive. It's a positive two meters per second. So that means that we are moving forward. Now over here, our velocity is zero meters per second. So this is neither. Now over here, our velocity is negative four meters per second. So one way to think about it is we're moving four meters per second backward."}, {"video_title": "Interpreting direction of motion from velocity-time graph AP Calculus AB Khan Academy.mp3", "Sentence": "Now over here, our velocity is zero meters per second. So this is neither. Now over here, our velocity is negative four meters per second. So one way to think about it is we're moving four meters per second backward. So I'll write backward. Now this is interesting, this last point. Because you might be tempted to say, all right, I'm oscillating, I'm going up, then I'm going down, then I'm going back up."}, {"video_title": "Interpreting direction of motion from velocity-time graph AP Calculus AB Khan Academy.mp3", "Sentence": "So one way to think about it is we're moving four meters per second backward. So I'll write backward. Now this is interesting, this last point. Because you might be tempted to say, all right, I'm oscillating, I'm going up, then I'm going down, then I'm going back up. Maybe I'm moving forward here. But remember, what we're thinking about here, this isn't position versus time, this is velocity versus time. So if our velocity is negative, we're moving backwards."}, {"video_title": "Interpreting direction of motion from velocity-time graph AP Calculus AB Khan Academy.mp3", "Sentence": "Because you might be tempted to say, all right, I'm oscillating, I'm going up, then I'm going down, then I'm going back up. Maybe I'm moving forward here. But remember, what we're thinking about here, this isn't position versus time, this is velocity versus time. So if our velocity is negative, we're moving backwards. And here, our velocity is still negative. It's becoming less negative, but it's still negative. So we are still moving, we are still moving backward."}, {"video_title": "Alternating series test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And I'll explain the alternating series test, and I'll apply it to an actual series while I do it to make the explanation of the alternating series test a little bit more concrete. So let's say that I have some series, some infinite series, let's say it goes from n equals k to infinity of a sub n. And let's say I can write it as, or I can rewrite a sub n. So let's say a sub n, I can write, so a sub n is equal to negative one to the n times b sub n, or a sub n is equal to negative one to the n plus one times b sub n, where b sub n is greater than or equal to zero for all the n's we care about, so for all of these integer n's greater than or equal to k. So if all of these things, if all of these things are true, and we know two more things, and we know, number one, the limit as n approaches infinity of b sub n is equal to zero, and number two, b sub n is a decreasing sequence, decreasing sequence, then that lets us know that the original infinite series, the original infinite series is going to converge. So this might seem a little bit abstract right now. Let's actually show, let's use this with an actual series to make it a little bit more concrete. So let's say that I had the series, let's say I had the series from n equals one to infinity of negative one to the n over n, and we could write it out just to make this series a little bit more concrete. When n is equal to one, this is gonna be negative one to the one power, actually let's just make this a little bit more interesting. Let's make this negative one to the n plus one."}, {"video_title": "Alternating series test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let's actually show, let's use this with an actual series to make it a little bit more concrete. So let's say that I had the series, let's say I had the series from n equals one to infinity of negative one to the n over n, and we could write it out just to make this series a little bit more concrete. When n is equal to one, this is gonna be negative one to the one power, actually let's just make this a little bit more interesting. Let's make this negative one to the n plus one. So when n is equal to one, this is gonna be negative one squared over one, which is going to be one, and then when n is two, it's gonna be negative one to the third power, which is gonna be negative 1 1 2, so it's minus 1 1 2 plus 1 3rd minus 1 4th plus minus, and it just keeps going on and on and on forever. Now, can we rewrite this a sub n like this? Well sure, the negative one to the n plus one is actually explicitly called out."}, {"video_title": "Alternating series test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let's make this negative one to the n plus one. So when n is equal to one, this is gonna be negative one squared over one, which is going to be one, and then when n is two, it's gonna be negative one to the third power, which is gonna be negative 1 1 2, so it's minus 1 1 2 plus 1 3rd minus 1 4th plus minus, and it just keeps going on and on and on forever. Now, can we rewrite this a sub n like this? Well sure, the negative one to the n plus one is actually explicitly called out. We can rewrite, we can rewrite our a sub n, so let me do that. So negative, so a sub n, which is equal to negative one to the n plus one over n. This is clearly the same thing as negative one to the n plus one times one over n, which is, which we can then say, this thing right over here could be our b sub n. So this right over here is our b sub n. And we can verify that our b sub n is going to be greater than or equal to zero for all the n's we care about. So our b sub n is equal to one over n. Now clearly this is going to be greater than or equal to zero for any positive n. Now what's the limit as b sub n, what's the limit of b sub n as n approaches infinity?"}, {"video_title": "Alternating series test Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well sure, the negative one to the n plus one is actually explicitly called out. We can rewrite, we can rewrite our a sub n, so let me do that. So negative, so a sub n, which is equal to negative one to the n plus one over n. This is clearly the same thing as negative one to the n plus one times one over n, which is, which we can then say, this thing right over here could be our b sub n. So this right over here is our b sub n. And we can verify that our b sub n is going to be greater than or equal to zero for all the n's we care about. So our b sub n is equal to one over n. Now clearly this is going to be greater than or equal to zero for any positive n. Now what's the limit as b sub n, what's the limit of b sub n as n approaches infinity? The limit of, let me just write one over n, one over n as n approaches infinity is going to be equal to zero, so we satisfy the first constraint. And then this is clearly a decreasing sequence. As n increases, the denominators are going to increase, and with a larger denominator, you're going to have a lower value."}, {"video_title": "Alternating series test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So our b sub n is equal to one over n. Now clearly this is going to be greater than or equal to zero for any positive n. Now what's the limit as b sub n, what's the limit of b sub n as n approaches infinity? The limit of, let me just write one over n, one over n as n approaches infinity is going to be equal to zero, so we satisfy the first constraint. And then this is clearly a decreasing sequence. As n increases, the denominators are going to increase, and with a larger denominator, you're going to have a lower value. So we can also say one over n is a decreasing, decreasing sequence for the n's that we care about. So this satisfies, this is satisfied as well. And so based on that, this thing, this thing is always, this thing right over here is always greater than or equal to zero."}, {"video_title": "Alternating series test Series AP Calculus BC Khan Academy.mp3", "Sentence": "As n increases, the denominators are going to increase, and with a larger denominator, you're going to have a lower value. So we can also say one over n is a decreasing, decreasing sequence for the n's that we care about. So this satisfies, this is satisfied as well. And so based on that, this thing, this thing is always, this thing right over here is always greater than or equal to zero. The limit as one over n or as our b sub n, as n approaches infinity is going to be zero. It's a decreasing sequence, therefore we can say that our original series actually converges. So n equals one to infinity of negative one to the n plus one over n. And that's kind of interesting, because we've already seen that if all of these were positive, if all of these terms were positive, we just have the harmonic series, and that one didn't converge."}, {"video_title": "Alternating series test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so based on that, this thing, this thing is always, this thing right over here is always greater than or equal to zero. The limit as one over n or as our b sub n, as n approaches infinity is going to be zero. It's a decreasing sequence, therefore we can say that our original series actually converges. So n equals one to infinity of negative one to the n plus one over n. And that's kind of interesting, because we've already seen that if all of these were positive, if all of these terms were positive, we just have the harmonic series, and that one didn't converge. But this one did, putting these negatives here do the trick. And actually we can prove this one over here converges using other techniques, and maybe if we have time, actually in particular the limit comparison test. I'll just throw that out there in case you are curious."}, {"video_title": "Alternating series test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So n equals one to infinity of negative one to the n plus one over n. And that's kind of interesting, because we've already seen that if all of these were positive, if all of these terms were positive, we just have the harmonic series, and that one didn't converge. But this one did, putting these negatives here do the trick. And actually we can prove this one over here converges using other techniques, and maybe if we have time, actually in particular the limit comparison test. I'll just throw that out there in case you are curious. So this is a pretty powerful tool. It looks a little bit like the divergence test, but remember the divergence test is really, is only useful if you wanna show something diverges. If the limit of, if the limit of your terms do not approach zero, then you say, okay, that thing is going to diverge."}, {"video_title": "Alternating series test Series AP Calculus BC Khan Academy.mp3", "Sentence": "I'll just throw that out there in case you are curious. So this is a pretty powerful tool. It looks a little bit like the divergence test, but remember the divergence test is really, is only useful if you wanna show something diverges. If the limit of, if the limit of your terms do not approach zero, then you say, okay, that thing is going to diverge. This thing is useful because you can actually prove convergence. Now once again, if something does not pass the alternating series test, that does not necessarily mean that it diverges. It just means that you couldn't use the alternating series test to prove that it converges."}, {"video_title": "Finding critical points Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "So assuming you've given a go at it, so let's just remind ourselves what a critical number is. So we would say c is a critical number of f if and only if, I'll write if with two f's, short for if and only if, f prime of c is equal to 0 or f prime of c is undefined. So if we look for the critical numbers for f, we want to figure out all the places where the derivative of this with respect to x is either equal to 0 or it is undefined. So let's think about how we can find the derivative of this. So let's see, f prime of x is going to be, well let's see, we're going to have to apply some combination of the product rule and the chain rule. So it's going to be the derivative with respect to x of x of x, so it's going to be that, times e to the negative 2x squared plus the derivative with respect to x of e to the negative 2x squared of e to the negative 2x squared times x. So this is just the product rule right over here, derivative of this, of the x, times e to the negative 2x squared plus the derivative of e to the negative 2x squared times x right over here."}, {"video_title": "Finding critical points Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "So let's think about how we can find the derivative of this. So let's see, f prime of x is going to be, well let's see, we're going to have to apply some combination of the product rule and the chain rule. So it's going to be the derivative with respect to x of x of x, so it's going to be that, times e to the negative 2x squared plus the derivative with respect to x of e to the negative 2x squared of e to the negative 2x squared times x. So this is just the product rule right over here, derivative of this, of the x, times e to the negative 2x squared plus the derivative of e to the negative 2x squared times x right over here. So what is this going to be? Well, all of this stuff in magenta, the derivative of x with respect to x, that's just going to be equal to 1. So this first part is going to be equal to e to the negative 2x squared."}, {"video_title": "Finding critical points Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "So this is just the product rule right over here, derivative of this, of the x, times e to the negative 2x squared plus the derivative of e to the negative 2x squared times x right over here. So what is this going to be? Well, all of this stuff in magenta, the derivative of x with respect to x, that's just going to be equal to 1. So this first part is going to be equal to e to the negative 2x squared. And now the derivative of e to the negative 2x squared over here, I'll do this in this pink color, so this part right over here, that is going to be equal to, or just apply the chain rule, derivative of e to the negative 2x squared with respect to negative 2x squared, well that's just going to be e to the negative 2x squared, and we're going to multiply that times the derivative of negative 2x squared with respect to x. And so that's going to be what? Negative 4x."}, {"video_title": "Finding critical points Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "So this first part is going to be equal to e to the negative 2x squared. And now the derivative of e to the negative 2x squared over here, I'll do this in this pink color, so this part right over here, that is going to be equal to, or just apply the chain rule, derivative of e to the negative 2x squared with respect to negative 2x squared, well that's just going to be e to the negative 2x squared, and we're going to multiply that times the derivative of negative 2x squared with respect to x. And so that's going to be what? Negative 4x. So times negative 4x, and of course we have this x over here. We have that x over there, and let's see, can we simplify it? Can we simplify it at all?"}, {"video_title": "Finding critical points Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "Negative 4x. So times negative 4x, and of course we have this x over here. We have that x over there, and let's see, can we simplify it? Can we simplify it at all? Well obviously both of these terms have an e to the negative 2x squared. So I'm going to try to figure out where this is either undefined or where this is equal to 0. So let's think about this a little bit."}, {"video_title": "Finding critical points Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "Can we simplify it at all? Well obviously both of these terms have an e to the negative 2x squared. So I'm going to try to figure out where this is either undefined or where this is equal to 0. So let's think about this a little bit. So let's see, if we factor in e to the negative 2x squared, I'll do that in green, we're going to have, this is equal to e to the negative 2x squared times, we have here 1 minus 4x squared. So this is the derivative of f. Now, where would this be undefined or equal to 0? Well, let's see, e to the negative 2x squared, this is going to be defined for any value of x."}, {"video_title": "Finding critical points Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "So let's think about this a little bit. So let's see, if we factor in e to the negative 2x squared, I'll do that in green, we're going to have, this is equal to e to the negative 2x squared times, we have here 1 minus 4x squared. So this is the derivative of f. Now, where would this be undefined or equal to 0? Well, let's see, e to the negative 2x squared, this is going to be defined for any value of x. This part is going to be defined, and this part is also going to be defined for any value of x. So there's no point where this is undefined, but let's think about when this is going to be equal to 0. So you have this product of these two expressions equaling 0. e to the negative 2x squared, that'll never be equal to 0."}, {"video_title": "Finding critical points Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "Well, let's see, e to the negative 2x squared, this is going to be defined for any value of x. This part is going to be defined, and this part is also going to be defined for any value of x. So there's no point where this is undefined, but let's think about when this is going to be equal to 0. So you have this product of these two expressions equaling 0. e to the negative 2x squared, that'll never be equal to 0. If you get this exponent to be a really, I guess you could say very negative number, you will approach 0, but you'll never get it to be, you'll never get it to be 0. So this part here can't be 0, but if the product of two things are 0, at least one of them has to be 0. So the only way that we can get f prime of x to be equal to 0 is when 1 minus 4x squared is equal to 0."}, {"video_title": "Finding critical points Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "So you have this product of these two expressions equaling 0. e to the negative 2x squared, that'll never be equal to 0. If you get this exponent to be a really, I guess you could say very negative number, you will approach 0, but you'll never get it to be, you'll never get it to be 0. So this part here can't be 0, but if the product of two things are 0, at least one of them has to be 0. So the only way that we can get f prime of x to be equal to 0 is when 1 minus 4x squared is equal to 0. So 1 minus 4x squared is equal to 0. Let me rewrite that. 1 minus 4x squared is equal to 0."}, {"video_title": "Finding critical points Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "So the only way that we can get f prime of x to be equal to 0 is when 1 minus 4x squared is equal to 0. So 1 minus 4x squared is equal to 0. Let me rewrite that. 1 minus 4x squared is equal to 0. When does that happen? And this one we can just solve. Add 4x squared to both sides, you get 1 is equal to 4x squared."}, {"video_title": "Finding critical points Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "1 minus 4x squared is equal to 0. When does that happen? And this one we can just solve. Add 4x squared to both sides, you get 1 is equal to 4x squared. Divide both sides by 4, you get 1 fourth is equal to x squared. And then what x values is this true at? Well, we just take the plus or minus square root of both sides and you get x is equal to plus or minus 1 half."}, {"video_title": "Finding critical points Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "Add 4x squared to both sides, you get 1 is equal to 4x squared. Divide both sides by 4, you get 1 fourth is equal to x squared. And then what x values is this true at? Well, we just take the plus or minus square root of both sides and you get x is equal to plus or minus 1 half. Negative 1 half squared is 1 fourth, positive 1 half squared is 1 fourth. So at x equals plus or minus 1 half, f prime or the derivative is equal to 0. So let me write it this way."}, {"video_title": "Finding critical points Using derivatives to analyze functions AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we just take the plus or minus square root of both sides and you get x is equal to plus or minus 1 half. Negative 1 half squared is 1 fourth, positive 1 half squared is 1 fourth. So at x equals plus or minus 1 half, f prime or the derivative is equal to 0. So let me write it this way. f prime of 1 half is equal to 0, and you can verify that right over here. And f prime of negative 1 half is equal to 0. So if someone asks what are the critical numbers here, critical numbers, they are 1 half and negative 1 half."}, {"video_title": "Horizontal tangent to implicit curve AP Calculus AB Khan Academy.mp3", "Sentence": "We've done that in other videos. Write the equation of the horizontal line that is tangent to the curve and is above the x-axis. Pause this video and see if you can have a go at it. So let's just make sure we're visualizing this right. Let me just draw a quick and dirty diagram. If that's my y-axis, this is my x-axis. I don't know exactly what that curve looks like, but imagine you have some type of a curve that looks something like this."}, {"video_title": "Horizontal tangent to implicit curve AP Calculus AB Khan Academy.mp3", "Sentence": "So let's just make sure we're visualizing this right. Let me just draw a quick and dirty diagram. If that's my y-axis, this is my x-axis. I don't know exactly what that curve looks like, but imagine you have some type of a curve that looks something like this. Well, there would be two tangent lines that are horizontal based on how I've drawn it. One might be right over there, so it might be like there, and then another one might be maybe right over here. And they want the equation of the horizontal line that is tangent to the curve and is above the x-axis."}, {"video_title": "Horizontal tangent to implicit curve AP Calculus AB Khan Academy.mp3", "Sentence": "I don't know exactly what that curve looks like, but imagine you have some type of a curve that looks something like this. Well, there would be two tangent lines that are horizontal based on how I've drawn it. One might be right over there, so it might be like there, and then another one might be maybe right over here. And they want the equation of the horizontal line that is tangent to the curve and is above the x-axis. So what do we know, what is true if this tangent line is horizontal? Well, that tells us that at this point, dy dx is equal to zero. In fact, that would be true at both of these points."}, {"video_title": "Horizontal tangent to implicit curve AP Calculus AB Khan Academy.mp3", "Sentence": "And they want the equation of the horizontal line that is tangent to the curve and is above the x-axis. So what do we know, what is true if this tangent line is horizontal? Well, that tells us that at this point, dy dx is equal to zero. In fact, that would be true at both of these points. And we know what dy dx is. We know that the derivative of y with respect to x is equal to negative two times x plus three over four y to the third power for any x and y. And so when will this equal zero?"}, {"video_title": "Horizontal tangent to implicit curve AP Calculus AB Khan Academy.mp3", "Sentence": "In fact, that would be true at both of these points. And we know what dy dx is. We know that the derivative of y with respect to x is equal to negative two times x plus three over four y to the third power for any x and y. And so when will this equal zero? Well, it's going to equal zero when our numerator is equal to zero and our denominator isn't. So when is our numerator going to be zero? When x is equal to negative three."}, {"video_title": "Horizontal tangent to implicit curve AP Calculus AB Khan Academy.mp3", "Sentence": "And so when will this equal zero? Well, it's going to equal zero when our numerator is equal to zero and our denominator isn't. So when is our numerator going to be zero? When x is equal to negative three. So when x is equal to negative three, the derivative is equal to zero. So what is going to be the corresponding y value when x is equal to negative three? And if we know that, well, this equation is just going to be y is equal to something."}, {"video_title": "Horizontal tangent to implicit curve AP Calculus AB Khan Academy.mp3", "Sentence": "When x is equal to negative three. So when x is equal to negative three, the derivative is equal to zero. So what is going to be the corresponding y value when x is equal to negative three? And if we know that, well, this equation is just going to be y is equal to something. It's going to be that y value. Well, to figure that out, we just take this x equals negative three, substitute it back into our original equation, and then solve for y. So let's do that."}, {"video_title": "Horizontal tangent to implicit curve AP Calculus AB Khan Academy.mp3", "Sentence": "And if we know that, well, this equation is just going to be y is equal to something. It's going to be that y value. Well, to figure that out, we just take this x equals negative three, substitute it back into our original equation, and then solve for y. So let's do that. So it's going to be negative three squared plus y to the fourth plus six times negative three is equal to seven. This is nine, this is negative 18. And so we're gonna get y to the fourth minus nine is equal to seven."}, {"video_title": "Horizontal tangent to implicit curve AP Calculus AB Khan Academy.mp3", "Sentence": "So let's do that. So it's going to be negative three squared plus y to the fourth plus six times negative three is equal to seven. This is nine, this is negative 18. And so we're gonna get y to the fourth minus nine is equal to seven. Or adding nine to both sides, we get y to the fourth power is equal to 16. And this would tell us that y is going to be equal to plus or minus two. Well, there would be then two horizontal lines."}, {"video_title": "Horizontal tangent to implicit curve AP Calculus AB Khan Academy.mp3", "Sentence": "And so we're gonna get y to the fourth minus nine is equal to seven. Or adding nine to both sides, we get y to the fourth power is equal to 16. And this would tell us that y is going to be equal to plus or minus two. Well, there would be then two horizontal lines. One would be y is equal to two. The other is y is equal to negative two. But they want us the equation of the horizontal line that is tangent to the curve and is above the x-axis."}, {"video_title": "Horizontal tangent to implicit curve AP Calculus AB Khan Academy.mp3", "Sentence": "Well, there would be then two horizontal lines. One would be y is equal to two. The other is y is equal to negative two. But they want us the equation of the horizontal line that is tangent to the curve and is above the x-axis. So only this one is going to be above the x-axis. And we're done. It's going to be y is equal to two."}, {"video_title": "_-substitution logarithmic function AP Calculus AB Khan Academy.mp3", "Sentence": "Is u substitution a possibility here? Well for u substitution we want to look for an expression and its derivative. Well what happens if we set u equal to the natural log of x? Now what would du be equal to in that scenario? Well du is going to be the derivative of the natural log of x with respect to x, which is just one over x dx. This is an equivalent statement to saying that du dx is equal to one over x. So do we see a one over x dx anywhere in this original expression?"}, {"video_title": "_-substitution logarithmic function AP Calculus AB Khan Academy.mp3", "Sentence": "Now what would du be equal to in that scenario? Well du is going to be the derivative of the natural log of x with respect to x, which is just one over x dx. This is an equivalent statement to saying that du dx is equal to one over x. So do we see a one over x dx anywhere in this original expression? Well it's kind of hiding, it's not so obvious, but this x in the denominator is essentially a one over x and then that's being multiplied by a dx. Let me rewrite this original expression to make a little bit more sense. So the first thing I'm going to do is I'm going to take the pi, I should do that in a different color since I've already used, let me take the pi and just stick it out front."}, {"video_title": "_-substitution logarithmic function AP Calculus AB Khan Academy.mp3", "Sentence": "So do we see a one over x dx anywhere in this original expression? Well it's kind of hiding, it's not so obvious, but this x in the denominator is essentially a one over x and then that's being multiplied by a dx. Let me rewrite this original expression to make a little bit more sense. So the first thing I'm going to do is I'm going to take the pi, I should do that in a different color since I've already used, let me take the pi and just stick it out front. So I'm going to stick the pi out in front of the integral and so this becomes the integral of, and let me write the one over natural log of x first, one over the natural log of x times one over x dx. Now it becomes a little bit clearer. These are completely equivalent statements."}, {"video_title": "_-substitution logarithmic function AP Calculus AB Khan Academy.mp3", "Sentence": "So the first thing I'm going to do is I'm going to take the pi, I should do that in a different color since I've already used, let me take the pi and just stick it out front. So I'm going to stick the pi out in front of the integral and so this becomes the integral of, and let me write the one over natural log of x first, one over the natural log of x times one over x dx. Now it becomes a little bit clearer. These are completely equivalent statements. But this makes it clear that yes, u substitution will work over here. If we set our u equal to natural log of x, then our du is one over x dx, one over x dx. Our du is one over x dx."}, {"video_title": "_-substitution logarithmic function AP Calculus AB Khan Academy.mp3", "Sentence": "These are completely equivalent statements. But this makes it clear that yes, u substitution will work over here. If we set our u equal to natural log of x, then our du is one over x dx, one over x dx. Our du is one over x dx. Let's rewrite this integral. It's going to be equal to pi times the indefinite integral of one over u, natural log of x is u, we set that equal to natural log of x, times du, times du. Now this becomes pretty straightforward."}, {"video_title": "_-substitution logarithmic function AP Calculus AB Khan Academy.mp3", "Sentence": "Our du is one over x dx. Let's rewrite this integral. It's going to be equal to pi times the indefinite integral of one over u, natural log of x is u, we set that equal to natural log of x, times du, times du. Now this becomes pretty straightforward. What is the antiderivative of all of this business? And we've done very similar things like this multiple times already. This is going to be equal to pi times the natural log, the natural log of the absolute value of u, so that we can handle even negative values of u, the natural log of the absolute value of u plus c, just in case we had a constant factor out here, plus c. And we're almost done."}, {"video_title": "_-substitution logarithmic function AP Calculus AB Khan Academy.mp3", "Sentence": "Now this becomes pretty straightforward. What is the antiderivative of all of this business? And we've done very similar things like this multiple times already. This is going to be equal to pi times the natural log, the natural log of the absolute value of u, so that we can handle even negative values of u, the natural log of the absolute value of u plus c, just in case we had a constant factor out here, plus c. And we're almost done. We just have to unsubstitute for the u. U is equal to natural log of x. So we end up with this kind of neat looking expression. This entire indefinite integral we have simplified, we have evaluated it."}, {"video_title": "_-substitution logarithmic function AP Calculus AB Khan Academy.mp3", "Sentence": "This is going to be equal to pi times the natural log, the natural log of the absolute value of u, so that we can handle even negative values of u, the natural log of the absolute value of u plus c, just in case we had a constant factor out here, plus c. And we're almost done. We just have to unsubstitute for the u. U is equal to natural log of x. So we end up with this kind of neat looking expression. This entire indefinite integral we have simplified, we have evaluated it. And it is now equal to pi times the natural log, the natural log of the absolute value of u, but u is just the natural log of x. The natural log of x, and then we have this plus c right over here. And we could have assumed that from the get-go, this original expression was only defined for positive values of x, because you have to take the natural log here and it wasn't an absolute value."}, {"video_title": "_-substitution logarithmic function AP Calculus AB Khan Academy.mp3", "Sentence": "This entire indefinite integral we have simplified, we have evaluated it. And it is now equal to pi times the natural log, the natural log of the absolute value of u, but u is just the natural log of x. The natural log of x, and then we have this plus c right over here. And we could have assumed that from the get-go, this original expression was only defined for positive values of x, because you have to take the natural log here and it wasn't an absolute value. But now, so we can leave this as just a natural log of x. But this also works for the situations now, because we're taking the absolute value of that, where the natural log of x might have been a negative number. For example, if it was a natural log of.5, or who knows, whatever it might be."}, {"video_title": "Worked example Parametric arc length AP Calculus BC Khan Academy.mp3", "Sentence": "Let's say that x is a function of the parameter t and it's equal to cosine of t and y is also defined as a function of t and it's equal to sine of t and we want to find the arc length of the curve traced out. So length of curve from t is equal to zero t is equal to zero to t is equal to pi over two. So pause this video and see if you can work that out based on formulas that we have seen in other videos. All right, so first I'm gonna look at the formula and then we're going to visualize it and appreciate why what we got from the formula actually makes sense. So the formula tells us that arc length of a parametric curve, arc length, is equal to the integral from our starting point of our parameter, t equals a, to our ending point of our parameter, t equals b, of the square root of the derivative of x with respect to t squared plus the derivative of y with respect to t squared dt. Dt. This could also be rewritten as this is equal to the integral from a to b of the square root of dx dt squared plus dy dt squared dt."}, {"video_title": "Worked example Parametric arc length AP Calculus BC Khan Academy.mp3", "Sentence": "All right, so first I'm gonna look at the formula and then we're going to visualize it and appreciate why what we got from the formula actually makes sense. So the formula tells us that arc length of a parametric curve, arc length, is equal to the integral from our starting point of our parameter, t equals a, to our ending point of our parameter, t equals b, of the square root of the derivative of x with respect to t squared plus the derivative of y with respect to t squared dt. Dt. This could also be rewritten as this is equal to the integral from a to b of the square root of dx dt squared plus dy dt squared dt. But either way, we can now apply it in this context. What is dx dt? So dx dt is equal to the derivative of cosine of t is equal to negative sine of t, negative sine of t. And what is dy dt?"}, {"video_title": "Worked example Parametric arc length AP Calculus BC Khan Academy.mp3", "Sentence": "This could also be rewritten as this is equal to the integral from a to b of the square root of dx dt squared plus dy dt squared dt. But either way, we can now apply it in this context. What is dx dt? So dx dt is equal to the derivative of cosine of t is equal to negative sine of t, negative sine of t. And what is dy dt? The derivative of y with respect to t. Derivative of sine of t is cosine of t. Cosine of t. So our arc length up here is going to be equal to the integral from t is equal to zero to pi over two. That's what we care about. Our parameter's going from zero to pi over two of the square root of the derivative of x with respect to t squared."}, {"video_title": "Worked example Parametric arc length AP Calculus BC Khan Academy.mp3", "Sentence": "So dx dt is equal to the derivative of cosine of t is equal to negative sine of t, negative sine of t. And what is dy dt? The derivative of y with respect to t. Derivative of sine of t is cosine of t. Cosine of t. So our arc length up here is going to be equal to the integral from t is equal to zero to pi over two. That's what we care about. Our parameter's going from zero to pi over two of the square root of the derivative of x with respect to t squared. That's negative sine of t squared. Well, if you square it, the negative's gonna go away. You're gonna, negative sine times negative sine is positive sine squared."}, {"video_title": "Worked example Parametric arc length AP Calculus BC Khan Academy.mp3", "Sentence": "Our parameter's going from zero to pi over two of the square root of the derivative of x with respect to t squared. That's negative sine of t squared. Well, if you square it, the negative's gonna go away. You're gonna, negative sine times negative sine is positive sine squared. So I could write this as sine squared of t and then dy dt squared, that's just cosine squared t, plus cosine squared t. And then we have our dt out here. Now, lucky for us, sine squared plus cosine squared of some variable is always going to be equal to one. So that's one of our most basic trig identities, it comes straight out of the unit circle definition of sine and cosine."}, {"video_title": "Worked example Parametric arc length AP Calculus BC Khan Academy.mp3", "Sentence": "You're gonna, negative sine times negative sine is positive sine squared. So I could write this as sine squared of t and then dy dt squared, that's just cosine squared t, plus cosine squared t. And then we have our dt out here. Now, lucky for us, sine squared plus cosine squared of some variable is always going to be equal to one. So that's one of our most basic trig identities, it comes straight out of the unit circle definition of sine and cosine. And so we have the square root of one, the principal root of one, which is just going to be one. So all of this thing, everything here has just simplified to the integral from zero to pi over two dt. Well, this is going to be equal to, so you could view this as a one here, the antiderivative of one with respect to t is just going to be t. We're gonna evaluate that from pi over two."}, {"video_title": "Worked example Parametric arc length AP Calculus BC Khan Academy.mp3", "Sentence": "So that's one of our most basic trig identities, it comes straight out of the unit circle definition of sine and cosine. And so we have the square root of one, the principal root of one, which is just going to be one. So all of this thing, everything here has just simplified to the integral from zero to pi over two dt. Well, this is going to be equal to, so you could view this as a one here, the antiderivative of one with respect to t is just going to be t. We're gonna evaluate that from pi over two. We're gonna evaluate that at pi over two and then subtract it evaluated at zero. So this is going to be equal to pi over two minus zero. That's going to be equal to pi over two."}, {"video_title": "Worked example Parametric arc length AP Calculus BC Khan Academy.mp3", "Sentence": "Well, this is going to be equal to, so you could view this as a one here, the antiderivative of one with respect to t is just going to be t. We're gonna evaluate that from pi over two. We're gonna evaluate that at pi over two and then subtract it evaluated at zero. So this is going to be equal to pi over two minus zero. That's going to be equal to pi over two. Now, let's think about why that actually does make sense. Let's plot this curve. So that is my y-axis."}, {"video_title": "Worked example Parametric arc length AP Calculus BC Khan Academy.mp3", "Sentence": "That's going to be equal to pi over two. Now, let's think about why that actually does make sense. Let's plot this curve. So that is my y-axis. This is my x-axis, right over here. When t is equal to zero, you have x of zero, cosine of zero, x is equal to one, x is equal to one, and y, sine of zero, is just zero. So y is equal to zero."}, {"video_title": "Worked example Parametric arc length AP Calculus BC Khan Academy.mp3", "Sentence": "So that is my y-axis. This is my x-axis, right over here. When t is equal to zero, you have x of zero, cosine of zero, x is equal to one, x is equal to one, and y, sine of zero, is just zero. So y is equal to zero. So we're at this point right over here at t is equal to zero and then as t increases up to pi over two, we trace out the top right corner of the unit circle and we end up right over here when t is equal to zero. So that is equal to pi over two. You could view t in this case as some type of an angle in radians."}, {"video_title": "Worked example Parametric arc length AP Calculus BC Khan Academy.mp3", "Sentence": "So y is equal to zero. So we're at this point right over here at t is equal to zero and then as t increases up to pi over two, we trace out the top right corner of the unit circle and we end up right over here when t is equal to zero. So that is equal to pi over two. You could view t in this case as some type of an angle in radians. And so the arc length is really just the length of a quarter of a unit circle. Well, we know what the circumference of a circle is. It is two pi r. In the unit circle case, the radius is one."}, {"video_title": "Worked example Parametric arc length AP Calculus BC Khan Academy.mp3", "Sentence": "You could view t in this case as some type of an angle in radians. And so the arc length is really just the length of a quarter of a unit circle. Well, we know what the circumference of a circle is. It is two pi r. In the unit circle case, the radius is one. So the circumference of the entire circle is two pi. 1 1\u2074 of that is going to be pi over two. So it's nice when this fancy thing that we feel good about in calculus is consistent with what we first learned in basic geometry."}, {"video_title": "Introduction to one-dimensional motion with calculus AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say in this situation, it is going to be our time to the third power minus three times our time squared plus five, and this is going to apply for our time being non-negative because the idea of negative time, at least for now, is a bit strange. So let's think about what this right over here is describing, and to help us do that, we could set up a little bit of a table to understand that depending on what time we are, let's say that time is in seconds, what is going to be our position along our x-axis? So at time equals zero, x of zero is just going to be five. At time one, you're gonna have one minus three plus five, so that is going to be, see, one minus three is negative two plus five is going to be, we're gonna be at position three, and then at time two, we are going to be at eight minus 12 plus five, so we're going to be at position one, and then at time t equals three, it's gonna be 27 minus 27 plus five. We're gonna be back at five, and so this can at least help us understand what's going on for the first three seconds. So let me draw our positive x-axis. So, say it looks something like that, and this is x equals zero, this is our x-axis, x equals one, two, three, four, and five."}, {"video_title": "Introduction to one-dimensional motion with calculus AP Calculus AB Khan Academy.mp3", "Sentence": "At time one, you're gonna have one minus three plus five, so that is going to be, see, one minus three is negative two plus five is going to be, we're gonna be at position three, and then at time two, we are going to be at eight minus 12 plus five, so we're going to be at position one, and then at time t equals three, it's gonna be 27 minus 27 plus five. We're gonna be back at five, and so this can at least help us understand what's going on for the first three seconds. So let me draw our positive x-axis. So, say it looks something like that, and this is x equals zero, this is our x-axis, x equals one, two, three, four, and five. And now let's play out how this particle that's being described is moving along the x-axis. So we're gonna start here, and we're gonna go one, two, three. Let's do it again."}, {"video_title": "Introduction to one-dimensional motion with calculus AP Calculus AB Khan Academy.mp3", "Sentence": "So, say it looks something like that, and this is x equals zero, this is our x-axis, x equals one, two, three, four, and five. And now let's play out how this particle that's being described is moving along the x-axis. So we're gonna start here, and we're gonna go one, two, three. Let's do it again. We're going to go one, two, three. The way I just moved my mouse, if we assume that I got the time roughly right, is how that particle would move. And we can graph this as well."}, {"video_title": "Introduction to one-dimensional motion with calculus AP Calculus AB Khan Academy.mp3", "Sentence": "Let's do it again. We're going to go one, two, three. The way I just moved my mouse, if we assume that I got the time roughly right, is how that particle would move. And we can graph this as well. So, for example, it would look like this. We are starting at time t equals zero. Our position, this is our vertical axis, our y-axis, but we're just saying y is going to be equal to our position along the x-axis."}, {"video_title": "Introduction to one-dimensional motion with calculus AP Calculus AB Khan Academy.mp3", "Sentence": "And we can graph this as well. So, for example, it would look like this. We are starting at time t equals zero. Our position, this is our vertical axis, our y-axis, but we're just saying y is going to be equal to our position along the x-axis. So that's a little bit counterintuitive, because we're talking about our position in the left-right dimension, and here you're seeing it start off in the vertical dimension. But you see the same thing. At time t equals one, our position has gone down to three, then it goes down further."}, {"video_title": "Introduction to one-dimensional motion with calculus AP Calculus AB Khan Academy.mp3", "Sentence": "Our position, this is our vertical axis, our y-axis, but we're just saying y is going to be equal to our position along the x-axis. So that's a little bit counterintuitive, because we're talking about our position in the left-right dimension, and here you're seeing it start off in the vertical dimension. But you see the same thing. At time t equals one, our position has gone down to three, then it goes down further. At time equals two, our position is down to one. And then we switch direction, and then over the next, if we say that time is in seconds, over the next second, we get back to five. Now, an interesting thing to think about in the context of calculus is, well, what is our velocity at any point in time?"}, {"video_title": "Introduction to one-dimensional motion with calculus AP Calculus AB Khan Academy.mp3", "Sentence": "At time t equals one, our position has gone down to three, then it goes down further. At time equals two, our position is down to one. And then we switch direction, and then over the next, if we say that time is in seconds, over the next second, we get back to five. Now, an interesting thing to think about in the context of calculus is, well, what is our velocity at any point in time? And velocity, as you might remember, is the derivative of position. So let me write that down. So we're gonna be thinking about velocity as a function of time."}, {"video_title": "Introduction to one-dimensional motion with calculus AP Calculus AB Khan Academy.mp3", "Sentence": "Now, an interesting thing to think about in the context of calculus is, well, what is our velocity at any point in time? And velocity, as you might remember, is the derivative of position. So let me write that down. So we're gonna be thinking about velocity as a function of time. And you could view velocity as the first derivative of position with respect to time, which is just going to be equal to, we're gonna apply the power rule and some derivative properties multiple times. If this is unfamiliar to you, I encourage you to review it. But this is going to be three t squared minus six t, and then plus zero, and we're going to restrict the domain as well, for t is greater than or equal to zero."}, {"video_title": "Introduction to one-dimensional motion with calculus AP Calculus AB Khan Academy.mp3", "Sentence": "So we're gonna be thinking about velocity as a function of time. And you could view velocity as the first derivative of position with respect to time, which is just going to be equal to, we're gonna apply the power rule and some derivative properties multiple times. If this is unfamiliar to you, I encourage you to review it. But this is going to be three t squared minus six t, and then plus zero, and we're going to restrict the domain as well, for t is greater than or equal to zero. And then we can plot that. It would look like that. Now, let's see if this curve makes intuitive sense."}, {"video_title": "Introduction to one-dimensional motion with calculus AP Calculus AB Khan Academy.mp3", "Sentence": "But this is going to be three t squared minus six t, and then plus zero, and we're going to restrict the domain as well, for t is greater than or equal to zero. And then we can plot that. It would look like that. Now, let's see if this curve makes intuitive sense. We mentioned that one second, two second, three seconds. So we're starting moving to the left. And the convention is, if you're moving to the left, you have negative velocity."}, {"video_title": "Introduction to one-dimensional motion with calculus AP Calculus AB Khan Academy.mp3", "Sentence": "Now, let's see if this curve makes intuitive sense. We mentioned that one second, two second, three seconds. So we're starting moving to the left. And the convention is, if you're moving to the left, you have negative velocity. And if you're moving to the right, you have positive velocity. And you can see here, our velocity immediately gets more and more negative until we get to one second. And then it stays negative, but it's getting less and less negative until we get to two seconds."}, {"video_title": "Introduction to one-dimensional motion with calculus AP Calculus AB Khan Academy.mp3", "Sentence": "And the convention is, if you're moving to the left, you have negative velocity. And if you're moving to the right, you have positive velocity. And you can see here, our velocity immediately gets more and more negative until we get to one second. And then it stays negative, but it's getting less and less negative until we get to two seconds. And at two seconds, our velocity becomes positive. And that makes sense, because at two seconds was when our velocity switched directions to the rightward direction. So our velocity is getting more and more negative, less and less negative, and then we switch directions, and we go just like that."}, {"video_title": "Introduction to one-dimensional motion with calculus AP Calculus AB Khan Academy.mp3", "Sentence": "And then it stays negative, but it's getting less and less negative until we get to two seconds. And at two seconds, our velocity becomes positive. And that makes sense, because at two seconds was when our velocity switched directions to the rightward direction. So our velocity is getting more and more negative, less and less negative, and then we switch directions, and we go just like that. And we see it right over here. Now, one thing to keep in mind when we're thinking about velocity as a function of time is that velocity and speed are two different things. Speed, speed, let me write it over here."}, {"video_title": "Introduction to one-dimensional motion with calculus AP Calculus AB Khan Academy.mp3", "Sentence": "So our velocity is getting more and more negative, less and less negative, and then we switch directions, and we go just like that. And we see it right over here. Now, one thing to keep in mind when we're thinking about velocity as a function of time is that velocity and speed are two different things. Speed, speed, let me write it over here. Speed is equal to, if you're thinking about it in one dimension, you could think about it as the absolute value of your velocity as a function of time, or your magnitude of velocity as a function of time. So in the beginning, even though your velocity is becoming more and more negative, your speed is actually increasing. Your speed is increasing to the left, then your speed is decreasing, you slow down, and then your speed is increasing as we go to the right."}, {"video_title": "Introduction to one-dimensional motion with calculus AP Calculus AB Khan Academy.mp3", "Sentence": "Speed, speed, let me write it over here. Speed is equal to, if you're thinking about it in one dimension, you could think about it as the absolute value of your velocity as a function of time, or your magnitude of velocity as a function of time. So in the beginning, even though your velocity is becoming more and more negative, your speed is actually increasing. Your speed is increasing to the left, then your speed is decreasing, you slow down, and then your speed is increasing as we go to the right. And we'll do some worked examples that work through that a little bit more. Now, the last concept we will touch on in this video is the idea of acceleration. And acceleration, you could view as the rate of change of velocity with respect to time."}, {"video_title": "Introduction to one-dimensional motion with calculus AP Calculus AB Khan Academy.mp3", "Sentence": "Your speed is increasing to the left, then your speed is decreasing, you slow down, and then your speed is increasing as we go to the right. And we'll do some worked examples that work through that a little bit more. Now, the last concept we will touch on in this video is the idea of acceleration. And acceleration, you could view as the rate of change of velocity with respect to time. So acceleration as a function of time is just going to be the first derivative of velocity with respect to time, which is equal to the second derivative of position with respect to time. It's just going to be the derivative of this expression. So once again, using the power rule here, that's going to be six t, and then using the power rule here minus six, and once again, we will restrict the domain."}, {"video_title": "Introduction to one-dimensional motion with calculus AP Calculus AB Khan Academy.mp3", "Sentence": "And acceleration, you could view as the rate of change of velocity with respect to time. So acceleration as a function of time is just going to be the first derivative of velocity with respect to time, which is equal to the second derivative of position with respect to time. It's just going to be the derivative of this expression. So once again, using the power rule here, that's going to be six t, and then using the power rule here minus six, and once again, we will restrict the domain. And we can graph that as well. And we would see right over here, this is y is equal to acceleration as a function of time. And you can see at time equals zero, our acceleration is quite negative, it is negative six."}, {"video_title": "Introduction to one-dimensional motion with calculus AP Calculus AB Khan Academy.mp3", "Sentence": "So once again, using the power rule here, that's going to be six t, and then using the power rule here minus six, and once again, we will restrict the domain. And we can graph that as well. And we would see right over here, this is y is equal to acceleration as a function of time. And you can see at time equals zero, our acceleration is quite negative, it is negative six. And then it becomes less and less and less negative. And then our acceleration actually becomes positive at t equals one. Now, does that make sense?"}, {"video_title": "Introduction to one-dimensional motion with calculus AP Calculus AB Khan Academy.mp3", "Sentence": "And you can see at time equals zero, our acceleration is quite negative, it is negative six. And then it becomes less and less and less negative. And then our acceleration actually becomes positive at t equals one. Now, does that make sense? Well, we're going one, two, three. You might say, wait, we didn't switch directions until we get to the second second. But remember, after we get to the first second, our velocity in the negative direction becomes less negative, which means that our acceleration is positive."}, {"video_title": "Introduction to one-dimensional motion with calculus AP Calculus AB Khan Academy.mp3", "Sentence": "Now, does that make sense? Well, we're going one, two, three. You might say, wait, we didn't switch directions until we get to the second second. But remember, after we get to the first second, our velocity in the negative direction becomes less negative, which means that our acceleration is positive. If that's a little confusing, pause the video and really think through that. So our acceleration is negative, then positive, and then positive continues. And so this is just to give you an intuition."}, {"video_title": "Defining a function at a point to make it continuous Limits Differential Calculus Khan Academy.mp3", "Sentence": "And we see why that is. If x is equal to positive or negative 2, then x squared is going to be equal to positive 4. And 4 minus 4 is 0. And then we're going to have a 0 in the denominator. And that's not defined. We don't know what happens when you divide. Or we've never defined what happens when you divide by 0."}, {"video_title": "Defining a function at a point to make it continuous Limits Differential Calculus Khan Academy.mp3", "Sentence": "And then we're going to have a 0 in the denominator. And that's not defined. We don't know what happens when you divide. Or we've never defined what happens when you divide by 0. So they say what value should be assigned to f of negative 2 to make f of x continuous at that point? So to think about that, let's try to actually simplify f of x. So f of x, I'll just rewrite it, is equal to."}, {"video_title": "Defining a function at a point to make it continuous Limits Differential Calculus Khan Academy.mp3", "Sentence": "Or we've never defined what happens when you divide by 0. So they say what value should be assigned to f of negative 2 to make f of x continuous at that point? So to think about that, let's try to actually simplify f of x. So f of x, I'll just rewrite it, is equal to. Actually, let me just start simplifying right from the get-go. So in the numerator, I can factor out a 6 out of every one of those terms. So it's 6 times x squared plus 3x plus 2 over."}, {"video_title": "Defining a function at a point to make it continuous Limits Differential Calculus Khan Academy.mp3", "Sentence": "So f of x, I'll just rewrite it, is equal to. Actually, let me just start simplifying right from the get-go. So in the numerator, I can factor out a 6 out of every one of those terms. So it's 6 times x squared plus 3x plus 2 over. And in the denominator, this is a difference of squares. This is x plus 2 times x minus 2. And then we can factor this expression up here."}, {"video_title": "Defining a function at a point to make it continuous Limits Differential Calculus Khan Academy.mp3", "Sentence": "So it's 6 times x squared plus 3x plus 2 over. And in the denominator, this is a difference of squares. This is x plus 2 times x minus 2. And then we can factor this expression up here. So this is going to be equal to 6 times. Let me do it a different color. So we think of two numbers that if I take their product, I get 2."}, {"video_title": "Defining a function at a point to make it continuous Limits Differential Calculus Khan Academy.mp3", "Sentence": "And then we can factor this expression up here. So this is going to be equal to 6 times. Let me do it a different color. So we think of two numbers that if I take their product, I get 2. If I take their sum, I get 3. The most obvious one is 2 and 1. So this is 6 times x plus 2 times x plus 1."}, {"video_title": "Defining a function at a point to make it continuous Limits Differential Calculus Khan Academy.mp3", "Sentence": "So we think of two numbers that if I take their product, I get 2. If I take their sum, I get 3. The most obvious one is 2 and 1. So this is 6 times x plus 2 times x plus 1. When you take the product there, you'll get x squared plus 3x plus 2, and then all of that over x plus 2 times x minus 2. Now, if we know that x does not equal negative 2, then we can divide both the numerator and the denominator by x plus 2. The reason why I'm making that constraint is that if x were to be equal to negative 2, then x plus 2 is going to be equal to 0."}, {"video_title": "Defining a function at a point to make it continuous Limits Differential Calculus Khan Academy.mp3", "Sentence": "So this is 6 times x plus 2 times x plus 1. When you take the product there, you'll get x squared plus 3x plus 2, and then all of that over x plus 2 times x minus 2. Now, if we know that x does not equal negative 2, then we can divide both the numerator and the denominator by x plus 2. The reason why I'm making that constraint is that if x were to be equal to negative 2, then x plus 2 is going to be equal to 0. And you won't be able to do that. You can't. We don't know what it means to divide something by 0."}, {"video_title": "Defining a function at a point to make it continuous Limits Differential Calculus Khan Academy.mp3", "Sentence": "The reason why I'm making that constraint is that if x were to be equal to negative 2, then x plus 2 is going to be equal to 0. And you won't be able to do that. You can't. We don't know what it means to divide something by 0. So we could say that this is going to be equal to. So we can divide the numerator and the denominator by x plus 2. But we have to assume that x is not equal to negative 2."}, {"video_title": "Defining a function at a point to make it continuous Limits Differential Calculus Khan Academy.mp3", "Sentence": "We don't know what it means to divide something by 0. So we could say that this is going to be equal to. So we can divide the numerator and the denominator by x plus 2. But we have to assume that x is not equal to negative 2. So this is equal to 6 times. So we're going to divide by x plus 2 in the numerator, x plus 2 in the denominator. So it's going to be 6 times x plus 1 over x minus 2 over x minus 2."}, {"video_title": "Defining a function at a point to make it continuous Limits Differential Calculus Khan Academy.mp3", "Sentence": "But we have to assume that x is not equal to negative 2. So this is equal to 6 times. So we're going to divide by x plus 2 in the numerator, x plus 2 in the denominator. So it's going to be 6 times x plus 1 over x minus 2 over x minus 2. And we have to put the constraint here because now we've changed it. Now, this expression over here is actually defined at x equals negative 2. But in order to be equivalent to the original function, we have to constrain it."}, {"video_title": "Defining a function at a point to make it continuous Limits Differential Calculus Khan Academy.mp3", "Sentence": "So it's going to be 6 times x plus 1 over x minus 2 over x minus 2. And we have to put the constraint here because now we've changed it. Now, this expression over here is actually defined at x equals negative 2. But in order to be equivalent to the original function, we have to constrain it. So we will say for x not equal to negative 2. And it's also obvious that x can't be equal to 2 here. This one also isn't defined at positive 2 because you're dividing by 0."}, {"video_title": "Defining a function at a point to make it continuous Limits Differential Calculus Khan Academy.mp3", "Sentence": "But in order to be equivalent to the original function, we have to constrain it. So we will say for x not equal to negative 2. And it's also obvious that x can't be equal to 2 here. This one also isn't defined at positive 2 because you're dividing by 0. So you could say for x does not equal to positive or negative 2 if you want to make it very explicit. But they ask us, what can we assign f of negative 2 to make the function continuous at the point? Well, the function is completely equivalent to this expression except that the function is not defined at x equals negative 2."}, {"video_title": "Defining a function at a point to make it continuous Limits Differential Calculus Khan Academy.mp3", "Sentence": "This one also isn't defined at positive 2 because you're dividing by 0. So you could say for x does not equal to positive or negative 2 if you want to make it very explicit. But they ask us, what can we assign f of negative 2 to make the function continuous at the point? Well, the function is completely equivalent to this expression except that the function is not defined at x equals negative 2. So that's why we had to put that constraint here if we wanted this to be the same thing as our original function. But if we wanted to re-engineer the function so it is continuous at that point, well, then we just have to set f of x equal to whatever this expression would have been when x is equal to negative 2. So let's think about that."}, {"video_title": "Defining a function at a point to make it continuous Limits Differential Calculus Khan Academy.mp3", "Sentence": "Well, the function is completely equivalent to this expression except that the function is not defined at x equals negative 2. So that's why we had to put that constraint here if we wanted this to be the same thing as our original function. But if we wanted to re-engineer the function so it is continuous at that point, well, then we just have to set f of x equal to whatever this expression would have been when x is equal to negative 2. So let's think about that. So 6 times negative 2 plus 1 over negative 2 minus 2 is equal to, this is 6 times negative 1, so it's negative 6 over negative 4, which is equal to 3 halves. So if we redefine f of x, if we say f of x is equal to 6x squared plus 18x plus 12 over x squared minus 4 for x not equal positive or negative 2, and it's equal to 3 halves for x equals negative 2, now this function is going to be the exact same thing as this right over here. This f of x, this new one, this new definition, this extended definition of our original one, is now equivalent to this expression."}, {"video_title": "Defining a function at a point to make it continuous Limits Differential Calculus Khan Academy.mp3", "Sentence": "So let's think about that. So 6 times negative 2 plus 1 over negative 2 minus 2 is equal to, this is 6 times negative 1, so it's negative 6 over negative 4, which is equal to 3 halves. So if we redefine f of x, if we say f of x is equal to 6x squared plus 18x plus 12 over x squared minus 4 for x not equal positive or negative 2, and it's equal to 3 halves for x equals negative 2, now this function is going to be the exact same thing as this right over here. This f of x, this new one, this new definition, this extended definition of our original one, is now equivalent to this expression. It is equal to 6 times x plus 1 over x minus 2. But just to answer their question, what value should be assigned to f of negative 2 to make f of x continuous at that point? Well, f of x should be, or f of negative 2 should be 3 halves."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So over here it says Nate tried to find the derivative of x squared plus five x times sine of x. Here is his work. Is Nate's work correct? If not, what's his mistake? So pause the video and see if you can answer this. Is Nate's work correct? And if not, what's his mistake?"}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "If not, what's his mistake? So pause the video and see if you can answer this. Is Nate's work correct? And if not, what's his mistake? All right, so I'm assuming you've had a go at it. So let's work through this step by step. So over here he's just trying to apply the derivative operator to the expression, which is exactly what he needed to do."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And if not, what's his mistake? All right, so I'm assuming you've had a go at it. So let's work through this step by step. So over here he's just trying to apply the derivative operator to the expression, which is exactly what he needed to do. He's trying to find the derivative of this thing. And he says, okay, this is a product of two expressions. And then he says, okay, well, this is gonna be the same thing as the derivative, or this is the same thing as the product of the derivatives."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So over here he's just trying to apply the derivative operator to the expression, which is exactly what he needed to do. He's trying to find the derivative of this thing. And he says, okay, this is a product of two expressions. And then he says, okay, well, this is gonna be the same thing as the derivative, or this is the same thing as the product of the derivatives. Now this is a problem. You are probably familiar, if I take the derivative of the sum of two things, so the derivative with respect to x of f of x plus g of x, that indeed is equal to the derivative of the first, f prime of x, plus the derivative of the second. But that is not true if we are dealing with the product of functions."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And then he says, okay, well, this is gonna be the same thing as the derivative, or this is the same thing as the product of the derivatives. Now this is a problem. You are probably familiar, if I take the derivative of the sum of two things, so the derivative with respect to x of f of x plus g of x, that indeed is equal to the derivative of the first, f prime of x, plus the derivative of the second. But that is not true if we are dealing with the product of functions. The derivative with respect to x of f of x, g of x, is not necessarily, maybe there's some very special circumstances, but in general, it's not going to be just the product of the derivative. It's not going to be just f prime of x, g prime of x. Here we would want to apply the product rule."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "But that is not true if we are dealing with the product of functions. The derivative with respect to x of f of x, g of x, is not necessarily, maybe there's some very special circumstances, but in general, it's not going to be just the product of the derivative. It's not going to be just f prime of x, g prime of x. Here we would want to apply the product rule. This is going to be equal to, this is going to be equal to the derivative of the first function times the second function, plus the derivative, well let me write it this way, plus the first function, not taking its derivative, times the derivative of the second function. So he should have applied the product rule here. And so let's do that, just to see what his answer should have been."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Here we would want to apply the product rule. This is going to be equal to, this is going to be equal to the derivative of the first function times the second function, plus the derivative, well let me write it this way, plus the first function, not taking its derivative, times the derivative of the second function. So he should have applied the product rule here. And so let's do that, just to see what his answer should have been. So what he should have done here, I'll get my correcting red pen out here, say no, that's not what he should have done. He says let's take the derivative of this first thing. So actually let me do it, color code it."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And so let's do that, just to see what his answer should have been. So what he should have done here, I'll get my correcting red pen out here, say no, that's not what he should have done. He says let's take the derivative of this first thing. So actually let me do it, color code it. So the derivative of this is two x plus five. So it should have been two x plus five, times the second thing, so times sine of x, times, let me do it in another color, times sine of x. And then to that, he would add the first thing, which is x squared plus five x, times the derivative of the second thing."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So actually let me do it, color code it. So the derivative of this is two x plus five. So it should have been two x plus five, times the second thing, so times sine of x, times, let me do it in another color, times sine of x. And then to that, he would add the first thing, which is x squared plus five x, times the derivative of the second thing. So the derivative of sine of x is cosine of x. So this is what we should have been seeing at this step right over here. He shouldn't have just taken the product of the derivatives, he should have applied the product rule."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And then to that, he would add the first thing, which is x squared plus five x, times the derivative of the second thing. So the derivative of sine of x is cosine of x. So this is what we should have been seeing at this step right over here. He shouldn't have just taken the product of the derivatives, he should have applied the product rule. So his work is not correct, and his mistake is that he didn't apply the product rule. He just assumed that the derivative of the products is the same thing as the product of the derivatives. Let's do more examples."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "He shouldn't have just taken the product of the derivatives, he should have applied the product rule. So his work is not correct, and his mistake is that he didn't apply the product rule. He just assumed that the derivative of the products is the same thing as the product of the derivatives. Let's do more examples. Okay, so let's see. It says, Katie tried to find the derivative of two x squared minus four, all of that to the third power. Here is her work."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Let's do more examples. Okay, so let's see. It says, Katie tried to find the derivative of two x squared minus four, all of that to the third power. Here is her work. Is Katie's work correct? If not, what is her mistake? So once again, pause the video, see if you can figure it out."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Here is her work. Is Katie's work correct? If not, what is her mistake? So once again, pause the video, see if you can figure it out. All right, now let's inspect Katie's work. So she's taking the derivative of this. And let's see, over here it looks like she's taking the derivative of the entire expression with respect to the inner expression."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So once again, pause the video, see if you can figure it out. All right, now let's inspect Katie's work. So she's taking the derivative of this. And let's see, over here it looks like she's taking the derivative of the entire expression with respect to the inner expression. And that is close to applying the chain rule properly, but it's not applying the chain rule properly. So her work is not correct, and her mistake is she's not correctly applying the chain rule. Just as a review, the chain rule says, look, if we're trying to take the derivative with respect to x of f of g of x, f of g of x, f of g of x, that this is going to be equal to, this is going to be equal to the derivative of the whole thing with respect to g of x."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And let's see, over here it looks like she's taking the derivative of the entire expression with respect to the inner expression. And that is close to applying the chain rule properly, but it's not applying the chain rule properly. So her work is not correct, and her mistake is she's not correctly applying the chain rule. Just as a review, the chain rule says, look, if we're trying to take the derivative with respect to x of f of g of x, f of g of x, f of g of x, that this is going to be equal to, this is going to be equal to the derivative of the whole thing with respect to g of x. So I could write that as f prime of g of x, f prime of g of x, times the derivative of the inner function with respect to x, times g prime of x. So over here, we could view our f function as a thing that takes its input and takes it to the third power. And so this right over here is f prime of g of x."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Just as a review, the chain rule says, look, if we're trying to take the derivative with respect to x of f of g of x, f of g of x, f of g of x, that this is going to be equal to, this is going to be equal to the derivative of the whole thing with respect to g of x. So I could write that as f prime of g of x, f prime of g of x, times the derivative of the inner function with respect to x, times g prime of x. So over here, we could view our f function as a thing that takes its input and takes it to the third power. And so this right over here is f prime of g of x. So this thing is the f prime of g of x, but she forgot to multiply it by the derivative of the inner function with respect to x. So she forgot to multiply this times the derivative of two x squared minus four with respect to x, which is going to be, let's see, the derivative of two x squared, power rule, two times two is four, so it's gonna be four x to the first. And then the derivative of negative four is just zero, so it's just gonna be times four x."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And so this right over here is f prime of g of x. So this thing is the f prime of g of x, but she forgot to multiply it by the derivative of the inner function with respect to x. So she forgot to multiply this times the derivative of two x squared minus four with respect to x, which is going to be, let's see, the derivative of two x squared, power rule, two times two is four, so it's gonna be four x to the first. And then the derivative of negative four is just zero, so it's just gonna be times four x. So that's what she needed to do in order for it to be correct. So she had to have this times four x here, times four x. So not correct, she didn't correctly apply the chain rule."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And then the derivative of negative four is just zero, so it's just gonna be times four x. So that's what she needed to do in order for it to be correct. So she had to have this times four x here, times four x. So not correct, she didn't correctly apply the chain rule. So let's do another one of these. So here it says Nejoman tried to find the derivative of sine of seven x squared plus four x. Here is his work."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So not correct, she didn't correctly apply the chain rule. So let's do another one of these. So here it says Nejoman tried to find the derivative of sine of seven x squared plus four x. Here is his work. Is Nejoman's work correct? If not, what is his mistake? Pause the video, see if you can figure it out."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Here is his work. Is Nejoman's work correct? If not, what is his mistake? Pause the video, see if you can figure it out. All right, so it's the derivative of sine of this expression. So you'd wanna use the chain rule. In fact, using the chain rule, you wanna find the derivative of the outside function with respect to the inside."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Pause the video, see if you can figure it out. All right, so it's the derivative of sine of this expression. So you'd wanna use the chain rule. In fact, using the chain rule, you wanna find the derivative of the outside function with respect to the inside. So the derivative of sine of something with respect to that something is gonna be cosine of that something. So that's right, that's right. And then you wanna multiply that times the derivative of the inside with respect to x."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "In fact, using the chain rule, you wanna find the derivative of the outside function with respect to the inside. So the derivative of sine of something with respect to that something is gonna be cosine of that something. So that's right, that's right. And then you wanna multiply that times the derivative of the inside with respect to x. So the derivative of seven x squared is 14 x. Derivative of four x is four. So this is actually, that step looks good."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And then you wanna multiply that times the derivative of the inside with respect to x. So the derivative of seven x squared is 14 x. Derivative of four x is four. So this is actually, that step looks good. But then Nejoman does something strange over here. This is the cosine of seven x squared plus four x, and then that whole thing times 14 x plus four. But they get confused where, just looking at these parentheses, and this tends to happen sometimes."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So this is actually, that step looks good. But then Nejoman does something strange over here. This is the cosine of seven x squared plus four x, and then that whole thing times 14 x plus four. But they get confused where, just looking at these parentheses, and this tends to happen sometimes. This is actually one of these key errors that the folks at the college board, the AP folks, told us about. Is that when dealing with these transcendental functions, cosine, sine, tangent, natural log, that are written like this, and people see the parentheses and then see another parentheses, their brain just says, oh, let me multiply these two expressions in parentheses. But that's not right, because if we were to add parentheses, this is what this is implying."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "But they get confused where, just looking at these parentheses, and this tends to happen sometimes. This is actually one of these key errors that the folks at the college board, the AP folks, told us about. Is that when dealing with these transcendental functions, cosine, sine, tangent, natural log, that are written like this, and people see the parentheses and then see another parentheses, their brain just says, oh, let me multiply these two expressions in parentheses. But that's not right, because if we were to add parentheses, this is what this is implying. So you can't just take the 14 x plus four and multiply it by this, and assuming you're taking the cosine of the whole thing. So this is where Nejoman makes the mistake. The work is not correct, and the mistake is trying to multiply these two expressions and taking the cosine of the whole thing."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "But that's not right, because if we were to add parentheses, this is what this is implying. So you can't just take the 14 x plus four and multiply it by this, and assuming you're taking the cosine of the whole thing. So this is where Nejoman makes the mistake. The work is not correct, and the mistake is trying to multiply these two expressions and taking the cosine of the whole thing. Let's do one more of these. I find these strangely fun. All right, this one is involved."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "The work is not correct, and the mistake is trying to multiply these two expressions and taking the cosine of the whole thing. Let's do one more of these. I find these strangely fun. All right, this one is involved. Tom tried to find the derivative of the square root of x over x to the fourth. Here is his work. Is Tom's work correct?"}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "All right, this one is involved. Tom tried to find the derivative of the square root of x over x to the fourth. Here is his work. Is Tom's work correct? If not, what's his mistake? Pause the video and see if you can figure that out. So it looks like he's trying to apply the quotient rule."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Is Tom's work correct? If not, what's his mistake? Pause the video and see if you can figure that out. So it looks like he's trying to apply the quotient rule. So applying the quotient rule, you would, in the numerator, you would take the derivative of the first expression times the second expression, and then minus the first expression times the derivative of the second expression, all of that over the, or I should say, the derivative of the numerator expression over the, or times the denominator expression, minus the numerator expression times the derivative of the denominator expression, all of that over the denominator expression squared. So this looks correct, actually. It's a correct application of the quotient rule."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So it looks like he's trying to apply the quotient rule. So applying the quotient rule, you would, in the numerator, you would take the derivative of the first expression times the second expression, and then minus the first expression times the derivative of the second expression, all of that over the, or I should say, the derivative of the numerator expression over the, or times the denominator expression, minus the numerator expression times the derivative of the denominator expression, all of that over the denominator expression squared. So this looks correct, actually. It's a correct application of the quotient rule. It looks like Tom is correctly simplifying. So the derivative of x to the 1 1\u20442 is 1 1\u20442 x to the negative 1 1\u20442, so that looks right. Derivative of x to the fourth is four x to the third, so that looks right."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "It's a correct application of the quotient rule. It looks like Tom is correctly simplifying. So the derivative of x to the 1 1\u20442 is 1 1\u20442 x to the negative 1 1\u20442, so that looks right. Derivative of x to the fourth is four x to the third, so that looks right. All of this looks algebraically right. And let's see, when you simplify this, so let's see, x to the negative 1 1\u20442 times x to the fourth is indeed x to, well, that's going to be x to the, oh, this correlates, so this simplifies to that, which looks correct, and that simplifies to that, which looks correct. We're just using exponent properties there."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Derivative of x to the fourth is four x to the third, so that looks right. All of this looks algebraically right. And let's see, when you simplify this, so let's see, x to the negative 1 1\u20442 times x to the fourth is indeed x to, well, that's going to be x to the, oh, this correlates, so this simplifies to that, which looks correct, and that simplifies to that, which looks correct. We're just using exponent properties there. And then divided everything by, let's see, oh, then everything is in terms of x to 3.5, so we're going to have negative 3.5 x to 3.5, and then you use exponent properties. So actually, it looks like he did everything correctly. This is the right answer."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "We're just using exponent properties there. And then divided everything by, let's see, oh, then everything is in terms of x to 3.5, so we're going to have negative 3.5 x to 3.5, and then you use exponent properties. So actually, it looks like he did everything correctly. This is the right answer. Now, so his work is correct. He did not make any mistakes, but I do have a bone to pick, so to speak, with Tom, because he didn't have to apply the quotient rule here. He did all of this hairy calculus and algebra, but there could have been a very simple simplification he could have made up here, and this is a key thing to realize."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "This is the right answer. Now, so his work is correct. He did not make any mistakes, but I do have a bone to pick, so to speak, with Tom, because he didn't have to apply the quotient rule here. He did all of this hairy calculus and algebra, but there could have been a very simple simplification he could have made up here, and this is a key thing to realize. He could have said, hey, you know what? This is the same thing as the derivative with respect to x of x to the 1.5, that's what the square root of x is, times x to the negative fourth power. That's what one over x to the fourth is."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "He did all of this hairy calculus and algebra, but there could have been a very simple simplification he could have made up here, and this is a key thing to realize. He could have said, hey, you know what? This is the same thing as the derivative with respect to x of x to the 1.5, that's what the square root of x is, times x to the negative fourth power. That's what one over x to the fourth is. And so let me color code it. So that is the same thing as that, and that is the same thing as that. And you wouldn't even have to use the product rule here."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "That's what one over x to the fourth is. And so let me color code it. So that is the same thing as that, and that is the same thing as that. And you wouldn't even have to use the product rule here. You could simplify this even further. This is the same thing as the derivative with respect to x of, just we have the same base. We can add the exponents."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And you wouldn't even have to use the product rule here. You could simplify this even further. This is the same thing as the derivative with respect to x of, just we have the same base. We can add the exponents. We're taking the product, so it's gonna be x to the negative 3.5. And so you could just use the power rule. So this is going to be equal to, bring the negative 3.5 out front, negative 3.5 x to the, and then we just decrement this by one."}, {"video_title": "Differentiating functions Find the error Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "We can add the exponents. We're taking the product, so it's gonna be x to the negative 3.5. And so you could just use the power rule. So this is going to be equal to, bring the negative 3.5 out front, negative 3.5 x to the, and then we just decrement this by one. We subtract one from that, negative 4.5 power. So as you can see, he could have gotten this answer much, much, much, much, much, much quicker, but he didn't make any mistakes. There's a little bit of a judgment error just immediately going forth with the quotient rule, which gets quite hairy quite fast."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And it's diving straight down near a streetlight. Let's get some information about what's going on. So the streetlight right over here is 20 feet high. So this is a 20 foot high street lamp. And right at this moment, and I haven't drawn it completely to scale, the owl is 15 feet above the mouse. So this distance right over here is 15 feet. And the mouse itself is 10 feet from the base of the lamp."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So this is a 20 foot high street lamp. And right at this moment, and I haven't drawn it completely to scale, the owl is 15 feet above the mouse. So this distance right over here is 15 feet. And the mouse itself is 10 feet from the base of the lamp. Let me draw that. So the mouse is 10 feet from the base of the lamp. And we also know, we have our little radar gun out, we know that this owl is driving straight down."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And the mouse itself is 10 feet from the base of the lamp. Let me draw that. So the mouse is 10 feet from the base of the lamp. And we also know, we have our little radar gun out, we know that this owl is driving straight down. And right now, it is going 20 feet per second. So right now, this is going down at 20 feet per second. Now, what we're curious about is we have the light over here."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And we also know, we have our little radar gun out, we know that this owl is driving straight down. And right now, it is going 20 feet per second. So right now, this is going down at 20 feet per second. Now, what we're curious about is we have the light over here. Light is coming from the street lamp in every direction. And it creates a shadow of the owl. So right now, the shadow is out here."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Now, what we're curious about is we have the light over here. Light is coming from the street lamp in every direction. And it creates a shadow of the owl. So right now, the shadow is out here. And as the owl goes further and further down, the shadow is going to move to the left like that. And so given everything that we've set up right over here, the question is, at what rate is the shadow moving? So let's think about what we know and what we don't know."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So right now, the shadow is out here. And as the owl goes further and further down, the shadow is going to move to the left like that. And so given everything that we've set up right over here, the question is, at what rate is the shadow moving? So let's think about what we know and what we don't know. And to do that, let's set up some variables. So let me draw the same thing a little bit more geometrically. So let's say that this right over here is the street light that is 20 feet tall."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let's think about what we know and what we don't know. And to do that, let's set up some variables. So let me draw the same thing a little bit more geometrically. So let's say that this right over here is the street light that is 20 feet tall. And then this right over here is the height of the owl right at this moment. So this is 15 feet. The distance between the base of the lamp and where the owl is going, where that mouse is right now, this is 10 feet."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say that this right over here is the street light that is 20 feet tall. And then this right over here is the height of the owl right at this moment. So this is 15 feet. The distance between the base of the lamp and where the owl is going, where that mouse is right now, this is 10 feet. And if I were to think about where the shadow is, well, the light's emitting from right over here. The light's emitting right over here. And so the owl blocks the light right over there."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "The distance between the base of the lamp and where the owl is going, where that mouse is right now, this is 10 feet. And if I were to think about where the shadow is, well, the light's emitting from right over here. The light's emitting right over here. And so the owl blocks the light right over there. So the shadow is going to be right over there. So if you just draw a straight line from the source of light through the owl, and you just keep going, and you hit the ground, you're going to figure out where the shadow is. So the shadow is going to be right over here."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And so the owl blocks the light right over there. So the shadow is going to be right over there. So if you just draw a straight line from the source of light through the owl, and you just keep going, and you hit the ground, you're going to figure out where the shadow is. So the shadow is going to be right over here. It's going to be right over there. And we need to figure out how quickly is that moving. And it's going to be moving in the leftward direction."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So the shadow is going to be right over here. It's going to be right over there. And we need to figure out how quickly is that moving. And it's going to be moving in the leftward direction. So let's set up some variables over here. So let's say, so what's changing? Well, we know that the height of the owl is changing."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And it's going to be moving in the leftward direction. So let's set up some variables over here. So let's say, so what's changing? Well, we know that the height of the owl is changing. So let's call that y. Right at this moment, it's equal to 15. But it is actually changing."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we know that the height of the owl is changing. So let's call that y. Right at this moment, it's equal to 15. But it is actually changing. And let's call the distance between the shadow and the mouse x. Now, given this setup, can we come up with a relationship between x and y? And then using that relationship, what we're really trying to come up with is what is the rate at which x is changing with respect to time?"}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "But it is actually changing. And let's call the distance between the shadow and the mouse x. Now, given this setup, can we come up with a relationship between x and y? And then using that relationship, what we're really trying to come up with is what is the rate at which x is changing with respect to time? We know what y is right at this moment. We know what dy dt is right at this moment. Can we come up with a relationship between x and y and maybe take the derivative with respect to t so we can figure out what dx dt is at a given moment in time?"}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And then using that relationship, what we're really trying to come up with is what is the rate at which x is changing with respect to time? We know what y is right at this moment. We know what dy dt is right at this moment. Can we come up with a relationship between x and y and maybe take the derivative with respect to t so we can figure out what dx dt is at a given moment in time? Well, both of these triangles, and when I say both of these triangles, let me be clear what I'm talking about. This triangle right over here, the smaller triangle in green, is a similar triangle to the larger triangle. It's a similar triangle to this larger triangle that I am tracing in blue."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Can we come up with a relationship between x and y and maybe take the derivative with respect to t so we can figure out what dx dt is at a given moment in time? Well, both of these triangles, and when I say both of these triangles, let me be clear what I'm talking about. This triangle right over here, the smaller triangle in green, is a similar triangle to the larger triangle. It's a similar triangle to this larger triangle that I am tracing in blue. It's similar to this larger one. How do I know that? Well, they both have a right angle right over here."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "It's a similar triangle to this larger triangle that I am tracing in blue. It's similar to this larger one. How do I know that? Well, they both have a right angle right over here. They both share this angle. So if they have two angles in common, then all three angles must be in common. So they are similar triangles, which means the ratio between corresponding sides must be the same."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, they both have a right angle right over here. They both share this angle. So if they have two angles in common, then all three angles must be in common. So they are similar triangles, which means the ratio between corresponding sides must be the same. So we know that the ratio of x to y must be the ratio of this entire base, which is x plus 10, to the height of the larger triangle, 220. And right there, we have a relationship between x and y. And if we take the derivative of both sides with respect to t, we're probably doing pretty well."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So they are similar triangles, which means the ratio between corresponding sides must be the same. So we know that the ratio of x to y must be the ratio of this entire base, which is x plus 10, to the height of the larger triangle, 220. And right there, we have a relationship between x and y. And if we take the derivative of both sides with respect to t, we're probably doing pretty well. Now, before taking the derivative with respect to t, I could do it right over here just to simplify things a little bit. Let me just cross multiply. So let me multiply both sides of this equation by 20 and y, just so that I don't have as many things in the denominator."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And if we take the derivative of both sides with respect to t, we're probably doing pretty well. Now, before taking the derivative with respect to t, I could do it right over here just to simplify things a little bit. Let me just cross multiply. So let me multiply both sides of this equation by 20 and y, just so that I don't have as many things in the denominator. So on the left-hand side, it simplifies to 20x. I don't want to write over it. Well, I'll just write 20x."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let me multiply both sides of this equation by 20 and y, just so that I don't have as many things in the denominator. So on the left-hand side, it simplifies to 20x. I don't want to write over it. Well, I'll just write 20x. And on the left-hand side, it is 20x. And then on the right-hand side, let's see, this cancels with that. We have xy plus 10y."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, I'll just write 20x. And on the left-hand side, it is 20x. And then on the right-hand side, let's see, this cancels with that. We have xy plus 10y. And now let me take the derivative of both sides with respect to time. So the derivative of 20 times something with respect to time is going to be the derivative of 20 times something with respect to the something, which is just 20. That's the derivative of 20x with respect to x times dx, the derivative of x with respect to t, is equal to."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We have xy plus 10y. And now let me take the derivative of both sides with respect to time. So the derivative of 20 times something with respect to time is going to be the derivative of 20 times something with respect to the something, which is just 20. That's the derivative of 20x with respect to x times dx, the derivative of x with respect to t, is equal to. Now, over here, we're going to have to break out a little bit of the product rule. So first, we want to figure out the derivative of x with respect to time. So the derivative of x with respect to time."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "That's the derivative of 20x with respect to x times dx, the derivative of x with respect to t, is equal to. Now, over here, we're going to have to break out a little bit of the product rule. So first, we want to figure out the derivative of x with respect to time. So the derivative of x with respect to time. So the derivative of the first thing times the second thing, times y, plus just the first thing, times the derivative of the second thing. The derivative of y with respect to t is just dy dt. dy dt."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So the derivative of x with respect to time. So the derivative of the first thing times the second thing, times y, plus just the first thing, times the derivative of the second thing. The derivative of y with respect to t is just dy dt. dy dt. And then finally, right over here, the derivative of 10y, with respect to t is the derivative of 10y with respect to y, which is just 10, times the derivative of y with respect to t, which is dy dy dt. And there you have it. You have your relationship between dx dt, dy dt, and x and y."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "dy dt. And then finally, right over here, the derivative of 10y, with respect to t is the derivative of 10y with respect to y, which is just 10, times the derivative of y with respect to t, which is dy dy dt. And there you have it. You have your relationship between dx dt, dy dt, and x and y. So let's just make sure we have everything. This is what we're trying to solve for, dx dt. And let's see, we have another dx dt here."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "You have your relationship between dx dt, dy dt, and x and y. So let's just make sure we have everything. This is what we're trying to solve for, dx dt. And let's see, we have another dx dt here. We're going to try to solve for that. We know what y is. y is equal to 15."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And let's see, we have another dx dt here. We're going to try to solve for that. We know what y is. y is equal to 15. We know what dy dt is. dy dt, if we make the convention since y is decreasing, we can say it's negative 20. So we know what this is."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "y is equal to 15. We know what dy dt is. dy dt, if we make the convention since y is decreasing, we can say it's negative 20. So we know what this is. And so if we just know what x is, we can solve for dx dt. So what is x right at this moment? Well, we can use this first equation."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So we know what this is. And so if we just know what x is, we can solve for dx dt. So what is x right at this moment? Well, we can use this first equation. We could actually use this one up here, but this one is simplified a little bit to actually solve for x. So let's do that, and then we'll substitute back into this thing where we've taken the derivative. So we get 20 times x is equal to x times y. y is 15."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we can use this first equation. We could actually use this one up here, but this one is simplified a little bit to actually solve for x. So let's do that, and then we'll substitute back into this thing where we've taken the derivative. So we get 20 times x is equal to x times y. y is 15. And just remember, I could have used this equation, but this is just one step further. We've already cross-multiplied. So it's x times y. y is 15."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So we get 20 times x is equal to x times y. y is 15. And just remember, I could have used this equation, but this is just one step further. We've already cross-multiplied. So it's x times y. y is 15. So it's x times 15 plus 10 times y. Plus 10 times 15. Did I do that right?"}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So it's x times y. y is 15. So it's x times 15 plus 10 times y. Plus 10 times 15. Did I do that right? 20x is equal to x times 15 plus 10 times 15. So let's see. If you subtract, so this is 20x is equal to 15x plus 150."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Did I do that right? 20x is equal to x times 15 plus 10 times 15. So let's see. If you subtract, so this is 20x is equal to 15x plus 150. Subtract 15x from both sides, you get 5x is equal to 30. 5x is equal to 150. My brain is getting ahead."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "If you subtract, so this is 20x is equal to 15x plus 150. Subtract 15x from both sides, you get 5x is equal to 30. 5x is equal to 150. My brain is getting ahead. 5x is equal to 150. Divide both sides by 5, you get x is equal to 30 feet. x is equal to 30 feet right at this moment."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "My brain is getting ahead. 5x is equal to 150. Divide both sides by 5, you get x is equal to 30 feet. x is equal to 30 feet right at this moment. So this distance, just going back to our original diagram, this distance right over here is 30 feet. So let's substitute all the values we know back into this equation to actually solve for dx dt. So we have, let me do it right over here."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "x is equal to 30 feet right at this moment. So this distance, just going back to our original diagram, this distance right over here is 30 feet. So let's substitute all the values we know back into this equation to actually solve for dx dt. So we have, let me do it right over here. We have 20 times dx dt. I'll do that in orange. We'll solve for that."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So we have, let me do it right over here. We have 20 times dx dt. I'll do that in orange. We'll solve for that. Actually, I already used orange. So let's say dx dt, I'll use this pink. 20 times dx dt is equal to dx dt times y. y right now is 15 feet."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We'll solve for that. Actually, I already used orange. So let's say dx dt, I'll use this pink. 20 times dx dt is equal to dx dt times y. y right now is 15 feet. So times 15 times, I didn't want to do that color, times 15 plus x. We already know that x is 30. Plus 30 times dy dt."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "20 times dx dt is equal to dx dt times y. y right now is 15 feet. So times 15 times, I didn't want to do that color, times 15 plus x. We already know that x is 30. Plus 30 times dy dt. What is dy dt? dy dt we could say is negative 20 feet per second. y is decreasing."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Plus 30 times dy dt. What is dy dt? dy dt we could say is negative 20 feet per second. y is decreasing. The bird is diving down to get its dinner. So times 20 feet per second, so that's that right over there, plus 10 times dy dt. So plus 10 times negative 20 feet per second."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "y is decreasing. The bird is diving down to get its dinner. So times 20 feet per second, so that's that right over there, plus 10 times dy dt. So plus 10 times negative 20 feet per second. And now we just solve for dx dt. So let's see, what do we have? We have 20 times, let's see, let me subtract 15 dx dt from both sides of this equation."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So plus 10 times negative 20 feet per second. And now we just solve for dx dt. So let's see, what do we have? We have 20 times, let's see, let me subtract 15 dx dt from both sides of this equation. And we get 5 dx dts. 5 dx dts, I just subtracted this from both sides of the equation. This is 15 dx dts, this is 20."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We have 20 times, let's see, let me subtract 15 dx dt from both sides of this equation. And we get 5 dx dts. 5 dx dts, I just subtracted this from both sides of the equation. This is 15 dx dts, this is 20. So we have 5 dx dts is equal to, this part right over here is negative 600. And this part right over here is negative 200. So it's equal to negative 800 feet per second."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "This is 15 dx dts, this is 20. So we have 5 dx dts is equal to, this part right over here is negative 600. And this part right over here is negative 200. So it's equal to negative 800 feet per second. Or negative 800, and actually this will actually be in feet per second. And so dx dt is equal to, dividing both sides by 5, 5 times 16 is 80. So this is negative 160 feet per second."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So it's equal to negative 800 feet per second. Or negative 800, and actually this will actually be in feet per second. And so dx dt is equal to, dividing both sides by 5, 5 times 16 is 80. So this is negative 160 feet per second. And we're done. And we see the shadow is moving very, very, very, very fast to the left. x is decreasing."}, {"video_title": "Related rates shadow Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So this is negative 160 feet per second. And we're done. And we see the shadow is moving very, very, very, very fast to the left. x is decreasing. And we see that. That's why we have this negative sign here. The value of x is decreasing."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "In the last video, we attempted to approximate the area under a curve by constructing four rectangles of equal width and using the left boundary of each rectangle, the function evaluated at the left boundary, to determine the height. And we came up with an approximation. What I want to do in this video is generalize things a bit, using the exact same method, but doing it for an arbitrary function with arbitrary boundaries and an arbitrary number of rectangles. So let's do it. I'm going to draw the diagram as large as I can to make things clear, to make things as clear as possible. So that's my y-axis. And this right over here is my x-axis."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So let's do it. I'm going to draw the diagram as large as I can to make things clear, to make things as clear as possible. So that's my y-axis. And this right over here is my x-axis. Let me draw an arbitrary function. So let's say my function looks something like that. So that is y is equal to f of x."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And this right over here is my x-axis. Let me draw an arbitrary function. So let's say my function looks something like that. So that is y is equal to f of x. And let me define my boundaries. So let's say this right over here is x equals a. This is x equals a."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So that is y is equal to f of x. And let me define my boundaries. So let's say this right over here is x equals a. This is x equals a. And this right over here is x. This is x equals b. So this is b."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "This is x equals a. And this right over here is x. This is x equals b. So this is b. And I'm going to use n rectangles. n rectangles. And I'm going to use the function evaluated at the left boundary of the rectangle to determine its height."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So this is b. And I'm going to use n rectangles. n rectangles. And I'm going to use the function evaluated at the left boundary of the rectangle to determine its height. So, for example, this will be rectangle 1. I'm going to evaluate what f of a is. I'm going to evaluate what f of a is."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And I'm going to use the function evaluated at the left boundary of the rectangle to determine its height. So, for example, this will be rectangle 1. I'm going to evaluate what f of a is. I'm going to evaluate what f of a is. So this right over here is f of a. And then I'm going to use that as the height of my first rectangle. The height of my first rectangle."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "I'm going to evaluate what f of a is. So this right over here is f of a. And then I'm going to use that as the height of my first rectangle. The height of my first rectangle. So just like that. So rectangle number 1 looks like this. And I'll even number it."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "The height of my first rectangle. So just like that. So rectangle number 1 looks like this. And I'll even number it. Rectangle 1 looks just like that. And just to have a convention here, because I'm going to want to label each of the boundaries of the left boundary. Each of the x values of the left boundary."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And I'll even number it. Rectangle 1 looks just like that. And just to have a convention here, because I'm going to want to label each of the boundaries of the left boundary. Each of the x values of the left boundary. So we'll say a is equal to x naught. a is equal to x naught. So we can also call this point right over here x naught."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Each of the x values of the left boundary. So we'll say a is equal to x naught. a is equal to x naught. So we can also call this point right over here x naught. That x value. And then we go to the next rectangle. And we can call this one right over here."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So we can also call this point right over here x naught. That x value. And then we go to the next rectangle. And we can call this one right over here. This x value, we'll call it x1. It's the left boundary of the next rectangle. If we evaluate f of x1, we get this value right over here."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And we can call this one right over here. This x value, we'll call it x1. It's the left boundary of the next rectangle. If we evaluate f of x1, we get this value right over here. This right over here is f of x1. So it tells us our height. And we want an equal width to the previous one."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "If we evaluate f of x1, we get this value right over here. This right over here is f of x1. So it tells us our height. And we want an equal width to the previous one. And we'll think about what the width is going to be in a second. So this right over here is our second rectangle. This right over here is our second rectangle that we're going to use to approximate the area under the curve."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And we want an equal width to the previous one. And we'll think about what the width is going to be in a second. So this right over here is our second rectangle. This right over here is our second rectangle that we're going to use to approximate the area under the curve. That's rectangle number 2. Let's do rectangle number 3. Well, rectangle number 3, the left boundary, we're just going to call it x sub 2."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "This right over here is our second rectangle that we're going to use to approximate the area under the curve. That's rectangle number 2. Let's do rectangle number 3. Well, rectangle number 3, the left boundary, we're just going to call it x sub 2. And its height is going to be f of x sub 2. And its width is going to be the same width as the other ones. I'm just eyeballing it right over here."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Well, rectangle number 3, the left boundary, we're just going to call it x sub 2. And its height is going to be f of x sub 2. And its width is going to be the same width as the other ones. I'm just eyeballing it right over here. So this is rectangle number 3. And we're going to continue this process all the way until we get to rectangle number n. So this is the nth rectangle. The nth rectangle right over here."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "I'm just eyeballing it right over here. So this is rectangle number 3. And we're going to continue this process all the way until we get to rectangle number n. So this is the nth rectangle. The nth rectangle right over here. And what am I going to label this point right over here? Well, we already see a pattern. The left boundary of the first rectangle is x sub 0."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "The nth rectangle right over here. And what am I going to label this point right over here? Well, we already see a pattern. The left boundary of the first rectangle is x sub 0. The left boundary of the second rectangle is x sub 1. The left boundary of the third rectangle is x sub 2. So the left boundary of the nth rectangle is going to be x sub n minus 1."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "The left boundary of the first rectangle is x sub 0. The left boundary of the second rectangle is x sub 1. The left boundary of the third rectangle is x sub 2. So the left boundary of the nth rectangle is going to be x sub n minus 1. Whatever the rectangle number is, the left boundary is x sub that number minus 1. And this is just based on the convention that we've defined. Now the next thing that we need to do in order to actually calculate this area is think about what is the width."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So the left boundary of the nth rectangle is going to be x sub n minus 1. Whatever the rectangle number is, the left boundary is x sub that number minus 1. And this is just based on the convention that we've defined. Now the next thing that we need to do in order to actually calculate this area is think about what is the width. So let's call the width of any of these rectangles. And for these purposes, or the purpose of this example, I'm going to assume that it's constant. Although you can do these sums where you actually vary the width of the rectangle, but then it gets a little bit fancier."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Now the next thing that we need to do in order to actually calculate this area is think about what is the width. So let's call the width of any of these rectangles. And for these purposes, or the purpose of this example, I'm going to assume that it's constant. Although you can do these sums where you actually vary the width of the rectangle, but then it gets a little bit fancier. So I want an equal width. So I want delta x to be equal width. And to think about what that has to be, we just have to think, what's the total width that we're covering?"}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Although you can do these sums where you actually vary the width of the rectangle, but then it gets a little bit fancier. So I want an equal width. So I want delta x to be equal width. And to think about what that has to be, we just have to think, what's the total width that we're covering? Well, the total distance here is going to be b minus a. And we're just going to divide by the number of rectangles that we want, the number of sections that we want. So we want to divide by n. So if we assume this is true, and then we assume that a is equal to x0, and then x1 is equal to x0 plus delta x, x2 is equal to x1 plus delta x, and we go all the way to xn is equal to xn minus 1 plus delta x, then we've essentially set up this diagram right over here."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "And to think about what that has to be, we just have to think, what's the total width that we're covering? Well, the total distance here is going to be b minus a. And we're just going to divide by the number of rectangles that we want, the number of sections that we want. So we want to divide by n. So if we assume this is true, and then we assume that a is equal to x0, and then x1 is equal to x0 plus delta x, x2 is equal to x1 plus delta x, and we go all the way to xn is equal to xn minus 1 plus delta x, then we've essentially set up this diagram right over here. b is actually going to be equal to xn. So this is xn. It's equal to xn minus 1 plus delta x."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So we want to divide by n. So if we assume this is true, and then we assume that a is equal to x0, and then x1 is equal to x0 plus delta x, x2 is equal to x1 plus delta x, and we go all the way to xn is equal to xn minus 1 plus delta x, then we've essentially set up this diagram right over here. b is actually going to be equal to xn. So this is xn. It's equal to xn minus 1 plus delta x. So now I think we've set up all of the notation and all of the conventions in order to actually calculate the area or our approximation of the area. So our approximation, approximate area, is going to be equal to what? Well, it's going to be the area of the first rectangle."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "It's equal to xn minus 1 plus delta x. So now I think we've set up all of the notation and all of the conventions in order to actually calculate the area or our approximation of the area. So our approximation, approximate area, is going to be equal to what? Well, it's going to be the area of the first rectangle. So let me write this down. So it's going to be rectangle 1, so the area of rectangle 1, so rectangle 1 plus the area of rectangle 2, plus the area of rectangle 2, plus the area of rectangle 3. I think you get the point here."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Well, it's going to be the area of the first rectangle. So let me write this down. So it's going to be rectangle 1, so the area of rectangle 1, so rectangle 1 plus the area of rectangle 2, plus the area of rectangle 2, plus the area of rectangle 3. I think you get the point here. Area of rectangle 3 all the way, plus all the way to the area of rectangle n. And so what are these going to be? Rectangle 1 is going to be its height, which is f of x0 or f of a. Either way, x0 and a are the same thing."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "I think you get the point here. Area of rectangle 3 all the way, plus all the way to the area of rectangle n. And so what are these going to be? Rectangle 1 is going to be its height, which is f of x0 or f of a. Either way, x0 and a are the same thing. So it's f of a times our delta x, times our width, our height times our width. So times delta, actually, let me write f of x0. I wanted to write f of x0 times delta x."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "Either way, x0 and a are the same thing. So it's f of a times our delta x, times our width, our height times our width. So times delta, actually, let me write f of x0. I wanted to write f of x0 times delta x. What is our height of rectangle 2? It's f of x1 times delta x. f of x1 times delta x. What's our area of rectangle 3?"}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "I wanted to write f of x0 times delta x. What is our height of rectangle 2? It's f of x1 times delta x. f of x1 times delta x. What's our area of rectangle 3? It's f of x2 times delta x. And then we go all the way to our area. We're taking all the sums all the way to rectangle n. What's its area?"}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "What's our area of rectangle 3? It's f of x2 times delta x. And then we go all the way to our area. We're taking all the sums all the way to rectangle n. What's its area? It's f of x sub n minus 1. Actually, that's a different shade of orange. We'll use that same shade."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "We're taking all the sums all the way to rectangle n. What's its area? It's f of x sub n minus 1. Actually, that's a different shade of orange. We'll use that same shade. It is f of x sub n minus 1 times delta x. And we're done. We've written it in a very general way."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "We'll use that same shade. It is f of x sub n minus 1 times delta x. And we're done. We've written it in a very general way. But to really make us comfortable with the various forms of notation, especially the types of notation you might see when people are talking about approximating areas or sums in general, I'm going to use the traditional sigma notation. So another way we could write this as the sum. This is equal to the sum from, and remember, this is just based on the conventions that I set up."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "We've written it in a very general way. But to really make us comfortable with the various forms of notation, especially the types of notation you might see when people are talking about approximating areas or sums in general, I'm going to use the traditional sigma notation. So another way we could write this as the sum. This is equal to the sum from, and remember, this is just based on the conventions that I set up. I'll let i count which rectangle we're in, from i equals 1 to n. And then we're going to look at each rectangle. So the first rectangle, that's rectangle 1. So it's going to be f of, well, if we're in the i-th rectangle, then we're going to take the left boundary is going to be x sub i minus 1 times delta x."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "This is equal to the sum from, and remember, this is just based on the conventions that I set up. I'll let i count which rectangle we're in, from i equals 1 to n. And then we're going to look at each rectangle. So the first rectangle, that's rectangle 1. So it's going to be f of, well, if we're in the i-th rectangle, then we're going to take the left boundary is going to be x sub i minus 1 times delta x. And so right over here is a general way of considering, of thinking about approximating the area under a curve using rectangles where the height of the rectangles are defined by the left boundary. And this tells us it's the left boundary. And we see for each, if this is the i-th rectangle right over here, if this is rectangle i, then this right over here is x sub i minus 1, and this height right over here is f of x sub i minus 1."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy.mp3", "Sentence": "So it's going to be f of, well, if we're in the i-th rectangle, then we're going to take the left boundary is going to be x sub i minus 1 times delta x. And so right over here is a general way of considering, of thinking about approximating the area under a curve using rectangles where the height of the rectangles are defined by the left boundary. And this tells us it's the left boundary. And we see for each, if this is the i-th rectangle right over here, if this is rectangle i, then this right over here is x sub i minus 1, and this height right over here is f of x sub i minus 1. So that's all we did right over there, times delta x. And then you sum all of these from the first rectangle all the way to the n-th. So hopefully that makes you a little bit more comfortable with this notation."}, {"video_title": "Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say that f of x is equal to x squared plus x minus 6 over x minus 2. And we're curious about what the limit of f of x as x approaches 2 is equal to. Now, the first attempt that you might want to do right when you see something like this is to see what happens. What is f of 2? Now, this won't always be the limit even if it's defined, but it's a good place to start just to see if something reasonable could pop out. So looking at it this way, if we just evaluate f of 2, on our numerator we're going to get 2 squared plus 2 minus 6. So that's going to be 4 plus 2, which is 6, minus 6."}, {"video_title": "Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "What is f of 2? Now, this won't always be the limit even if it's defined, but it's a good place to start just to see if something reasonable could pop out. So looking at it this way, if we just evaluate f of 2, on our numerator we're going to get 2 squared plus 2 minus 6. So that's going to be 4 plus 2, which is 6, minus 6. You're going to get 0 in the numerator, and you're going to get 0 in the denominator. So we don't have the function is not defined. So not defined at x is equal to 2. f not defined."}, {"video_title": "Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So that's going to be 4 plus 2, which is 6, minus 6. You're going to get 0 in the numerator, and you're going to get 0 in the denominator. So we don't have the function is not defined. So not defined at x is equal to 2. f not defined. So there's no simple thing there. Even if this did evaluate, if it was a continuous function, then actually the limit would be whatever the function is, but that doesn't necessarily mean the case. But we see very clearly that the function is not defined here."}, {"video_title": "Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So not defined at x is equal to 2. f not defined. So there's no simple thing there. Even if this did evaluate, if it was a continuous function, then actually the limit would be whatever the function is, but that doesn't necessarily mean the case. But we see very clearly that the function is not defined here. So let's see if we can simplify this, and we'll also try to graph it in some way. So one thing that might have jumped out at your head is you might want to factor this expression on top. So if we want to rewrite this, we could rewrite the top expression, and this just goes back to your algebra 1, two numbers whose product is negative 6, whose sum is positive 3."}, {"video_title": "Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "But we see very clearly that the function is not defined here. So let's see if we can simplify this, and we'll also try to graph it in some way. So one thing that might have jumped out at your head is you might want to factor this expression on top. So if we want to rewrite this, we could rewrite the top expression, and this just goes back to your algebra 1, two numbers whose product is negative 6, whose sum is positive 3. Well, that could be positive 3 and negative 2. So this could be x plus 3 times x minus 2, all of that, over x minus 2. So as long as x does not equal 2, these two things will cancel out."}, {"video_title": "Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So if we want to rewrite this, we could rewrite the top expression, and this just goes back to your algebra 1, two numbers whose product is negative 6, whose sum is positive 3. Well, that could be positive 3 and negative 2. So this could be x plus 3 times x minus 2, all of that, over x minus 2. So as long as x does not equal 2, these two things will cancel out. So we could say this is equal to x plus 3 for all x's, except for x is equal to 2, as long as x does not equal 2. So that's another way of looking at it. Another way we could rewrite our f of x. I'm doing it in blue just to change the colors."}, {"video_title": "Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So as long as x does not equal 2, these two things will cancel out. So we could say this is equal to x plus 3 for all x's, except for x is equal to 2, as long as x does not equal 2. So that's another way of looking at it. Another way we could rewrite our f of x. I'm doing it in blue just to change the colors. We could rewrite f of x. This is the exact same function. f of x is equal to x plus 3 when x does not equal 2."}, {"video_title": "Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Another way we could rewrite our f of x. I'm doing it in blue just to change the colors. We could rewrite f of x. This is the exact same function. f of x is equal to x plus 3 when x does not equal 2. And we could even say it's undefined when x is equal to 2. So given this definition, it becomes much clearer to us of how we can actually graph f of x. So let's try to do it."}, {"video_title": "Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "f of x is equal to x plus 3 when x does not equal 2. And we could even say it's undefined when x is equal to 2. So given this definition, it becomes much clearer to us of how we can actually graph f of x. So let's try to do it. So that is not anywhere near being a straight line. That is much better. So let's call this the y-axis."}, {"video_title": "Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So let's try to do it. So that is not anywhere near being a straight line. That is much better. So let's call this the y-axis. I'll call it y equals f of x. And then let's, over here, let me make a horizontal line. That is my x-axis."}, {"video_title": "Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So let's call this the y-axis. I'll call it y equals f of x. And then let's, over here, let me make a horizontal line. That is my x-axis. So defined this way, f of x is equal to x plus 3. So if this is 1, 2, 3, we have a y-intercept at 3, and then the slope is 1. The slope is 1."}, {"video_title": "Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "That is my x-axis. So defined this way, f of x is equal to x plus 3. So if this is 1, 2, 3, we have a y-intercept at 3, and then the slope is 1. The slope is 1. But it's defined for all x's, except for x is equal to 2. So this is x is equal to 1, x is equal to 2. So when x is equal to 2, it is undefined."}, {"video_title": "Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "The slope is 1. But it's defined for all x's, except for x is equal to 2. So this is x is equal to 1, x is equal to 2. So when x is equal to 2, it is undefined. So let me make sure I can. So it's undefined right over there. It's undefined right over there."}, {"video_title": "Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So when x is equal to 2, it is undefined. So let me make sure I can. So it's undefined right over there. It's undefined right over there. So this is what f of x looks like. Now, given this, let's try to answer our question. What is the limit of f of x as x approaches 2?"}, {"video_title": "Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "It's undefined right over there. So this is what f of x looks like. Now, given this, let's try to answer our question. What is the limit of f of x as x approaches 2? Well, we can look at this graphically. As x approaches 2 from lower values than 2, so this right over here is x is equal to 2. If we get to maybe, let's say this is 1.7, we see that our f of x is right over there."}, {"video_title": "Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "What is the limit of f of x as x approaches 2? Well, we can look at this graphically. As x approaches 2 from lower values than 2, so this right over here is x is equal to 2. If we get to maybe, let's say this is 1.7, we see that our f of x is right over there. If we get to 1.9, our f of x is right over there. So it seems to be approaching this value right over there. Similarly, as we approach 2 from values greater than it, if we're at, I don't know, this could be like 2.5, 2.5, our f of x is right over there."}, {"video_title": "Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "If we get to maybe, let's say this is 1.7, we see that our f of x is right over there. If we get to 1.9, our f of x is right over there. So it seems to be approaching this value right over there. Similarly, as we approach 2 from values greater than it, if we're at, I don't know, this could be like 2.5, 2.5, our f of x is right over there. If we get even closer to 2, our f of x is right over there. Once again, we look like we are approaching this value. Or another way of thinking about it, if we ride this line from the positive direction, we seem to be approaching this value for f of x."}, {"video_title": "Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Similarly, as we approach 2 from values greater than it, if we're at, I don't know, this could be like 2.5, 2.5, our f of x is right over there. If we get even closer to 2, our f of x is right over there. Once again, we look like we are approaching this value. Or another way of thinking about it, if we ride this line from the positive direction, we seem to be approaching this value for f of x. If we ride this line from the negative direction, from values less than 2, we seem to be approaching this value right over here. This is essentially the value of x plus 3 if we set x is equal to 2. So this is essentially going to be this value right over here is equal to 5."}, {"video_title": "Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Or another way of thinking about it, if we ride this line from the positive direction, we seem to be approaching this value for f of x. If we ride this line from the negative direction, from values less than 2, we seem to be approaching this value right over here. This is essentially the value of x plus 3 if we set x is equal to 2. So this is essentially going to be this value right over here is equal to 5. If we just look at it visually, if we just graphed a line with slope 1 with a y-intercept of 3, this value right over here is 5. Now, we can also try to do this numerically. So let's try to do that."}, {"video_title": "Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So this is essentially going to be this value right over here is equal to 5. If we just look at it visually, if we just graphed a line with slope 1 with a y-intercept of 3, this value right over here is 5. Now, we can also try to do this numerically. So let's try to do that. So if this is our function definition, completely identical to our original definition, then we just try values as x gets closer and closer to 2. So let's try values less than 2. So 1.9999, this is almost obvious, 1.9999 plus 3, well, that gets you pretty darn close to 5."}, {"video_title": "Limits by factoring Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So let's try to do that. So if this is our function definition, completely identical to our original definition, then we just try values as x gets closer and closer to 2. So let's try values less than 2. So 1.9999, this is almost obvious, 1.9999 plus 3, well, that gets you pretty darn close to 5. If I put even more 9s here, got even closer to 2, we'd get even closer to 5 here. If we approach 2 from the positive direction, and then we once again, we're getting closer and closer to 5 from the positive direction. If we were even closer to 2, we'd be even closer to 5."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "And we came up with an approximation. What I want to do in this video is generalize things a bit, using the exact same method, but doing it for an arbitrary function with arbitrary boundaries and an arbitrary number of rectangles. So let's do it. I'm going to draw the diagram as large as I can to make things clear, to make things as clear as possible. So that's my y-axis. And this right over here is my x-axis. Let me draw an arbitrary function."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "I'm going to draw the diagram as large as I can to make things clear, to make things as clear as possible. So that's my y-axis. And this right over here is my x-axis. Let me draw an arbitrary function. So let's say my function looks something like that. So that is y is equal to f of x. And let me define my boundaries."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "Let me draw an arbitrary function. So let's say my function looks something like that. So that is y is equal to f of x. And let me define my boundaries. So let's say this right over here is x equals a. This is x equals a. And this right over here is x."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "And let me define my boundaries. So let's say this right over here is x equals a. This is x equals a. And this right over here is x. This is x equals b. So this is b. And I'm going to use n rectangles."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "And this right over here is x. This is x equals b. So this is b. And I'm going to use n rectangles. n rectangles. And I'm going to use the function evaluated at the left boundary of the rectangle to determine its height. So, for example, this will be rectangle 1."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "And I'm going to use n rectangles. n rectangles. And I'm going to use the function evaluated at the left boundary of the rectangle to determine its height. So, for example, this will be rectangle 1. I'm going to evaluate what f of a is. I'm going to evaluate what f of a is. So this right over here is f of a."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "So, for example, this will be rectangle 1. I'm going to evaluate what f of a is. I'm going to evaluate what f of a is. So this right over here is f of a. And then I'm going to use that as the height of my first rectangle. The height of my first rectangle. So just like that."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "So this right over here is f of a. And then I'm going to use that as the height of my first rectangle. The height of my first rectangle. So just like that. So rectangle number 1 looks like this. And I'll even number it. Rectangle 1 looks just like that."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "So just like that. So rectangle number 1 looks like this. And I'll even number it. Rectangle 1 looks just like that. And just to have a convention here, because I'm going to want to label each of the boundaries of the left boundary. Each of the x values of the left boundary. So we'll say a is equal to x naught."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "Rectangle 1 looks just like that. And just to have a convention here, because I'm going to want to label each of the boundaries of the left boundary. Each of the x values of the left boundary. So we'll say a is equal to x naught. a is equal to x naught. So we can also call this point right over here x naught. That x value."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "So we'll say a is equal to x naught. a is equal to x naught. So we can also call this point right over here x naught. That x value. And then we go to the next rectangle. And we can call this one right over here. This x value, we'll call it x1."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "That x value. And then we go to the next rectangle. And we can call this one right over here. This x value, we'll call it x1. It's the left boundary of the next rectangle. If we evaluate f of x1, we get this value right over here. This right over here is f of x1."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "This x value, we'll call it x1. It's the left boundary of the next rectangle. If we evaluate f of x1, we get this value right over here. This right over here is f of x1. So it tells us our height. And we want an equal width to the previous one. And we'll think about what the width is going to be in a second."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "This right over here is f of x1. So it tells us our height. And we want an equal width to the previous one. And we'll think about what the width is going to be in a second. So this right over here is our second rectangle. This right over here is our second rectangle that we're going to use to approximate the area under the curve. That's rectangle number 2."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "And we'll think about what the width is going to be in a second. So this right over here is our second rectangle. This right over here is our second rectangle that we're going to use to approximate the area under the curve. That's rectangle number 2. Let's do rectangle number 3. Well, rectangle number 3, the left boundary, we're just going to call it x sub 2. And its height is going to be f of x sub 2."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "That's rectangle number 2. Let's do rectangle number 3. Well, rectangle number 3, the left boundary, we're just going to call it x sub 2. And its height is going to be f of x sub 2. And its width is going to be the same width as the other ones. I'm just eyeballing it right over here. So this is rectangle number 3."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "And its height is going to be f of x sub 2. And its width is going to be the same width as the other ones. I'm just eyeballing it right over here. So this is rectangle number 3. And we're going to continue this process all the way until we get to rectangle number n. So this is the nth rectangle. The nth rectangle right over here. And what am I going to label this point right over here?"}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "So this is rectangle number 3. And we're going to continue this process all the way until we get to rectangle number n. So this is the nth rectangle. The nth rectangle right over here. And what am I going to label this point right over here? Well, we already see a pattern. The left boundary of the first rectangle is x sub 0. The left boundary of the second rectangle is x sub 1."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "And what am I going to label this point right over here? Well, we already see a pattern. The left boundary of the first rectangle is x sub 0. The left boundary of the second rectangle is x sub 1. The left boundary of the third rectangle is x sub 2. So the left boundary of the nth rectangle is going to be x sub n minus 1. Whatever the rectangle number is, the left boundary is x sub that number minus 1."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "The left boundary of the second rectangle is x sub 1. The left boundary of the third rectangle is x sub 2. So the left boundary of the nth rectangle is going to be x sub n minus 1. Whatever the rectangle number is, the left boundary is x sub that number minus 1. And this is just based on the convention that we've defined. Now the next thing that we need to do in order to actually calculate this area is think about what is the width. So let's call the width of any of these rectangles."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "Whatever the rectangle number is, the left boundary is x sub that number minus 1. And this is just based on the convention that we've defined. Now the next thing that we need to do in order to actually calculate this area is think about what is the width. So let's call the width of any of these rectangles. And for these purposes, or the purpose of this example, I'm going to assume that it's constant. Although you can do these sums where you actually vary the width of the rectangle, but then it gets a little bit fancier. So I want an equal width."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "So let's call the width of any of these rectangles. And for these purposes, or the purpose of this example, I'm going to assume that it's constant. Although you can do these sums where you actually vary the width of the rectangle, but then it gets a little bit fancier. So I want an equal width. So I want delta x to be equal width. And to think about what that has to be, we just have to think, what's the total width that we're covering? Well, the total distance here is going to be b minus a."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "So I want an equal width. So I want delta x to be equal width. And to think about what that has to be, we just have to think, what's the total width that we're covering? Well, the total distance here is going to be b minus a. And we're just going to divide by the number of rectangles that we want, the number of sections that we want. So we want to divide by n. So if we assume this is true, and then we assume that a is equal to x0, and then x1 is equal to x0 plus delta x, x2 is equal to x1 plus delta x, and we go all the way to xn is equal to xn minus 1 plus delta x, then we've essentially set up this diagram right over here. b is actually going to be equal to xn."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "Well, the total distance here is going to be b minus a. And we're just going to divide by the number of rectangles that we want, the number of sections that we want. So we want to divide by n. So if we assume this is true, and then we assume that a is equal to x0, and then x1 is equal to x0 plus delta x, x2 is equal to x1 plus delta x, and we go all the way to xn is equal to xn minus 1 plus delta x, then we've essentially set up this diagram right over here. b is actually going to be equal to xn. So this is xn. It's equal to xn minus 1 plus delta x. So now I think we've set up all of the notation and all of the conventions in order to actually calculate the area or our approximation of the area."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "b is actually going to be equal to xn. So this is xn. It's equal to xn minus 1 plus delta x. So now I think we've set up all of the notation and all of the conventions in order to actually calculate the area or our approximation of the area. So our approximation, approximate area, is going to be equal to what? Well, it's going to be the area of the first rectangle. So let me write this down."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "So now I think we've set up all of the notation and all of the conventions in order to actually calculate the area or our approximation of the area. So our approximation, approximate area, is going to be equal to what? Well, it's going to be the area of the first rectangle. So let me write this down. So it's going to be rectangle 1, so the area of rectangle 1, so rectangle 1 plus the area of rectangle 2, plus the area of rectangle 2, plus the area of rectangle 3. I think you get the point here. Area of rectangle 3 all the way, plus all the way to the area of rectangle n. And so what are these going to be?"}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "So let me write this down. So it's going to be rectangle 1, so the area of rectangle 1, so rectangle 1 plus the area of rectangle 2, plus the area of rectangle 2, plus the area of rectangle 3. I think you get the point here. Area of rectangle 3 all the way, plus all the way to the area of rectangle n. And so what are these going to be? Rectangle 1 is going to be its height, which is f of x0 or f of a. Either way, x0 and a are the same thing. So it's f of a times our delta x, times our width, our height times our width."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "Area of rectangle 3 all the way, plus all the way to the area of rectangle n. And so what are these going to be? Rectangle 1 is going to be its height, which is f of x0 or f of a. Either way, x0 and a are the same thing. So it's f of a times our delta x, times our width, our height times our width. So times delta, actually, let me write f of x0. I wanted to write f of x0 times delta x. What is our height of rectangle 2?"}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "So it's f of a times our delta x, times our width, our height times our width. So times delta, actually, let me write f of x0. I wanted to write f of x0 times delta x. What is our height of rectangle 2? It's f of x1 times delta x. f of x1 times delta x. What's our area of rectangle 3? It's f of x2 times delta x."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "What is our height of rectangle 2? It's f of x1 times delta x. f of x1 times delta x. What's our area of rectangle 3? It's f of x2 times delta x. And then we go all the way to our area. We're taking all the sums all the way to rectangle n. What's its area? It's f of x sub n minus 1."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "It's f of x2 times delta x. And then we go all the way to our area. We're taking all the sums all the way to rectangle n. What's its area? It's f of x sub n minus 1. Actually, that's a different shade of orange. We'll use that same shade. It is f of x sub n minus 1 times delta x."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "It's f of x sub n minus 1. Actually, that's a different shade of orange. We'll use that same shade. It is f of x sub n minus 1 times delta x. And we're done. We've written it in a very general way. But to really make us comfortable with the various forms of notation, especially the types of notation you might see when people are talking about approximating areas or sums in general, I'm going to use the traditional sigma notation."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "It is f of x sub n minus 1 times delta x. And we're done. We've written it in a very general way. But to really make us comfortable with the various forms of notation, especially the types of notation you might see when people are talking about approximating areas or sums in general, I'm going to use the traditional sigma notation. So another way we could write this as the sum. This is equal to the sum from, and remember, this is just based on the conventions that I set up. I'll let i count which rectangle we're in, from i equals 1 to n. And then we're going to look at each rectangle."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "But to really make us comfortable with the various forms of notation, especially the types of notation you might see when people are talking about approximating areas or sums in general, I'm going to use the traditional sigma notation. So another way we could write this as the sum. This is equal to the sum from, and remember, this is just based on the conventions that I set up. I'll let i count which rectangle we're in, from i equals 1 to n. And then we're going to look at each rectangle. So the first rectangle, that's rectangle 1. So it's going to be f of, well, if we're in the i-th rectangle, then we're going to take the left boundary is going to be x sub i minus 1 times delta x. And so right over here is a general way of considering, of thinking about approximating the area under a curve using rectangles where the height of the rectangles are defined by the left boundary."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "I'll let i count which rectangle we're in, from i equals 1 to n. And then we're going to look at each rectangle. So the first rectangle, that's rectangle 1. So it's going to be f of, well, if we're in the i-th rectangle, then we're going to take the left boundary is going to be x sub i minus 1 times delta x. And so right over here is a general way of considering, of thinking about approximating the area under a curve using rectangles where the height of the rectangles are defined by the left boundary. And this tells us it's the left boundary. And we see for each, if this is the i-th rectangle right over here, if this is rectangle i, then this right over here is x sub i minus 1, and this height right over here is f of x sub i minus 1. So that's all we did right over there, times delta x."}, {"video_title": "Riemann sums in summation notation Accumulation and Riemann sums AP Calculus AB Khan Academy (2).mp3", "Sentence": "And so right over here is a general way of considering, of thinking about approximating the area under a curve using rectangles where the height of the rectangles are defined by the left boundary. And this tells us it's the left boundary. And we see for each, if this is the i-th rectangle right over here, if this is rectangle i, then this right over here is x sub i minus 1, and this height right over here is f of x sub i minus 1. So that's all we did right over there, times delta x. And then you sum all of these from the first rectangle all the way to the n-th. So hopefully that makes you a little bit more comfortable with this notation. We're not doing anything different than we did in this first video, which was hopefully fairly straightforward for you."}, {"video_title": "Worked example arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "And what I want to do is find the arc length of this curve from when x equals zero to when x is equal to, and I'm going to pick a strange number here, and I picked the strange number because it makes the numbers work out very well, to x is equal to 32 over nine. 32 over nine is, let's see, that's 3 5 9ths. So it's going to be right around, so that's 3 1 half, so it's going to be a little bit past 3 1 half, so it's going to be like right over there. So we want to find this arc length right over here, this thing that I have depicted in yellow. So it's from zero to 32 over nine. And I encourage you to pause the video and try this out on your own. So I'm assuming you've had a go at it, and if at any point while I'm working through it you feel inspired, always feel free to pause the video and continue working with it."}, {"video_title": "Worked example arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So we want to find this arc length right over here, this thing that I have depicted in yellow. So it's from zero to 32 over nine. And I encourage you to pause the video and try this out on your own. So I'm assuming you've had a go at it, and if at any point while I'm working through it you feel inspired, always feel free to pause the video and continue working with it. So let's just apply the arc length formula that we got kind of a conceptual proof for in the previous video. So we know that the arc length, let me write this, the arc length is going to be equal to the definite integral from zero to 32 over nine of the square root, actually let me just write it in general terms first so that you can kind of see the formula and then how we apply it. So it's the square root of one plus f prime of x squared dx."}, {"video_title": "Worked example arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So I'm assuming you've had a go at it, and if at any point while I'm working through it you feel inspired, always feel free to pause the video and continue working with it. So let's just apply the arc length formula that we got kind of a conceptual proof for in the previous video. So we know that the arc length, let me write this, the arc length is going to be equal to the definite integral from zero to 32 over nine of the square root, actually let me just write it in general terms first so that you can kind of see the formula and then how we apply it. So it's the square root of one plus f prime of x squared dx. And in this case, that's going to be, it's going to be the definite integral from zero to 32 over nine of, I should say, the square root of one plus, now what's the derivative? If f of x is x to the three halves, then f prime of x is going to be three halves x to the one half. And I picked this particular function because it simplifies quite well when we put it under the radical."}, {"video_title": "Worked example arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So it's the square root of one plus f prime of x squared dx. And in this case, that's going to be, it's going to be the definite integral from zero to 32 over nine of, I should say, the square root of one plus, now what's the derivative? If f of x is x to the three halves, then f prime of x is going to be three halves x to the one half. And I picked this particular function because it simplifies quite well when we put it under the radical. It's fairly straightforward to find the antiderivative. So we've done a lot of engineering of this problem to make the numbers work out well. But let's just go through it."}, {"video_title": "Worked example arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "And I picked this particular function because it simplifies quite well when we put it under the radical. It's fairly straightforward to find the antiderivative. So we've done a lot of engineering of this problem to make the numbers work out well. But let's just go through it. So this is f prime of x. f prime of x squared is going to be this quantity squared. It's going to be nine over four. x to the one half squared is x."}, {"video_title": "Worked example arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "But let's just go through it. So this is f prime of x. f prime of x squared is going to be this quantity squared. It's going to be nine over four. x to the one half squared is x. So one plus nine fourths x dx. And so now we just have a definite integral that we know how to solve this type of thing. And you might be able to even do this in your head, essentially do the u substitution, say, okay, I have one plus nine fourths x."}, {"video_title": "Worked example arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "x to the one half squared is x. So one plus nine fourths x dx. And so now we just have a definite integral that we know how to solve this type of thing. And you might be able to even do this in your head, essentially do the u substitution, say, okay, I have one plus nine fourths x. Its derivative is nine fourths. I can kind of engineer that if I want. But instead, I'm just going to do straight up u substitution."}, {"video_title": "Worked example arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "And you might be able to even do this in your head, essentially do the u substitution, say, okay, I have one plus nine fourths x. Its derivative is nine fourths. I can kind of engineer that if I want. But instead, I'm just going to do straight up u substitution. So u substitution, so if I say u is equal to one plus nine over four x, then we know, let's see, du dx would be du dx is going to be equal to nine fourths. Or we could say du is equal to nine fourths dx. Or we could say dx, or we could say, let me scroll down a little bit."}, {"video_title": "Worked example arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "But instead, I'm just going to do straight up u substitution. So u substitution, so if I say u is equal to one plus nine over four x, then we know, let's see, du dx would be du dx is going to be equal to nine fourths. Or we could say du is equal to nine fourths dx. Or we could say dx, or we could say, let me scroll down a little bit. We could say dx is equal to, I'm just going to multiply both sides times four ninths, is equal to four ninths du. And then we just have to change the bounds of integration. When x is equal to zero, then u is going to be equal to, nine fourths times zero is zero, so u is going to be equal to one."}, {"video_title": "Worked example arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "Or we could say dx, or we could say, let me scroll down a little bit. We could say dx is equal to, I'm just going to multiply both sides times four ninths, is equal to four ninths du. And then we just have to change the bounds of integration. When x is equal to zero, then u is going to be equal to, nine fourths times zero is zero, so u is going to be equal to one. And when x is equal to 32 over nine, 32 over nine, and this is why that number was picked, what's u going to be equal to? 32 over nine times nine fourths is going to be 32 over four, which is going to be eight plus one. So that worked out very nicely, imagine that."}, {"video_title": "Worked example arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "When x is equal to zero, then u is going to be equal to, nine fourths times zero is zero, so u is going to be equal to one. And when x is equal to 32 over nine, 32 over nine, and this is why that number was picked, what's u going to be equal to? 32 over nine times nine fourths is going to be 32 over four, which is going to be eight plus one. So that worked out very nicely, imagine that. So there we have it. So this is going to be equal to, this is going to be equal to the definite integral, actually let me make it clear, this is what is equal to this. The definite integral from u is equal to, u is equal to one to u is equal to nine, I'm going to make it very explicit that I'm dealing with u now, of the square root of u, instead of dx we have dx is four ninths du."}, {"video_title": "Worked example arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "So that worked out very nicely, imagine that. So there we have it. So this is going to be equal to, this is going to be equal to the definite integral, actually let me make it clear, this is what is equal to this. The definite integral from u is equal to, u is equal to one to u is equal to nine, I'm going to make it very explicit that I'm dealing with u now, of the square root of u, instead of dx we have dx is four ninths du. Let me do it this way. Square root, whoops, that's not the right color. Square root of u, instead of dx, we have times four ninths du, and I'm just going to take the four ninths and stick it out here."}, {"video_title": "Worked example arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "The definite integral from u is equal to, u is equal to one to u is equal to nine, I'm going to make it very explicit that I'm dealing with u now, of the square root of u, instead of dx we have dx is four ninths du. Let me do it this way. Square root, whoops, that's not the right color. Square root of u, instead of dx, we have times four ninths du, and I'm just going to take the four ninths and stick it out here. Four ninths du, and we know how to apply the fundamental, or the second fundamental theorem of calculus here to evaluate this definite integral. This is going to be four ninths times the antiderivative of u, of the square root of u, which is the same thing as u to the one half, is going to be u to the three halves, and then we divide by three halves, which is the same thing as multiplying by two thirds, and we're going to evaluate that at u equals nine, and u is equal to one. We're in the home stretch here."}, {"video_title": "Worked example arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "Square root of u, instead of dx, we have times four ninths du, and I'm just going to take the four ninths and stick it out here. Four ninths du, and we know how to apply the fundamental, or the second fundamental theorem of calculus here to evaluate this definite integral. This is going to be four ninths times the antiderivative of u, of the square root of u, which is the same thing as u to the one half, is going to be u to the three halves, and then we divide by three halves, which is the same thing as multiplying by two thirds, and we're going to evaluate that at u equals nine, and u is equal to one. We're in the home stretch here. This is going to be equal to four ninths times, let's see, two thirds times nine to the three halves, minus two thirds times one to the three half. Nine to the three halves, that is, let's see, the square root of nine is three to the third power is 27, and this, of course, is one. We are left with two thirds of, actually, let's just factor out the two thirds."}, {"video_title": "Worked example arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "We're in the home stretch here. This is going to be equal to four ninths times, let's see, two thirds times nine to the three halves, minus two thirds times one to the three half. Nine to the three halves, that is, let's see, the square root of nine is three to the third power is 27, and this, of course, is one. We are left with two thirds of, actually, let's just factor out the two thirds. That makes it easier. We're going to be left with, this is going to be equal to two thirds times four ninths is equal to eight over 27. I've just factored out the two thirds, and then we're going to have 27 minus one inside, I guess you could say, the brackets now."}, {"video_title": "Worked example arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "We are left with two thirds of, actually, let's just factor out the two thirds. That makes it easier. We're going to be left with, this is going to be equal to two thirds times four ninths is equal to eight over 27. I've just factored out the two thirds, and then we're going to have 27 minus one inside, I guess you could say, the brackets now. 27 minus one is just going to be 26. It's going to be 26 times 26. We could obviously simplify this more if we want."}, {"video_title": "Worked example arc length Applications of definite integrals AP Calculus BC Khan Academy.mp3", "Sentence": "I've just factored out the two thirds, and then we're going to have 27 minus one inside, I guess you could say, the brackets now. 27 minus one is just going to be 26. It's going to be 26 times 26. We could obviously simplify this more if we want. We could do whatever eight times 26 is. Actually, let's just figure that out just for fun. Eight times 26 is going to be, this is going to be 160 plus eight times six is 48, so it's 208 over 27."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "That rate will increase as the population increases. And when you actually try to solve this differential equation you try to find an n of t that satisfies this, we found that an exponential would work. An exponential satisfies this differential equation. And it would look like this visually. It would look like this visually, where you're starting at a population of n naught. This is the time axis, this is the population axis. And as time increases, population increases exponentially."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "And it would look like this visually. It would look like this visually, where you're starting at a population of n naught. This is the time axis, this is the population axis. And as time increases, population increases exponentially. Now we said there's an issue there. What if Thomas Malthus is right? That the environment can't support, let's say that the environment can't support, let me do this in a new color, let's say that the environment really can't support more than k, more than a population, more than a population of k. Then clearly the population can't just go and go right through the ceiling."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "And as time increases, population increases exponentially. Now we said there's an issue there. What if Thomas Malthus is right? That the environment can't support, let's say that the environment can't support, let me do this in a new color, let's say that the environment really can't support more than k, more than a population, more than a population of k. Then clearly the population can't just go and go right through the ceiling. They're not going to be able to have food or water or resources or whatever it might be. They might generate too much pollution. Who knows what it might be."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "That the environment can't support, let's say that the environment can't support, let me do this in a new color, let's say that the environment really can't support more than k, more than a population, more than a population of k. Then clearly the population can't just go and go right through the ceiling. They're not going to be able to have food or water or resources or whatever it might be. They might generate too much pollution. Who knows what it might be. And so this first stab at modeling population doesn't quite do the trick, especially if you are kind of in Malthus' camp. And that's where P.F., and once again, I'm sure I'm mispronouncing the name, Verhulst, who is going to come into the picture, because he read Malthus' work and said, well, yeah, I think I can do a pretty good job of modeling the type of behavior that Malthus is talking about. And he says, you know, what we really want, what we really want is something, let me write it, so the rate, let's try to model, let's set up another differential equation."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "Who knows what it might be. And so this first stab at modeling population doesn't quite do the trick, especially if you are kind of in Malthus' camp. And that's where P.F., and once again, I'm sure I'm mispronouncing the name, Verhulst, who is going to come into the picture, because he read Malthus' work and said, well, yeah, I think I can do a pretty good job of modeling the type of behavior that Malthus is talking about. And he says, you know, what we really want, what we really want is something, let me write it, so the rate, let's try to model, let's set up another differential equation. Let's set up another differential equation. And now let's say, okay, instead of, you know, if N is substantially smaller than what the environment can support, then yeah, that makes sense to have exponential growth. That makes sense to have exponential growth."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "And he says, you know, what we really want, what we really want is something, let me write it, so the rate, let's try to model, let's set up another differential equation. Let's set up another differential equation. And now let's say, okay, instead of, you know, if N is substantially smaller than what the environment can support, then yeah, that makes sense to have exponential growth. That makes sense to have exponential growth. But maybe we can dampen this, or maybe we can bring this growth to zero as N approaches, as N approaches K. And so how can we actually modify this? Maybe we can multiply it by something that for when N is small, when N is much smaller than K, this term right over here is going to be close to one. And when N is close to K, this term is close to zero."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "That makes sense to have exponential growth. But maybe we can dampen this, or maybe we can bring this growth to zero as N approaches, as N approaches K. And so how can we actually modify this? Maybe we can multiply it by something that for when N is small, when N is much smaller than K, this term right over here is going to be close to one. And when N is close to K, this term is close to zero. So let me write that. When N, so these are our goals for this term right over here. When N is much smaller, so much smaller, much smaller than K, so now the population is not constrained at all."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "And when N is close to K, this term is close to zero. So let me write that. When N, so these are our goals for this term right over here. When N is much smaller, so much smaller, much smaller than K, so now the population is not constrained at all. People can have babies, and those babies can be fed, and then they can have babies, et cetera, et cetera. Then this thing should be close to one. And so then you have essentially our old model."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "When N is much smaller, so much smaller, much smaller than K, so now the population is not constrained at all. People can have babies, and those babies can be fed, and then they can have babies, et cetera, et cetera. Then this thing should be close to one. And so then you have essentially our old model. But then as N approaches K, when N, as N approaches K, then this thing should approach, then this term, this term or this expression, should approach zero. And what that does is as N approaches the natural limit, the ceiling to population, then no matter what this is doing, if this thing is approaching zero, that's going to make the actual rate of growth approach zero. So food is going to be more scarce."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "And so then you have essentially our old model. But then as N approaches K, when N, as N approaches K, then this thing should approach, then this term, this term or this expression, should approach zero. And what that does is as N approaches the natural limit, the ceiling to population, then no matter what this is doing, if this thing is approaching zero, that's going to make the actual rate of growth approach zero. So food is going to be more scarce. It's going to be harder to find things. And so what can I construct here dealing with N and K that will have these properties? And for fun, you might actually want to pause the video and see if you can construct a fairly simple algebraic statement using N and K, and maybe the number one if you find the need, for an expression that has these properties."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "So food is going to be more scarce. It's going to be harder to find things. And so what can I construct here dealing with N and K that will have these properties? And for fun, you might actually want to pause the video and see if you can construct a fairly simple algebraic statement using N and K, and maybe the number one if you find the need, for an expression that has these properties. Well, let's see. What if we start with a one, we start with a one, and we subtract N over K? We subtract N over, my K is in pink, over K. Does this have those properties?"}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "And for fun, you might actually want to pause the video and see if you can construct a fairly simple algebraic statement using N and K, and maybe the number one if you find the need, for an expression that has these properties. Well, let's see. What if we start with a one, we start with a one, and we subtract N over K? We subtract N over, my K is in pink, over K. Does this have those properties? Well, yeah, sure it does. When N is really small, or I should say when it's a small fraction of K, then one mind, this is going to be a small fraction, then this whole thing is going to be pretty close to one. It's going to be a little bit less than one."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "We subtract N over, my K is in pink, over K. Does this have those properties? Well, yeah, sure it does. When N is really small, or I should say when it's a small fraction of K, then one mind, this is going to be a small fraction, then this whole thing is going to be pretty close to one. It's going to be a little bit less than one. And when N approaches K, as N gets closer and closer and closer to K, then this thing right over here is going to approach one, which means this whole expression is going to approach zero, which is exactly what we wanted. And this thing right over here is actually, and this is used in tons of applications, not just in population modeling, but that's kind of one of its first reasons or motivations. This differential equation right over here is actually quite famous."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "It's going to be a little bit less than one. And when N approaches K, as N gets closer and closer and closer to K, then this thing right over here is going to approach one, which means this whole expression is going to approach zero, which is exactly what we wanted. And this thing right over here is actually, and this is used in tons of applications, not just in population modeling, but that's kind of one of its first reasons or motivations. This differential equation right over here is actually quite famous. It's called the logistic differential equation. Logistic differential equation. Logistic differential equation."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "This differential equation right over here is actually quite famous. It's called the logistic differential equation. Logistic differential equation. Logistic differential equation. And in the next video, we're actually going to solve this. And this is a separable differential equation. You can actually solve it just using standard techniques of integration."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "Logistic differential equation. And in the next video, we're actually going to solve this. And this is a separable differential equation. You can actually solve it just using standard techniques of integration. It's a little bit hairier than this one, so we're going to work through it together. And we're going to look at the solution. The solution to the logistic differential equation is a logistic function, which, once again, is really, essentially models population in this way."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "You can actually solve it just using standard techniques of integration. It's a little bit hairier than this one, so we're going to work through it together. And we're going to look at the solution. The solution to the logistic differential equation is a logistic function, which, once again, is really, essentially models population in this way. But before we actually solve for it, let's just try to interpret this differential equation and think about what the shape of this function might look like. And to do that, actually, let me, it's nice to see the faces. So let me draw some axes here."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "The solution to the logistic differential equation is a logistic function, which, once again, is really, essentially models population in this way. But before we actually solve for it, let's just try to interpret this differential equation and think about what the shape of this function might look like. And to do that, actually, let me, it's nice to see the faces. So let me draw some axes here. Let me draw some axes here. So that's my time axis. That is my population axis."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "So let me draw some axes here. Let me draw some axes here. So that's my time axis. That is my population axis. Let me scroll up a little bit, because sometimes the subtitles show up around here, and then people can't see what's going on. So let's think about a couple of permutations, a couple of situations. So if our initial, if our initial, if our n at time equals zero, remember, n is a function of t. If at time equals zero, n is equal to zero."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "That is my population axis. Let me scroll up a little bit, because sometimes the subtitles show up around here, and then people can't see what's going on. So let's think about a couple of permutations, a couple of situations. So if our initial, if our initial, if our n at time equals zero, remember, n is a function of t. If at time equals zero, n is equal to zero. So if n is equal to zero, then this term is going to be zero, and then your rate of change is going to be zero, and so you're not going to add any population. And that's good, because if your population is zero, how are you going to actually be able to add population? There's no one there to have children."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "So if our initial, if our initial, if our n at time equals zero, remember, n is a function of t. If at time equals zero, n is equal to zero. So if n is equal to zero, then this term is going to be zero, and then your rate of change is going to be zero, and so you're not going to add any population. And that's good, because if your population is zero, how are you going to actually be able to add population? There's no one there to have children. So there's actually one constant solution to this differential equation, which is just n of t, that is n of t, is equal to zero. And that's neat that this satisfies the logistic differential equation. Hey, if your population starts at zero, if n sub naught is zero, then you're just going to be at zero forever."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "There's no one there to have children. So there's actually one constant solution to this differential equation, which is just n of t, that is n of t, is equal to zero. And that's neat that this satisfies the logistic differential equation. Hey, if your population starts at zero, if n sub naught is zero, then you're just going to be at zero forever. Well, that's actually what would happen in real life. There's no one there to have kids. Now let's think about another situation."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "Hey, if your population starts at zero, if n sub naught is zero, then you're just going to be at zero forever. Well, that's actually what would happen in real life. There's no one there to have kids. Now let's think about another situation. What if our population, what if n naught is equal to k? What happens if n naught is equal to, what happens if n naught is equal, so that's k right over there. What happens if at time equals zero, this is our population?"}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "Now let's think about another situation. What if our population, what if n naught is equal to k? What happens if n naught is equal to, what happens if n naught is equal, so that's k right over there. What happens if at time equals zero, this is our population? Well, if n is equal to k, then this is one minus one, then this thing is zero, and so our rate of population change is going to be zero. So essentially, if my population is zero, then after a little bit of time, my population is still the same k. If my rate of change of population is zero, that means my population is staying constant. And so my population is just going to stay there at k. And that's actually believable."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "What happens if at time equals zero, this is our population? Well, if n is equal to k, then this is one minus one, then this thing is zero, and so our rate of population change is going to be zero. So essentially, if my population is zero, then after a little bit of time, my population is still the same k. If my rate of change of population is zero, that means my population is staying constant. And so my population is just going to stay there at k. And that's actually believable. Malthus would actually probably say that you're going to have, maybe it grows a little bit beyond the capacity of the environment, and then you have some flood, or some hurricane, or some famine, and it goes around. But for our purposes, you can never model anything perfectly, for our purposes, that's pretty good. You're at the limit of what the environment can handle, so you just kind of stay there."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "And so my population is just going to stay there at k. And that's actually believable. Malthus would actually probably say that you're going to have, maybe it grows a little bit beyond the capacity of the environment, and then you have some flood, or some hurricane, or some famine, and it goes around. But for our purposes, you can never model anything perfectly, for our purposes, that's pretty good. You're at the limit of what the environment can handle, so you just kind of stay there. So that's actually another constant solution, that n of t, if it starts, and now you can kind of appreciate why initial conditions are important. If you start at zero, you're going to stay at zero. If you start at k, you're going to stay at k. So that is n of t just stays at k. But now let's think of a more interesting scenario."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "You're at the limit of what the environment can handle, so you just kind of stay there. So that's actually another constant solution, that n of t, if it starts, and now you can kind of appreciate why initial conditions are important. If you start at zero, you're going to stay at zero. If you start at k, you're going to stay at k. So that is n of t just stays at k. But now let's think of a more interesting scenario. Let's assume an initial population that's someplace between zero and k. So this is going to be, I'm going to assume initial population that is someplace, it's greater than zero, so there are people to actually have children, and it is less than k, so we aren't fully maxing out the environment, or the land, or the food, or the water, or whatever it might be. So what's going to happen? So, and once again, I'm just going to kind of sketch it, and then we're going to actually solve it in the next video."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "If you start at k, you're going to stay at k. So that is n of t just stays at k. But now let's think of a more interesting scenario. Let's assume an initial population that's someplace between zero and k. So this is going to be, I'm going to assume initial population that is someplace, it's greater than zero, so there are people to actually have children, and it is less than k, so we aren't fully maxing out the environment, or the land, or the food, or the water, or whatever it might be. So what's going to happen? So, and once again, I'm just going to kind of sketch it, and then we're going to actually solve it in the next video. So when n is a lot less than k, when it's a small fraction of k, you're going to, it's going to, you know, this term is going to be the main one that's influencing it, because this is a small fraction of k. I mean, even the way I drew it, it looks like it's about a, I don't know, it looks like it's about a sixth, or a seventh, or an eighth of k. So it's one minus one eighth. So it's seven eighths. It's going to be seven eighths times this."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "So, and once again, I'm just going to kind of sketch it, and then we're going to actually solve it in the next video. So when n is a lot less than k, when it's a small fraction of k, you're going to, it's going to, you know, this term is going to be the main one that's influencing it, because this is a small fraction of k. I mean, even the way I drew it, it looks like it's about a, I don't know, it looks like it's about a sixth, or a seventh, or an eighth of k. So it's one minus one eighth. So it's seven eighths. It's going to be seven eighths times this. So really, this is what's going to dictate, this is what's going to dictate what our rate of growth is. And if this is kind of dictating it, we're kind of looking more of a, well, let's just think of it this way. As the population grows, the rate of change is going to grow."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "It's going to be seven eighths times this. So really, this is what's going to dictate, this is what's going to dictate what our rate of growth is. And if this is kind of dictating it, we're kind of looking more of a, well, let's just think of it this way. As the population grows, the rate of change is going to grow. So it's going to look, it's going to look something, it's going to look something like this. As our population gets larger, our slope is getting higher, and it's getting steeper and steeper. But then as n approaches k, then this thing is going to become, this is going to be close to one minus, close to one, and so this is going to become a very small number."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "As the population grows, the rate of change is going to grow. So it's going to look, it's going to look something, it's going to look something like this. As our population gets larger, our slope is getting higher, and it's getting steeper and steeper. But then as n approaches k, then this thing is going to become, this is going to be close to one minus, close to one, and so this is going to become a very small number. It's going to make this whole thing approach zero. So as n approaches k, the whole thing, the rate of change, is going to flatten out, and we're going to asymptote towards k. And so the solution, the solution to the logistic differential equation should look something like this, depending on what your initial conditions are. If your initial condition is here, maybe it does something like this."}, {"video_title": "Logistic differential equation intuition First order differential equations Khan Academy.mp3", "Sentence": "But then as n approaches k, then this thing is going to become, this is going to be close to one minus, close to one, and so this is going to become a very small number. It's going to make this whole thing approach zero. So as n approaches k, the whole thing, the rate of change, is going to flatten out, and we're going to asymptote towards k. And so the solution, the solution to the logistic differential equation should look something like this, depending on what your initial conditions are. If your initial condition is here, maybe it does something like this. If your initial condition's here, maybe it does something like, something like this. And once again, and this is what's fun about differential equations, before even doing the fancy math, you can kind of get an intuition, just by thinking through the differential equation, of what it is likely, what it is likely to be. Down here, or when n is much smaller than k, its rate of increase is increasing as n increases, and over here, as n gets close to k, its rate of increase is decreasing."}, {"video_title": "Justification with the mean value theorem table AP Calculus AB Khan Academy.mp3", "Sentence": "Well, to use the mean value theorem, you have to be differentiable over the open interval and continuous over the closed interval. So it seems like we've met that because if you're differentiable over an interval, you're definitely continuous over that interval. I'm not saying that it's just a generally differentiable function f, I guess, over any interval. But the next part is to say, all right, that if that condition is met, then the slope of the secant line between four comma f of four and six comma f of six, that some, at least one point in between four and six, will have a derivative that is equal to the slope of the secant line. And so let's figure out what the slope of the secant line is between four comma f of four and six comma f of six. And if it's equal to five, then we could use the mean value theorem. If it's not equal to five, then the mean value theorem would not apply."}, {"video_title": "Justification with the mean value theorem table AP Calculus AB Khan Academy.mp3", "Sentence": "But the next part is to say, all right, that if that condition is met, then the slope of the secant line between four comma f of four and six comma f of six, that some, at least one point in between four and six, will have a derivative that is equal to the slope of the secant line. And so let's figure out what the slope of the secant line is between four comma f of four and six comma f of six. And if it's equal to five, then we could use the mean value theorem. If it's not equal to five, then the mean value theorem would not apply. And so let's do that. F of six minus f of four, all of that over six minus four is equal to seven minus three over two, which is equal to two. So two not equal to five, so mean value theorem doesn't apply."}, {"video_title": "Justification with the mean value theorem table AP Calculus AB Khan Academy.mp3", "Sentence": "If it's not equal to five, then the mean value theorem would not apply. And so let's do that. F of six minus f of four, all of that over six minus four is equal to seven minus three over two, which is equal to two. So two not equal to five, so mean value theorem doesn't apply. Apply. All right, let's, I'll put an exclamation mark there for emphasis. All right, let's do the next part."}, {"video_title": "Justification with the mean value theorem table AP Calculus AB Khan Academy.mp3", "Sentence": "So two not equal to five, so mean value theorem doesn't apply. Apply. All right, let's, I'll put an exclamation mark there for emphasis. All right, let's do the next part. Can we use the mean value theorem to say that the equation f prime of x is equal to negative one has a solution, and now the interval is from zero to two. If so, write a justification. All right, so let's see this."}, {"video_title": "Justification with the mean value theorem table AP Calculus AB Khan Academy.mp3", "Sentence": "All right, let's do the next part. Can we use the mean value theorem to say that the equation f prime of x is equal to negative one has a solution, and now the interval is from zero to two. If so, write a justification. All right, so let's see this. So if we were to take the slope of the secant line, so f of two minus f of zero, all that over two minus zero, this is equal to negative two minus zero. All of that over two, which is equal to negative two over two, which is equal to negative one. And so, and we also know that we meet the continuity and differentiability conditions."}, {"video_title": "Justification with the mean value theorem table AP Calculus AB Khan Academy.mp3", "Sentence": "All right, so let's see this. So if we were to take the slope of the secant line, so f of two minus f of zero, all that over two minus zero, this is equal to negative two minus zero. All of that over two, which is equal to negative two over two, which is equal to negative one. And so, and we also know that we meet the continuity and differentiability conditions. And so we could say, and since f is generally differentiable, generally differentiable, differentiable, it will be differentiable, differentiable, and continuous over the interval from zero to two. And I say the closed interval. You just have to be differentiable over the open interval, but it's even better, I guess, if you're differentiable over the closed interval, because you have to be continuous over the closed interval."}, {"video_title": "Justification with the mean value theorem table AP Calculus AB Khan Academy.mp3", "Sentence": "And so, and we also know that we meet the continuity and differentiability conditions. And so we could say, and since f is generally differentiable, generally differentiable, differentiable, it will be differentiable, differentiable, and continuous over the interval from zero to two. And I say the closed interval. You just have to be differentiable over the open interval, but it's even better, I guess, if you're differentiable over the closed interval, because you have to be continuous over the closed interval. And so, and since f is generally differentiable, it will be differentiable and continuous over zero, two. So the mean value theorem tells us, tells us, that there is an x in that interval from zero to two such that f prime of x is equal to that secant slope, or you could say that average rate of change is equal to negative one. And so I could write yes, yes, and then this would be my justification."}, {"video_title": "Justification with the mean value theorem table AP Calculus AB Khan Academy.mp3", "Sentence": "You just have to be differentiable over the open interval, but it's even better, I guess, if you're differentiable over the closed interval, because you have to be continuous over the closed interval. And so, and since f is generally differentiable, it will be differentiable and continuous over zero, two. So the mean value theorem tells us, tells us, that there is an x in that interval from zero to two such that f prime of x is equal to that secant slope, or you could say that average rate of change is equal to negative one. And so I could write yes, yes, and then this would be my justification. This is the slope of the secant line or the average rate of change. And since f is generally differentiable, it will be differentiable and continuous over the closed interval. So the mean value theorem tells us that there is an x in this interval such that f prime of x is equal to negative one."}, {"video_title": "Derivatives of inverse functions from equation AP Calculus AB Khan Academy.mp3", "Sentence": "Let h be the inverse of f. Notice that f of negative two is equal to negative 14. And then they're asking us, what is h prime of negative 14? And if you're not familiar with how functions and their derivatives relate to their inverses and the derivatives of the inverse, well this will seem like a very hard thing to do. Because if you were tempted to take the inverse of f to figure out what h is, well it's tough to figure out the inverse of a third degree polynomial defined function like this. So the key, I guess, property to realize or the key truth to realize is that if f and h are inverses, then h prime of x, h prime of x, is going to be equal to, it's going to be equal to one over f prime of h of x. One over f prime of h of x. And you could now use this in order to figure out what h prime of negative 14 is."}, {"video_title": "Derivatives of inverse functions from equation AP Calculus AB Khan Academy.mp3", "Sentence": "Because if you were tempted to take the inverse of f to figure out what h is, well it's tough to figure out the inverse of a third degree polynomial defined function like this. So the key, I guess, property to realize or the key truth to realize is that if f and h are inverses, then h prime of x, h prime of x, is going to be equal to, it's going to be equal to one over f prime of h of x. One over f prime of h of x. And you could now use this in order to figure out what h prime of negative 14 is. Now I know what some of you are thinking because it's exactly what I would be thinking if someone just sprung this on me, is where does this come from? And I would tell you this comes straight out of the chain rule. We know that if a function and its inverse, we know that if we have a function and its inverse, that f of the inverse of our function, so f of h of x, f of h of x, we know that this is going to be equal to x."}, {"video_title": "Derivatives of inverse functions from equation AP Calculus AB Khan Academy.mp3", "Sentence": "And you could now use this in order to figure out what h prime of negative 14 is. Now I know what some of you are thinking because it's exactly what I would be thinking if someone just sprung this on me, is where does this come from? And I would tell you this comes straight out of the chain rule. We know that if a function and its inverse, we know that if we have a function and its inverse, that f of the inverse of our function, so f of h of x, f of h of x, we know that this is going to be equal to x. This literally, this comes out of them being each other's inverses. We could have also said h of f of x will also be equal to x. Remember, f is going to map, or h is going to map from some x to h of x, and then f is going to map back to that original x."}, {"video_title": "Derivatives of inverse functions from equation AP Calculus AB Khan Academy.mp3", "Sentence": "We know that if a function and its inverse, we know that if we have a function and its inverse, that f of the inverse of our function, so f of h of x, f of h of x, we know that this is going to be equal to x. This literally, this comes out of them being each other's inverses. We could have also said h of f of x will also be equal to x. Remember, f is going to map, or h is going to map from some x to h of x, and then f is going to map back to that original x. That's what inverses do. So that's because they are inverses. This is by definition, this is what inverses do to each other."}, {"video_title": "Derivatives of inverse functions from equation AP Calculus AB Khan Academy.mp3", "Sentence": "Remember, f is going to map, or h is going to map from some x to h of x, and then f is going to map back to that original x. That's what inverses do. So that's because they are inverses. This is by definition, this is what inverses do to each other. But then if you took the derivative of both sides of this, what would you get? Let me do that. So if we take the derivative of both sides of this, d dx on the left-hand side, d dx on the right-hand side, and I think you see where this is going."}, {"video_title": "Derivatives of inverse functions from equation AP Calculus AB Khan Academy.mp3", "Sentence": "This is by definition, this is what inverses do to each other. But then if you took the derivative of both sides of this, what would you get? Let me do that. So if we take the derivative of both sides of this, d dx on the left-hand side, d dx on the right-hand side, and I think you see where this is going. You're essentially going to get a version of that. The left-hand side, use the chain rule. You're going to get f prime of h of x, f prime of h of x times h prime of x, comes straight out of the chain rule, is equal to, is equal to the derivative of x, is just going to be equal to one."}, {"video_title": "Derivatives of inverse functions from equation AP Calculus AB Khan Academy.mp3", "Sentence": "So if we take the derivative of both sides of this, d dx on the left-hand side, d dx on the right-hand side, and I think you see where this is going. You're essentially going to get a version of that. The left-hand side, use the chain rule. You're going to get f prime of h of x, f prime of h of x times h prime of x, comes straight out of the chain rule, is equal to, is equal to the derivative of x, is just going to be equal to one. And then you divide both sides by f prime of h of x, and you get our original property there. So now with that out of the way, let's just actually apply this. So we want to evaluate h prime of 14."}, {"video_title": "Derivatives of inverse functions from equation AP Calculus AB Khan Academy.mp3", "Sentence": "You're going to get f prime of h of x, f prime of h of x times h prime of x, comes straight out of the chain rule, is equal to, is equal to the derivative of x, is just going to be equal to one. And then you divide both sides by f prime of h of x, and you get our original property there. So now with that out of the way, let's just actually apply this. So we want to evaluate h prime of 14. Or sorry, h prime of negative 14 is going to be equal to one over f prime of h of negative 14. H of negative 14. Now have they given us h of negative 14?"}, {"video_title": "Derivatives of inverse functions from equation AP Calculus AB Khan Academy.mp3", "Sentence": "So we want to evaluate h prime of 14. Or sorry, h prime of negative 14 is going to be equal to one over f prime of h of negative 14. H of negative 14. Now have they given us h of negative 14? Well, they didn't give it to us explicitly, but we have to remember that f and h are inverses of each other. So if f of negative two is negative 14, well, h is gonna go from the other way around. If you input negative 14 into h, you're going to get negative two."}, {"video_title": "Derivatives of inverse functions from equation AP Calculus AB Khan Academy.mp3", "Sentence": "Now have they given us h of negative 14? Well, they didn't give it to us explicitly, but we have to remember that f and h are inverses of each other. So if f of negative two is negative 14, well, h is gonna go from the other way around. If you input negative 14 into h, you're going to get negative two. So h of negative 14, well, this is going to be equal to negative two. Once again, they are inverses of each other. So h of negative 14 is equal to negative negative two."}, {"video_title": "Derivatives of inverse functions from equation AP Calculus AB Khan Academy.mp3", "Sentence": "If you input negative 14 into h, you're going to get negative two. So h of negative 14, well, this is going to be equal to negative two. Once again, they are inverses of each other. So h of negative 14 is equal to negative negative two. And once again, I just swapped these two around. That's what the inverse function will do. If you're mapping from, if f goes from negative two to negative 14, h is going to go from negative 14 back to negative two."}, {"video_title": "Derivatives of inverse functions from equation AP Calculus AB Khan Academy.mp3", "Sentence": "So h of negative 14 is equal to negative negative two. And once again, I just swapped these two around. That's what the inverse function will do. If you're mapping from, if f goes from negative two to negative 14, h is going to go from negative 14 back to negative two. So now we want to evaluate f prime of negative two. Well, let's figure out what f prime of x is. So f prime of x is equal to, see, we're just gonna leverage the power rule."}, {"video_title": "Derivatives of inverse functions from equation AP Calculus AB Khan Academy.mp3", "Sentence": "If you're mapping from, if f goes from negative two to negative 14, h is going to go from negative 14 back to negative two. So now we want to evaluate f prime of negative two. Well, let's figure out what f prime of x is. So f prime of x is equal to, see, we're just gonna leverage the power rule. So three times 1 1\u20442 is 3 1\u20442 times x to the three minus one power, which is just the second power, plus the derivative of three x with respect to x. Well, that's just going to be three. And you could view that as just the power rule."}, {"video_title": "Derivatives of inverse functions from equation AP Calculus AB Khan Academy.mp3", "Sentence": "So f prime of x is equal to, see, we're just gonna leverage the power rule. So three times 1 1\u20442 is 3 1\u20442 times x to the three minus one power, which is just the second power, plus the derivative of three x with respect to x. Well, that's just going to be three. And you could view that as just the power rule. If this was x to the first power, one times three x to the zeroth power. Well, x to the zero is just one, so you're just left with three. And the derivative of a constant, that's just gonna be zero."}, {"video_title": "Derivatives of inverse functions from equation AP Calculus AB Khan Academy.mp3", "Sentence": "And you could view that as just the power rule. If this was x to the first power, one times three x to the zeroth power. Well, x to the zero is just one, so you're just left with three. And the derivative of a constant, that's just gonna be zero. So that's f prime of x. So f prime of negative two is going to be 3 1\u20442 times negative two squared is four, positive four. So plus three."}, {"video_title": "Derivatives of inverse functions from equation AP Calculus AB Khan Academy.mp3", "Sentence": "And the derivative of a constant, that's just gonna be zero. So that's f prime of x. So f prime of negative two is going to be 3 1\u20442 times negative two squared is four, positive four. So plus three. So this is going to be equal to two times three plus three. So six plus three is equal to nine. So this denominator right here is going to be equal to nine."}, {"video_title": "Derivatives of inverse functions from equation AP Calculus AB Khan Academy.mp3", "Sentence": "So plus three. So this is going to be equal to two times three plus three. So six plus three is equal to nine. So this denominator right here is going to be equal to nine. So this whole thing is equal to one over nine. So this involved, this isn't something you're gonna see every day. This isn't a typical problem in your calculus class."}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "So let's say I have the differential equation, the derivative of y with respect to x is equal to two y squared. And let's say that the graph of a particular solution to this the graph of a particular solution passes through the point one comma negative one. So my question to you is, what is y, what is y when x is equal to three for this particular solution? So the particular solution to the differential equation that passes through the point one comma negative one, what is y when x is equal to three? And I encourage you to pause the video and try to work through it on your own. So I'm assuming you had a go at it and the key with a separable differential equation, and that's a big clue, that even calling it a separable differential equation is that you separate the x's from the y's, or all the x's and the dx's from the y's and dy's. So how do you do that here?"}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "So the particular solution to the differential equation that passes through the point one comma negative one, what is y when x is equal to three? And I encourage you to pause the video and try to work through it on your own. So I'm assuming you had a go at it and the key with a separable differential equation, and that's a big clue, that even calling it a separable differential equation is that you separate the x's from the y's, or all the x's and the dx's from the y's and dy's. So how do you do that here? Well, what I could do, let me just rewrite it. So it's gonna be dy dx is equal to two y squared. Is equal to two y, equal to two y squared."}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "So how do you do that here? Well, what I could do, let me just rewrite it. So it's gonna be dy dx is equal to two y squared. Is equal to two y, equal to two y squared. So let's see, we can multiply both sides by dx. And let's see, so then we're gonna have, that cancels with that if we treat it as just a value or as a variable, we're gonna have dy is equal to two y squared dx. Well, we're not quite done yet."}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "Is equal to two y, equal to two y squared. So let's see, we can multiply both sides by dx. And let's see, so then we're gonna have, that cancels with that if we treat it as just a value or as a variable, we're gonna have dy is equal to two y squared dx. Well, we're not quite done yet. We need to get this two y squared on the left-hand side. So we can divide both sides by two y squared. So if we divide both sides by two y squared, two y squared, the left-hand side, we could rewrite this as 1 1 2 y to the negative two power is going to be equal to dy."}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "Well, we're not quite done yet. We need to get this two y squared on the left-hand side. So we can divide both sides by two y squared. So if we divide both sides by two y squared, two y squared, the left-hand side, we could rewrite this as 1 1 2 y to the negative two power is going to be equal to dy. Let me, dy is equal to dx. And now we can integrate both sides. So we can integrate both sides."}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "So if we divide both sides by two y squared, two y squared, the left-hand side, we could rewrite this as 1 1 2 y to the negative two power is going to be equal to dy. Let me, dy is equal to dx. And now we can integrate both sides. So we can integrate both sides. Let me get myself a little bit more space. And so what is, what is this left-hand side going to be? Well, we increment the exponent and then divide by that value."}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "So we can integrate both sides. Let me get myself a little bit more space. And so what is, what is this left-hand side going to be? Well, we increment the exponent and then divide by that value. So y to the negative two, if you increment it to y to the negative one and then divide by negative one, so this is going to be negative 1 1 2 y to the negative one power. And we could do a plus c like we did in the previous video, but we're gonna have a plus c on both sides. And you could subtract, or you know, you have different arbitrary constants on both sides and you could subtract them from each other."}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "Well, we increment the exponent and then divide by that value. So y to the negative two, if you increment it to y to the negative one and then divide by negative one, so this is going to be negative 1 1 2 y to the negative one power. And we could do a plus c like we did in the previous video, but we're gonna have a plus c on both sides. And you could subtract, or you know, you have different arbitrary constants on both sides and you could subtract them from each other. So I'm just gonna write the constant only on one side. So you have that is equal to, well, if I integrate just dx, that's just going to give me x. That's just gonna give me x."}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "And you could subtract, or you know, you have different arbitrary constants on both sides and you could subtract them from each other. So I'm just gonna write the constant only on one side. So you have that is equal to, well, if I integrate just dx, that's just going to give me x. That's just gonna give me x. So this right over here is x, and of course I can have a plus c over there. And if I want, I can solve for y. If I multiply, let's see, I can multiply both sides by negative two, and then I'm gonna have, the left-hand side you're just gonna have y to the negative one, or one over y."}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "That's just gonna give me x. So this right over here is x, and of course I can have a plus c over there. And if I want, I can solve for y. If I multiply, let's see, I can multiply both sides by negative two, and then I'm gonna have, the left-hand side you're just gonna have y to the negative one, or one over y. Is equal to, if I multiply the right-hand side times negative two, I'm gonna have negative two times x plus, well, it's some arbitrary constant. It's still going to be negative two times this arbitrary constant, but I could still just call it some arbitrary constant. And then if we want, we can take the reciprocal of both sides."}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "If I multiply, let's see, I can multiply both sides by negative two, and then I'm gonna have, the left-hand side you're just gonna have y to the negative one, or one over y. Is equal to, if I multiply the right-hand side times negative two, I'm gonna have negative two times x plus, well, it's some arbitrary constant. It's still going to be negative two times this arbitrary constant, but I could still just call it some arbitrary constant. And then if we want, we can take the reciprocal of both sides. And so we will get y is equal to, is equal to one over negative two x plus c. And now we can use, we can use the information they gave us right over here, the fact that our particular solution needs to go through this point to solve for c. So when x is negative one, so when x is negative one, or sorry, when x is one, when x is one, y is negative one. So we get negative one is equal to one over negative two plus c. Or we could say c minus two. We could multiply both sides times c minus two."}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "And then if we want, we can take the reciprocal of both sides. And so we will get y is equal to, is equal to one over negative two x plus c. And now we can use, we can use the information they gave us right over here, the fact that our particular solution needs to go through this point to solve for c. So when x is negative one, so when x is negative one, or sorry, when x is one, when x is one, y is negative one. So we get negative one is equal to one over negative two plus c. Or we could say c minus two. We could multiply both sides times c minus two. If then we will get, actually let me just scroll down a little bit. So if you multiply both sides times c minus two, negative one times c minus two is gonna be negative c plus two, or two minus c is equal to one. All I did is I multiplied c minus two times both sides."}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "We could multiply both sides times c minus two. If then we will get, actually let me just scroll down a little bit. So if you multiply both sides times c minus two, negative one times c minus two is gonna be negative c plus two, or two minus c is equal to one. All I did is I multiplied c minus two times both sides. And then let's see, I can subtract two from both sides. So negative c is equal to negative one. And then if I multiply both sides by negative one, we get c is equal to one."}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "All I did is I multiplied c minus two times both sides. And then let's see, I can subtract two from both sides. So negative c is equal to negative one. And then if I multiply both sides by negative one, we get c is equal to one. So our particular solution is y is equal to one over negative two x plus one. And we are almost done. They didn't just ask for, we didn't just ask for the particular solution."}, {"video_title": "Particular solution to differential equation example Khan Academy.mp3", "Sentence": "And then if I multiply both sides by negative one, we get c is equal to one. So our particular solution is y is equal to one over negative two x plus one. And we are almost done. They didn't just ask for, we didn't just ask for the particular solution. We asked what is y when x is equal to three. So y is going to be equal to one over, three times negative two is negative six plus one, which is equal to negative, is going to be equal to one over negative five, or negative one fifth. And we are done."}, {"video_title": "Limits by direct substitution Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And when you think about this, I'm not doing a rigorous proof here, a parabola would look something like this, and this would be an upward opening parabola, looks something like this. It is a, this graph visually is continuous, you don't see any jumps or gaps in it. And in general, a quadratic like this is going to be defined for all values of x, for all real numbers, and it's going to be continuous for all real numbers. And so if something is continuous for all real numbers, well then the limit as x approaches some real number is going to be the same thing, it's just evaluating the expression at that real number. So what am I saying? I'm just gonna say it another way. We know that some function is continuous, is continuous at some x value, at x equals a, if and only if, I'll write that as if, or if, if and only if the limit as x approaches a of f of x is equal to f of, is equal to f of a."}, {"video_title": "Limits by direct substitution Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And so if something is continuous for all real numbers, well then the limit as x approaches some real number is going to be the same thing, it's just evaluating the expression at that real number. So what am I saying? I'm just gonna say it another way. We know that some function is continuous, is continuous at some x value, at x equals a, if and only if, I'll write that as if, or if, if and only if the limit as x approaches a of f of x is equal to f of, is equal to f of a. So I didn't do a rigorous proof here, but just, it's conceptually not a big jump to say okay, well this is just a standard quadratic right over here, it's defined for all real numbers, and in fact it's continuous for all real numbers, and so we know that this expression, it could define a continuous function, so that means that the limit as x approaches a for this expression is just the same thing as evaluating this expression at a. And in this case our a is negative one. So all I have to do is evaluate this at negative one."}, {"video_title": "Limits by direct substitution Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "We know that some function is continuous, is continuous at some x value, at x equals a, if and only if, I'll write that as if, or if, if and only if the limit as x approaches a of f of x is equal to f of, is equal to f of a. So I didn't do a rigorous proof here, but just, it's conceptually not a big jump to say okay, well this is just a standard quadratic right over here, it's defined for all real numbers, and in fact it's continuous for all real numbers, and so we know that this expression, it could define a continuous function, so that means that the limit as x approaches a for this expression is just the same thing as evaluating this expression at a. And in this case our a is negative one. So all I have to do is evaluate this at negative one. This is going to be six times negative one squared plus five times negative one minus one. So that's just one, this is negative five. So it's six minus five minus one, which is equal to zero."}, {"video_title": "Area between a curve and and the _-axis AP Calculus AB Khan Academy.mp3", "Sentence": "And what I'm curious about in this video is I wanna find the area, not between this curve and the positive x-axis, I wanna find the area between the curve and the y-axis, bounded not by two x values, but bounded by two y values. So with the bottom bound of the horizontal line y is equal to e, and an upper bound with y is equal to e to the third power. So pause this video and see if you can work through it. So one way to think about it, this is just like definite integrals we've done where we're looking between the curve and the x-axis, but now it looks like things are swapped around. We now care about the y-axis. So let's just rewrite our function here, and let's rewrite it in terms of x. So if y is equal to 15 over x, that means if we multiply both sides by x, xy is equal to 15, and if we divide both sides by y, we get x is equal to 15 over y."}, {"video_title": "Area between a curve and and the _-axis AP Calculus AB Khan Academy.mp3", "Sentence": "So one way to think about it, this is just like definite integrals we've done where we're looking between the curve and the x-axis, but now it looks like things are swapped around. We now care about the y-axis. So let's just rewrite our function here, and let's rewrite it in terms of x. So if y is equal to 15 over x, that means if we multiply both sides by x, xy is equal to 15, and if we divide both sides by y, we get x is equal to 15 over y. These right over here are all going to be equivalent. Now how does this right over here help you? Well, think about the area, think about estimating the area as a bunch of little rectangles here."}, {"video_title": "Area between a curve and and the _-axis AP Calculus AB Khan Academy.mp3", "Sentence": "So if y is equal to 15 over x, that means if we multiply both sides by x, xy is equal to 15, and if we divide both sides by y, we get x is equal to 15 over y. These right over here are all going to be equivalent. Now how does this right over here help you? Well, think about the area, think about estimating the area as a bunch of little rectangles here. So that's one rectangle, and then another rectangle right over there, and then another rectangle right over there. So what's the area of each of those rectangles? So the width here, that is going to be x, but we can express x as a function of y."}, {"video_title": "Area between a curve and and the _-axis AP Calculus AB Khan Academy.mp3", "Sentence": "Well, think about the area, think about estimating the area as a bunch of little rectangles here. So that's one rectangle, and then another rectangle right over there, and then another rectangle right over there. So what's the area of each of those rectangles? So the width here, that is going to be x, but we can express x as a function of y. So that's the width right over there, and we know that that's going to be 15 over y. And then what's the height gonna be? Well, that's gonna be a very small change in y."}, {"video_title": "Area between a curve and and the _-axis AP Calculus AB Khan Academy.mp3", "Sentence": "So the width here, that is going to be x, but we can express x as a function of y. So that's the width right over there, and we know that that's going to be 15 over y. And then what's the height gonna be? Well, that's gonna be a very small change in y. The height is going to be dy. So the area of one of those little rectangles right over there, say the area of that one right over there, you could view as, let me do it over here, as 15 over y dy. And then we wanna sum all of these little rectangles from y is equal to e all the way to y is equal to e to the third power."}, {"video_title": "Area between a curve and and the _-axis AP Calculus AB Khan Academy.mp3", "Sentence": "Well, that's gonna be a very small change in y. The height is going to be dy. So the area of one of those little rectangles right over there, say the area of that one right over there, you could view as, let me do it over here, as 15 over y dy. And then we wanna sum all of these little rectangles from y is equal to e all the way to y is equal to e to the third power. So that's what our definite integral does. We go from y is equal to e to y is equal to e to the third power. So all we did, we're used to seeing things like this, where this would be 15 over x dx."}, {"video_title": "Area between a curve and and the _-axis AP Calculus AB Khan Academy.mp3", "Sentence": "And then we wanna sum all of these little rectangles from y is equal to e all the way to y is equal to e to the third power. So that's what our definite integral does. We go from y is equal to e to y is equal to e to the third power. So all we did, we're used to seeing things like this, where this would be 15 over x dx. All we're doing here is, this is 15 over y dy. So let's evaluate this. So we take the antiderivative of 15 over y, and then evaluate at these two points."}, {"video_title": "Area between a curve and and the _-axis AP Calculus AB Khan Academy.mp3", "Sentence": "So all we did, we're used to seeing things like this, where this would be 15 over x dx. All we're doing here is, this is 15 over y dy. So let's evaluate this. So we take the antiderivative of 15 over y, and then evaluate at these two points. So this is going to be equal to, antiderivative of one over y is going to be the natural log of the absolute value of y. So it's 15 times the natural log of the absolute value of y. And then we're going to evaluate that at our endpoints."}, {"video_title": "Area between a curve and and the _-axis AP Calculus AB Khan Academy.mp3", "Sentence": "So we take the antiderivative of 15 over y, and then evaluate at these two points. So this is going to be equal to, antiderivative of one over y is going to be the natural log of the absolute value of y. So it's 15 times the natural log of the absolute value of y. And then we're going to evaluate that at our endpoints. So we're gonna evaluate it at e to the third and at e. So let's first evaluate at e to the third. So that's 15 times the natural log, the absolute time, the natural log of the absolute value of e to the third power minus 15 times the natural log of the absolute value of e. So what does this simplify to? The natural log of e to the third power, what power do I have to raise e to to get to e to the third?"}, {"video_title": "Area between a curve and and the _-axis AP Calculus AB Khan Academy.mp3", "Sentence": "And then we're going to evaluate that at our endpoints. So we're gonna evaluate it at e to the third and at e. So let's first evaluate at e to the third. So that's 15 times the natural log, the absolute time, the natural log of the absolute value of e to the third power minus 15 times the natural log of the absolute value of e. So what does this simplify to? The natural log of e to the third power, what power do I have to raise e to to get to e to the third? Well, that's just going to be three. And then the natural log of e, what power do I have to raise e to to get e? Well, that's just one."}, {"video_title": "Area between a curve and and the _-axis AP Calculus AB Khan Academy.mp3", "Sentence": "The natural log of e to the third power, what power do I have to raise e to to get to e to the third? Well, that's just going to be three. And then the natural log of e, what power do I have to raise e to to get e? Well, that's just one. So this is 15 times three minus 15. So that is all going to get us to 30. And we are done, 45 minus 15."}, {"video_title": "Worked example alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "Now, when we've looked at convergence tests for infinite series, we saw things like this. This passes the alternating series test, and so we know that this converges. Let's say it converges to some value s. But what we're concerned about in this video is not whether or not this converges, but estimating what this actually converges to. We know that we can estimate this by taking a partial sum, let's say s sub k, this is the first k terms right over here, and then you're going to have a remainder. So plus the remainder after you've taken the first k terms. So this actually starts with the k plus first term. And so what I'm concerned about or what I wanna figure out is how many, what's the minimum number of terms?"}, {"video_title": "Worked example alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "We know that we can estimate this by taking a partial sum, let's say s sub k, this is the first k terms right over here, and then you're going to have a remainder. So plus the remainder after you've taken the first k terms. So this actually starts with the k plus first term. And so what I'm concerned about or what I wanna figure out is how many, what's the minimum number of terms? What's the minimum k here so that my remainder, so that the absolute value of my remainder is less than or equal to 0.001. And I encourage you to pause the video based on what we've seen in previous alternating series estimation situations to see if you can figure this out. All right, I'm assuming you've had a go at it."}, {"video_title": "Worked example alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so what I'm concerned about or what I wanna figure out is how many, what's the minimum number of terms? What's the minimum k here so that my remainder, so that the absolute value of my remainder is less than or equal to 0.001. And I encourage you to pause the video based on what we've seen in previous alternating series estimation situations to see if you can figure this out. All right, I'm assuming you've had a go at it. So let's just remind ourselves what r sub k looks like. So r sub k, r sub k is going to be, it's gonna start with a k plus first term. So it's gonna be negative one, k plus one plus one, so it's negative one to the k plus two over the square root of k, of k plus, k plus one, k plus one."}, {"video_title": "Worked example alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "All right, I'm assuming you've had a go at it. So let's just remind ourselves what r sub k looks like. So r sub k, r sub k is going to be, it's gonna start with a k plus first term. So it's gonna be negative one, k plus one plus one, so it's negative one to the k plus two over the square root of k, of k plus, k plus one, k plus one. And then the next term, I can just write plus negative one to the k plus three over the square root of k plus two. And it's just gonna go on and on and on like that. Now what we know already or what we've seen an example of and we'll verify it or at least get a conceptual verification for us at the end of this video is, well, the absolute value of this entire sum, the absolute value of this entire sum is going to be less than the absolute value of the first term."}, {"video_title": "Worked example alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "So it's gonna be negative one, k plus one plus one, so it's negative one to the k plus two over the square root of k, of k plus, k plus one, k plus one. And then the next term, I can just write plus negative one to the k plus three over the square root of k plus two. And it's just gonna go on and on and on like that. Now what we know already or what we've seen an example of and we'll verify it or at least get a conceptual verification for us at the end of this video is, well, the absolute value of this entire sum, the absolute value of this entire sum is going to be less than the absolute value of the first term. So let me write that down. The absolute value of this entire sum, of this entire remainder, r sub k, and this is, you know, some people refer to this too as kind of the alternating series remainder property or whatever you wanna call it. But the absolute value of this entire thing is going to be less than or equal to the absolute value of the first term, negative one to the k plus two over the square root of k plus one."}, {"video_title": "Worked example alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "Now what we know already or what we've seen an example of and we'll verify it or at least get a conceptual verification for us at the end of this video is, well, the absolute value of this entire sum, the absolute value of this entire sum is going to be less than the absolute value of the first term. So let me write that down. The absolute value of this entire sum, of this entire remainder, r sub k, and this is, you know, some people refer to this too as kind of the alternating series remainder property or whatever you wanna call it. But the absolute value of this entire thing is going to be less than or equal to the absolute value of the first term, negative one to the k plus two over the square root of k plus one. And of course, we want that to be less than 1,000th. So less than, that needs to be less than or equal to 0.001. And so now this setup right over here inspires you."}, {"video_title": "Worked example alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "But the absolute value of this entire thing is going to be less than or equal to the absolute value of the first term, negative one to the k plus two over the square root of k plus one. And of course, we want that to be less than 1,000th. So less than, that needs to be less than or equal to 0.001. And so now this setup right over here inspires you. I once again encourage you to pause the video and see if you can figure out what is the minimum k, what is the minimum k that satisfies this inequality? Well, once again, I'm assuming you've had a go at it. And the key realization here is that this negative one to the k plus two, this makes this whole thing either positive or negative."}, {"video_title": "Worked example alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so now this setup right over here inspires you. I once again encourage you to pause the video and see if you can figure out what is the minimum k, what is the minimum k that satisfies this inequality? Well, once again, I'm assuming you've had a go at it. And the key realization here is that this negative one to the k plus two, this makes this whole thing either positive or negative. The denominator is going to be positive for, well, it's definitely going to be positive for all the k's, frankly, for which the principal root is going to be defined, but we obviously, these are all positive k's as well. So the numerator here just flips the sign. We flip between negative one or positive one, negative one or positive one."}, {"video_title": "Worked example alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "And the key realization here is that this negative one to the k plus two, this makes this whole thing either positive or negative. The denominator is going to be positive for, well, it's definitely going to be positive for all the k's, frankly, for which the principal root is going to be defined, but we obviously, these are all positive k's as well. So the numerator here just flips the sign. We flip between negative one or positive one, negative one or positive one. But if we take the absolute value, then we know whatever, whether it's negative or positive, it's going to end up being positive. So this first, this thing on the left-hand side, this thing on the left-hand side, that's the same thing as just one over the square root of k, the square root of k plus one, and then that's going to be less than or equal to, that needs to be less than or equal to 0.001. And now we just have to solve, we have to just solve this inequality, find out which k's satisfy this inequality."}, {"video_title": "Worked example alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "We flip between negative one or positive one, negative one or positive one. But if we take the absolute value, then we know whatever, whether it's negative or positive, it's going to end up being positive. So this first, this thing on the left-hand side, this thing on the left-hand side, that's the same thing as just one over the square root of k, the square root of k plus one, and then that's going to be less than or equal to, that needs to be less than or equal to 0.001. And now we just have to solve, we have to just solve this inequality, find out which k's satisfy this inequality. And so let's see, we can multiply both sides by the square root of k plus one. So square root of k plus one, so we can get this out of the denominator. And let's actually multiply both sides times 1,000, because this is 1,000th, and so we'll end up with a one on the right-hand side."}, {"video_title": "Worked example alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "And now we just have to solve, we have to just solve this inequality, find out which k's satisfy this inequality. And so let's see, we can multiply both sides by the square root of k plus one. So square root of k plus one, so we can get this out of the denominator. And let's actually multiply both sides times 1,000, because this is 1,000th, and so we'll end up with a one on the right-hand side. So times 1,000, times 1,000. And what we're going to be left with is, these cancel out, on the left-hand side, we don't have to flip the sign, because we just multiplied by positive things. So we have 1,000 is less than or equal to the square root of k plus one."}, {"video_title": "Worked example alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "And let's actually multiply both sides times 1,000, because this is 1,000th, and so we'll end up with a one on the right-hand side. So times 1,000, times 1,000. And what we're going to be left with is, these cancel out, on the left-hand side, we don't have to flip the sign, because we just multiplied by positive things. So we have 1,000 is less than or equal to the square root of k plus one. We could square both sides, and we get one million, one million is less than or equal to k plus one. And so, and then we can just subtract one from both sides, and we have 999,999 is going to be less than or equal to k. So k just has to be greater than or equal to 999,999. And remember, we want the smallest k that satisfies these conditions."}, {"video_title": "Worked example alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "So we have 1,000 is less than or equal to the square root of k plus one. We could square both sides, and we get one million, one million is less than or equal to k plus one. And so, and then we can just subtract one from both sides, and we have 999,999 is going to be less than or equal to k. So k just has to be greater than or equal to 999,999. And remember, we want the smallest k that satisfies these conditions. So the smallest k, if k has to be greater than or equal to this, the smallest k that satisfies this is k is equal to 999,999. So let's write that down. So the smallest k for which this is true is going to be k is equal to 999,999."}, {"video_title": "Worked example alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "And remember, we want the smallest k that satisfies these conditions. So the smallest k, if k has to be greater than or equal to this, the smallest k that satisfies this is k is equal to 999,999. So let's write that down. So the smallest k for which this is true is going to be k is equal to 999,999. Now let's convince ourselves that that remainder, that the absolute value of the remainder is definitely going to be less, is definitely, if our k, if we take the partial sum of the first 999,999 terms, that this remainder is actually going to be less than this. I'm telling you, so far we've just kind of worked from the premise that it will be, but let's actually look at that and feel good about it. And once again, I encourage you to pause the video and try it on your own."}, {"video_title": "Worked example alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "So the smallest k for which this is true is going to be k is equal to 999,999. Now let's convince ourselves that that remainder, that the absolute value of the remainder is definitely going to be less, is definitely, if our k, if we take the partial sum of the first 999,999 terms, that this remainder is actually going to be less than this. I'm telling you, so far we've just kind of worked from the premise that it will be, but let's actually look at that and feel good about it. And once again, I encourage you to pause the video and try it on your own. So let's just rewrite this again, but I'm gonna expand out r sub k. So we're saying that it's all gonna converge to s, and we're gonna take the partial sum, the partial sum of the first 999,999 terms, and we're claiming that this is gonna be within 1,000th of this right over here. So this is gonna be plus, and so the first term is going to be the, of the remainder of r sub k, is going to be our millionth, is going to be our millionth, our millionth term. And so the millionth term, we just have to remind ourselves, it's gonna be negative 1 to the millionth and onenth power."}, {"video_title": "Worked example alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "And once again, I encourage you to pause the video and try it on your own. So let's just rewrite this again, but I'm gonna expand out r sub k. So we're saying that it's all gonna converge to s, and we're gonna take the partial sum, the partial sum of the first 999,999 terms, and we're claiming that this is gonna be within 1,000th of this right over here. So this is gonna be plus, and so the first term is going to be the, of the remainder of r sub k, is going to be our millionth, is going to be our millionth, our millionth term. And so the millionth term, we just have to remind ourselves, it's gonna be negative 1 to the millionth and onenth power. So negative 1 to the millionth and oneth power, that's going to be a negative. That's gonna be negative 1, cause that's an odd number. So it's gonna be negative 1 over the square root of a millionth."}, {"video_title": "Worked example alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so the millionth term, we just have to remind ourselves, it's gonna be negative 1 to the millionth and onenth power. So negative 1 to the millionth and oneth power, that's going to be a negative. That's gonna be negative 1, cause that's an odd number. So it's gonna be negative 1 over the square root of a millionth. Let me do it like this. It's gonna be, it's gonna be minus 1 over the square root of a million. A million."}, {"video_title": "Worked example alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "So it's gonna be negative 1 over the square root of a millionth. Let me do it like this. It's gonna be, it's gonna be minus 1 over the square root of a million. A million. And then the next term is gonna be plus 1 over the square root of a million, of a million and one, and it's positive now, because this is going to be negative one to the million and two second power. And then it's going to be minus one over the square root of one million and two, and then plus, I'll just do two more terms, one over the square root of a million and three, and then minus one over the square root of a million and four, and then plus minus, it just keeps going on and on, on and on and on forever. Now, I want to convince ourselves that this thing is going to be, the absolute value of this thing is going to be less than 1,000."}, {"video_title": "Worked example alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "A million. And then the next term is gonna be plus 1 over the square root of a million, of a million and one, and it's positive now, because this is going to be negative one to the million and two second power. And then it's going to be minus one over the square root of one million and two, and then plus, I'll just do two more terms, one over the square root of a million and three, and then minus one over the square root of a million and four, and then plus minus, it just keeps going on and on, on and on and on forever. Now, I want to convince ourselves that this thing is going to be, the absolute value of this thing is going to be less than 1,000. It's going to be essentially less than the absolute value of this first term. So how do we think about it? So this first term, we already know, this right over here is, this first term right over here is negative negative 0.001, because it's one over 1,000, and we have this negative right over here."}, {"video_title": "Worked example alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "Now, I want to convince ourselves that this thing is going to be, the absolute value of this thing is going to be less than 1,000. It's going to be essentially less than the absolute value of this first term. So how do we think about it? So this first term, we already know, this right over here is, this first term right over here is negative negative 0.001, because it's one over 1,000, and we have this negative right over here. Now, the first thing we could realize, if we just put parentheses like this, if we just put parentheses like this, we see that this is going to be positive. This term is larger than this term. This term is larger than this term, and we could keep going."}, {"video_title": "Worked example alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "So this first term, we already know, this right over here is, this first term right over here is negative negative 0.001, because it's one over 1,000, and we have this negative right over here. Now, the first thing we could realize, if we just put parentheses like this, if we just put parentheses like this, we see that this is going to be positive. This term is larger than this term. This term is larger than this term, and we could keep going. So we're going to have, the remainder kind of starts at negative 1,000, and then it just adds a bunch of positive terms. So it's not gonna get any more negative than this. So we know that the remainder can't get more negative than this, but let's also verify that it can't, you know, get more than a positive thousandth either."}, {"video_title": "Worked example alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "This term is larger than this term, and we could keep going. So we're going to have, the remainder kind of starts at negative 1,000, and then it just adds a bunch of positive terms. So it's not gonna get any more negative than this. So we know that the remainder can't get more negative than this, but let's also verify that it can't, you know, get more than a positive thousandth either. And the way that we can do that is to just look at the parentheses in a different way. If we do it this way, if we do it this way, that's going to be, this is going to be a negative value. This right over here is going to be a negative value, and we can just keep, so we could say plus that, plus that, and so we're gonna have a bunch of negative values together."}, {"video_title": "Worked example alternating series remainder Series AP Calculus BC Khan Academy.mp3", "Sentence": "So we know that the remainder can't get more negative than this, but let's also verify that it can't, you know, get more than a positive thousandth either. And the way that we can do that is to just look at the parentheses in a different way. If we do it this way, if we do it this way, that's going to be, this is going to be a negative value. This right over here is going to be a negative value, and we can just keep, so we could say plus that, plus that, and so we're gonna have a bunch of negative values together. So just like that, by really looking at the parentheses a little bit differently, we're able to say this is definitely going to be negative. This remainder is definitely going to be a negative value, but it can't be any more negative than our first term. So that tells us that our remainder is not gonna get any, is not gonna be any more, can't get any, the absolute value of the remainder can't get any larger than 1,000."}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And like always, pause the video and see if you can work through this. All right, well, our first temptation is to say, well, this is going to be the same thing as the limit of one minus cosine theta as x approaches, or not x, as theta approaches zero, of theta as theta approaches zero over the limit as theta approaches zero of two sine squared theta. Now, both of these expressions, which could be used to define a function, they'd be continuous if you graphed them. They'd be continuous at theta equals zero, so the limit is going to be the same thing as just evaluating them at theta equals zero. So this is going to be equal to one minus cosine of zero over two sine squared of zero. Now, cosine of zero is one, and then one minus one is zero, and sine of zero is zero, and you square it, you still got zero, and you multiply it times two, you still got zero, so you got zero over zero. So once again, we have that indeterminate form."}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "They'd be continuous at theta equals zero, so the limit is going to be the same thing as just evaluating them at theta equals zero. So this is going to be equal to one minus cosine of zero over two sine squared of zero. Now, cosine of zero is one, and then one minus one is zero, and sine of zero is zero, and you square it, you still got zero, and you multiply it times two, you still got zero, so you got zero over zero. So once again, we have that indeterminate form. And once again, this indeterminate form, when you have zero over zero, doesn't mean to give up, it doesn't mean that the limit doesn't exist, it just means, well, maybe there's some other approaches here to work on. If you got some non-zero number divided by zero, then you say, okay, that limit doesn't exist, and you'd say, well, you'd just say it doesn't exist. But let's see what we can do to maybe think about this expression in a different way."}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So once again, we have that indeterminate form. And once again, this indeterminate form, when you have zero over zero, doesn't mean to give up, it doesn't mean that the limit doesn't exist, it just means, well, maybe there's some other approaches here to work on. If you got some non-zero number divided by zero, then you say, okay, that limit doesn't exist, and you'd say, well, you'd just say it doesn't exist. But let's see what we can do to maybe think about this expression in a different way. So if we said, so let's just say that this, let me use some other colors here. Let's say that this right over here is f of x. So f of x is equal to one minus cosine theta over two sine squared theta."}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "But let's see what we can do to maybe think about this expression in a different way. So if we said, so let's just say that this, let me use some other colors here. Let's say that this right over here is f of x. So f of x is equal to one minus cosine theta over two sine squared theta. And let's see if we can rewrite it in some way, that at least the limit as theta approaches zero isn't going to, we're not gonna get the same zero over zero. Well, we can, we got some trig functions here, so maybe we can use some of our trig identities to simplify this. And the one that jumps out at me is that we have the sine squared of theta, and we know from the Pythagorean identity in trigonometries, comes straight out of the unit circle definition of sine and cosine, we know that sine squared theta plus cosine squared theta is equal to one, or we know that sine squared theta is one minus cosine squared theta."}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So f of x is equal to one minus cosine theta over two sine squared theta. And let's see if we can rewrite it in some way, that at least the limit as theta approaches zero isn't going to, we're not gonna get the same zero over zero. Well, we can, we got some trig functions here, so maybe we can use some of our trig identities to simplify this. And the one that jumps out at me is that we have the sine squared of theta, and we know from the Pythagorean identity in trigonometries, comes straight out of the unit circle definition of sine and cosine, we know that sine squared theta plus cosine squared theta is equal to one, or we know that sine squared theta is one minus cosine squared theta. One minus cosine squared theta. So we could rewrite this. This is equal to one minus cosine theta over two times one minus cosine squared theta."}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And the one that jumps out at me is that we have the sine squared of theta, and we know from the Pythagorean identity in trigonometries, comes straight out of the unit circle definition of sine and cosine, we know that sine squared theta plus cosine squared theta is equal to one, or we know that sine squared theta is one minus cosine squared theta. One minus cosine squared theta. So we could rewrite this. This is equal to one minus cosine theta over two times one minus cosine squared theta. Now, this is a one minus cosine theta, this is a one minus cosine squared theta, so it's not completely obvious yet of how you can simplify it, until you realize that this could be viewed as a difference of squares. If you view this as, if you view this as a squared minus b squared, we know that this can be factored as a plus b times a minus b. So I could rewrite this."}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "This is equal to one minus cosine theta over two times one minus cosine squared theta. Now, this is a one minus cosine theta, this is a one minus cosine squared theta, so it's not completely obvious yet of how you can simplify it, until you realize that this could be viewed as a difference of squares. If you view this as, if you view this as a squared minus b squared, we know that this can be factored as a plus b times a minus b. So I could rewrite this. This is equal to one minus cosine theta over two times, I could write this as one plus cosine theta times one minus cosine theta. One plus cosine theta times one minus, one minus cosine theta. And now this is interesting."}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So I could rewrite this. This is equal to one minus cosine theta over two times, I could write this as one plus cosine theta times one minus cosine theta. One plus cosine theta times one minus, one minus cosine theta. And now this is interesting. I have a one minus cosine theta in the numerator, and I have a one minus cosine theta in the denominator. Now we might be tempted to say, oh, well, let's just cross that out with that, and we would get, we would simplify it and get f of x is equal to one over, and we could distribute this two now, we could say two plus two cosine theta. We could say, well, aren't these the same thing?"}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And now this is interesting. I have a one minus cosine theta in the numerator, and I have a one minus cosine theta in the denominator. Now we might be tempted to say, oh, well, let's just cross that out with that, and we would get, we would simplify it and get f of x is equal to one over, and we could distribute this two now, we could say two plus two cosine theta. We could say, well, aren't these the same thing? And we would be almost right, because f of x, this one right over here, this is defined, this right over here is defined when theta is equal to zero, while this one is not defined when theta is equal to zero. When theta is equal to zero, you have a zero in the denominator. And so what we need to do in order for this f of x, or in order for this to be the same thing, we have to say theta cannot be equal to zero."}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "We could say, well, aren't these the same thing? And we would be almost right, because f of x, this one right over here, this is defined, this right over here is defined when theta is equal to zero, while this one is not defined when theta is equal to zero. When theta is equal to zero, you have a zero in the denominator. And so what we need to do in order for this f of x, or in order for this to be the same thing, we have to say theta cannot be equal to zero. But now let's think about the limit again. Essentially what we want to do is we want to find the limit as theta approaches zero of f of x. And we can't just do direct substitution into, if we really take this seriously, because we're going to like, oh, well, if I try to put zero here, it says theta cannot be equal to zero."}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And so what we need to do in order for this f of x, or in order for this to be the same thing, we have to say theta cannot be equal to zero. But now let's think about the limit again. Essentially what we want to do is we want to find the limit as theta approaches zero of f of x. And we can't just do direct substitution into, if we really take this seriously, because we're going to like, oh, well, if I try to put zero here, it says theta cannot be equal to zero. f of x is not defined at zero. This expression is defined at zero, but this tells me, well, I really shouldn't apply zero to this function. But we know that if we can find another function that is defined, that is the exact same thing as f of x, except at zero, and it is continuous at zero."}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And we can't just do direct substitution into, if we really take this seriously, because we're going to like, oh, well, if I try to put zero here, it says theta cannot be equal to zero. f of x is not defined at zero. This expression is defined at zero, but this tells me, well, I really shouldn't apply zero to this function. But we know that if we can find another function that is defined, that is the exact same thing as f of x, except at zero, and it is continuous at zero. And so we could say g of x is equal to one over two plus two cosine theta. Well, then we know this limit is going to be the exact same thing as the limit of g of x as theta approaches zero. Once again, these two functions are identical, except f of x is not defined at theta equals zero, while g of x is."}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "But we know that if we can find another function that is defined, that is the exact same thing as f of x, except at zero, and it is continuous at zero. And so we could say g of x is equal to one over two plus two cosine theta. Well, then we know this limit is going to be the exact same thing as the limit of g of x as theta approaches zero. Once again, these two functions are identical, except f of x is not defined at theta equals zero, while g of x is. But the limits as theta approaches zero are going to be the same, and we've seen that in previous videos. And I know what a lot of you are thinking, Sal, this seems like a very, you know, why don't I just, you know, do this algebra here, cross these things out, and just substitute zero for theta? Well, you could do that, and you would get the answer, but you need to be clear, or it's important to be mathematically clear of what you are doing."}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Once again, these two functions are identical, except f of x is not defined at theta equals zero, while g of x is. But the limits as theta approaches zero are going to be the same, and we've seen that in previous videos. And I know what a lot of you are thinking, Sal, this seems like a very, you know, why don't I just, you know, do this algebra here, cross these things out, and just substitute zero for theta? Well, you could do that, and you would get the answer, but you need to be clear, or it's important to be mathematically clear of what you are doing. If you do that, if you just cross these two out, and all of a sudden, your expression becomes defined at zero, you are now dealing with a different expression, or a different function definition. So to be clear, if you want to say this is the function you're finding the limit of, you have to put this constraint in to make sure it has the exact same domain. But lucky for us, we can say if we found another function that's continuous at that point that doesn't have that gap there, that doesn't have that point discontinuity out, the limits are going to be equivalent."}, {"video_title": "Trig limit using pythagorean identity Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Well, you could do that, and you would get the answer, but you need to be clear, or it's important to be mathematically clear of what you are doing. If you do that, if you just cross these two out, and all of a sudden, your expression becomes defined at zero, you are now dealing with a different expression, or a different function definition. So to be clear, if you want to say this is the function you're finding the limit of, you have to put this constraint in to make sure it has the exact same domain. But lucky for us, we can say if we found another function that's continuous at that point that doesn't have that gap there, that doesn't have that point discontinuity out, the limits are going to be equivalent. So the limit as theta approaches zero of g of x, well, that's just going to be, since it's continuous at zero, we could say that's just going to be, we can just substitute, that's going to be equal to g of zero, which is equal to one over two plus cosine two, one over two plus two times cosine of zero. Cosine of zero is one, so it's just one over two plus two, which is equal to, deserve a little bit of a drum roll here, which is equal to 1 4th. And we are done."}, {"video_title": "Formal definition of limits Part 1 intuition review AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say this is my y-axis. So try to draw a vertical line. So that right over there is my y-axis. And then let's say this is my x-axis. I'll focus on the first quadrant, although I don't have to. So let's say this right over here is my x-axis. And then let me draw a function."}, {"video_title": "Formal definition of limits Part 1 intuition review AP Calculus AB Khan Academy.mp3", "Sentence": "And then let's say this is my x-axis. I'll focus on the first quadrant, although I don't have to. So let's say this right over here is my x-axis. And then let me draw a function. So let's say my function looks something like that. It could look like anything, but that seems suitable. So this is the function y is equal to f of x."}, {"video_title": "Formal definition of limits Part 1 intuition review AP Calculus AB Khan Academy.mp3", "Sentence": "And then let me draw a function. So let's say my function looks something like that. It could look like anything, but that seems suitable. So this is the function y is equal to f of x. And just for the sake of conceptual understanding, I'm going to say it's not defined at a point. I didn't have to do this. You can find the limit as x approaches a point where the function actually is defined, but it becomes that much more interesting, at least for me, where you start to understand why a limit might be relevant, where a function is not defined at some point."}, {"video_title": "Formal definition of limits Part 1 intuition review AP Calculus AB Khan Academy.mp3", "Sentence": "So this is the function y is equal to f of x. And just for the sake of conceptual understanding, I'm going to say it's not defined at a point. I didn't have to do this. You can find the limit as x approaches a point where the function actually is defined, but it becomes that much more interesting, at least for me, where you start to understand why a limit might be relevant, where a function is not defined at some point. So the way I've drawn it, this function is not defined when x is equal to c. Now, the way that we've thought about a limit is what does f of x approach as x approaches c? So let's think about that a little bit. When x is a reasonable bit lower than c, f of x, for our function that we just drew, is right over here."}, {"video_title": "Formal definition of limits Part 1 intuition review AP Calculus AB Khan Academy.mp3", "Sentence": "You can find the limit as x approaches a point where the function actually is defined, but it becomes that much more interesting, at least for me, where you start to understand why a limit might be relevant, where a function is not defined at some point. So the way I've drawn it, this function is not defined when x is equal to c. Now, the way that we've thought about a limit is what does f of x approach as x approaches c? So let's think about that a little bit. When x is a reasonable bit lower than c, f of x, for our function that we just drew, is right over here. That's what f of x is going to be equal. y is equal to f of x. When x gets a little bit closer, then our f of x is right over there."}, {"video_title": "Formal definition of limits Part 1 intuition review AP Calculus AB Khan Academy.mp3", "Sentence": "When x is a reasonable bit lower than c, f of x, for our function that we just drew, is right over here. That's what f of x is going to be equal. y is equal to f of x. When x gets a little bit closer, then our f of x is right over there. When x gets even closer, maybe really almost at c, but not quite at c, then our f of x is right over here. And the way we see it, we see that our f of x seems to be, as x gets closer and closer to c, it looks like our f of x is getting closer and closer to some value. It's getting closer and closer to some value right over there."}, {"video_title": "Formal definition of limits Part 1 intuition review AP Calculus AB Khan Academy.mp3", "Sentence": "When x gets a little bit closer, then our f of x is right over there. When x gets even closer, maybe really almost at c, but not quite at c, then our f of x is right over here. And the way we see it, we see that our f of x seems to be, as x gets closer and closer to c, it looks like our f of x is getting closer and closer to some value. It's getting closer and closer to some value right over there. I'll even draw it, maybe I'll draw it with a more solid line. And that was actually only the case when x was getting closer to c from the left, from values of x less than c. But what happens as we get closer and closer to c from values of x that are larger than c? Well, when x is over here, f of x is right over here."}, {"video_title": "Formal definition of limits Part 1 intuition review AP Calculus AB Khan Academy.mp3", "Sentence": "It's getting closer and closer to some value right over there. I'll even draw it, maybe I'll draw it with a more solid line. And that was actually only the case when x was getting closer to c from the left, from values of x less than c. But what happens as we get closer and closer to c from values of x that are larger than c? Well, when x is over here, f of x is right over here. And so that's what f of x is, is right over there. When x gets a little bit closer to c, our f of x is right over there. When x is just very slightly larger than c, then our f of x is right over there."}, {"video_title": "Formal definition of limits Part 1 intuition review AP Calculus AB Khan Academy.mp3", "Sentence": "Well, when x is over here, f of x is right over here. And so that's what f of x is, is right over there. When x gets a little bit closer to c, our f of x is right over there. When x is just very slightly larger than c, then our f of x is right over there. And you see, once again, it seems to be approaching that same value. And we call that value, that value that f of x seems to be approaching as x approaches c, we call that value L, or the limit. And so the way we would denote it, is we would call that the limit."}, {"video_title": "Formal definition of limits Part 1 intuition review AP Calculus AB Khan Academy.mp3", "Sentence": "When x is just very slightly larger than c, then our f of x is right over there. And you see, once again, it seems to be approaching that same value. And we call that value, that value that f of x seems to be approaching as x approaches c, we call that value L, or the limit. And so the way we would denote it, is we would call that the limit. We don't have to call it L all the time, but it's referred to as the limit. And the way that we would kind of write that mathematically, is we would say the limit of f of x as x approaches c is equal to L. And this is a fine conceptual understanding of limits. And it really will take you pretty far."}, {"video_title": "Formal definition of limits Part 1 intuition review AP Calculus AB Khan Academy.mp3", "Sentence": "And so the way we would denote it, is we would call that the limit. We don't have to call it L all the time, but it's referred to as the limit. And the way that we would kind of write that mathematically, is we would say the limit of f of x as x approaches c is equal to L. And this is a fine conceptual understanding of limits. And it really will take you pretty far. And you're ready to progress and start thinking about taking a lot of limits. But this isn't a very mathematically rigorous definition of limits. And so this sets us up for the intuition."}, {"video_title": "Sum of an infinite geometric series Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "What I want to do in this video is now think about the sum of an infinite geometric series. And I've always found this mildly mind-blowing, or actually more than mildly mind-blowing, because you're taking the sum of an infinite things, but as we see, you can actually get a finite value, depending on what your common ratio is. So there's a couple of ways to think about it. One is, you could say that the sum of an infinite geometric series is just the limit of this as n approaches infinity. So we could say, what is the limit as n approaches infinity of this business, of the sum from k equals 0 to n of a times r to the k, which would be the same thing as taking the limit as n approaches infinity right over here. So that would be the same thing as the limit as n approaches infinity of all of this business. Let me just copy and paste that so I don't start switching colors."}, {"video_title": "Sum of an infinite geometric series Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "One is, you could say that the sum of an infinite geometric series is just the limit of this as n approaches infinity. So we could say, what is the limit as n approaches infinity of this business, of the sum from k equals 0 to n of a times r to the k, which would be the same thing as taking the limit as n approaches infinity right over here. So that would be the same thing as the limit as n approaches infinity of all of this business. Let me just copy and paste that so I don't start switching colors. So copy and then paste. So what's the limit as n approaches infinity here? Let's think about that for a second."}, {"video_title": "Sum of an infinite geometric series Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "Let me just copy and paste that so I don't start switching colors. So copy and then paste. So what's the limit as n approaches infinity here? Let's think about that for a second. I encourage you to pause the video, and I'll give you one hint. Think about it for r is greater than 1, for r is equal to 1. And actually, let me make it clear."}, {"video_title": "Sum of an infinite geometric series Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "Let's think about that for a second. I encourage you to pause the video, and I'll give you one hint. Think about it for r is greater than 1, for r is equal to 1. And actually, let me make it clear. But think about it for the absolute values of r is greater than 1, the absolute values of r equal to 1, and then the absolute value of r less than 1. Well, I'm assuming you've given a go at it. So if the absolute value of r is greater than 1, as this exponent explodes, as it approaches infinity, this number is just going to become massively, massively huge."}, {"video_title": "Sum of an infinite geometric series Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "And actually, let me make it clear. But think about it for the absolute values of r is greater than 1, the absolute values of r equal to 1, and then the absolute value of r less than 1. Well, I'm assuming you've given a go at it. So if the absolute value of r is greater than 1, as this exponent explodes, as it approaches infinity, this number is just going to become massively, massively huge. And so the whole thing is just going to become, or at least you could think of the absolute value of the whole thing, is just going to become a very, very, very large number. If r was equal to 1, then the denominator is going to become 0. And we're going to be dividing by that denominator."}, {"video_title": "Sum of an infinite geometric series Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "So if the absolute value of r is greater than 1, as this exponent explodes, as it approaches infinity, this number is just going to become massively, massively huge. And so the whole thing is just going to become, or at least you could think of the absolute value of the whole thing, is just going to become a very, very, very large number. If r was equal to 1, then the denominator is going to become 0. And we're going to be dividing by that denominator. And this formula just breaks down. But where this formula can be helpful and where we can get this to actually give us a sensical result is when the absolute value of r is between 0 and 1. We've already talked about, we're not even dealing with the geometric, we're not even talking about a geometric series if r is equal to 0."}, {"video_title": "Sum of an infinite geometric series Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "And we're going to be dividing by that denominator. And this formula just breaks down. But where this formula can be helpful and where we can get this to actually give us a sensical result is when the absolute value of r is between 0 and 1. We've already talked about, we're not even dealing with the geometric, we're not even talking about a geometric series if r is equal to 0. So let's think about the case where the absolute value of r is greater than 0 and it is less than 1. What's going to happen in that case? Well, the denominator is going to make sense right over here."}, {"video_title": "Sum of an infinite geometric series Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "We've already talked about, we're not even dealing with the geometric, we're not even talking about a geometric series if r is equal to 0. So let's think about the case where the absolute value of r is greater than 0 and it is less than 1. What's going to happen in that case? Well, the denominator is going to make sense right over here. And then up here, what's going to happen? Well, if you take something with an absolute value less than 1 and you take it to higher and higher and higher exponents, every time you multiply it by itself, you're going to get a number with a smaller absolute value. So this term right over here, this entire term, is going to go to 0 as n approaches infinity."}, {"video_title": "Sum of an infinite geometric series Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "Well, the denominator is going to make sense right over here. And then up here, what's going to happen? Well, if you take something with an absolute value less than 1 and you take it to higher and higher and higher exponents, every time you multiply it by itself, you're going to get a number with a smaller absolute value. So this term right over here, this entire term, is going to go to 0 as n approaches infinity. Imagine if r was 1 half. You're talking about 1 half to the 100th power, 1 half to the 1,000th power, 1 half to the millionth power, 1 half to the billionth power. That quickly approaches 0."}, {"video_title": "Sum of an infinite geometric series Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "So this term right over here, this entire term, is going to go to 0 as n approaches infinity. Imagine if r was 1 half. You're talking about 1 half to the 100th power, 1 half to the 1,000th power, 1 half to the millionth power, 1 half to the billionth power. That quickly approaches 0. So this goes to 0 if the absolute value of r is less than 1. So this, we could argue, would be equal to a over 1 minus r. So for example, if I had the infinite geometric series, let's just have a simple one. Let's say that my first term is 1."}, {"video_title": "Sum of an infinite geometric series Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "That quickly approaches 0. So this goes to 0 if the absolute value of r is less than 1. So this, we could argue, would be equal to a over 1 minus r. So for example, if I had the infinite geometric series, let's just have a simple one. Let's say that my first term is 1. And then each successive term, I'm going to multiply by 1 third. So it's 1 plus 1 third plus 1 third squared plus 1 third to the third plus. And I were to just keep on going forever, this is telling us that that sum, this infinite sum, I have an infinite number of terms here."}, {"video_title": "Sum of an infinite geometric series Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "Let's say that my first term is 1. And then each successive term, I'm going to multiply by 1 third. So it's 1 plus 1 third plus 1 third squared plus 1 third to the third plus. And I were to just keep on going forever, this is telling us that that sum, this infinite sum, I have an infinite number of terms here. This is a pretty fascinating concept here. We'll come out to this. It's going to be my first term."}, {"video_title": "Sum of an infinite geometric series Sequences, series and induction Precalculus Khan Academy.mp3", "Sentence": "And I were to just keep on going forever, this is telling us that that sum, this infinite sum, I have an infinite number of terms here. This is a pretty fascinating concept here. We'll come out to this. It's going to be my first term. It's going to be my first term, 1 over 1 minus my common ratio. My common ratio in this case is 1 third. 1 minus 1 third, which is the same thing as 1 over 2 thirds, which is equal to 3 halves."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "We've got the function f of x is equal to x to the third power minus 12x plus 2. And what I want to do in this video is think about at what points does my function f take on minimum or maximum values. And to figure that out, I have to first figure out what are the critical points for my function f, and then which of those critical points do we achieve a minimum or maximum value. And to determine the critical points, we have to find the derivative of our function, because our critical points are just the points at which our derivative is either equal to 0 or undefined. So the derivative of this thing right over here, we're just going to use the power rule several times, and then I guess you could call it the constant rule. But the derivative of x to the third is 3x squared. Derivative of negative 12x is negative 12."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "And to determine the critical points, we have to find the derivative of our function, because our critical points are just the points at which our derivative is either equal to 0 or undefined. So the derivative of this thing right over here, we're just going to use the power rule several times, and then I guess you could call it the constant rule. But the derivative of x to the third is 3x squared. Derivative of negative 12x is negative 12. And the derivative of a constant, it doesn't change with respect to x, so it's just going to be equal to 0. So we're going to get a critical point when this thing right over here, for some value of x, is either undefined or 0. Well, this thing is defined for all values of x."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "Derivative of negative 12x is negative 12. And the derivative of a constant, it doesn't change with respect to x, so it's just going to be equal to 0. So we're going to get a critical point when this thing right over here, for some value of x, is either undefined or 0. Well, this thing is defined for all values of x. So the only places we're going to find critical points is when this thing is equal to 0. So let's set it equal to 0. When does 3x squared minus 12 equal 0?"}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "Well, this thing is defined for all values of x. So the only places we're going to find critical points is when this thing is equal to 0. So let's set it equal to 0. When does 3x squared minus 12 equal 0? So let's add 12 to both sides. You get 3x squared is equal to 12. Divide both sides by 3."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "When does 3x squared minus 12 equal 0? So let's add 12 to both sides. You get 3x squared is equal to 12. Divide both sides by 3. You get x squared is equal to 4. Well, this is going to happen when x is equal to 2 and x is equal to negative 2. Just to be clear, f prime of 2, you get 3 times 4 minus 12, which is equal to 0."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "Divide both sides by 3. You get x squared is equal to 4. Well, this is going to happen when x is equal to 2 and x is equal to negative 2. Just to be clear, f prime of 2, you get 3 times 4 minus 12, which is equal to 0. And f prime of negative 2 is also, same exact reason, is also equal to 0. So we can say, and I'll switch colors here, that f has critical points at x equals 2 and x equals negative 2. Well, that's fair enough."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "Just to be clear, f prime of 2, you get 3 times 4 minus 12, which is equal to 0. And f prime of negative 2 is also, same exact reason, is also equal to 0. So we can say, and I'll switch colors here, that f has critical points at x equals 2 and x equals negative 2. Well, that's fair enough. But we still don't know whether the function takes on minimum values at those points, maximum values at those points, or neither. To figure that out, we have to figure out whether the derivative changes signs around these points. So let's actually try to graph the derivative to think about this."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "Well, that's fair enough. But we still don't know whether the function takes on minimum values at those points, maximum values at those points, or neither. To figure that out, we have to figure out whether the derivative changes signs around these points. So let's actually try to graph the derivative to think about this. So let's graph. So I'll draw an axis right over here. I'll do it down here, because maybe we can use that information later on to graph f of x."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "So let's actually try to graph the derivative to think about this. So let's graph. So I'll draw an axis right over here. I'll do it down here, because maybe we can use that information later on to graph f of x. So let's say this is my x-axis. This is my y-axis. And so we have critical points at x is equal to positive 2."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "I'll do it down here, because maybe we can use that information later on to graph f of x. So let's say this is my x-axis. This is my y-axis. And so we have critical points at x is equal to positive 2. So it's 1, 2. And x is equal to negative 2. 1, 2. x is equal to negative 2."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "And so we have critical points at x is equal to positive 2. So it's 1, 2. And x is equal to negative 2. 1, 2. x is equal to negative 2. So what does this derivative look like if we wanted to graph it? Well, when x is equal to 0 for the derivative, we're at negative 12. So this is a point y is equal to negative 12."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "1, 2. x is equal to negative 2. So what does this derivative look like if we wanted to graph it? Well, when x is equal to 0 for the derivative, we're at negative 12. So this is a point y is equal to negative 12. So we're graphing y is equal to f prime of x. So it'll look something like this. It will look something like this."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "So this is a point y is equal to negative 12. So we're graphing y is equal to f prime of x. So it'll look something like this. It will look something like this. These are obviously the 0's of our derivative. So it has to move up to cross the x-axis there and over here. So what is the derivative doing at each of these critical points?"}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "It will look something like this. These are obviously the 0's of our derivative. So it has to move up to cross the x-axis there and over here. So what is the derivative doing at each of these critical points? Well, over here, our derivative is crossing from being positive. We have a positive derivative to being a negative derivative. So we're crossing from being a positive derivative to being a negative derivative."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "So what is the derivative doing at each of these critical points? Well, over here, our derivative is crossing from being positive. We have a positive derivative to being a negative derivative. So we're crossing from being a positive derivative to being a negative derivative. That was our criteria for a critical point to be a maximum point. Over here, we're crossing from a negative derivative to a positive derivative, which is our criteria for a critical point for the function to have a minimum value at a critical point. So a minimum."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "So we're crossing from being a positive derivative to being a negative derivative. That was our criteria for a critical point to be a maximum point. Over here, we're crossing from a negative derivative to a positive derivative, which is our criteria for a critical point for the function to have a minimum value at a critical point. So a minimum. And I just want to make sure we have the correct intuition. If some function is increasing going into some point, and at that point, we see actually we have a derivative of 0. The derivative could also be undefined."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "So a minimum. And I just want to make sure we have the correct intuition. If some function is increasing going into some point, and at that point, we see actually we have a derivative of 0. The derivative could also be undefined. We have a derivative of 0. And then the function begins decreasing. That's why this would be a maximum point."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "The derivative could also be undefined. We have a derivative of 0. And then the function begins decreasing. That's why this would be a maximum point. Similarly, if we have a situation where the function is decreasing going into a point, the derivative is negative. Remember, this is the graph of the derivative. Let me make this clear."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "That's why this would be a maximum point. Similarly, if we have a situation where the function is decreasing going into a point, the derivative is negative. Remember, this is the graph of the derivative. Let me make this clear. This is the graph of y is equal to not f of x, but f prime of x. So if we have a situation where going into the point, the function has a negative slope. We see we have a negative slope here."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "Let me make this clear. This is the graph of y is equal to not f of x, but f prime of x. So if we have a situation where going into the point, the function has a negative slope. We see we have a negative slope here. So the function might look something like this. And then right at this point, the function is either undefined or has 0 slope. So in this case, it has 0 slope."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "We see we have a negative slope here. So the function might look something like this. And then right at this point, the function is either undefined or has 0 slope. So in this case, it has 0 slope. And then after that point, let me do it right under it. So going into it, we have a negative slope. And then right over here, we have a 0 slope."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "So in this case, it has 0 slope. And then after that point, let me do it right under it. So going into it, we have a negative slope. And then right over here, we have a 0 slope. Actually, I could draw it even better than that. So if we were to imagine going into it, we have a negative slope. Right at that point, we have a 0 slope."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "And then right over here, we have a 0 slope. Actually, I could draw it even better than that. So if we were to imagine going into it, we have a negative slope. Right at that point, we have a 0 slope. And then we have a positive slope. So the function begins increasing. That's why we say we have a minimum point right over there."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "Right at that point, we have a 0 slope. And then we have a positive slope. So the function begins increasing. That's why we say we have a minimum point right over there. So what I did right over here is to try to conceptualize what the function itself could look like given the derivative, in this case, switching from a positive derivative to a negative derivative across that critical point or going from a negative derivative to a positive derivative. That's why this is a criteria for a maximum point. This is a criteria for a minimum point."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "That's why we say we have a minimum point right over there. So what I did right over here is to try to conceptualize what the function itself could look like given the derivative, in this case, switching from a positive derivative to a negative derivative across that critical point or going from a negative derivative to a positive derivative. That's why this is a criteria for a maximum point. This is a criteria for a minimum point. Well, with that out of the way, can we use this intuition that we just talked about to at least try to sketch the graph of f of x? So let's try to do it. And it's just going to be a sketch."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "This is a criteria for a minimum point. Well, with that out of the way, can we use this intuition that we just talked about to at least try to sketch the graph of f of x? So let's try to do it. And it's just going to be a sketch. It's not going to be very exact. But at least it will give us a sense of the shape of what f of x looks like. So my best attempt."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "And it's just going to be a sketch. It's not going to be very exact. But at least it will give us a sense of the shape of what f of x looks like. So my best attempt. So it might not be drawn completely to scale. So it's my x-axis. This is my y-axis."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "So my best attempt. So it might not be drawn completely to scale. So it's my x-axis. This is my y-axis. We know we have a critical point at x is equal to positive 2. And we have a critical point at x is equal to negative 2. We know just from inspection that the y-intercept right here of the graph of y is equal to f of x, when x is 0, f of x is 2."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "This is my y-axis. We know we have a critical point at x is equal to positive 2. And we have a critical point at x is equal to negative 2. We know just from inspection that the y-intercept right here of the graph of y is equal to f of x, when x is 0, f of x is 2. So we're going to hit right over. Let's say we're going to hit right over. I don't want to draw this completely to the same scale as the x-axis."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "We know just from inspection that the y-intercept right here of the graph of y is equal to f of x, when x is 0, f of x is 2. So we're going to hit right over. Let's say we're going to hit right over. I don't want to draw this completely to the same scale as the x-axis. So let's say that this is 2 right over here. So this is where we're going to cross. This is going to be our y-intercept."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "I don't want to draw this completely to the same scale as the x-axis. So let's say that this is 2 right over here. So this is where we're going to cross. This is going to be our y-intercept. And so we said already that we have a maximum point at x is equal to negative 2. So what is f of negative 2? f of negative 2 is equal to 8."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "This is going to be our y-intercept. And so we said already that we have a maximum point at x is equal to negative 2. So what is f of negative 2? f of negative 2 is equal to 8. Or negative 8. Let me be careful. It's negative 8."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "f of negative 2 is equal to 8. Or negative 8. Let me be careful. It's negative 8. And then we're going to have 12 times negative 2, which is negative 24. But then we're going to add it. So we're subtracting negative 24."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "It's negative 8. And then we're going to have 12 times negative 2, which is negative 24. But then we're going to add it. So we're subtracting negative 24. So this is plus 24. And then we finally have plus 2. So negative 8 plus 24 plus 2."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "So we're subtracting negative 24. So this is plus 24. And then we finally have plus 2. So negative 8 plus 24 plus 2. So that's going to be negative 8 plus 24 is 16 plus 2 is 18. So f of negative 2 is equal to 18. And I'm not drawing it completely to scale."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "So negative 8 plus 24 plus 2. So that's going to be negative 8 plus 24 is 16 plus 2 is 18. So f of negative 2 is equal to 18. And I'm not drawing it completely to scale. Let's say that this is 18 right over here. So this is the function. This is the point negative 2 comma 18."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "And I'm not drawing it completely to scale. Let's say that this is 18 right over here. So this is the function. This is the point negative 2 comma 18. And we know that it's a maximum point. The derivative going into that point is negative. The derivative going into that point is positive."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "This is the point negative 2 comma 18. And we know that it's a maximum point. The derivative going into that point is negative. The derivative going into that point is positive. So we are increasing. The slope is positive. And then after we cross that point, the slope becomes negative."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "The derivative going into that point is positive. So we are increasing. The slope is positive. And then after we cross that point, the slope becomes negative. The derivative crossed the x-axis. The slope becomes negative. Actually, I want to use that same color."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "And then after we cross that point, the slope becomes negative. The derivative crossed the x-axis. The slope becomes negative. Actually, I want to use that same color. It looks like this. And then, of course, the graph is going to have a y-intercept, something like that. And then as we approach 2, we are approaching another critical point."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "Actually, I want to use that same color. It looks like this. And then, of course, the graph is going to have a y-intercept, something like that. And then as we approach 2, we are approaching another critical point. Now what is f of 2? f of 2 is going to be equal to positive 8 minus 24 plus 2. So this is 10 minus 24, which is equal to negative 14."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "And then as we approach 2, we are approaching another critical point. Now what is f of 2? f of 2 is going to be equal to positive 8 minus 24 plus 2. So this is 10 minus 24, which is equal to negative 14. So let's say that this is a point negative 14 right over here. Actually, I could draw it a little bit. Let's say this is negative 14."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "So this is 10 minus 24, which is equal to negative 14. So let's say that this is a point negative 14 right over here. Actually, I could draw it a little bit. Let's say this is negative 14. So this is f of 2 right over there. And we saw already that the slope is negative as we approach it. So our function is decreasing as we approach it."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say this is negative 14. So this is f of 2 right over there. And we saw already that the slope is negative as we approach it. So our function is decreasing as we approach it. And then right there, the slope is 0. We figured that out earlier. That's how we identified it being a critical point."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "So our function is decreasing as we approach it. And then right there, the slope is 0. We figured that out earlier. That's how we identified it being a critical point. And then the slope is increasing after that. The derivative is positive. The slope is increasing."}, {"video_title": "Analyzing a function with its derivative AP Calculus AB Khan Academy.mp3", "Sentence": "That's how we identified it being a critical point. And then the slope is increasing after that. The derivative is positive. The slope is increasing. So this is our sketch of what f of x could look like given that these are the critical points. And we were able to identify 2 as a minimum point. So this was a minimum value."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "You've opened up a shoe factory, and you're trying to figure out how many thousands of shoes to produce in order to optimize your profit. And so let's let x equal the thousands of pairs produced. Thousands of pairs produced. Now let's think about how much money you're going to make per pair. Actually, let me say how much revenue, which is how much do you actually get to sell those shoes for? So let's write a function right here. Revenue as a function of x."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's think about how much money you're going to make per pair. Actually, let me say how much revenue, which is how much do you actually get to sell those shoes for? So let's write a function right here. Revenue as a function of x. Well, you have a wholesaler who's willing to pay you $10 per pair for as many pairs as you're willing to give him. So your revenue as a function of x is going to be 10 times x. And since x is in thousands of pairs produced, if x is one, that means 1,000 pairs produced times 10, which means $10,000."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Revenue as a function of x. Well, you have a wholesaler who's willing to pay you $10 per pair for as many pairs as you're willing to give him. So your revenue as a function of x is going to be 10 times x. And since x is in thousands of pairs produced, if x is one, that means 1,000 pairs produced times 10, which means $10,000. But this will just give you 10. So this right over here is in thousands of dollars. Thousands of dollars."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And since x is in thousands of pairs produced, if x is one, that means 1,000 pairs produced times 10, which means $10,000. But this will just give you 10. So this right over here is in thousands of dollars. Thousands of dollars. So if x is one, that means 1,000 pairs produced. 10 times one says r is equal to 10, but that really means $10,000. Now, it would be a nice business if all you had was revenue and no costs, but you do have costs."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Thousands of dollars. So if x is one, that means 1,000 pairs produced. 10 times one says r is equal to 10, but that really means $10,000. Now, it would be a nice business if all you had was revenue and no costs, but you do have costs. You have materials, you have to build your factory, you have to pay your employees, you have to pay the electricity bill. And so you hire a bunch of consultants to come up with what your cost is as a function of x. What your cost is as a function of x."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Now, it would be a nice business if all you had was revenue and no costs, but you do have costs. You have materials, you have to build your factory, you have to pay your employees, you have to pay the electricity bill. And so you hire a bunch of consultants to come up with what your cost is as a function of x. What your cost is as a function of x. And they come up with a function. They say it is the number of the thousands of pairs you produce cubed minus six times the thousands of pairs you produce squared plus 15 times the thousands of pairs that you produce. And once again, this is going to be, this is also going to be in thousands of dollars."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "What your cost is as a function of x. And they come up with a function. They say it is the number of the thousands of pairs you produce cubed minus six times the thousands of pairs you produce squared plus 15 times the thousands of pairs that you produce. And once again, this is going to be, this is also going to be in thousands of dollars. Now, given these functions of x for revenue and cost, what is profit as a function of x going to be? Well, your profit as a function of x is just going to be equal to your revenue as a function of x minus your cost as a function of x. Minus your cost as a function of x."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And once again, this is going to be, this is also going to be in thousands of dollars. Now, given these functions of x for revenue and cost, what is profit as a function of x going to be? Well, your profit as a function of x is just going to be equal to your revenue as a function of x minus your cost as a function of x. Minus your cost as a function of x. If you produce a certain amount, and let's say you bring in, I don't know, $10,000 of revenue and it costs you $5,000 to produce those shoes, you'll have $5,000 in profit. Those numbers aren't the ones that would actually, you would get from this right here. I'm just giving you an example."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Minus your cost as a function of x. If you produce a certain amount, and let's say you bring in, I don't know, $10,000 of revenue and it costs you $5,000 to produce those shoes, you'll have $5,000 in profit. Those numbers aren't the ones that would actually, you would get from this right here. I'm just giving you an example. So this is what you want to optimize. You want to optimize p as a function, optimize p as a function of x. So what is it?"}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "I'm just giving you an example. So this is what you want to optimize. You want to optimize p as a function, optimize p as a function of x. So what is it? I've just said it here in abstract terms, but we know what r of x is and what c of x. This is 10x, this is 10x minus all of this business. So minus x to the third plus 6x squared minus 15x."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So what is it? I've just said it here in abstract terms, but we know what r of x is and what c of x. This is 10x, this is 10x minus all of this business. So minus x to the third plus 6x squared minus 15x. I just subtracted x squared. You subtract a 6x squared, it becomes positive. You subtract a 15x, it becomes negative 15x."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So minus x to the third plus 6x squared minus 15x. I just subtracted x squared. You subtract a 6x squared, it becomes positive. You subtract a 15x, it becomes negative 15x. And then we can simplify this as, let's see, we have negative x to the third plus 6x squared minus 15x plus 10x, so that is minus 5x. Now if we want to optimize this profit function analytically, the easiest way is to think about what are the critical points of this profit function and are any of those critical points minimum points or maximum points? And if one of them is a maximum point, then we can say, well, let's produce that many."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "You subtract a 15x, it becomes negative 15x. And then we can simplify this as, let's see, we have negative x to the third plus 6x squared minus 15x plus 10x, so that is minus 5x. Now if we want to optimize this profit function analytically, the easiest way is to think about what are the critical points of this profit function and are any of those critical points minimum points or maximum points? And if one of them is a maximum point, then we can say, well, let's produce that many. That is going to be, we will have optimized or we'll figure out the quantity we need to produce in order to optimize our profit. So to figure out critical points, we essentially have to find the derivative of our function and figure out when does that derivative equal zero or when is that derivative undefined? That's the definition of critical points."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And if one of them is a maximum point, then we can say, well, let's produce that many. That is going to be, we will have optimized or we'll figure out the quantity we need to produce in order to optimize our profit. So to figure out critical points, we essentially have to find the derivative of our function and figure out when does that derivative equal zero or when is that derivative undefined? That's the definition of critical points. So p prime of x is going to be equal to negative 3x squared plus 12x minus five. And so this thing is going to be defined for all x. So the only critical points we're going to have is when the first derivative right over here is equal to zero."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "That's the definition of critical points. So p prime of x is going to be equal to negative 3x squared plus 12x minus five. And so this thing is going to be defined for all x. So the only critical points we're going to have is when the first derivative right over here is equal to zero. So negative 3x squared plus 12x minus five needs to be equal to zero in order for x to be a critical point. So now we just have to solve for x. And so we just are essentially solving a quadratic equation just so that I don't have as many negatives."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So the only critical points we're going to have is when the first derivative right over here is equal to zero. So negative 3x squared plus 12x minus five needs to be equal to zero in order for x to be a critical point. So now we just have to solve for x. And so we just are essentially solving a quadratic equation just so that I don't have as many negatives. Let's multiply both sides by negative one. I just like to have a clean first coefficient. So if we multiply both sides by negative one, we get 3x squared minus 12x plus five is equal to zero."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And so we just are essentially solving a quadratic equation just so that I don't have as many negatives. Let's multiply both sides by negative one. I just like to have a clean first coefficient. So if we multiply both sides by negative one, we get 3x squared minus 12x plus five is equal to zero. And now we can use the quadratic formula to solve for x. So x is going to be equal to negative b, which is 12 plus or minus the square root. I always need to make my radical signs wide enough."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So if we multiply both sides by negative one, we get 3x squared minus 12x plus five is equal to zero. And now we can use the quadratic formula to solve for x. So x is going to be equal to negative b, which is 12 plus or minus the square root. I always need to make my radical signs wide enough. The square root of b squared, which is 144, minus four times a, which is three, times c, which is five, times five. All of that, all of that over 2a. So two times three is six."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "I always need to make my radical signs wide enough. The square root of b squared, which is 144, minus four times a, which is three, times c, which is five, times five. All of that, all of that over 2a. So two times three is six. So x is equal to 12 plus or minus the square root of, let's see, four times three is 12, times five is 60. 144 minus 60 is 84. All of that over six."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So two times three is six. So x is equal to 12 plus or minus the square root of, let's see, four times three is 12, times five is 60. 144 minus 60 is 84. All of that over six. So x could be equal to 12 plus the square root of 84 over six or x could be equal to 12 minus the square root of 84 over six. So let's figure out what these two are. And I'll use the calculator."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "All of that over six. So x could be equal to 12 plus the square root of 84 over six or x could be equal to 12 minus the square root of 84 over six. So let's figure out what these two are. And I'll use the calculator. I'll use the calculator for this one. So I get, let's see, 12 plus the square root of 84 divided by six gives me 3.5, I'll just say 3.53. So approximately, actually let me go one more digit because I'm talking about thousands."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And I'll use the calculator. I'll use the calculator for this one. So I get, let's see, 12 plus the square root of 84 divided by six gives me 3.5, I'll just say 3.53. So approximately, actually let me go one more digit because I'm talking about thousands. So let me say 3.528. 3.528, 3.528. So this would literally be 3,528 shoes because this is in thousands."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So approximately, actually let me go one more digit because I'm talking about thousands. So let me say 3.528. 3.528, 3.528. So this would literally be 3,528 shoes because this is in thousands. And let's do our pairs of shoes. And then let's do the situation where we subtract. And actually we can look at our previous entry and just change this to a subtraction."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So this would literally be 3,528 shoes because this is in thousands. And let's do our pairs of shoes. And then let's do the situation where we subtract. And actually we can look at our previous entry and just change this to a subtraction. Change that to not a negative sign, a subtraction. There you go. And we get.4725,.47, let me remember that,.4725."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And actually we can look at our previous entry and just change this to a subtraction. Change that to not a negative sign, a subtraction. There you go. And we get.4725,.47, let me remember that,.4725. Approximately equal to 0.4725. I have a horrible memory so let me review that I wrote the same thing. 47, 4725, yep, all right."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And we get.4725,.47, let me remember that,.4725. Approximately equal to 0.4725. I have a horrible memory so let me review that I wrote the same thing. 47, 4725, yep, all right. Now, all we know about these are these are both critical points. These are points at which our derivative is equal to zero. But we don't know whether they're minimum points."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "47, 4725, yep, all right. Now, all we know about these are these are both critical points. These are points at which our derivative is equal to zero. But we don't know whether they're minimum points. They're points at which the function takes on a minimum value, a maximum value, or neither. To do that, I'll use the second derivative test to figure out if our function is concave upwards or concave downwards or neither at one of these points. So let's look at the second derivative."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "But we don't know whether they're minimum points. They're points at which the function takes on a minimum value, a maximum value, or neither. To do that, I'll use the second derivative test to figure out if our function is concave upwards or concave downwards or neither at one of these points. So let's look at the second derivative. So P prime prime of x is going to be equal to negative six x plus negative six x plus 12. And so if we look at, let me make sure I have enough space. So if we look at P prime prime, P prime prime of 3.528."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let's look at the second derivative. So P prime prime of x is going to be equal to negative six x plus negative six x plus 12. And so if we look at, let me make sure I have enough space. So if we look at P prime prime, P prime prime of 3.528. So let's see if I can think this through. So this is between three and four. So if we take the lower value, three times negative six is negative 18 plus 12."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So if we look at P prime prime, P prime prime of 3.528. So let's see if I can think this through. So this is between three and four. So if we take the lower value, three times negative six is negative 18 plus 12. So that's going to be less than zero. And if this was four, it would be even more negative. So this thing is going to be less than zero."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So if we take the lower value, three times negative six is negative 18 plus 12. So that's going to be less than zero. And if this was four, it would be even more negative. So this thing is going to be less than zero. I don't even have to use my calculator to evaluate it. Now what about this thing right over here,.47? Well,.47, that's gonna get us, that's roughly.5."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So this thing is going to be less than zero. I don't even have to use my calculator to evaluate it. Now what about this thing right over here,.47? Well,.47, that's gonna get us, that's roughly.5. So negative six times.5 is negative three. This is going to be nowhere close to being negative. This is definitely going to be positive."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well,.47, that's gonna get us, that's roughly.5. So negative six times.5 is negative three. This is going to be nowhere close to being negative. This is definitely going to be positive. So P prime prime of 0.4725 is greater than zero. So the fact that the second derivative is less than zero, that means that my derivative is decreasing, my derivative, my first derivative is decreasing when x is equal to this value, which means that our graph, our function is concave downwards here. Concave, concave downwards."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "This is definitely going to be positive. So P prime prime of 0.4725 is greater than zero. So the fact that the second derivative is less than zero, that means that my derivative is decreasing, my derivative, my first derivative is decreasing when x is equal to this value, which means that our graph, our function is concave downwards here. Concave, concave downwards. And concave downwards means it looks something like this. And so, and you can see when it looks something like that, the slope is constantly decreasing. So if you have an interval where the slope is decreasing and you know the point where the slope is exactly zero, which is where x is equal to 3.528, it must be a maximum."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Concave, concave downwards. And concave downwards means it looks something like this. And so, and you can see when it looks something like that, the slope is constantly decreasing. So if you have an interval where the slope is decreasing and you know the point where the slope is exactly zero, which is where x is equal to 3.528, it must be a maximum. It must be a maximum. So we actually do take on a maximum value when x is 3.528. On the other side, we see that over here we are concave upwards, concave upwards."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So if you have an interval where the slope is decreasing and you know the point where the slope is exactly zero, which is where x is equal to 3.528, it must be a maximum. It must be a maximum. So we actually do take on a maximum value when x is 3.528. On the other side, we see that over here we are concave upwards, concave upwards. Concave upwards, the graph will look something like this over here. And if the slope is zero where the graph looks like that, we see that that is a local minimum. That is a local minimum."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "On the other side, we see that over here we are concave upwards, concave upwards. Concave upwards, the graph will look something like this over here. And if the slope is zero where the graph looks like that, we see that that is a local minimum. That is a local minimum. And so we definitely don't want to do this. We would produce 472 and a half units if we were looking to minimize our profit, maximize our loss. So we definitely don't want to do this."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "That is a local minimum. And so we definitely don't want to do this. We would produce 472 and a half units if we were looking to minimize our profit, maximize our loss. So we definitely don't want to do this. But let's actually think about what our profit is going to be if we produce 3.528 thousands of shoes, or 3,528 shoes. Well, to do that, we just have to input it back into our original profit function right over here. So let's do that."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So we definitely don't want to do this. But let's actually think about what our profit is going to be if we produce 3.528 thousands of shoes, or 3,528 shoes. Well, to do that, we just have to input it back into our original profit function right over here. So let's do that. So get my calculator out. So my original profit function is right over there. So I want to be able to see that and that."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let's do that. So get my calculator out. So my original profit function is right over there. So I want to be able to see that and that. So I get negative 3.528 to the third power plus 6 times 3.528 squared minus 5 times 3.528 gives me, and we get a drum roll now, gives me a profit of 13.128. So let me write this down. The profit when I produce 3,528 shoes is approximately equal to, or it is equal to, if I produce exactly that many shoes, it's equal to 13.128."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So I want to be able to see that and that. So I get negative 3.528 to the third power plus 6 times 3.528 squared minus 5 times 3.528 gives me, and we get a drum roll now, gives me a profit of 13.128. So let me write this down. The profit when I produce 3,528 shoes is approximately equal to, or it is equal to, if I produce exactly that many shoes, it's equal to 13.128. Or actually, it's approximately because I'm still rounding. 13.128. So if I produce 3,528 shoes in a given period, I am going to have a profit of $13,128."}, {"video_title": "Optimization profit Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "The profit when I produce 3,528 shoes is approximately equal to, or it is equal to, if I produce exactly that many shoes, it's equal to 13.128. Or actually, it's approximately because I'm still rounding. 13.128. So if I produce 3,528 shoes in a given period, I am going to have a profit of $13,128. Remember, this right over here is in thousands. This right over here is 13.128 thousands of dollars in profit, which is $13,128. Anyway, we are now going to be rich shoe manufacturers."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Differentiability. So let's think about that first. It's always helpful to draw ourselves a function. So that's our y-axis. This is our x-axis. And let's just draw some function here. So let's say my function looks like this."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So that's our y-axis. This is our x-axis. And let's just draw some function here. So let's say my function looks like this. And we care about the point x equals C, which is right over here. So that's the point x equals C. And then this value, of course, is going to be f of C. And one way that we can find the derivative at x equals C, or the slope of the tangent line at x equals C, is we could start with some other point, say some arbitrary x out here. So let's say this is some arbitrary x out here."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say my function looks like this. And we care about the point x equals C, which is right over here. So that's the point x equals C. And then this value, of course, is going to be f of C. And one way that we can find the derivative at x equals C, or the slope of the tangent line at x equals C, is we could start with some other point, say some arbitrary x out here. So let's say this is some arbitrary x out here. So then this point right over there, this value, this y value, would be f of x. This graph, of course, is a graph of y equals f of x. And we can think about finding the slope of this line, this secant line between these two points."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say this is some arbitrary x out here. So then this point right over there, this value, this y value, would be f of x. This graph, of course, is a graph of y equals f of x. And we can think about finding the slope of this line, this secant line between these two points. But then we can find the limit as x approaches C. And as x approaches C, this secant, the slope of the secant line is going to approach the slope of the tangent line, or it's going to be the derivative. And so we could take the limit as x approaches C, of the slope of this secant line. So what's the slope?"}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And we can think about finding the slope of this line, this secant line between these two points. But then we can find the limit as x approaches C. And as x approaches C, this secant, the slope of the secant line is going to approach the slope of the tangent line, or it's going to be the derivative. And so we could take the limit as x approaches C, of the slope of this secant line. So what's the slope? Well, it's going to be change in y over change in x. The change in y is f of x minus f of C. That's our change in y right over here. And this is all a review."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So what's the slope? Well, it's going to be change in y over change in x. The change in y is f of x minus f of C. That's our change in y right over here. And this is all a review. This is just one definition of the derivative, or one way to think about the derivative. So it's going to be f of x minus f of C, that's our change in y, over our change in x, which is x minus C. It is x minus C. So if this limit exists, then we're able to find the slope of the tangent line at this point. And we call that slope of the tangent line, we call that the derivative at x equals C. We say that this is going to be equal to f prime of C. All of this is review."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And this is all a review. This is just one definition of the derivative, or one way to think about the derivative. So it's going to be f of x minus f of C, that's our change in y, over our change in x, which is x minus C. It is x minus C. So if this limit exists, then we're able to find the slope of the tangent line at this point. And we call that slope of the tangent line, we call that the derivative at x equals C. We say that this is going to be equal to f prime of C. All of this is review. So if we're saying, one way to think about it, if we're saying that the function f is differentiable at x equals C, we're really just saying that this limit right over here actually exists. And if this limit actually exists, we just call that value f prime of C. So that's just a review of differentiability. Now let's give ourselves a review of continuity."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And we call that slope of the tangent line, we call that the derivative at x equals C. We say that this is going to be equal to f prime of C. All of this is review. So if we're saying, one way to think about it, if we're saying that the function f is differentiable at x equals C, we're really just saying that this limit right over here actually exists. And if this limit actually exists, we just call that value f prime of C. So that's just a review of differentiability. Now let's give ourselves a review of continuity. Con-ti-nuity. So the definition for continuity is if the limit as x approaches C of f of x is equal to f of C. Now this might seem a little bit, you know, well, it might pop out to you as being intuitive, or it might seem a little, well, where did this come from? Well, let's visualize it, and then hopefully it'll make some intuitive sense."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's give ourselves a review of continuity. Con-ti-nuity. So the definition for continuity is if the limit as x approaches C of f of x is equal to f of C. Now this might seem a little bit, you know, well, it might pop out to you as being intuitive, or it might seem a little, well, where did this come from? Well, let's visualize it, and then hopefully it'll make some intuitive sense. So if you have a function, so let's actually look at some cases where you're not continuous. And that actually might make it a little bit more clear. So if you had a point discontinuity at x equals C, so this is x equals C. So if you had a point discontinuity, so let me draw it like this actually."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, let's visualize it, and then hopefully it'll make some intuitive sense. So if you have a function, so let's actually look at some cases where you're not continuous. And that actually might make it a little bit more clear. So if you had a point discontinuity at x equals C, so this is x equals C. So if you had a point discontinuity, so let me draw it like this actually. So you have a gap here, and x equals, when x equals C, f of C is actually way up here. So this is f of C, and then the function continues like this. The limit as x approaches C of f of x is going to be this value, which is clearly different than f of C, this value right over here."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So if you had a point discontinuity at x equals C, so this is x equals C. So if you had a point discontinuity, so let me draw it like this actually. So you have a gap here, and x equals, when x equals C, f of C is actually way up here. So this is f of C, and then the function continues like this. The limit as x approaches C of f of x is going to be this value, which is clearly different than f of C, this value right over here. If you take the limit, if you take the limit as x approaches C of f of x, you're approaching this value. This right over here is the limit as x approaches C of f of x, which is different than f of C. So this definition of continuity seems to be good at least for this case, because this is not a continuous function. You have a point discontinuity."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "The limit as x approaches C of f of x is going to be this value, which is clearly different than f of C, this value right over here. If you take the limit, if you take the limit as x approaches C of f of x, you're approaching this value. This right over here is the limit as x approaches C of f of x, which is different than f of C. So this definition of continuity seems to be good at least for this case, because this is not a continuous function. You have a point discontinuity. So for at least in this case, this definition of continuity would properly identify this as not a continuous function. Now you could also think about a jump discontinuity. You could also think about a jump discontinuity."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "You have a point discontinuity. So for at least in this case, this definition of continuity would properly identify this as not a continuous function. Now you could also think about a jump discontinuity. You could also think about a jump discontinuity. So let's look at this. And all of this is hopefully a little bit of review. So a jump discontinuity at C, at x equals C, might look like this."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "You could also think about a jump discontinuity. So let's look at this. And all of this is hopefully a little bit of review. So a jump discontinuity at C, at x equals C, might look like this. Might look like this. So this is at x equals C. So this is x equals C right over here. This would be f of C. But if you tried to evaluate the limit as x approaches C of f of x, you'd get a different value as you approach C from the negative side."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So a jump discontinuity at C, at x equals C, might look like this. Might look like this. So this is at x equals C. So this is x equals C right over here. This would be f of C. But if you tried to evaluate the limit as x approaches C of f of x, you'd get a different value as you approach C from the negative side. You would approach this value. And as you approach C from the positive side, you would approach f of C. And so the limit wouldn't exist. So this limit right over here wouldn't exist in the case of this type of a jump discontinuity."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "This would be f of C. But if you tried to evaluate the limit as x approaches C of f of x, you'd get a different value as you approach C from the negative side. You would approach this value. And as you approach C from the positive side, you would approach f of C. And so the limit wouldn't exist. So this limit right over here wouldn't exist in the case of this type of a jump discontinuity. So once again, this definition would properly say that this is not, this one right over here is not continuous. This limit actually would not even exist. And then you could even look at a, you could look at a function that is truly continuous."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So this limit right over here wouldn't exist in the case of this type of a jump discontinuity. So once again, this definition would properly say that this is not, this one right over here is not continuous. This limit actually would not even exist. And then you could even look at a, you could look at a function that is truly continuous. If you look at a function that is truly continuous, so something like this. Something like this. That is x equals C. Well, this is f of C. This is f of C. And if you were to take the limit as x approaches C, as x approaches C from either side of f of x, you're going to approach f of C. So here you have the limit as x approaches C of f of x indeed is equal to f of C. So it's what you would expect for a continuous function."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And then you could even look at a, you could look at a function that is truly continuous. If you look at a function that is truly continuous, so something like this. Something like this. That is x equals C. Well, this is f of C. This is f of C. And if you were to take the limit as x approaches C, as x approaches C from either side of f of x, you're going to approach f of C. So here you have the limit as x approaches C of f of x indeed is equal to f of C. So it's what you would expect for a continuous function. So now that we've done that review of differentiability and continuity, let's prove that differentiability actually implies continuity. And I think it's important to kind of do this review just so that you can really visualize things. So differentiability implies this, this limit right over here exists."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "That is x equals C. Well, this is f of C. This is f of C. And if you were to take the limit as x approaches C, as x approaches C from either side of f of x, you're going to approach f of C. So here you have the limit as x approaches C of f of x indeed is equal to f of C. So it's what you would expect for a continuous function. So now that we've done that review of differentiability and continuity, let's prove that differentiability actually implies continuity. And I think it's important to kind of do this review just so that you can really visualize things. So differentiability implies this, this limit right over here exists. So let's start with a slightly different limit. Let me draw a line here actually. Let me draw a line just so we're doing something different."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So differentiability implies this, this limit right over here exists. So let's start with a slightly different limit. Let me draw a line here actually. Let me draw a line just so we're doing something different. So let's take, let us take the limit as x approaches C of f of x, of f of x minus f of C. Of f of x minus f of C. Well can we rewrite this? Well we could rewrite this as the limit as x approaches C. And we can essentially take this expression and multiply and divide it by x minus C. So let's multiply it times x minus C. x minus C and divide it by x minus C. So we have f of x minus f of C. All of that over x minus C. So all I did is I multiplied and I divided by x minus C. Well what's this limit going to be equal to? This is going to be equal to, it's going to be the limit, and I'm just applying the property of limit, property, I'm applying a property of limits here."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Let me draw a line just so we're doing something different. So let's take, let us take the limit as x approaches C of f of x, of f of x minus f of C. Of f of x minus f of C. Well can we rewrite this? Well we could rewrite this as the limit as x approaches C. And we can essentially take this expression and multiply and divide it by x minus C. So let's multiply it times x minus C. x minus C and divide it by x minus C. So we have f of x minus f of C. All of that over x minus C. So all I did is I multiplied and I divided by x minus C. Well what's this limit going to be equal to? This is going to be equal to, it's going to be the limit, and I'm just applying the property of limit, property, I'm applying a property of limits here. So the limit of the product is equal to the same thing as the product of the limits. So it's the limit as x approaches C of x minus C times the limit, let me write it this way, times the limit as x approaches C of f of x minus f of C. All of that over x minus C. Now what is this thing right over here? Well if we assume that f is differentiable at C, and we're going to do that, actually I should have started off there."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "This is going to be equal to, it's going to be the limit, and I'm just applying the property of limit, property, I'm applying a property of limits here. So the limit of the product is equal to the same thing as the product of the limits. So it's the limit as x approaches C of x minus C times the limit, let me write it this way, times the limit as x approaches C of f of x minus f of C. All of that over x minus C. Now what is this thing right over here? Well if we assume that f is differentiable at C, and we're going to do that, actually I should have started off there. Let's assume, let's assume, because we wanted to show that differentiability improves continuity. If we assume f differentiable, differentiable at C, well then this right over here is just going to be f prime of C. This right over here, we just saw it right over here, that's this exact same thing. This is f prime, f prime of C. And what is this thing right over here?"}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well if we assume that f is differentiable at C, and we're going to do that, actually I should have started off there. Let's assume, let's assume, because we wanted to show that differentiability improves continuity. If we assume f differentiable, differentiable at C, well then this right over here is just going to be f prime of C. This right over here, we just saw it right over here, that's this exact same thing. This is f prime, f prime of C. And what is this thing right over here? The limit as x approaches C of x minus C? Well that's just going to be zero. As x approaches C, it's going to approach C minus C, it's just going to be zero."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "This is f prime, f prime of C. And what is this thing right over here? The limit as x approaches C of x minus C? Well that's just going to be zero. As x approaches C, it's going to approach C minus C, it's just going to be zero. So what's zero times f prime of C? Well f prime of C is just going to be some value, so zero times anything is just going to be zero. So I did all that work to get a zero."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "As x approaches C, it's going to approach C minus C, it's just going to be zero. So what's zero times f prime of C? Well f prime of C is just going to be some value, so zero times anything is just going to be zero. So I did all that work to get a zero. Now why is this interesting? Well we just said, we just assumed that if f is differentiable at C, and we evaluate this limit, we get zero. So if we assume f is differentiable at C, we can write, we can write the limit, I'm just rewriting it, the limit as x approaches C of f of x minus f of C, and I could even put parentheses around it like that, which I already did up here, is equal to zero."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So I did all that work to get a zero. Now why is this interesting? Well we just said, we just assumed that if f is differentiable at C, and we evaluate this limit, we get zero. So if we assume f is differentiable at C, we can write, we can write the limit, I'm just rewriting it, the limit as x approaches C of f of x minus f of C, and I could even put parentheses around it like that, which I already did up here, is equal to zero. Well this is the same thing, I could use limit properties again, this is the same thing as saying, and I'll do it over here, actually let me do it down here. The limit as x approaches C of f of x minus the limit as x approaches C of f of C, of f of C, is equal to zero. The limit of the difference is the same thing as the difference of the limits."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So if we assume f is differentiable at C, we can write, we can write the limit, I'm just rewriting it, the limit as x approaches C of f of x minus f of C, and I could even put parentheses around it like that, which I already did up here, is equal to zero. Well this is the same thing, I could use limit properties again, this is the same thing as saying, and I'll do it over here, actually let me do it down here. The limit as x approaches C of f of x minus the limit as x approaches C of f of C, of f of C, is equal to zero. The limit of the difference is the same thing as the difference of the limits. Well what's this thing over here going to be? Well f of C is just a number, it's not a function of x anymore, it's just f of C is going to evaluate to something. So this is just going to be f of C. This is just going to be f of C. So if the limit of f of x as x approaches C minus f of C is equal to zero."}, {"video_title": "Proof Differentiability implies continuity Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "The limit of the difference is the same thing as the difference of the limits. Well what's this thing over here going to be? Well f of C is just a number, it's not a function of x anymore, it's just f of C is going to evaluate to something. So this is just going to be f of C. This is just going to be f of C. So if the limit of f of x as x approaches C minus f of C is equal to zero. Well just add f of C to both sides and what do you get? Well you get the limit as x approaches C of f of x is equal to f of C. And this is the definition of continuity, the limit of my function as x approaches C is equal to the function, is equal to the value of the function at C. This is, this means that our function is continuous. Continuous at C. So just a reminder, we started assuming f differentiable at C, we use that fact to evaluate this limit right over here, which we got to be equal to zero."}, {"video_title": "Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So we're told this table gives select values of the differentiable function f. So it gives us the value of the function at a few values for x, in particular five different values for x. It tells us what the corresponding f of x is. And they say what is the best estimate for f prime of four? So this is the derivative of our function f when x is equal to four. Or another way to think about it, what is the slope of the tangent line when x is equal to four for f of x? So what is the best estimate for f prime of four we can make based on this table? So let's just visualize what's going on before we even look at the choices."}, {"video_title": "Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So this is the derivative of our function f when x is equal to four. Or another way to think about it, what is the slope of the tangent line when x is equal to four for f of x? So what is the best estimate for f prime of four we can make based on this table? So let's just visualize what's going on before we even look at the choices. So let me draw some axes here. And let me plot these points. We know that these would sit on the curve of y is equal to f of x."}, {"video_title": "Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So let's just visualize what's going on before we even look at the choices. So let me draw some axes here. And let me plot these points. We know that these would sit on the curve of y is equal to f of x. When x is zero, f of x is 72. So this is the point zero, 72. This is the point three, 95."}, {"video_title": "Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "We know that these would sit on the curve of y is equal to f of x. When x is zero, f of x is 72. So this is the point zero, 72. This is the point three, 95. Clearly two different scales on the x and y-axis. This is the point five, 112. This is 677."}, {"video_title": "Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "This is the point three, 95. Clearly two different scales on the x and y-axis. This is the point five, 112. This is 677. This is 954. Actually, let me write out the, this is one, two, three, four, five, six, seven, eight, nine, and 10. Now they want us to know, they wanna know what is the derivative of our function when f is equal to four."}, {"video_title": "Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "This is 677. This is 954. Actually, let me write out the, this is one, two, three, four, five, six, seven, eight, nine, and 10. Now they want us to know, they wanna know what is the derivative of our function when f is equal to four. Well, they haven't told us even what the value of f is at four. We don't know what that point is. But what they're trying to do is, well, we're trying to make a best estimate."}, {"video_title": "Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Now they want us to know, they wanna know what is the derivative of our function when f is equal to four. Well, they haven't told us even what the value of f is at four. We don't know what that point is. But what they're trying to do is, well, we're trying to make a best estimate. And using these points, we don't even know exactly what the curve looks like. It could look like all sorts of things. We could try to fit a reasonably smooth curve."}, {"video_title": "Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "But what they're trying to do is, well, we're trying to make a best estimate. And using these points, we don't even know exactly what the curve looks like. It could look like all sorts of things. We could try to fit a reasonably smooth curve. The curve might look something like that. But it might be wackier. It might do something like this."}, {"video_title": "Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "We could try to fit a reasonably smooth curve. The curve might look something like that. But it might be wackier. It might do something like this. Well, let me try to do it. It might look something like this. So we don't know for sure."}, {"video_title": "Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "It might do something like this. Well, let me try to do it. It might look something like this. So we don't know for sure. All we know is that it needs to go through those points because they've just sampled the function at those points. But let's just, for the sake of this exercise, let's assume the simplest, let's say it's a nice, smooth curve without too many twists and turns that goes through these points just like that. So what they're asking, okay, when x is equal to four, if this yellow curve were the actual curve, then what is the slope of the tangent line at that point?"}, {"video_title": "Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So we don't know for sure. All we know is that it needs to go through those points because they've just sampled the function at those points. But let's just, for the sake of this exercise, let's assume the simplest, let's say it's a nice, smooth curve without too many twists and turns that goes through these points just like that. So what they're asking, okay, when x is equal to four, if this yellow curve were the actual curve, then what is the slope of the tangent line at that point? So we would be visualizing that. Now to be clear, this tangent line that I just drew, this would be for this version of our function that I did connecting these points. That does not have to be the actual function."}, {"video_title": "Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So what they're asking, okay, when x is equal to four, if this yellow curve were the actual curve, then what is the slope of the tangent line at that point? So we would be visualizing that. Now to be clear, this tangent line that I just drew, this would be for this version of our function that I did connecting these points. That does not have to be the actual function. We know that the actual function has to go through those points. But I'm just doing this for visualization purposes. One of the whole ideas here is that all we do have is a sample and we're trying to get a best estimate."}, {"video_title": "Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "That does not have to be the actual function. We know that the actual function has to go through those points. But I'm just doing this for visualization purposes. One of the whole ideas here is that all we do have is a sample and we're trying to get a best estimate. We don't know if it's even gonna be a good estimate. It's just going to be a best estimate. So what we generally do when we just have some data around a point is let's use the data points that are closest to that point and find slopes of secant lines pretty close around that point."}, {"video_title": "Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "One of the whole ideas here is that all we do have is a sample and we're trying to get a best estimate. We don't know if it's even gonna be a good estimate. It's just going to be a best estimate. So what we generally do when we just have some data around a point is let's use the data points that are closest to that point and find slopes of secant lines pretty close around that point. And that's going to give us our best estimate for the slope of the tangent line. So what points do we have near F of four or near the point four comma F of four? Well, they give us what F is equal to when x is equal to three."}, {"video_title": "Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So what we generally do when we just have some data around a point is let's use the data points that are closest to that point and find slopes of secant lines pretty close around that point. And that's going to give us our best estimate for the slope of the tangent line. So what points do we have near F of four or near the point four comma F of four? Well, they give us what F is equal to when x is equal to three. They give us this point right over here. Let me do this in another color. So three comma 95, that is that right over there."}, {"video_title": "Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Well, they give us what F is equal to when x is equal to three. They give us this point right over here. Let me do this in another color. So three comma 95, that is that right over there. And they also give us five comma 112. That is that point right over there. And so what we could do, we could say, well, what is the average rate of change between these two points?"}, {"video_title": "Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "So three comma 95, that is that right over there. And they also give us five comma 112. That is that point right over there. And so what we could do, we could say, well, what is the average rate of change between these two points? Another way to think about it is what is the slope of the secant line between those two points? And that would be our best estimate for the slope of the tangent line at x equals four. Do we know that it's a good estimate?"}, {"video_title": "Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And so what we could do, we could say, well, what is the average rate of change between these two points? Another way to think about it is what is the slope of the secant line between those two points? And that would be our best estimate for the slope of the tangent line at x equals four. Do we know that it's a good estimate? Do we know that it's even close? No, we don't know for sure, but that would be the best estimate. It would be better than trying to take the average rate of change between when x equals three and x equals six, or between when x equals zero and x equals nine."}, {"video_title": "Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Do we know that it's a good estimate? Do we know that it's even close? No, we don't know for sure, but that would be the best estimate. It would be better than trying to take the average rate of change between when x equals three and x equals six, or between when x equals zero and x equals nine. These are pretty close around four. And so let's do that. Let's find the average rate of change between when x goes from three to five."}, {"video_title": "Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "It would be better than trying to take the average rate of change between when x equals three and x equals six, or between when x equals zero and x equals nine. These are pretty close around four. And so let's do that. Let's find the average rate of change between when x goes from three to five. So we can see here our change in x, let me do this in a new color. So our change in x here is equal to plus two, and I can draw that out. My change in x here is plus two, and my change in y is going to be, when my x increased by two, my change in y is plus, let's see, this is, if I add 10, I get to 105, and then I add another seven, so this is plus 17."}, {"video_title": "Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "Let's find the average rate of change between when x goes from three to five. So we can see here our change in x, let me do this in a new color. So our change in x here is equal to plus two, and I can draw that out. My change in x here is plus two, and my change in y is going to be, when my x increased by two, my change in y is plus, let's see, this is, if I add 10, I get to 105, and then I add another seven, so this is plus 17. So this is plus 17 right over here. Plus 17. And so my change in y over change in x, change in y over my change in x, for this secant line between when x is equaling three and x is equaling five, is going to be equal to 17 over two."}, {"video_title": "Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "My change in x here is plus two, and my change in y is going to be, when my x increased by two, my change in y is plus, let's see, this is, if I add 10, I get to 105, and then I add another seven, so this is plus 17. So this is plus 17 right over here. Plus 17. And so my change in y over change in x, change in y over my change in x, for this secant line between when x is equaling three and x is equaling five, is going to be equal to 17 over two. 17 over two, which is equal to 8.5. So the slope of this green line here is 8.5, and that would be our best estimate for the slope of the tangent line when x equals four of the curve y is equal to f of x. And so lucky for us, the people who wrote this question had the exact same logic, and they did it right over there."}, {"video_title": "Estimating derivatives Derivatives introduction AP Calculus AB Khan Academy.mp3", "Sentence": "And so my change in y over change in x, change in y over my change in x, for this secant line between when x is equaling three and x is equaling five, is going to be equal to 17 over two. 17 over two, which is equal to 8.5. So the slope of this green line here is 8.5, and that would be our best estimate for the slope of the tangent line when x equals four of the curve y is equal to f of x. And so lucky for us, the people who wrote this question had the exact same logic, and they did it right over there. So you wouldn't have to graph it the way I did. I did it just to help us visualize what's going on. In general, when you see a question like this, they're really saying, look, you don't have all the data you need to figure out exactly what f prime of four is, but if you can find points close to or around f prime of four and find the secant line, the average rate of, the slope of the secant line or the average rate of change between those points, that's going to be our best estimate for the instantaneous rate of change when x equals four or the derivative when x equals four."}, {"video_title": "Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3", "Sentence": "This one is actually pretty straightforward to define explicitly in terms of x, to solve for y. So if we divide both sides by x, we get square root of y is equal to 1 over x. And then if you square both sides, you get y is equal to 1 over x squared, which is the same thing as x to the negative 2 power. And so if you want the derivative of y with respect to x, this is pretty straightforward. This is just an application of the chain rule. We get dy dx is equal to negative 2 x to the negative 2 minus 1, x to the negative 3 power. So that's pretty straightforward."}, {"video_title": "Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3", "Sentence": "And so if you want the derivative of y with respect to x, this is pretty straightforward. This is just an application of the chain rule. We get dy dx is equal to negative 2 x to the negative 2 minus 1, x to the negative 3 power. So that's pretty straightforward. But what I want to see is if we get the same exact result when we differentiate implicitly. So let's apply our derivative operator to both sides of this equation. And so let me make it clear what we're doing."}, {"video_title": "Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3", "Sentence": "So that's pretty straightforward. But what I want to see is if we get the same exact result when we differentiate implicitly. So let's apply our derivative operator to both sides of this equation. And so let me make it clear what we're doing. x times the square root of y and 1 right over there. When you apply the derivative operator to the expression on the left-hand side, we are just going to have to apply, well, actually we're going to apply both the product rule and the chain rule. The product rule tells us, so we have the product of two functions of x."}, {"video_title": "Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3", "Sentence": "And so let me make it clear what we're doing. x times the square root of y and 1 right over there. When you apply the derivative operator to the expression on the left-hand side, we are just going to have to apply, well, actually we're going to apply both the product rule and the chain rule. The product rule tells us, so we have the product of two functions of x. You could view it that way. So the product rule tells us this is going to be the derivative with respect to x of x times the square root of y plus x, not taking its derivative, plus x times the derivative with respect to x of the square root of y. Let me make it clear this bracket, of the square root of y."}, {"video_title": "Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3", "Sentence": "The product rule tells us, so we have the product of two functions of x. You could view it that way. So the product rule tells us this is going to be the derivative with respect to x of x times the square root of y plus x, not taking its derivative, plus x times the derivative with respect to x of the square root of y. Let me make it clear this bracket, of the square root of y. And on the right-hand side, right over here, the derivative with respect to x of this constant, that's just going to be equal to 0. So what does this simplify to? Well, the derivative with respect to x of x is just 1."}, {"video_title": "Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3", "Sentence": "Let me make it clear this bracket, of the square root of y. And on the right-hand side, right over here, the derivative with respect to x of this constant, that's just going to be equal to 0. So what does this simplify to? Well, the derivative with respect to x of x is just 1. So we're just going to be left, this simplifies to 1, so we're just going to be left with the square root of y right over here. So we're just going to be, this is going to simplify to a square root of y. And what does this over here simplify to?"}, {"video_title": "Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the derivative with respect to x of x is just 1. So we're just going to be left, this simplifies to 1, so we're just going to be left with the square root of y right over here. So we're just going to be, this is going to simplify to a square root of y. And what does this over here simplify to? Well, the derivative with respect to x of the square root of y, here we want to apply the chain rule, so let me make it clear. So we have plus this x, plus this x, plus whatever business this is. And I'm going to do this in blue."}, {"video_title": "Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3", "Sentence": "And what does this over here simplify to? Well, the derivative with respect to x of the square root of y, here we want to apply the chain rule, so let me make it clear. So we have plus this x, plus this x, plus whatever business this is. And I'm going to do this in blue. Well, it's going to be the derivative of square root of something with respect to that something. Well, the derivative of square root of something with respect to that something, or the derivative of something to the one half with respect to that something is going to be one half times that something to the negative one half power. Once again, this right over here is the derivative of the square root of y with respect to y."}, {"video_title": "Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3", "Sentence": "And I'm going to do this in blue. Well, it's going to be the derivative of square root of something with respect to that something. Well, the derivative of square root of something with respect to that something, or the derivative of something to the one half with respect to that something is going to be one half times that something to the negative one half power. Once again, this right over here is the derivative of the square root of y with respect to y. We've seen this multiple times. If I were to say the derivative of the square root of x with respect to x, you would get 1 half x to the negative 1 half. Now I'm just doing it with y's."}, {"video_title": "Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3", "Sentence": "Once again, this right over here is the derivative of the square root of y with respect to y. We've seen this multiple times. If I were to say the derivative of the square root of x with respect to x, you would get 1 half x to the negative 1 half. Now I'm just doing it with y's. But we're not done yet. Remember, our derivative operator wasn't to say with respect to y. It's with respect to x."}, {"video_title": "Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3", "Sentence": "Now I'm just doing it with y's. But we're not done yet. Remember, our derivative operator wasn't to say with respect to y. It's with respect to x. So this only gets us with respect to y. We need to apply the entire chain rule. We have to multiply that."}, {"video_title": "Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3", "Sentence": "It's with respect to x. So this only gets us with respect to y. We need to apply the entire chain rule. We have to multiply that. We have to multiply that times the derivative of y with respect to x in order to get the real derivative of this expression with respect to x. So let's multiply times the derivative of y with respect to x. We don't know what that is."}, {"video_title": "Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3", "Sentence": "We have to multiply that. We have to multiply that times the derivative of y with respect to x in order to get the real derivative of this expression with respect to x. So let's multiply times the derivative of y with respect to x. We don't know what that is. That's actually what we're trying to solve for. But to use the chain rule, we just have to say it's the derivative of the square root of y with respect to y times the derivative of y with respect to x. This is the derivative of this thing with respect to x."}, {"video_title": "Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3", "Sentence": "We don't know what that is. That's actually what we're trying to solve for. But to use the chain rule, we just have to say it's the derivative of the square root of y with respect to y times the derivative of y with respect to x. This is the derivative of this thing with respect to x. So we get this on the left-hand side, on the right-hand side, we just have a 0. And now once again, we can attempt to solve for the derivative of y with respect to x. And maybe the easiest first step is to subtract the square root of y from both sides of this equation."}, {"video_title": "Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3", "Sentence": "This is the derivative of this thing with respect to x. So we get this on the left-hand side, on the right-hand side, we just have a 0. And now once again, we can attempt to solve for the derivative of y with respect to x. And maybe the easiest first step is to subtract the square root of y from both sides of this equation. And actually, let me move all of this stuff over so I have, once again, more room to work with. Let me cut it, actually. And then let me paste it."}, {"video_title": "Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3", "Sentence": "And maybe the easiest first step is to subtract the square root of y from both sides of this equation. And actually, let me move all of this stuff over so I have, once again, more room to work with. Let me cut it, actually. And then let me paste it. Let me move it over right over here. So we went from there to there. I didn't gain a lot of real estate, but hopefully this helps a little bit."}, {"video_title": "Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3", "Sentence": "And then let me paste it. Let me move it over right over here. So we went from there to there. I didn't gain a lot of real estate, but hopefully this helps a little bit. And actually, I don't even like that. Let me leave it where it was before. So then if we subtract the square root of y from both sides, we get, and I'll try to simplify as I go, we get this thing, which I can rewrite as x times, well, it's just going to be x in the numerator, divided by 2 times the square root of y. y to the negative 1 half is just the square root of y in the denominator."}, {"video_title": "Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3", "Sentence": "I didn't gain a lot of real estate, but hopefully this helps a little bit. And actually, I don't even like that. Let me leave it where it was before. So then if we subtract the square root of y from both sides, we get, and I'll try to simplify as I go, we get this thing, which I can rewrite as x times, well, it's just going to be x in the numerator, divided by 2 times the square root of y. y to the negative 1 half is just the square root of y in the denominator. And 1 half, I just put the 2 in the denominator there, times dy dx times the derivative of y with respect to x is going to be equal to the negative square root of y. I just subtracted the square root of y from both sides. And actually, this is something that I might actually want to copy and paste up here. So copy and then paste."}, {"video_title": "Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3", "Sentence": "So then if we subtract the square root of y from both sides, we get, and I'll try to simplify as I go, we get this thing, which I can rewrite as x times, well, it's just going to be x in the numerator, divided by 2 times the square root of y. y to the negative 1 half is just the square root of y in the denominator. And 1 half, I just put the 2 in the denominator there, times dy dx times the derivative of y with respect to x is going to be equal to the negative square root of y. I just subtracted the square root of y from both sides. And actually, this is something that I might actually want to copy and paste up here. So copy and then paste. So let's go back up here just to continue our simplification solving for dy dx. Well, to solve for dy dx, we just have to divide both sides by x over 2 times the square root of y. So we're left with dy dx is equal to, or dividing both sides by this is the same thing as multiplying by the reciprocal of this, is equal to 2 times the square root of y over x, over my yellow x, times the negative square root of y."}, {"video_title": "Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3", "Sentence": "So copy and then paste. So let's go back up here just to continue our simplification solving for dy dx. Well, to solve for dy dx, we just have to divide both sides by x over 2 times the square root of y. So we're left with dy dx is equal to, or dividing both sides by this is the same thing as multiplying by the reciprocal of this, is equal to 2 times the square root of y over x, over my yellow x, times the negative square root of y. Well, what's this going to simplify to? This is going to be equal to the square root of y times the square root of y is just y. The negative times the 2, you get negative 2."}, {"video_title": "Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3", "Sentence": "So we're left with dy dx is equal to, or dividing both sides by this is the same thing as multiplying by the reciprocal of this, is equal to 2 times the square root of y over x, over my yellow x, times the negative square root of y. Well, what's this going to simplify to? This is going to be equal to the square root of y times the square root of y is just y. The negative times the 2, you get negative 2. So you get negative 2y over x is equal to the derivative of y with respect to x. Now, you might be saying, look, we just figured out the derivative implicitly, and it looks very different than what we did right over here. When we just used the power rule, we got negative 2x to the negative third power."}, {"video_title": "Showing explicit and implicit differentiation give same result AP Calculus AB Khan Academy.mp3", "Sentence": "The negative times the 2, you get negative 2. So you get negative 2y over x is equal to the derivative of y with respect to x. Now, you might be saying, look, we just figured out the derivative implicitly, and it looks very different than what we did right over here. When we just used the power rule, we got negative 2x to the negative third power. x to the negative 3 power. The key here is to realize that this thing right over here, we could solve explicitly in terms of, we could solve for y. So we could just make this substitution back here to see that these are the exact same thing."}, {"video_title": "Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So this car right over here is approaching an intersection at 60 miles per hour. And right now, right at this moment, it is 0.8 miles from the intersection. Now we have this truck over here. It's approaching the same intersection on a street that is perpendicular to the street that the car is on. And right now it is 0.6 miles. So that is 0.6 miles from the intersection and is approaching the intersection at 30 miles per hour. Now my question to you is, what is the rate at which the distance between the car and the truck is changing?"}, {"video_title": "Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "It's approaching the same intersection on a street that is perpendicular to the street that the car is on. And right now it is 0.6 miles. So that is 0.6 miles from the intersection and is approaching the intersection at 30 miles per hour. Now my question to you is, what is the rate at which the distance between the car and the truck is changing? Well, to think about that, let's first just think about what we're asking. So we're asking about the distance between the car and the truck. So right at this moment when the car is 0.8 miles from the intersection, the truck is 0.6 miles from the intersection, the truck is traveling at 30 miles per hour towards the intersection, the car is traveling 60 miles per hour towards the intersection, right at this moment, what is the rate at which this distance right over here is changing?"}, {"video_title": "Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Now my question to you is, what is the rate at which the distance between the car and the truck is changing? Well, to think about that, let's first just think about what we're asking. So we're asking about the distance between the car and the truck. So right at this moment when the car is 0.8 miles from the intersection, the truck is 0.6 miles from the intersection, the truck is traveling at 30 miles per hour towards the intersection, the car is traveling 60 miles per hour towards the intersection, right at this moment, what is the rate at which this distance right over here is changing? And just so that we have some variables in place, let's call this distance s. So what we really are trying to figure out is right at this moment, what is ds dt going to be equal to? Well, let's think about what we know that we could use to somehow come to terms or figure out what ds dt is. Well, we know the distance of the car and the intersection."}, {"video_title": "Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So right at this moment when the car is 0.8 miles from the intersection, the truck is 0.6 miles from the intersection, the truck is traveling at 30 miles per hour towards the intersection, the car is traveling 60 miles per hour towards the intersection, right at this moment, what is the rate at which this distance right over here is changing? And just so that we have some variables in place, let's call this distance s. So what we really are trying to figure out is right at this moment, what is ds dt going to be equal to? Well, let's think about what we know that we could use to somehow come to terms or figure out what ds dt is. Well, we know the distance of the car and the intersection. And let's just call that distance y. So y is equal to 0.8 miles. We also know that d, so let me write this."}, {"video_title": "Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we know the distance of the car and the intersection. And let's just call that distance y. So y is equal to 0.8 miles. We also know that d, so let me write this. We know that y is 0.8 miles right now. 0.8 miles. We also know that dy dt, the rate at which y is changing with respect to time, is what?"}, {"video_title": "Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We also know that d, so let me write this. We know that y is 0.8 miles right now. 0.8 miles. We also know that dy dt, the rate at which y is changing with respect to time, is what? Well, y is decreasing by 60 miles per hour. So let me write it as negative 60 miles per hour. Now, similarly, let's say that this distance right over here is x. x is 0.6 miles right at this moment."}, {"video_title": "Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We also know that dy dt, the rate at which y is changing with respect to time, is what? Well, y is decreasing by 60 miles per hour. So let me write it as negative 60 miles per hour. Now, similarly, let's say that this distance right over here is x. x is 0.6 miles right at this moment. So we know that x is equal to 0.6 miles. What is the rate at which x is changing with respect to time? Well, we know it's 30 miles per hour is how fast we're approaching the intersection."}, {"video_title": "Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Now, similarly, let's say that this distance right over here is x. x is 0.6 miles right at this moment. So we know that x is equal to 0.6 miles. What is the rate at which x is changing with respect to time? Well, we know it's 30 miles per hour is how fast we're approaching the intersection. But x is decreasing by 30 miles every hour. So we should say it's negative 30 miles per hour. So we know what y is."}, {"video_title": "Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we know it's 30 miles per hour is how fast we're approaching the intersection. But x is decreasing by 30 miles every hour. So we should say it's negative 30 miles per hour. So we know what y is. We know what x is. We know how fast y is changing, how fast x is changing with respect to time. So what we could try to do here is come up with a relationship between x, y, and s, and then differentiate that relationship with respect to time."}, {"video_title": "Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So we know what y is. We know what x is. We know how fast y is changing, how fast x is changing with respect to time. So what we could try to do here is come up with a relationship between x, y, and s, and then differentiate that relationship with respect to time. And it seems like we have pretty much everything we need to solve for this. So what's the relationship between x, y, and s? Well, we know that this is a right triangle."}, {"video_title": "Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So what we could try to do here is come up with a relationship between x, y, and s, and then differentiate that relationship with respect to time. And it seems like we have pretty much everything we need to solve for this. So what's the relationship between x, y, and s? Well, we know that this is a right triangle. The streets are perpendicular to each other. So we can use the Pythagorean theorem. We know that x squared plus y squared is going to be equal to s squared."}, {"video_title": "Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we know that this is a right triangle. The streets are perpendicular to each other. So we can use the Pythagorean theorem. We know that x squared plus y squared is going to be equal to s squared. And then we can take the derivative of both sides of this with respect to time to get a relationship between all the things that we care about. So what's the derivative of x squared with respect to time? Well, it's going to be the derivative of x squared with respect to x, which is just 2x, times the derivative of x with respect to time, times dx dt."}, {"video_title": "Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We know that x squared plus y squared is going to be equal to s squared. And then we can take the derivative of both sides of this with respect to time to get a relationship between all the things that we care about. So what's the derivative of x squared with respect to time? Well, it's going to be the derivative of x squared with respect to x, which is just 2x, times the derivative of x with respect to time, times dx dt. Once again, just a chain rule. Derivative of something squared with respect to the something times the derivative of the something with respect to time. And we use similar logic right over here when we want to take the derivative of y squared with respect to time."}, {"video_title": "Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, it's going to be the derivative of x squared with respect to x, which is just 2x, times the derivative of x with respect to time, times dx dt. Once again, just a chain rule. Derivative of something squared with respect to the something times the derivative of the something with respect to time. And we use similar logic right over here when we want to take the derivative of y squared with respect to time. Derivative of y squared with respect to y times the derivative of y with respect to time. Now on the right-hand side of this equation, we, once again, take the derivative with respect to time, so it's the derivative of s squared with respect to s, which is just 2s, times the derivative of s with respect to time. Once again, this is all just an application of the chain rule."}, {"video_title": "Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And we use similar logic right over here when we want to take the derivative of y squared with respect to time. Derivative of y squared with respect to y times the derivative of y with respect to time. Now on the right-hand side of this equation, we, once again, take the derivative with respect to time, so it's the derivative of s squared with respect to s, which is just 2s, times the derivative of s with respect to time. Once again, this is all just an application of the chain rule. Now it looks like we know what x is, we know what dx dt is, we know what y is, we know what dy dt is. All we need to figure out is what s and then what ds dt is, the rate at which this distance is changing with respect to time. Well, what's s right now?"}, {"video_title": "Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Once again, this is all just an application of the chain rule. Now it looks like we know what x is, we know what dx dt is, we know what y is, we know what dy dt is. All we need to figure out is what s and then what ds dt is, the rate at which this distance is changing with respect to time. Well, what's s right now? Well, we can actually use the Pythagorean theorem at this exact moment. We know that x squared, so x is 0.6, we know 0.6 squared plus y squared, 0.8 squared, is equal to s squared. Well, this is 0.36 plus 0.64 is equal to s squared."}, {"video_title": "Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, what's s right now? Well, we can actually use the Pythagorean theorem at this exact moment. We know that x squared, so x is 0.6, we know 0.6 squared plus y squared, 0.8 squared, is equal to s squared. Well, this is 0.36 plus 0.64 is equal to s squared. This is 1 is equal to s squared. And we only care about positive distances, so we have s is equal to 1 right now. So we also know what s is."}, {"video_title": "Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, this is 0.36 plus 0.64 is equal to s squared. This is 1 is equal to s squared. And we only care about positive distances, so we have s is equal to 1 right now. So we also know what s is. So let's substitute all of these numbers in and then try to solve for what we came here to do, solve for ds dt. So the rate at which 2 times x, maybe I'll do that in yellow, 2 times x, x is 0.6, is going to be 1.2 times dx dt. So that's negative 30 miles per hour."}, {"video_title": "Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So we also know what s is. So let's substitute all of these numbers in and then try to solve for what we came here to do, solve for ds dt. So the rate at which 2 times x, maybe I'll do that in yellow, 2 times x, x is 0.6, is going to be 1.2 times dx dt. So that's negative 30 miles per hour. So times negative 30 miles per hour plus 2 times y is 1.6, times dy dt is negative 60 miles per hour. And I'm not writing the units here, but if you were to write the units, you will see that all of our distances are in miles and all of our time is within hours. So we're going to get an answer when we solve for ds dt."}, {"video_title": "Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So that's negative 30 miles per hour. So times negative 30 miles per hour plus 2 times y is 1.6, times dy dt is negative 60 miles per hour. And I'm not writing the units here, but if you were to write the units, you will see that all of our distances are in miles and all of our time is within hours. So we're going to get an answer when we solve for ds dt. That's miles per hour. But I encourage you, if you want to, to actually write out the units and see how they work out. And so this is going to be equal to 2 times s. Well, s is 1 mile, so it's just going to be 2 times ds dt, which is what we're trying to solve for."}, {"video_title": "Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So we're going to get an answer when we solve for ds dt. That's miles per hour. But I encourage you, if you want to, to actually write out the units and see how they work out. And so this is going to be equal to 2 times s. Well, s is 1 mile, so it's just going to be 2 times ds dt, which is what we're trying to solve for. So what do we get here on the left-hand side? So 1.2 times negative 30, that's negative 36. 1 over 5 of 30 is 6."}, {"video_title": "Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And so this is going to be equal to 2 times s. Well, s is 1 mile, so it's just going to be 2 times ds dt, which is what we're trying to solve for. So what do we get here on the left-hand side? So 1.2 times negative 30, that's negative 36. 1 over 5 of 30 is 6. Yep, that's right. And then 1.6 times negative 60, that's going to be negative 96, is equal to 2 times ds dt, is equal to 2 times the rate at which our distance is changing with respect to time. On the left-hand side right over here, this is negative 132."}, {"video_title": "Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "1 over 5 of 30 is 6. Yep, that's right. And then 1.6 times negative 60, that's going to be negative 96, is equal to 2 times ds dt, is equal to 2 times the rate at which our distance is changing with respect to time. On the left-hand side right over here, this is negative 132. Negative 132 is equal to 2 times ds dt. Divide both sides by 2. We get negative 66."}, {"video_title": "Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "On the left-hand side right over here, this is negative 132. Negative 132 is equal to 2 times ds dt. Divide both sides by 2. We get negative 66. And now we can put our units if we want. Miles per hour is the rate at which our distance is changing with respect to time. So ds dt is negative 66 miles per hour."}, {"video_title": "Related rates Approaching cars Applications of derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "We get negative 66. And now we can put our units if we want. Miles per hour is the rate at which our distance is changing with respect to time. So ds dt is negative 66 miles per hour. Does it make sense that we got a negative number here? Well, sure. This distance is decreasing right at this moment as they approach the intersection."}, {"video_title": "Limits of combined functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And we know from our limit properties that this is going to be the same thing as the limit as x approaches zero of f of x times, times the limit limit as x approaches zero of h of x. And let's think about what each of these are. So let's first think about f of x right over here. So on f of x, as x approaches zero, notice the function itself isn't defined there, but we see when we approach from the left, we are approaching, the function seems to be approaching the value of negative one right over here. And as we approach from the right, the function seems to be approaching the value of negative one. So the limit here, this limit here is negative one. As we approach from the left, we're approaching negative one."}, {"video_title": "Limits of combined functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So on f of x, as x approaches zero, notice the function itself isn't defined there, but we see when we approach from the left, we are approaching, the function seems to be approaching the value of negative one right over here. And as we approach from the right, the function seems to be approaching the value of negative one. So the limit here, this limit here is negative one. As we approach from the left, we're approaching negative one. As we approach from the right, the value of the function seems to be approaching negative one. Now what about h of x? Well, h of x we have down here."}, {"video_title": "Limits of combined functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "As we approach from the left, we're approaching negative one. As we approach from the right, the value of the function seems to be approaching negative one. Now what about h of x? Well, h of x we have down here. As x approaches zero, as x approaches zero, the function is defined at x equals zero. It looks like it is equal to one. And the limit is also equal to one."}, {"video_title": "Limits of combined functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Well, h of x we have down here. As x approaches zero, as x approaches zero, the function is defined at x equals zero. It looks like it is equal to one. And the limit is also equal to one. We can see that as we approach it from the left, we are approaching one. As we approach from the right, we are approaching one. As we approach x equals sex from the left, the function approaches one."}, {"video_title": "Limits of combined functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And the limit is also equal to one. We can see that as we approach it from the left, we are approaching one. As we approach from the right, we are approaching one. As we approach x equals sex from the left, the function approaches one. As we approach x equals zero from the right, the function itself is approaching one. And it makes sense that the function is defined there, is defined at x equals zero, and the limit as x approaches zero is equal to the value of the function at that point, because this is a continuous function. So this is, this is one, and so negative one times one is going to be equal to, is equal to negative one."}, {"video_title": "Limits of combined functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "As we approach x equals sex from the left, the function approaches one. As we approach x equals zero from the right, the function itself is approaching one. And it makes sense that the function is defined there, is defined at x equals zero, and the limit as x approaches zero is equal to the value of the function at that point, because this is a continuous function. So this is, this is one, and so negative one times one is going to be equal to, is equal to negative one. So that is equal to negative one. Let's do one more. All right, so these are both, looks like continuous functions."}, {"video_title": "Limits of combined functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So this is, this is one, and so negative one times one is going to be equal to, is equal to negative one. So that is equal to negative one. Let's do one more. All right, so these are both, looks like continuous functions. So we have the limit as x approaches zero of h of x over g of x. So once again, using our limit properties, this is going to be the same as the limit of h of x as x approaches zero over the limit of g of x as x approaches zero. Now what's the limit of h of x as x approaches zero?"}, {"video_title": "Limits of combined functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "All right, so these are both, looks like continuous functions. So we have the limit as x approaches zero of h of x over g of x. So once again, using our limit properties, this is going to be the same as the limit of h of x as x approaches zero over the limit of g of x as x approaches zero. Now what's the limit of h of x as x approaches zero? This is, let's see, as we approach zero from the left, as we approach x equals zero from the left, our function seems to be approaching four. And as we approach z equals zero from the right, our function seems to be approaching four, And that's also what the value of the function is at x equals zero. That makes sense because this is a continuous function."}, {"video_title": "Limits of combined functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Now what's the limit of h of x as x approaches zero? This is, let's see, as we approach zero from the left, as we approach x equals zero from the left, our function seems to be approaching four. And as we approach z equals zero from the right, our function seems to be approaching four, And that's also what the value of the function is at x equals zero. That makes sense because this is a continuous function. So the limit as we approach x equals zero should be the same as the value of the function at x equals zero. So this top, this is going to be four. Now let's think about the limit of g of x as x approaches zero."}, {"video_title": "Limits of combined functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "That makes sense because this is a continuous function. So the limit as we approach x equals zero should be the same as the value of the function at x equals zero. So this top, this is going to be four. Now let's think about the limit of g of x as x approaches zero. So from the left, it looks like as x approaches zero, the value of the function is approaching zero. And as x approaches zero from the right, the value of the function is also approaching zero, which happens to also be g of zero. G of zero is also zero."}, {"video_title": "Limits of combined functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Now let's think about the limit of g of x as x approaches zero. So from the left, it looks like as x approaches zero, the value of the function is approaching zero. And as x approaches zero from the right, the value of the function is also approaching zero, which happens to also be g of zero. G of zero is also zero. And that makes sense that the limit and the actual value of the function at that point is the same because it's continuous. So this also is zero. But now we're in a strange situation."}, {"video_title": "Limits of combined functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "G of zero is also zero. And that makes sense that the limit and the actual value of the function at that point is the same because it's continuous. So this also is zero. But now we're in a strange situation. We have to take four and divide it by zero. So this limit will not exist because we can't take four and divide it by zero. So even though the limit of h of x is x equals as x approaches zero exists, and the limit of g of x as x approaches zero exists, we can't divide four by zero, so this whole entire limit does not exist."}, {"video_title": "Limits of combined functions Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "But now we're in a strange situation. We have to take four and divide it by zero. So this limit will not exist because we can't take four and divide it by zero. So even though the limit of h of x is x equals as x approaches zero exists, and the limit of g of x as x approaches zero exists, we can't divide four by zero, so this whole entire limit does not exist. Does not exist. And actually, if you were to plot h of x over g of x, if you were to plot that graph, you would see it even clearer that that limit does not exist. You would actually be able to see it graphically."}, {"video_title": "Continuity over an interval Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So we say that f is continuous when x is equal to c if and only if, so I'm gonna make these two-way arrows right over here, the limit of f of x as x approaches c is equal to f of c. And when we first introduced this, we said, hey, this looks a little bit technical, but it's actually pretty intuitive. Think about what's happening. The limit as x approaches c of f of x, so let's say that f of x as x approaches c is approaching some value. So if we approach from the left, we're getting to this value. If we approach from the right, we're getting this value. Well, in order for the function to be continuous, if I had to draw this function without picking up my pen, well, the value of the function at that point should be the same as the limit. This is really just a more rigorous way of describing this notion of not having to pick up your pencil, this notion of connectedness, that you don't have any jumps or any discontinuities of any kind."}, {"video_title": "Continuity over an interval Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So if we approach from the left, we're getting to this value. If we approach from the right, we're getting this value. Well, in order for the function to be continuous, if I had to draw this function without picking up my pen, well, the value of the function at that point should be the same as the limit. This is really just a more rigorous way of describing this notion of not having to pick up your pencil, this notion of connectedness, that you don't have any jumps or any discontinuities of any kind. So with that out of the way, let's discuss continuity over intervals. Let me delete this really fast so I have space to work with. So we say, so I'm gonna first talk about an open interval, and then we're gonna talk about a closed interval, because a closed interval gets a little bit more involved."}, {"video_title": "Continuity over an interval Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "This is really just a more rigorous way of describing this notion of not having to pick up your pencil, this notion of connectedness, that you don't have any jumps or any discontinuities of any kind. So with that out of the way, let's discuss continuity over intervals. Let me delete this really fast so I have space to work with. So we say, so I'm gonna first talk about an open interval, and then we're gonna talk about a closed interval, because a closed interval gets a little bit more involved. So we say f is continuous over an open interval from a to b. So the parentheses instead of brackets, this shows that we're not including the endpoints. So this would be all of the points between x equals a and x equals b, but not equaling x equals a and x equals b."}, {"video_title": "Continuity over an interval Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So we say, so I'm gonna first talk about an open interval, and then we're gonna talk about a closed interval, because a closed interval gets a little bit more involved. So we say f is continuous over an open interval from a to b. So the parentheses instead of brackets, this shows that we're not including the endpoints. So this would be all of the points between x equals a and x equals b, but not equaling x equals a and x equals b. So f is continuous over this open interval if and only if, if and only if, f is continuous, f is continuous over every point in, over every point in the interval. So let's do a couple of examples of that. So let's say we're talking about the open interval from negative seven to negative five."}, {"video_title": "Continuity over an interval Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So this would be all of the points between x equals a and x equals b, but not equaling x equals a and x equals b. So f is continuous over this open interval if and only if, if and only if, f is continuous, f is continuous over every point in, over every point in the interval. So let's do a couple of examples of that. So let's say we're talking about the open interval from negative seven to negative five. Is f continuous over that interval? Let's see, we're going from negative seven to negative five, and there's a couple of ways you could do it. There's the not-so-mathematically-rigorous way, where you could say, hey, look, if I start here, I can get all the way to negative five without having to pick up my pencil."}, {"video_title": "Continuity over an interval Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say we're talking about the open interval from negative seven to negative five. Is f continuous over that interval? Let's see, we're going from negative seven to negative five, and there's a couple of ways you could do it. There's the not-so-mathematically-rigorous way, where you could say, hey, look, if I start here, I can get all the way to negative five without having to pick up my pencil. If you wanted to do more rigorously, and you actually had the definition of the function, you might be able to do a proof that for any of these points over the interval, that the limit as x approaches any one of these points of f of x is equal to the value of the function at that point. It's harder to do when you only have a graph. When you only have a graph, you can only just do it by inspection and say, okay, I can go from that point to that point without picking up my pencil, so I feel pretty good about it."}, {"video_title": "Continuity over an interval Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "There's the not-so-mathematically-rigorous way, where you could say, hey, look, if I start here, I can get all the way to negative five without having to pick up my pencil. If you wanted to do more rigorously, and you actually had the definition of the function, you might be able to do a proof that for any of these points over the interval, that the limit as x approaches any one of these points of f of x is equal to the value of the function at that point. It's harder to do when you only have a graph. When you only have a graph, you can only just do it by inspection and say, okay, I can go from that point to that point without picking up my pencil, so I feel pretty good about it. Now let's do another interval. Let's say the, so let me put a check mark here. That is continuous."}, {"video_title": "Continuity over an interval Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "When you only have a graph, you can only just do it by inspection and say, okay, I can go from that point to that point without picking up my pencil, so I feel pretty good about it. Now let's do another interval. Let's say the, so let me put a check mark here. That is continuous. Let's think about the interval from negative two to positive one, the open interval. So this is interesting, because the function at negative two is up here, and so if you really wanted to start at negative two, you would have to start here, and then jump immediately down as soon as you get slightly larger than negative two, and then keep going, but this is an open interval, so we're not actually concerned with what exactly happens at negative two. We're concerned what happens when we are all the numbers larger than negative two, so we would actually start right over here, and then we would go to one, and once again, based on the intuitive, I didn't have to pick up my pen idea, this function would be continuous over this interval."}, {"video_title": "Continuity over an interval Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "That is continuous. Let's think about the interval from negative two to positive one, the open interval. So this is interesting, because the function at negative two is up here, and so if you really wanted to start at negative two, you would have to start here, and then jump immediately down as soon as you get slightly larger than negative two, and then keep going, but this is an open interval, so we're not actually concerned with what exactly happens at negative two. We're concerned what happens when we are all the numbers larger than negative two, so we would actually start right over here, and then we would go to one, and once again, based on the intuitive, I didn't have to pick up my pen idea, this function would be continuous over this interval. So what's an example of an interval where the function would not be continuous? Well, think about the interval from, well, this is a pretty straightforward one, the open interval from three to five. The function is here when x is equal to three, but if we wanted to get to five, it looks like we're asymptoting, it looks like we're asymptoting up towards infinity, and we just keep on going for a very long time, and then we would have to pick up our pencil and jump over, and then we would come back down right over here, and so here, we are not continuous over that interval."}, {"video_title": "Continuity over an interval Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "We're concerned what happens when we are all the numbers larger than negative two, so we would actually start right over here, and then we would go to one, and once again, based on the intuitive, I didn't have to pick up my pen idea, this function would be continuous over this interval. So what's an example of an interval where the function would not be continuous? Well, think about the interval from, well, this is a pretty straightforward one, the open interval from three to five. The function is here when x is equal to three, but if we wanted to get to five, it looks like we're asymptoting, it looks like we're asymptoting up towards infinity, and we just keep on going for a very long time, and then we would have to pick up our pencil and jump over, and then we would come back down right over here, and so here, we are not continuous over that interval. So now let's think about the more, the slightly more involved interval, the slightly more involved case is when you have a closed interval. F is continuous over the closed interval from a to b, so this includes not just the points between a and b, but the endpoints as well, if and only if f is continuous over the open interval and the one-sided limits, let me write this, and the limit as x approaches a from the right of f of x is equal to f of a, and the limit as x approaches b from the left of f of x is equal to f of b. Now what's going on here?"}, {"video_title": "Continuity over an interval Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "The function is here when x is equal to three, but if we wanted to get to five, it looks like we're asymptoting, it looks like we're asymptoting up towards infinity, and we just keep on going for a very long time, and then we would have to pick up our pencil and jump over, and then we would come back down right over here, and so here, we are not continuous over that interval. So now let's think about the more, the slightly more involved interval, the slightly more involved case is when you have a closed interval. F is continuous over the closed interval from a to b, so this includes not just the points between a and b, but the endpoints as well, if and only if f is continuous over the open interval and the one-sided limits, let me write this, and the limit as x approaches a from the right of f of x is equal to f of a, and the limit as x approaches b from the left of f of x is equal to f of b. Now what's going on here? Well, it's just saying that the one-sided limit when you're operating within the interval has to approach the same value as the function. So for example, if we said the closed interval from negative seven to negative five, well, this one is still reasonable, just based on the picking up your pencil thing, you don't have to pick up your pencil, and what you would do is at the endpoint, and at negative seven, this function is just plain old continuous, but if it wasn't defined over here, it could still be continuous because you would do the right-handed limit towards it, and you'd say, okay, the right-handed limit is equal to the value of the function, and then at this endpoint, at the second endpoint, you'd say, okay, the left-handed limit is equal to the function, even if it wasn't defined here, even if the two-sided limit were not defined. And so we could actually look at an example of that."}, {"video_title": "Continuity over an interval Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Now what's going on here? Well, it's just saying that the one-sided limit when you're operating within the interval has to approach the same value as the function. So for example, if we said the closed interval from negative seven to negative five, well, this one is still reasonable, just based on the picking up your pencil thing, you don't have to pick up your pencil, and what you would do is at the endpoint, and at negative seven, this function is just plain old continuous, but if it wasn't defined over here, it could still be continuous because you would do the right-handed limit towards it, and you'd say, okay, the right-handed limit is equal to the value of the function, and then at this endpoint, at the second endpoint, you'd say, okay, the left-handed limit is equal to the function, even if it wasn't defined here, even if the two-sided limit were not defined. And so we could actually look at an example of that. If we were looking at the interval from the closed, and you could have one side open, one side closed, but let's just do the closed interval from negative three to negative two. So notice I did not have to pick up my pencil. I'm including negative three, and I'm getting all the way to negative two."}, {"video_title": "Continuity over an interval Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "And so we could actually look at an example of that. If we were looking at the interval from the closed, and you could have one side open, one side closed, but let's just do the closed interval from negative three to negative two. So notice I did not have to pick up my pencil. I'm including negative three, and I'm getting all the way to negative two. If you knew the analytic definition of this function, you could prove that, hey, the limit at any of these points inside between negative three and negative two is equal to the value of the function. Negative three, the function is clearly, at negative three, the function is just plain old continuous. The two-sided limit approaches the value of the function, but at negative two, the two-sided limit does not exist."}, {"video_title": "Continuity over an interval Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "I'm including negative three, and I'm getting all the way to negative two. If you knew the analytic definition of this function, you could prove that, hey, the limit at any of these points inside between negative three and negative two is equal to the value of the function. Negative three, the function is clearly, at negative three, the function is just plain old continuous. The two-sided limit approaches the value of the function, but at negative two, the two-sided limit does not exist. When you approach from the left, looks like you're approaching zero, f of x is equal to zero. When you approach from the right, it looks like f of x is approaching negative three. So even though the two-sided limit does not exist, we can still be good because the left-handed limit does exist, and the left-handed limit is approaching the value of the function."}, {"video_title": "Continuity over an interval Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "The two-sided limit approaches the value of the function, but at negative two, the two-sided limit does not exist. When you approach from the left, looks like you're approaching zero, f of x is equal to zero. When you approach from the right, it looks like f of x is approaching negative three. So even though the two-sided limit does not exist, we can still be good because the left-handed limit does exist, and the left-handed limit is approaching the value of the function. So we actually are continuous over that interval. But then if we did the interval, if we did the closed interval from negative two to negative two to one, pause the video and think about, based on what we just talked about, are we continuous over this interval? Well, we're going from negative two to one, and negative two is the lower bound."}, {"video_title": "Continuity over an interval Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "So even though the two-sided limit does not exist, we can still be good because the left-handed limit does exist, and the left-handed limit is approaching the value of the function. So we actually are continuous over that interval. But then if we did the interval, if we did the closed interval from negative two to negative two to one, pause the video and think about, based on what we just talked about, are we continuous over this interval? Well, we're going from negative two to one, and negative two is the lower bound. So is this right over here, is this right over here true? Is the limit as we approach negative two from the right, is that the same thing as f of negative two? Well, the limit as we approach from the right seems to be approaching negative three, and f of negative two is zero."}, {"video_title": "Continuity over an interval Limits and continuity AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we're going from negative two to one, and negative two is the lower bound. So is this right over here, is this right over here true? Is the limit as we approach negative two from the right, is that the same thing as f of negative two? Well, the limit as we approach from the right seems to be approaching negative three, and f of negative two is zero. So this limit does not, these two things, the limit as we approach from the right and the value of the function are not the same, and so we do not have that, I guess you could say that one-sided continuity at negative two. And that also makes sense. If I start at negative two, let me do this in a color you can see."}, {"video_title": "Definite integrals reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "Let's evaluate the definite integral from negative three to five of four dx. What is this going to be equal to? And I encourage you to pause the video and try to figure it out on your own. All right, so in order to evaluate this, we need to remember the fundamental theorem of calculus, which connects the notion of a definite integral and an antiderivative. So the fundamental theorem of calculus tells us that our definite integral from a to b of f of x dx is going to be equal to the antiderivative of our function f, which we denote with a capital F, evaluated at the upper bound minus our antiderivative evaluated at the lower bound. So we just have to do that right over here. So this is going to be equal to, well, what is the antiderivative of four?"}, {"video_title": "Definite integrals reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "All right, so in order to evaluate this, we need to remember the fundamental theorem of calculus, which connects the notion of a definite integral and an antiderivative. So the fundamental theorem of calculus tells us that our definite integral from a to b of f of x dx is going to be equal to the antiderivative of our function f, which we denote with a capital F, evaluated at the upper bound minus our antiderivative evaluated at the lower bound. So we just have to do that right over here. So this is going to be equal to, well, what is the antiderivative of four? Well, you might immediately say, well, that's just going to be four x. You could even think of it in terms of reverse power rule. Four is the same thing as four x to the zero."}, {"video_title": "Definite integrals reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be equal to, well, what is the antiderivative of four? Well, you might immediately say, well, that's just going to be four x. You could even think of it in terms of reverse power rule. Four is the same thing as four x to the zero. So you increase zero by one, so it's going to be four x to the first, and then you divide by that new exponent. Four x to the first divided by one, well, that's just going to be four x. So the antiderivative is four x."}, {"video_title": "Definite integrals reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "Four is the same thing as four x to the zero. So you increase zero by one, so it's going to be four x to the first, and then you divide by that new exponent. Four x to the first divided by one, well, that's just going to be four x. So the antiderivative is four x. This is, you could say, our capital F of x. And we're going to evaluate that at five and at negative three and we're gonna find the difference between these two. So what we have right over here, evaluating the antiderivative at our upper bound, that is going to be four x five."}, {"video_title": "Definite integrals reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "So the antiderivative is four x. This is, you could say, our capital F of x. And we're going to evaluate that at five and at negative three and we're gonna find the difference between these two. So what we have right over here, evaluating the antiderivative at our upper bound, that is going to be four x five. And then from that, we're going to subtract evaluating our antiderivative at the lower bound. So that's four x negative three. Four x negative three."}, {"video_title": "Definite integrals reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "So what we have right over here, evaluating the antiderivative at our upper bound, that is going to be four x five. And then from that, we're going to subtract evaluating our antiderivative at the lower bound. So that's four x negative three. Four x negative three. And what is that going to be equal to? So this is 20 and then minus negative 12. So this is going to be plus 12, which is going to be equal to 32."}, {"video_title": "Definite integrals reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "Four x negative three. And what is that going to be equal to? So this is 20 and then minus negative 12. So this is going to be plus 12, which is going to be equal to 32. Let's do another example where we're going to do the reverse power rule. So let's say that we want to find the indefinite, or we want to find the definite integral going from negative one to three of seven x squared dx. What is this going to be equal to?"}, {"video_title": "Definite integrals reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "So this is going to be plus 12, which is going to be equal to 32. Let's do another example where we're going to do the reverse power rule. So let's say that we want to find the indefinite, or we want to find the definite integral going from negative one to three of seven x squared dx. What is this going to be equal to? Well, what we want to do is evaluate what is the antiderivative of this? Or you could say, if this is lowercase f of x, what is capital F of x? Well, the reverse power rule, we increase this exponent by one."}, {"video_title": "Definite integrals reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "What is this going to be equal to? Well, what we want to do is evaluate what is the antiderivative of this? Or you could say, if this is lowercase f of x, what is capital F of x? Well, the reverse power rule, we increase this exponent by one. So we're going to have seven times x to the third and then we divide by that increased exponent. So seven x to the third divided by three. And we want to evaluate that at our upper bound and then subtract from that and evaluate it at our lower bound."}, {"video_title": "Definite integrals reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the reverse power rule, we increase this exponent by one. So we're going to have seven times x to the third and then we divide by that increased exponent. So seven x to the third divided by three. And we want to evaluate that at our upper bound and then subtract from that and evaluate it at our lower bound. So this is going to be equal to, so evaluating it at our upper bound, it's going to be seven times three to the third. I'll just write that, three to the third over three. And then from that, we are going to subtract this capital F of x, the antiderivative, evaluated at the lower bound."}, {"video_title": "Definite integrals reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "And we want to evaluate that at our upper bound and then subtract from that and evaluate it at our lower bound. So this is going to be equal to, so evaluating it at our upper bound, it's going to be seven times three to the third. I'll just write that, three to the third over three. And then from that, we are going to subtract this capital F of x, the antiderivative, evaluated at the lower bound. So that is going to be seven times negative one to the third, all of that over three. And so this first expression, let's see, this is going to be seven times three to the third over three this is 27 over three, this is going to be the same thing as seven times nine. So this is going to be 63."}, {"video_title": "Definite integrals reverse power rule AP Calculus AB Khan Academy.mp3", "Sentence": "And then from that, we are going to subtract this capital F of x, the antiderivative, evaluated at the lower bound. So that is going to be seven times negative one to the third, all of that over three. And so this first expression, let's see, this is going to be seven times three to the third over three this is 27 over three, this is going to be the same thing as seven times nine. So this is going to be 63. And this over here, negative one to the third power is negative one, but then we're subtracting a negative, so this is just going to be adding. And so this is just going to be plus seven over three. Plus seven over three, if we wanted to express this as a mixed number, seven over three is the same thing as two and 1 3rd, so when we add everything together, we are going to get 65 and 1 3rd."}, {"video_title": "_-substitution rational function AP Calculus AB Khan Academy.mp3", "Sentence": "So how can we tackle this? This seems like a hairy integral. Now the key insight here is to realize that you have this expression x to the fourth plus 7 and you also have its derivative up here. The derivative of x to the fourth plus 7 is equal to 4x to the third. The derivative of x to the fourth is 4x to the third. The derivative of 7 is just 0. So that's a big clue that u-substitution might be the tool of choice here."}, {"video_title": "_-substitution rational function AP Calculus AB Khan Academy.mp3", "Sentence": "The derivative of x to the fourth plus 7 is equal to 4x to the third. The derivative of x to the fourth is 4x to the third. The derivative of 7 is just 0. So that's a big clue that u-substitution might be the tool of choice here. I'll just write u. I'll write the whole thing. U-substitution could be the tool of choice. So given that, what would you want to set your u equal to?"}, {"video_title": "_-substitution rational function AP Calculus AB Khan Academy.mp3", "Sentence": "So that's a big clue that u-substitution might be the tool of choice here. I'll just write u. I'll write the whole thing. U-substitution could be the tool of choice. So given that, what would you want to set your u equal to? I'll let you think about that because if you can figure out this part, then the rest will just boil down to a fairly straightforward integral. You want to set u to be equal to the expression that you have its derivative laying around. So we could set u equal to x to the fourth plus 7."}, {"video_title": "_-substitution rational function AP Calculus AB Khan Academy.mp3", "Sentence": "So given that, what would you want to set your u equal to? I'll let you think about that because if you can figure out this part, then the rest will just boil down to a fairly straightforward integral. You want to set u to be equal to the expression that you have its derivative laying around. So we could set u equal to x to the fourth plus 7. Now what is du going to be equal to? I will do it in magenta. du is just going to be the derivative of x to the fourth plus 7 with respect to x, so 4x to the third plus 0 times dx."}, {"video_title": "_-substitution rational function AP Calculus AB Khan Academy.mp3", "Sentence": "So we could set u equal to x to the fourth plus 7. Now what is du going to be equal to? I will do it in magenta. du is just going to be the derivative of x to the fourth plus 7 with respect to x, so 4x to the third plus 0 times dx. I wrote it in differential form right over here, but it's a completely equivalent statement to saying that du, the derivative of u with respect to x, is equal to 4x to the third power. When someone writes du over dx like this, this is really a notation to say the derivative of u with respect to x. It really isn't a fraction in a very formal way, but oftentimes you can kind of pseudo-manipulate them like fractions."}, {"video_title": "_-substitution rational function AP Calculus AB Khan Academy.mp3", "Sentence": "du is just going to be the derivative of x to the fourth plus 7 with respect to x, so 4x to the third plus 0 times dx. I wrote it in differential form right over here, but it's a completely equivalent statement to saying that du, the derivative of u with respect to x, is equal to 4x to the third power. When someone writes du over dx like this, this is really a notation to say the derivative of u with respect to x. It really isn't a fraction in a very formal way, but oftentimes you can kind of pseudo-manipulate them like fractions. So if you wanted to go from here to there, you could kind of pretend that you're multiplying both sides by dx. But these are equivalent statements, and we want to get it in differential form in order to do proper u-substitution. The reason why this is useful, and I'll just rewrite it up here so that it becomes pretty obvious, our original integral we can rewrite as 4x to the third dx over x to the fourth plus 7."}, {"video_title": "_-substitution rational function AP Calculus AB Khan Academy.mp3", "Sentence": "It really isn't a fraction in a very formal way, but oftentimes you can kind of pseudo-manipulate them like fractions. So if you wanted to go from here to there, you could kind of pretend that you're multiplying both sides by dx. But these are equivalent statements, and we want to get it in differential form in order to do proper u-substitution. The reason why this is useful, and I'll just rewrite it up here so that it becomes pretty obvious, our original integral we can rewrite as 4x to the third dx over x to the fourth plus 7. Then it's pretty clear what's du and what's u. u, which we set to be equal to x to the fourth plus 7, and then du is equal to this. It's equal to 4x to the third dx. We saw it right over here."}, {"video_title": "_-substitution rational function AP Calculus AB Khan Academy.mp3", "Sentence": "The reason why this is useful, and I'll just rewrite it up here so that it becomes pretty obvious, our original integral we can rewrite as 4x to the third dx over x to the fourth plus 7. Then it's pretty clear what's du and what's u. u, which we set to be equal to x to the fourth plus 7, and then du is equal to this. It's equal to 4x to the third dx. We saw it right over here. So we can rewrite this integral, and I'll try to stay consistent with the colors, as the indefinite integral. What we have in magenta right over here, that's du over x to the fourth plus 7, which is just u. Or we could rewrite this entire thing as the integral of 1 over u du."}, {"video_title": "_-substitution rational function AP Calculus AB Khan Academy.mp3", "Sentence": "We saw it right over here. So we can rewrite this integral, and I'll try to stay consistent with the colors, as the indefinite integral. What we have in magenta right over here, that's du over x to the fourth plus 7, which is just u. Or we could rewrite this entire thing as the integral of 1 over u du. What is the indefinite integral of 1 over u du? That's just going to be equal to the natural log of the absolute value, and we use the absolute value so it'll be defined even for negative u's. It actually does work out, and I'll do another video where I show you that it definitely does."}, {"video_title": "_-substitution rational function AP Calculus AB Khan Academy.mp3", "Sentence": "Or we could rewrite this entire thing as the integral of 1 over u du. What is the indefinite integral of 1 over u du? That's just going to be equal to the natural log of the absolute value, and we use the absolute value so it'll be defined even for negative u's. It actually does work out, and I'll do another video where I show you that it definitely does. The natural log of the absolute value of u, and then we might have had a constant there that was lost when we took the derivative. That's essentially our answer in terms of u, but now we need to unsubstitute the u. What happens when we unsubstitute the u?"}, {"video_title": "_-substitution rational function AP Calculus AB Khan Academy.mp3", "Sentence": "It actually does work out, and I'll do another video where I show you that it definitely does. The natural log of the absolute value of u, and then we might have had a constant there that was lost when we took the derivative. That's essentially our answer in terms of u, but now we need to unsubstitute the u. What happens when we unsubstitute the u? Then we are left with, this is going to be equal to the natural log of the absolute value of, well u is x to the fourth plus 7. Then we can't forget our plus c out here. And we are done."}, {"video_title": "L'H\u00f4pital's rule example 2 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "You can't just plug in infinity and see what happens. But if you wanted to evaluate this limit, what you might try to do is just evaluate, if you want to find the limit as this numerator approaches infinity, you put in really large numbers there, you're going to see that it approaches infinity, that the numerator approaches infinity as x approaches infinity. And if you put really large numbers in the denominator, you're going to see that that also, well, not quite infinity, 3x squared will approach infinity, but we're subtracting it. So if you subtract infinity from some non-infinite number, this is going to be negative infinity. So if you were to just kind of evaluate it at infinity, the numerator, you would get positive infinity, the denominator, you would get negative infinity. And that's one of the indeterminate forms that L'Hopital's Rule can be applied to. And you're probably saying, hey, Sal, why are we even using L'Hopital's Rule?"}, {"video_title": "L'H\u00f4pital's rule example 2 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So if you subtract infinity from some non-infinite number, this is going to be negative infinity. So if you were to just kind of evaluate it at infinity, the numerator, you would get positive infinity, the denominator, you would get negative infinity. And that's one of the indeterminate forms that L'Hopital's Rule can be applied to. And you're probably saying, hey, Sal, why are we even using L'Hopital's Rule? I know how to do this without L'Hopital's Rule, and you probably do, or you should, and we'll do that in a second. But I just want to show you that L'Hopital's Rule also works for this type of problems, and I really just want to show you an example that had an infinity over a negative or positive infinity indeterminate form. But let's apply L'Hopital's Rule here."}, {"video_title": "L'H\u00f4pital's rule example 2 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And you're probably saying, hey, Sal, why are we even using L'Hopital's Rule? I know how to do this without L'Hopital's Rule, and you probably do, or you should, and we'll do that in a second. But I just want to show you that L'Hopital's Rule also works for this type of problems, and I really just want to show you an example that had an infinity over a negative or positive infinity indeterminate form. But let's apply L'Hopital's Rule here. So if this limit exists, or if the limit of their derivatives exists, then this limit's going to be equal to the limit as x approaches infinity of the derivative of the numerator. So the derivative of the numerator is the derivative of 4x squared is 8x minus 5 over, the derivative of the denominator is, well, the derivative of 1 is 0, the derivative of negative 3x squared is negative 6x. And once again, when you evaluate at infinity, the numerator is going to approach infinity, and the denominator is approaching negative infinity."}, {"video_title": "L'H\u00f4pital's rule example 2 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "But let's apply L'Hopital's Rule here. So if this limit exists, or if the limit of their derivatives exists, then this limit's going to be equal to the limit as x approaches infinity of the derivative of the numerator. So the derivative of the numerator is the derivative of 4x squared is 8x minus 5 over, the derivative of the denominator is, well, the derivative of 1 is 0, the derivative of negative 3x squared is negative 6x. And once again, when you evaluate at infinity, the numerator is going to approach infinity, and the denominator is approaching negative infinity. Negative 6 times infinity is negative infinity. So this is negative infinity. So let's apply L'Hopital's Rule again."}, {"video_title": "L'H\u00f4pital's rule example 2 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And once again, when you evaluate at infinity, the numerator is going to approach infinity, and the denominator is approaching negative infinity. Negative 6 times infinity is negative infinity. So this is negative infinity. So let's apply L'Hopital's Rule again. So if the limit of these guys' derivatives exist, or the rational function of the derivative of this guy divided by the derivative of that guy, if that exists, then this limit's going to be equal to the limit as x approaches infinity of arbitrary switch colors. Derivative of 8x minus 5 is just 8. Derivative of negative 6x is negative 6."}, {"video_title": "L'H\u00f4pital's rule example 2 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So let's apply L'Hopital's Rule again. So if the limit of these guys' derivatives exist, or the rational function of the derivative of this guy divided by the derivative of that guy, if that exists, then this limit's going to be equal to the limit as x approaches infinity of arbitrary switch colors. Derivative of 8x minus 5 is just 8. Derivative of negative 6x is negative 6. And this is just going to be, I mean, this is just a constant here, so it doesn't matter what limit you're approaching. This is just going to equal this value, which is what? If we put it in lowest common form, simplified form, it's negative 4 over 3."}, {"video_title": "L'H\u00f4pital's rule example 2 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Derivative of negative 6x is negative 6. And this is just going to be, I mean, this is just a constant here, so it doesn't matter what limit you're approaching. This is just going to equal this value, which is what? If we put it in lowest common form, simplified form, it's negative 4 over 3. So this limit exists. This was an indeterminate form, and the limit of this function's derivative over this function's derivative exists. So this limit must also equal negative 4 over 3."}, {"video_title": "L'H\u00f4pital's rule example 2 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "If we put it in lowest common form, simplified form, it's negative 4 over 3. So this limit exists. This was an indeterminate form, and the limit of this function's derivative over this function's derivative exists. So this limit must also equal negative 4 over 3. And by that same argument, that limit also must be equal to negative 4 over 3. And for those of you who say, hey, we already knew how to do this. We could just factor out an x squared."}, {"video_title": "L'H\u00f4pital's rule example 2 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "So this limit must also equal negative 4 over 3. And by that same argument, that limit also must be equal to negative 4 over 3. And for those of you who say, hey, we already knew how to do this. We could just factor out an x squared. You are absolutely right, and I'll show you that right here. Just to show you that it's not the only, L'Hopital's Rule isn't the only game in town. And frankly, for this type of problem, my first reaction probably wouldn't have been to use L'Hopital's Rule first."}, {"video_title": "L'H\u00f4pital's rule example 2 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "We could just factor out an x squared. You are absolutely right, and I'll show you that right here. Just to show you that it's not the only, L'Hopital's Rule isn't the only game in town. And frankly, for this type of problem, my first reaction probably wouldn't have been to use L'Hopital's Rule first. You could have said that that first limit, so the limit as x approaches infinity of 4x squared minus 5x over 1 minus 3x squared is equal to the limit as x approaches infinity. Let me draw a little line here to show you that this is equal to that, not to this thing over here. This is equal to the limit as x approaches infinity."}, {"video_title": "L'H\u00f4pital's rule example 2 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "And frankly, for this type of problem, my first reaction probably wouldn't have been to use L'Hopital's Rule first. You could have said that that first limit, so the limit as x approaches infinity of 4x squared minus 5x over 1 minus 3x squared is equal to the limit as x approaches infinity. Let me draw a little line here to show you that this is equal to that, not to this thing over here. This is equal to the limit as x approaches infinity. Let's factor out an x squared out of the numerator and the denominator. So you have an x squared times 4 minus 5 over x. x squared times 5 over x is going to be 5x. Divided by, let's factor an x squared out of the numerator."}, {"video_title": "L'H\u00f4pital's rule example 2 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "This is equal to the limit as x approaches infinity. Let's factor out an x squared out of the numerator and the denominator. So you have an x squared times 4 minus 5 over x. x squared times 5 over x is going to be 5x. Divided by, let's factor an x squared out of the numerator. So x squared times 1 over x squared minus 3. And then these x squareds cancel out. This is going to be equal to the limit as x approaches infinity of 4 minus 5 over x over 1 over x squared minus 3."}, {"video_title": "L'H\u00f4pital's rule example 2 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "Divided by, let's factor an x squared out of the numerator. So x squared times 1 over x squared minus 3. And then these x squareds cancel out. This is going to be equal to the limit as x approaches infinity of 4 minus 5 over x over 1 over x squared minus 3. And what's this going to be equal to? Well, as x approaches infinity, 5 divided by infinity, this term is going to be 0. Super duper infinitely large denominator, this is going to be 0."}, {"video_title": "L'H\u00f4pital's rule example 2 Derivative applications Differential Calculus Khan Academy.mp3", "Sentence": "This is going to be equal to the limit as x approaches infinity of 4 minus 5 over x over 1 over x squared minus 3. And what's this going to be equal to? Well, as x approaches infinity, 5 divided by infinity, this term is going to be 0. Super duper infinitely large denominator, this is going to be 0. That is going to approach 0. And same argument, this right here is going to approach 0. So all you're left with is the 4 and the negative 3."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "Alright, let's see if we can find the limit of one over square root of two sine of theta over cosine of two theta as theta approaches negative pi over four. And like always, try to give it a shot before we go through it together. Well, one take on it is, well, let's just say that this is going to be the same thing as the limit as theta approaches negative pi over four of one plus square root of two sine theta theta over the limit as theta approaches negative pi over four, make sure we can see that negative there, of cosine of two theta. And both of these expressions are, if these were function definitions, or if we were to graph y equals one plus square root of two times sine theta, or y equals cosine of two theta, we would get continuous functions, especially at theta is equal to negative pi over four, so we could just substitute in. We say, well, this is going to be equal to, this expression evaluated at negative pi over four, so one plus square root of two times sine of negative pi over four over cosine of two times negative pi over four. Now, negative pi over four, sine of negative pi over four is going to be negative square root of two over two. So this is negative square root of two over two."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "And both of these expressions are, if these were function definitions, or if we were to graph y equals one plus square root of two times sine theta, or y equals cosine of two theta, we would get continuous functions, especially at theta is equal to negative pi over four, so we could just substitute in. We say, well, this is going to be equal to, this expression evaluated at negative pi over four, so one plus square root of two times sine of negative pi over four over cosine of two times negative pi over four. Now, negative pi over four, sine of negative pi over four is going to be negative square root of two over two. So this is negative square root of two over two. We're assuming this is in radians. If we're thinking in degrees, this would be a negative 45 degree angle, so this is one of the trig values that it's good to know. And so if you have one, so let's see."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "So this is negative square root of two over two. We're assuming this is in radians. If we're thinking in degrees, this would be a negative 45 degree angle, so this is one of the trig values that it's good to know. And so if you have one, so let's see. Well, actually, let me just rewrite it. So this is going to be equal to one plus square root of two times that is going to be negative two over two. So this is going to be minus one."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "And so if you have one, so let's see. Well, actually, let me just rewrite it. So this is going to be equal to one plus square root of two times that is going to be negative two over two. So this is going to be minus one. That's the numerator over here. All of this stuff simplifies to negative one over, this is going to be cosine of negative pi over two. All right, this is negative pi over two."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "So this is going to be minus one. That's the numerator over here. All of this stuff simplifies to negative one over, this is going to be cosine of negative pi over two. All right, this is negative pi over two. Cosine of negative pi over two, if you thought in degrees, that's gonna be negative 90 degrees. Well, cosine of that is just going to be zero. So what we end up with is equal to zero over zero."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "All right, this is negative pi over two. Cosine of negative pi over two, if you thought in degrees, that's gonna be negative 90 degrees. Well, cosine of that is just going to be zero. So what we end up with is equal to zero over zero. And as we've talked about before, if we had something non-zero divided by zero, we'd say, okay, that's undefined. We might as well give up. But when we have this indeterminate form, it does not mean that the limit does not exist."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "So what we end up with is equal to zero over zero. And as we've talked about before, if we had something non-zero divided by zero, we'd say, okay, that's undefined. We might as well give up. But when we have this indeterminate form, it does not mean that the limit does not exist. It's usually a clue that we should use some tools in our toolkit, one of which is to do some manipulation here to get an expression that maybe is defined at theta is equal to, or is not an indeterminate form, at theta is equal to negative pi over four, and I will see other tools in our toolkit in the future. So let me algebraically manipulate this a little bit. So if I have one plus the square root of two sine theta over cosine two theta, as you can imagine, the things that might be useful here are our trig identities."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "But when we have this indeterminate form, it does not mean that the limit does not exist. It's usually a clue that we should use some tools in our toolkit, one of which is to do some manipulation here to get an expression that maybe is defined at theta is equal to, or is not an indeterminate form, at theta is equal to negative pi over four, and I will see other tools in our toolkit in the future. So let me algebraically manipulate this a little bit. So if I have one plus the square root of two sine theta over cosine two theta, as you can imagine, the things that might be useful here are our trig identities. And in particular, cosine of two theta seems interesting. Let me write some trig identities involving cosine of two theta. I'll write it over here."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "So if I have one plus the square root of two sine theta over cosine two theta, as you can imagine, the things that might be useful here are our trig identities. And in particular, cosine of two theta seems interesting. Let me write some trig identities involving cosine of two theta. I'll write it over here. So we know that cosine of two theta is equal to cosine squared of theta minus sine squared of theta, which is equal to one minus two sine squared of theta, which is equal to two cosine squared theta minus one. And you can go from this one to this one to this one just using the Pythagorean identity, and we proved that in earlier videos in trigonometry on Khan Academy. Now, do any of these look useful?"}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "I'll write it over here. So we know that cosine of two theta is equal to cosine squared of theta minus sine squared of theta, which is equal to one minus two sine squared of theta, which is equal to two cosine squared theta minus one. And you can go from this one to this one to this one just using the Pythagorean identity, and we proved that in earlier videos in trigonometry on Khan Academy. Now, do any of these look useful? Well, all of these three are gonna be differences of squares so we can factor them in interesting ways. And remember, our goal at the end of the day is maybe cancel things out that are making us get this zero over zero. And if I could factor this into something that involves a one plus square root of two sine theta, then I'm gonna be in business."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "Now, do any of these look useful? Well, all of these three are gonna be differences of squares so we can factor them in interesting ways. And remember, our goal at the end of the day is maybe cancel things out that are making us get this zero over zero. And if I could factor this into something that involves a one plus square root of two sine theta, then I'm gonna be in business. And it looks like this right over here, that can be factored as one plus square root of two sine theta times one minus square root of two sine theta. So let me use this. Cosine of two theta is the same thing, cosine of two theta is the same thing as one minus two sine squared theta, which is just a difference of squares."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "And if I could factor this into something that involves a one plus square root of two sine theta, then I'm gonna be in business. And it looks like this right over here, that can be factored as one plus square root of two sine theta times one minus square root of two sine theta. So let me use this. Cosine of two theta is the same thing, cosine of two theta is the same thing as one minus two sine squared theta, which is just a difference of squares. We can rewrite that as, if this is a squared minus b squared, this is a plus b times a minus b. So I can just replace this with one plus square root of two sine theta times one minus square root of two sine theta. And now we have some nice canceling, or potential canceling that can occur."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "Cosine of two theta is the same thing, cosine of two theta is the same thing as one minus two sine squared theta, which is just a difference of squares. We can rewrite that as, if this is a squared minus b squared, this is a plus b times a minus b. So I can just replace this with one plus square root of two sine theta times one minus square root of two sine theta. And now we have some nice canceling, or potential canceling that can occur. So we could say that cancels with that. And we could say that that is going to be equal, and let me do this in a new color, this is going to be equal to, in the numerator we just have one, in the denominator we just are left with one minus square root of two sine theta. And if we want these expressions to truly be equal, we would have to have them to have the same, if you view them as function definitions, as having the same domain."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "And now we have some nice canceling, or potential canceling that can occur. So we could say that cancels with that. And we could say that that is going to be equal, and let me do this in a new color, this is going to be equal to, in the numerator we just have one, in the denominator we just are left with one minus square root of two sine theta. And if we want these expressions to truly be equal, we would have to have them to have the same, if you view them as function definitions, as having the same domain. So this one right over here, this one we already saw is not defined at theta is equal to negative pi over four. And so this one, in order for these to be equivalent, we have to say that this one is also not. And actually other places, but let's just say theta does not equal negative pi over four."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "And if we want these expressions to truly be equal, we would have to have them to have the same, if you view them as function definitions, as having the same domain. So this one right over here, this one we already saw is not defined at theta is equal to negative pi over four. And so this one, in order for these to be equivalent, we have to say that this one is also not. And actually other places, but let's just say theta does not equal negative pi over four. And we could think about all of this happening in some type of an open interval around negative pi over four if we wanted to get very precise. But if we, for this particular case, well let's just say everything we're doing is in the open interval. So in open interval, in open interval between theta, or say negative one and one."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "And actually other places, but let's just say theta does not equal negative pi over four. And we could think about all of this happening in some type of an open interval around negative pi over four if we wanted to get very precise. But if we, for this particular case, well let's just say everything we're doing is in the open interval. So in open interval, in open interval between theta, or say negative one and one. And I think that covers it, because if we have pi over four, that is not going to get us the zero over zero form. And pi over four would make this denominator equal to zero, and it also makes, let's see, pi over four also will make this denominator equal to zero. Because we would get one minus one."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "So in open interval, in open interval between theta, or say negative one and one. And I think that covers it, because if we have pi over four, that is not going to get us the zero over zero form. And pi over four would make this denominator equal to zero, and it also makes, let's see, pi over four also will make this denominator equal to zero. Because we would get one minus one. So I think we're good if we're just assuming, if we're restricting to this open interval, and that's okay because we're taking the limit and say it approaches something within this open interval. And I'm being extra precise, because I'm trying to explain it to you, and it's important to be precise. But obviously if you're working this out on a test or notebook, you wouldn't be taking, or taking as much trouble to be putting all of these caveats in."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "Because we would get one minus one. So I think we're good if we're just assuming, if we're restricting to this open interval, and that's okay because we're taking the limit and say it approaches something within this open interval. And I'm being extra precise, because I'm trying to explain it to you, and it's important to be precise. But obviously if you're working this out on a test or notebook, you wouldn't be taking, or taking as much trouble to be putting all of these caveats in. So what we've now realized is that, okay, this expression, actually let's think about this. Let's think about the limit, the limit as theta approaches negative pi over four of this thing, without the restriction, of one over one minus the square root of two sine of theta. If we're dealing with this over, you know, in this open interval, or actually even disregarding that, this theta, or this expression is continuous at, it is defined and it is continuous at theta is equal to negative pi over four."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "But obviously if you're working this out on a test or notebook, you wouldn't be taking, or taking as much trouble to be putting all of these caveats in. So what we've now realized is that, okay, this expression, actually let's think about this. Let's think about the limit, the limit as theta approaches negative pi over four of this thing, without the restriction, of one over one minus the square root of two sine of theta. If we're dealing with this over, you know, in this open interval, or actually even disregarding that, this theta, or this expression is continuous at, it is defined and it is continuous at theta is equal to negative pi over four. So this is just going to be equal to one over one minus the square root of two times sine of negative pi over four. Sine of negative pi over four. Sine of negative pi over four, which we've already seen is negative square root of two over two."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "If we're dealing with this over, you know, in this open interval, or actually even disregarding that, this theta, or this expression is continuous at, it is defined and it is continuous at theta is equal to negative pi over four. So this is just going to be equal to one over one minus the square root of two times sine of negative pi over four. Sine of negative pi over four. Sine of negative pi over four, which we've already seen is negative square root of two over two. And so this is going to be equal to one over one minus square root of two times the negative square root of two over two. So negative negative, you get a positive. Square root of two times square root of two is two over two is gonna be one."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "Sine of negative pi over four, which we've already seen is negative square root of two over two. And so this is going to be equal to one over one minus square root of two times the negative square root of two over two. So negative negative, you get a positive. Square root of two times square root of two is two over two is gonna be one. So this is going to be equal to 1 1 2. And so, I wanna be very clear. This expression is not the same thing as this expression."}, {"video_title": "Trig limit using double angle identity Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "Square root of two times square root of two is two over two is gonna be one. So this is going to be equal to 1 1 2. And so, I wanna be very clear. This expression is not the same thing as this expression. Where the same thing at all values of theta, especially if we're dealing in this open interval, except at theta equals negative pi over four, this one is not defined, and this one is defined. But as we've seen multiple times before, if we find a function that is equal to our original, or an expression that's equal to our original expression, at all values of theta, except where the original one was not defined at a certain point, but this new one is defined and is continuous there, well then these two limits are going to be equal. So if this limit is 1 1 2, then this limit is going to be 1 1 2."}, {"video_title": "Harmonic series and \ud835\udc5d-series AP\u00ae\ufe0e Calculus BC Khan Academy.mp3", "Sentence": "For many hundreds of years, mathematicians have been fascinated by the infinite sum, which we would call a series, of one plus 1 1\u20442 plus 1 3rd plus 1 4th, and you just keep adding on and on and on forever. And this is interesting on many layers. One, it just feels like something that it would be interesting to explore. It's one over one plus one over two plus one over three, that each of these terms are getting smaller and smaller. They're approaching zero, but when you add them all together, these infinite number of terms, do you get a finite number, or does it diverge? Do you not get a finite number? This also shows up in music, and this actually might have been one of the early motivations for studying this series, where if you have a fundamental note, a fundamental frequency in music, and the point of this video isn't to teach you too much about music, but if you have a fundamental note, that might be a pure A or something like that."}, {"video_title": "Harmonic series and \ud835\udc5d-series AP\u00ae\ufe0e Calculus BC Khan Academy.mp3", "Sentence": "It's one over one plus one over two plus one over three, that each of these terms are getting smaller and smaller. They're approaching zero, but when you add them all together, these infinite number of terms, do you get a finite number, or does it diverge? Do you not get a finite number? This also shows up in music, and this actually might have been one of the early motivations for studying this series, where if you have a fundamental note, a fundamental frequency in music, and the point of this video isn't to teach you too much about music, but if you have a fundamental note, that might be a pure A or something like that. I'm just showing you one of its wavelengths. Obviously, you would keep going like that, and this is a hand-drawn version, so it's not perfect. The harmonics are the frequencies, the overtones, that at least to our ear reinforce that A."}, {"video_title": "Harmonic series and \ud835\udc5d-series AP\u00ae\ufe0e Calculus BC Khan Academy.mp3", "Sentence": "This also shows up in music, and this actually might have been one of the early motivations for studying this series, where if you have a fundamental note, a fundamental frequency in music, and the point of this video isn't to teach you too much about music, but if you have a fundamental note, that might be a pure A or something like that. I'm just showing you one of its wavelengths. Obviously, you would keep going like that, and this is a hand-drawn version, so it's not perfect. The harmonics are the frequencies, the overtones, that at least to our ear reinforce that A. And what's true about the harmonics are that they will be one half of the wavelength of A, in which case it might look something like this. So this would be a harmonic of A. It has half of the wavelength of A."}, {"video_title": "Harmonic series and \ud835\udc5d-series AP\u00ae\ufe0e Calculus BC Khan Academy.mp3", "Sentence": "The harmonics are the frequencies, the overtones, that at least to our ear reinforce that A. And what's true about the harmonics are that they will be one half of the wavelength of A, in which case it might look something like this. So this would be a harmonic of A. It has half of the wavelength of A. Notice, when it finishes its second full waveform, it ends again right at the same time that the wavelength of A ends. And then another harmonic would be something that has a third the wavelength of an A, and a fourth of the wavelength of A. And if you look at a lot of musical instruments or what sounds good to our ears, they're playing not just the fundamental tone, but a lot of the harmonics."}, {"video_title": "Harmonic series and \ud835\udc5d-series AP\u00ae\ufe0e Calculus BC Khan Academy.mp3", "Sentence": "It has half of the wavelength of A. Notice, when it finishes its second full waveform, it ends again right at the same time that the wavelength of A ends. And then another harmonic would be something that has a third the wavelength of an A, and a fourth of the wavelength of A. And if you look at a lot of musical instruments or what sounds good to our ears, they're playing not just the fundamental tone, but a lot of the harmonics. But anyway, that was a long-winded way of justifying why this is called the harmonic series. Harmonic series. And in a future video, we will prove that, and I don't wanna ruin the punchline, but this actually diverges."}, {"video_title": "Harmonic series and \ud835\udc5d-series AP\u00ae\ufe0e Calculus BC Khan Academy.mp3", "Sentence": "And if you look at a lot of musical instruments or what sounds good to our ears, they're playing not just the fundamental tone, but a lot of the harmonics. But anyway, that was a long-winded way of justifying why this is called the harmonic series. Harmonic series. And in a future video, we will prove that, and I don't wanna ruin the punchline, but this actually diverges. And I will come up with general rules for when things that look like this might converge or diverge. But the harmonic series in particular diverges. So if we were to write it, so in sigma form, we would write it like this."}, {"video_title": "Harmonic series and \ud835\udc5d-series AP\u00ae\ufe0e Calculus BC Khan Academy.mp3", "Sentence": "And in a future video, we will prove that, and I don't wanna ruin the punchline, but this actually diverges. And I will come up with general rules for when things that look like this might converge or diverge. But the harmonic series in particular diverges. So if we were to write it, so in sigma form, we would write it like this. We're going from n equals one to infinity of one over n. Now, another interesting thing is, well, what if we were to throw in some exponents here? So we already said, and I'll just rewrite it. It doesn't hurt to rewrite it and get more familiar with it."}, {"video_title": "Harmonic series and \ud835\udc5d-series AP\u00ae\ufe0e Calculus BC Khan Academy.mp3", "Sentence": "So if we were to write it, so in sigma form, we would write it like this. We're going from n equals one to infinity of one over n. Now, another interesting thing is, well, what if we were to throw in some exponents here? So we already said, and I'll just rewrite it. It doesn't hurt to rewrite it and get more familiar with it. This right over here is the harmonic series. One over one, which is just one, plus one over two, plus one over three, so on and so forth. But what if we were to raise each of these denominators to, say, the second power?"}, {"video_title": "Harmonic series and \ud835\udc5d-series AP\u00ae\ufe0e Calculus BC Khan Academy.mp3", "Sentence": "It doesn't hurt to rewrite it and get more familiar with it. This right over here is the harmonic series. One over one, which is just one, plus one over two, plus one over three, so on and so forth. But what if we were to raise each of these denominators to, say, the second power? So you might have something that looks like this, where you have from n equals one to infinity of one over n to the second power. Well, then it would look like this. It would be one over one squared, which is one."}, {"video_title": "Harmonic series and \ud835\udc5d-series AP\u00ae\ufe0e Calculus BC Khan Academy.mp3", "Sentence": "But what if we were to raise each of these denominators to, say, the second power? So you might have something that looks like this, where you have from n equals one to infinity of one over n to the second power. Well, then it would look like this. It would be one over one squared, which is one. And we could just write that first term as one, plus one over two squared, which would be 1 1\u2074, plus one over three squared, which is 1 9th. And then you could go on and on forever, forever. And then you could generalize it."}, {"video_title": "Harmonic series and \ud835\udc5d-series AP\u00ae\ufe0e Calculus BC Khan Academy.mp3", "Sentence": "It would be one over one squared, which is one. And we could just write that first term as one, plus one over two squared, which would be 1 1\u2074, plus one over three squared, which is 1 9th. And then you could go on and on forever, forever. And then you could generalize it. You could say, hey, all right, what if we wanted to have a general class of series that we were to describe like this, going from n equals one to infinity of one over n to the p, where p could be any exponent. So for example, well, the way this would play out is this would be one plus one over two to the p plus one over three to the p plus one over four to the p. And it doesn't just have to be an integer value. It could be, some p could be 1\u20442, in which case you would have one plus one over the square root of two plus one over the square root of three."}, {"video_title": "Harmonic series and \ud835\udc5d-series AP\u00ae\ufe0e Calculus BC Khan Academy.mp3", "Sentence": "And then you could generalize it. You could say, hey, all right, what if we wanted to have a general class of series that we were to describe like this, going from n equals one to infinity of one over n to the p, where p could be any exponent. So for example, well, the way this would play out is this would be one plus one over two to the p plus one over three to the p plus one over four to the p. And it doesn't just have to be an integer value. It could be, some p could be 1\u20442, in which case you would have one plus one over the square root of two plus one over the square root of three. This entire class of series, and of course harmonic series is a special case where p is equal to one, this is known as p-series. So these are known as p-series. And I tried to remember it because it's p for the power that you are raising this denominator to."}, {"video_title": "Harmonic series and \ud835\udc5d-series AP\u00ae\ufe0e Calculus BC Khan Academy.mp3", "Sentence": "It could be, some p could be 1\u20442, in which case you would have one plus one over the square root of two plus one over the square root of three. This entire class of series, and of course harmonic series is a special case where p is equal to one, this is known as p-series. So these are known as p-series. And I tried to remember it because it's p for the power that you are raising this denominator to. You could also view it as you're raising the whole expression to it because one to any exponent is still going to be one. But I hinted a little bit that maybe some of these converge and some of these diverge. And we're going to prove it in future videos, but the general principle is if p is greater than one, then we are going to converge."}, {"video_title": "Harmonic series and \ud835\udc5d-series AP\u00ae\ufe0e Calculus BC Khan Academy.mp3", "Sentence": "And I tried to remember it because it's p for the power that you are raising this denominator to. You could also view it as you're raising the whole expression to it because one to any exponent is still going to be one. But I hinted a little bit that maybe some of these converge and some of these diverge. And we're going to prove it in future videos, but the general principle is if p is greater than one, then we are going to converge. And that makes sense intuitively because that means that the terms are getting smaller and smaller fast enough because the larger the exponent for that denominator, that means that the denominator is going to get bigger faster, which means that the fraction is going to get smaller faster. And if p is less than or equal to one, and of course when p is equal to one, we're dealing with the famous harmonic series, that's a situation in which we diverge. And we will prove these things in future videos."}, {"video_title": "Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3", "Sentence": "For each point on the graph, is the object speeding up, slowing down, or neither? So pause this video and see if you can figure that out. All right, now let's do it together. And first, we need to make sure we're reading this carefully, because they're not asking is the velocity increasing, decreasing, or neither. They're saying is the object speeding up, slowing down, or neither? So they're talking about speed, which is the magnitude of velocity. You could think of it as the absolute value of velocity, especially when we're thinking about it in one dimension here."}, {"video_title": "Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3", "Sentence": "And first, we need to make sure we're reading this carefully, because they're not asking is the velocity increasing, decreasing, or neither. They're saying is the object speeding up, slowing down, or neither? So they're talking about speed, which is the magnitude of velocity. You could think of it as the absolute value of velocity, especially when we're thinking about it in one dimension here. So even though they're not asking about velocity, I'm actually going to answer both, so that we can see how sometimes they move together, velocity and speed, but sometimes one might be increasing while the other might be decreasing. So if we look at this point right over here, where our velocity is two meters per second, the speed is the absolute value of velocity, which would also be two meters per second, and we can see that the slope of the velocity-time graph is positive, and so our velocity is increasing, and the absolute value of our velocity, which is speed, is also increasing. A moment later, our velocity might be 2.1 meters per second, and our speed would also be 2.1 meters per second."}, {"video_title": "Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3", "Sentence": "You could think of it as the absolute value of velocity, especially when we're thinking about it in one dimension here. So even though they're not asking about velocity, I'm actually going to answer both, so that we can see how sometimes they move together, velocity and speed, but sometimes one might be increasing while the other might be decreasing. So if we look at this point right over here, where our velocity is two meters per second, the speed is the absolute value of velocity, which would also be two meters per second, and we can see that the slope of the velocity-time graph is positive, and so our velocity is increasing, and the absolute value of our velocity, which is speed, is also increasing. A moment later, our velocity might be 2.1 meters per second, and our speed would also be 2.1 meters per second. That seems intuitive enough. And we get the other scenario if we go to this point right over here. Our velocity is still positive, but we see that our velocity-time graph is now downward sloping."}, {"video_title": "Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3", "Sentence": "A moment later, our velocity might be 2.1 meters per second, and our speed would also be 2.1 meters per second. That seems intuitive enough. And we get the other scenario if we go to this point right over here. Our velocity is still positive, but we see that our velocity-time graph is now downward sloping. So our velocity is decreasing because of that downward slope, and the absolute value of our velocity is also decreasing. Right at that moment, our speed is two meters per second, and then a moment later, it might be 1.9 meters per second. All right, now let's go to this point."}, {"video_title": "Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3", "Sentence": "Our velocity is still positive, but we see that our velocity-time graph is now downward sloping. So our velocity is decreasing because of that downward slope, and the absolute value of our velocity is also decreasing. Right at that moment, our speed is two meters per second, and then a moment later, it might be 1.9 meters per second. All right, now let's go to this point. So this point is really interesting. Here we see that our velocity, the slope of the tangent line, is still negative, so our velocity is still decreasing. But what about the absolute value of our velocity, which is speed?"}, {"video_title": "Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3", "Sentence": "All right, now let's go to this point. So this point is really interesting. Here we see that our velocity, the slope of the tangent line, is still negative, so our velocity is still decreasing. But what about the absolute value of our velocity, which is speed? Well, if you think about it, a moment before this, we were slowing down to get to a zero velocity, and a moment after this, we're going to be speeding up to start having negative velocity. You might say, wait, speeding up for negative velocity? Remember, speed is the absolute value."}, {"video_title": "Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3", "Sentence": "But what about the absolute value of our velocity, which is speed? Well, if you think about it, a moment before this, we were slowing down to get to a zero velocity, and a moment after this, we're going to be speeding up to start having negative velocity. You might say, wait, speeding up for negative velocity? Remember, speed is the absolute value. So if your velocity goes from zero to negative one meters per second, your speed just went from zero to one meter per second. So we're slowing down here, and we're speeding up here, but right at this moment, neither is happening. We are neither speeding up nor slowing down."}, {"video_title": "Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3", "Sentence": "Remember, speed is the absolute value. So if your velocity goes from zero to negative one meters per second, your speed just went from zero to one meter per second. So we're slowing down here, and we're speeding up here, but right at this moment, neither is happening. We are neither speeding up nor slowing down. Now what about this point? Here, the slope of our velocity time graph, or the slope of the tangent line, is still negative, so our velocity is still decreasing. But what about speed?"}, {"video_title": "Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3", "Sentence": "We are neither speeding up nor slowing down. Now what about this point? Here, the slope of our velocity time graph, or the slope of the tangent line, is still negative, so our velocity is still decreasing. But what about speed? Well, our velocity is already negative, and it's becoming more negative. So the absolute value of velocity, which is two meters per second, that is increasing at that moment in time. So our speed is actually increasing."}, {"video_title": "Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3", "Sentence": "But what about speed? Well, our velocity is already negative, and it's becoming more negative. So the absolute value of velocity, which is two meters per second, that is increasing at that moment in time. So our speed is actually increasing. So notice here, you see a difference. Now what about this point? Well, the slope of the tangent line here, of our velocity time graph, is zero right at that point."}, {"video_title": "Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3", "Sentence": "So our speed is actually increasing. So notice here, you see a difference. Now what about this point? Well, the slope of the tangent line here, of our velocity time graph, is zero right at that point. So that means that our velocity is not changing. So you could say velocity not changing. And if speed is the absolute value, or the magnitude of velocity, well, that will also be not changing."}, {"video_title": "Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3", "Sentence": "Well, the slope of the tangent line here, of our velocity time graph, is zero right at that point. So that means that our velocity is not changing. So you could say velocity not changing. And if speed is the absolute value, or the magnitude of velocity, well, that will also be not changing. So we would say speed is, I'll just say, neither slowing down nor speeding up. Last but not least, this point right over here, the slope of the tangent line is positive, so our velocity is increasing. What about speed?"}, {"video_title": "Interpreting change in speed from velocity-time graph Differential Calculus Khan Academy.mp3", "Sentence": "And if speed is the absolute value, or the magnitude of velocity, well, that will also be not changing. So we would say speed is, I'll just say, neither slowing down nor speeding up. Last but not least, this point right over here, the slope of the tangent line is positive, so our velocity is increasing. What about speed? Well, the speed here is two meters per second. Remember, it would be the absolute value of the velocity. And the absolute value is actually going down if we forward in time a little bit."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "We've already seen that if we were to start with a differential equation, the derivative of y with respect to x is equal to y, and we had the initial condition that y of zero is equal to one, that the particular solution to this, given this initial conditions, is y of x is equal to e to the x, or I guess we could just say y is equal to e to the x if we didn't want to write it with the function notation. And that's all fair and well, and this works out well. This is a separable differential equation, and we can integrate things quite easily. But as you will see as you go further in the world of differential equations, most differential equations are not so easy to solve. In fact, many of them are impossible to solve using analytic methods. And so given that, what do you do? You've nicely described some phenomena, modeled some phenomena using differential equations, but if you can't solve it analytically, do you just give up?"}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "But as you will see as you go further in the world of differential equations, most differential equations are not so easy to solve. In fact, many of them are impossible to solve using analytic methods. And so given that, what do you do? You've nicely described some phenomena, modeled some phenomena using differential equations, but if you can't solve it analytically, do you just give up? And the answer to that question is no. You do not just give up, because we now have computers, and computers are really good at numerical methods, numerical methods for approximating and giving us a sense of what the solution to a differential equation might look like. And so how do we do that?"}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "You've nicely described some phenomena, modeled some phenomena using differential equations, but if you can't solve it analytically, do you just give up? And the answer to that question is no. You do not just give up, because we now have computers, and computers are really good at numerical methods, numerical methods for approximating and giving us a sense of what the solution to a differential equation might look like. And so how do we do that? Well, in this video, we can explore one of the most straightforward numerical methods for approximating a particular solution. So what we do is, so let me draw a little table here. So a little table here."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "And so how do we do that? Well, in this video, we can explore one of the most straightforward numerical methods for approximating a particular solution. So what we do is, so let me draw a little table here. So a little table here. And so, actually let me give myself some, I'm gonna do it over here on the left-hand side. So a little table. So x and then y, with x, y, and then dy, dx."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So a little table here. And so, actually let me give myself some, I'm gonna do it over here on the left-hand side. So a little table. So x and then y, with x, y, and then dy, dx. And you could set up a table like this to create a slope field. You could just pick all the, you could sample x's and y's in the xy plane, and then figure out, for a first-order differential equation like this, what is the slope going to be at that point, and you could construct a slope field. And we're gonna do something kind of related, but instead of trying to construct a slope field, we're gonna start with this initial condition."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So x and then y, with x, y, and then dy, dx. And you could set up a table like this to create a slope field. You could just pick all the, you could sample x's and y's in the xy plane, and then figure out, for a first-order differential equation like this, what is the slope going to be at that point, and you could construct a slope field. And we're gonna do something kind of related, but instead of trying to construct a slope field, we're gonna start with this initial condition. We know that y of zero is equal to one. We know that the particular solution of this differential equation contains this point. So we're gonna start with that point."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "And we're gonna do something kind of related, but instead of trying to construct a slope field, we're gonna start with this initial condition. We know that y of zero is equal to one. We know that the particular solution of this differential equation contains this point. So we're gonna start with that point. So we're gonna start with x is equal to zero, and let me do this in a different color. We're gonna start with x is equal to zero, y is equal to one, which is that point right over there. And we're gonna say, well, okay, what is the derivative at that point?"}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So we're gonna start with that point. So we're gonna start with x is equal to zero, and let me do this in a different color. We're gonna start with x is equal to zero, y is equal to one, which is that point right over there. And we're gonna say, well, okay, what is the derivative at that point? Well, we know the derivative at any point or any solution to this differential equation, the derivative is going to be equal to the y value. So in this case, the derivative is going to be equal to y. It's going to be equal to one."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "And we're gonna say, well, okay, what is the derivative at that point? Well, we know the derivative at any point or any solution to this differential equation, the derivative is going to be equal to the y value. So in this case, the derivative is going to be equal to y. It's going to be equal to one. And in general, if the derivative, just like what we saw in the case of slope fields, as long as the derivative is expressed as a function of x's and y of x's, then you can figure out what the slope of the tangent line will be at that point. And so you say, okay, there's a slope of one at that point. So I can depict it like that."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "It's going to be equal to one. And in general, if the derivative, just like what we saw in the case of slope fields, as long as the derivative is expressed as a function of x's and y of x's, then you can figure out what the slope of the tangent line will be at that point. And so you say, okay, there's a slope of one at that point. So I can depict it like that. And instead of just keep doing that at a bunch of points, you say, okay, well, let's just, we know that the slope is changing, or it's probably changing for most cases, but let's just assume it's fixed until our next x, and then use that assumption to estimate what the next y would be. So what am I talking about here? So when I talk about the next x, we're talking about, well, let's just step, let's say for the sake of simplicity, we're gonna have a delta x of one, a change in x of one."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So I can depict it like that. And instead of just keep doing that at a bunch of points, you say, okay, well, let's just, we know that the slope is changing, or it's probably changing for most cases, but let's just assume it's fixed until our next x, and then use that assumption to estimate what the next y would be. So what am I talking about here? So when I talk about the next x, we're talking about, well, let's just step, let's say for the sake of simplicity, we're gonna have a delta x of one, a change in x of one. So we're gonna step from x equals zero now. We're gonna now step from that to x is equal to one. So we're now gonna go to, actually, let me not use that, I used that yellow color already for the actual graph, or for the actual e to the x."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So when I talk about the next x, we're talking about, well, let's just step, let's say for the sake of simplicity, we're gonna have a delta x of one, a change in x of one. So we're gonna step from x equals zero now. We're gonna now step from that to x is equal to one. So we're now gonna go to, actually, let me not use that, I used that yellow color already for the actual graph, or for the actual e to the x. So now let's say x is equal to one. So we've, our delta x is one. So we've just added one here."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So we're now gonna go to, actually, let me not use that, I used that yellow color already for the actual graph, or for the actual e to the x. So now let's say x is equal to one. So we've, our delta x is one. So we've just added one here. And what we can do in our little approximation scheme here is let's just assume that that slope was constant over that interval. So where does that get us to? Well, if y was at one, and if I have a slope of one, for one more, for one increase in x, I'm gonna increase by y by one."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So we've just added one here. And what we can do in our little approximation scheme here is let's just assume that that slope was constant over that interval. So where does that get us to? Well, if y was at one, and if I have a slope of one, for one more, for one increase in x, I'm gonna increase by y by one. So then y is going to increase by one, and is going to get to two. And we see that point right over there. And you already might see where this is going."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "Well, if y was at one, and if I have a slope of one, for one more, for one increase in x, I'm gonna increase by y by one. So then y is going to increase by one, and is going to get to two. And we see that point right over there. And you already might see where this is going. Now, if this were actually a point on the curve, on the solution, and if it was satisfying this, what would then the derivative be? Well, the derivative is equal to y. The slope of the tangent line is going to be equal to y."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "And you already might see where this is going. Now, if this were actually a point on the curve, on the solution, and if it was satisfying this, what would then the derivative be? Well, the derivative is equal to y. The slope of the tangent line is going to be equal to y. So in this case, the slope of the tangent line is now going to be equal to two. And we could depict that, let me depict that in magenta here. So it is going to be, it is going to be two."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "The slope of the tangent line is going to be equal to y. So in this case, the slope of the tangent line is now going to be equal to two. And we could depict that, let me depict that in magenta here. So it is going to be, it is going to be two. So it's gonna look, the slope of the tangent line there is going to be two. And so what does that tell us? Well, if we step, if we step by our delta x one more, so now our x is equal to two, what should the corresponding, what should the corresponding y be?"}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So it is going to be, it is going to be two. So it's gonna look, the slope of the tangent line there is going to be two. And so what does that tell us? Well, if we step, if we step by our delta x one more, so now our x is equal to two, what should the corresponding, what should the corresponding y be? Well, let's see. Now, for every one that we increase in the x direction, we should increase two in the y direction, because the slope is two. So the very next one should be four."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "Well, if we step, if we step by our delta x one more, so now our x is equal to two, what should the corresponding, what should the corresponding y be? Well, let's see. Now, for every one that we increase in the x direction, we should increase two in the y direction, because the slope is two. So the very next one should be four. Y is equal to four. So we could imagine, we have now kind of had a constant slope, and we get to that point right over there. And now we can do the same thing."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So the very next one should be four. Y is equal to four. So we could imagine, we have now kind of had a constant slope, and we get to that point right over there. And now we can do the same thing. Well, if we assume dy dx, based on the differential equation, has to be equal to y, we say, okay, the slope of the tangent line there is going to be the same thing as y. It's going to be four. And so if we step our x up by one, if we increment our x by one again, and once again, we just decided to increment by one."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "And now we can do the same thing. Well, if we assume dy dx, based on the differential equation, has to be equal to y, we say, okay, the slope of the tangent line there is going to be the same thing as y. It's going to be four. And so if we step our x up by one, if we increment our x by one again, and once again, we just decided to increment by one. We could have incremented by 10. We could have incremented by.01. And you could guess which one's going to give you a more accurate result."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "And so if we step our x up by one, if we increment our x by one again, and once again, we just decided to increment by one. We could have incremented by 10. We could have incremented by.01. And you could guess which one's going to give you a more accurate result. But if we step up by one now, and our slope is four, well, we're going to increase by, if we increased x by one, we're going to increase y by four. So we are going to get, we are going to get to eight. And so we are at the.3,8, which is right over here."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "And you could guess which one's going to give you a more accurate result. But if we step up by one now, and our slope is four, well, we're going to increase by, if we increased x by one, we're going to increase y by four. So we are going to get, we are going to get to eight. And so we are at the.3,8, which is right over here. And so for this next stretch, the next stretch is going to look like that. And as you can see, just by doing this, we have been able to approximate what the particular solution looks like. And you might say, hey, Sal, well, you know, that's not so good of an approximation."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "And so we are at the.3,8, which is right over here. And so for this next stretch, the next stretch is going to look like that. And as you can see, just by doing this, we have been able to approximate what the particular solution looks like. And you might say, hey, Sal, well, you know, that's not so good of an approximation. And my reply to you is, well, yeah, I mean, depends on what your goals are. But I did this by hand. I didn't even do this using a computer."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "And you might say, hey, Sal, well, you know, that's not so good of an approximation. And my reply to you is, well, yeah, I mean, depends on what your goals are. But I did this by hand. I didn't even do this using a computer. And because I wanted to do it by hand, I took fairly large delta x steps. If I wanted a better approximation, I could have lowered the delta x. And let's do that."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "I didn't even do this using a computer. And because I wanted to do it by hand, I took fairly large delta x steps. If I wanted a better approximation, I could have lowered the delta x. And let's do that. So let's take another scenario. So let's do another scenario where instead of delta x equals one, let's say delta x equals 1 1.5. So once again, x, y, and the derivative of y with respect to x."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "And let's do that. So let's take another scenario. So let's do another scenario where instead of delta x equals one, let's say delta x equals 1 1.5. So once again, x, y, and the derivative of y with respect to x. So now let's say I want to take, so we know this first point, we're given this initial condition. When x is zero, y is one. And so the slope of the tangent line is going to be one."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So once again, x, y, and the derivative of y with respect to x. So now let's say I want to take, so we know this first point, we're given this initial condition. When x is zero, y is one. And so the slope of the tangent line is going to be one. But then if we're incrementing by 1 1.5, so then when x is, I'll just write it as 0.5, 0.5, what is our new y going to be? Well, we're gonna assume that our slope from this to this is this slope right over here. So our slope is one."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "And so the slope of the tangent line is going to be one. But then if we're incrementing by 1 1.5, so then when x is, I'll just write it as 0.5, 0.5, what is our new y going to be? Well, we're gonna assume that our slope from this to this is this slope right over here. So our slope is one. So if we increase x by 0.5, we're gonna increase y by 0.5. And we're going to get to 1.5. So we're gonna, we get 0.5, 1.5."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So our slope is one. So if we increase x by 0.5, we're gonna increase y by 0.5. And we're going to get to 1.5. So we're gonna, we get 0.5, 1.5. We get to that point right over there. Actually, you're having trouble seeing that. This stuff right over here is this point right over here."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "So we're gonna, we get 0.5, 1.5. We get to that point right over there. Actually, you're having trouble seeing that. This stuff right over here is this point right over here. And now our new slope is going to be 1.5, which is going to look, which is going to look like, which is going to look like, actually not quite that steep. I don't want to overstate how good of an approximation it is and it's starting to get a little bit messy. But it's gonna look something like that."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "This stuff right over here is this point right over here. And now our new slope is going to be 1.5, which is going to look, which is going to look like, which is going to look like, actually not quite that steep. I don't want to overstate how good of an approximation it is and it's starting to get a little bit messy. But it's gonna look something like that. And what you would see if you kept doing this process, so if your slope is now 1.5, when you increment x by another 0.5, we get to one. So now if you increment by, if you increment by 0.5, and your slope is 1.5, your y is gonna increment by half of that, by 0.75. And so you're gonna get to 2.25."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "But it's gonna look something like that. And what you would see if you kept doing this process, so if your slope is now 1.5, when you increment x by another 0.5, we get to one. So now if you increment by, if you increment by 0.5, and your slope is 1.5, your y is gonna increment by half of that, by 0.75. And so you're gonna get to 2.25. So now you get to one, 2.25, which is this point right over here. And once again, this is a better approximation. Remember in the original one, y of one, you know, should be equal to e. y of one in the actual solution should be equal to e, 2.7, on and on and on and on and on."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "And so you're gonna get to 2.25. So now you get to one, 2.25, which is this point right over here. And once again, this is a better approximation. Remember in the original one, y of one, you know, should be equal to e. y of one in the actual solution should be equal to e, 2.7, on and on and on and on and on. Now in this one, y of, y of, y of one got us to two. In this one, y of one got us to 2.25. Once again, closer to the actual reality, closer to e. Instead of stepping by 0.5, if we stepped by 0.1, we would get even closer."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "Remember in the original one, y of one, you know, should be equal to e. y of one in the actual solution should be equal to e, 2.7, on and on and on and on and on. Now in this one, y of, y of, y of one got us to two. In this one, y of one got us to 2.25. Once again, closer to the actual reality, closer to e. Instead of stepping by 0.5, if we stepped by 0.1, we would get even closer. If we stepped by, if we stepped by 0.0001, we would get even closer and closer and closer. So there's a bunch of interesting things here. This is actually how most differential equations or techniques that are derived from this or that are based on numerical methods similar to this are how most differential equations get solved."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "Once again, closer to the actual reality, closer to e. Instead of stepping by 0.5, if we stepped by 0.1, we would get even closer. If we stepped by, if we stepped by 0.0001, we would get even closer and closer and closer. So there's a bunch of interesting things here. This is actually how most differential equations or techniques that are derived from this or that are based on numerical methods similar to this are how most differential equations get solved. And even if it's not the exact same solution or the same method, the idea that most differential equations are actually solved or, I guess you could say simulated with a numerical method because most of them actually cannot be solved in analytical form. Now you might be saying, hey, well what method is this one right over here called? Well this right over here is called Euler's, Euler's method after the famous Leonard Euler."}, {"video_title": "Euler's method Differential equations AP Calculus BC Khan Academy.mp3", "Sentence": "This is actually how most differential equations or techniques that are derived from this or that are based on numerical methods similar to this are how most differential equations get solved. And even if it's not the exact same solution or the same method, the idea that most differential equations are actually solved or, I guess you could say simulated with a numerical method because most of them actually cannot be solved in analytical form. Now you might be saying, hey, well what method is this one right over here called? Well this right over here is called Euler's, Euler's method after the famous Leonard Euler. Euler's, Euler's method. And not only is actually this one a good way of approximating what the solution to this or any differential equation is, but actually for this differential equation in particular, you can actually even use this to find, to find E with more and more and more precision. Anyway, hopefully you found that exciting."}, {"video_title": "Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3", "Sentence": "If we were to take the derivative with respect to time, so if we were to take the derivative with respect to time of this function s, what are we going to get? Well, we're going to get ds dt, or the rate at which position changes with respect to time, and what's another word for that? The rate at which position changes with respect to time? Well, that's just velocity. So that we could write as velocity as a function of time. Now, what if we were to take the derivative of that with respect to time? So we could either view this as the second derivative, we're taking the derivative not once, but twice of our position function, or you could say that we're taking the derivative with respect to time of our velocity function."}, {"video_title": "Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3", "Sentence": "Well, that's just velocity. So that we could write as velocity as a function of time. Now, what if we were to take the derivative of that with respect to time? So we could either view this as the second derivative, we're taking the derivative not once, but twice of our position function, or you could say that we're taking the derivative with respect to time of our velocity function. So this is going to be, we can write this as dv dt, the rate at which velocity is changing with respect to time, and what's another word for that? Well, that's also called acceleration. This is going to be our acceleration as a function of time."}, {"video_title": "Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3", "Sentence": "So we could either view this as the second derivative, we're taking the derivative not once, but twice of our position function, or you could say that we're taking the derivative with respect to time of our velocity function. So this is going to be, we can write this as dv dt, the rate at which velocity is changing with respect to time, and what's another word for that? Well, that's also called acceleration. This is going to be our acceleration as a function of time. So you start with the position as a function of time, take its derivative with respect to time, you get velocity. Take that derivative with respect to time, you get acceleration. Well, you could go the other way around."}, {"video_title": "Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3", "Sentence": "This is going to be our acceleration as a function of time. So you start with the position as a function of time, take its derivative with respect to time, you get velocity. Take that derivative with respect to time, you get acceleration. Well, you could go the other way around. If you started with acceleration, and you were to take the antiderivative of it, an antiderivative of it is going to be, actually, let me just write it this way. So an antiderivative, I'll just use the integral symbol to show that I'm taking the antiderivative, is going to be the integral of the antiderivative of a of t, and this is going to give you some expression with a plus c, and we could say, well, that's a general form of our velocity function. This is going to be equal to our velocity function, and to find the particular velocity function, we would have to know what the velocity is at a particular time that we could solve for our c. But then if we're able to do that, and we were to take the antiderivative again, then now we're taking the antiderivative of our velocity function, which would give us some expression as a function of t, and then some other constant, and if we could solve for that constant, then we know what the position is going to be."}, {"video_title": "Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3", "Sentence": "Well, you could go the other way around. If you started with acceleration, and you were to take the antiderivative of it, an antiderivative of it is going to be, actually, let me just write it this way. So an antiderivative, I'll just use the integral symbol to show that I'm taking the antiderivative, is going to be the integral of the antiderivative of a of t, and this is going to give you some expression with a plus c, and we could say, well, that's a general form of our velocity function. This is going to be equal to our velocity function, and to find the particular velocity function, we would have to know what the velocity is at a particular time that we could solve for our c. But then if we're able to do that, and we were to take the antiderivative again, then now we're taking the antiderivative of our velocity function, which would give us some expression as a function of t, and then some other constant, and if we could solve for that constant, then we know what the position is going to be. The position is a function of time. Just like this, we would have some plus c here, but if we know our position at a given time, we could solve for that c. Now that we've reviewed it a little bit, but we've rewritten it in, I guess you could say, thinking of it not just from the differential point of view or from the derivative point of view, but thinking of it from the antiderivative point of view, let's see if we can solve an interesting problem. Let's say that we know that the acceleration of a particle as a function of time is equal to 1."}, {"video_title": "Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3", "Sentence": "This is going to be equal to our velocity function, and to find the particular velocity function, we would have to know what the velocity is at a particular time that we could solve for our c. But then if we're able to do that, and we were to take the antiderivative again, then now we're taking the antiderivative of our velocity function, which would give us some expression as a function of t, and then some other constant, and if we could solve for that constant, then we know what the position is going to be. The position is a function of time. Just like this, we would have some plus c here, but if we know our position at a given time, we could solve for that c. Now that we've reviewed it a little bit, but we've rewritten it in, I guess you could say, thinking of it not just from the differential point of view or from the derivative point of view, but thinking of it from the antiderivative point of view, let's see if we can solve an interesting problem. Let's say that we know that the acceleration of a particle as a function of time is equal to 1. It's always accelerating at 1 unit per, and I'm not giving you time, but let's just say that we're thinking in terms of meters and seconds. This is 1 meter per second squared right over here. That's our acceleration as a function of time."}, {"video_title": "Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say that we know that the acceleration of a particle as a function of time is equal to 1. It's always accelerating at 1 unit per, and I'm not giving you time, but let's just say that we're thinking in terms of meters and seconds. This is 1 meter per second squared right over here. That's our acceleration as a function of time. Let's say we don't know the velocity expressions, but we know the velocity at a particular time, and we don't know the position expressions, but we know the position at a particular time. Let's say we know that the velocity at time 3, let's say 3 seconds, is negative 3 meters per second. Actually, I'm going to write the units here and make it a little bit."}, {"video_title": "Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3", "Sentence": "That's our acceleration as a function of time. Let's say we don't know the velocity expressions, but we know the velocity at a particular time, and we don't know the position expressions, but we know the position at a particular time. Let's say we know that the velocity at time 3, let's say 3 seconds, is negative 3 meters per second. Actually, I'm going to write the units here and make it a little bit. This is meters per second squared. That's going to be our unit for acceleration. This is our unit for velocity."}, {"video_title": "Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3", "Sentence": "Actually, I'm going to write the units here and make it a little bit. This is meters per second squared. That's going to be our unit for acceleration. This is our unit for velocity. Let's say that we know that the position at time 2, at 2 seconds, is equal to negative 10 meters. If we're thinking in one dimension, if we're just moving along the number line, it's 10 to the left of the origin. Given this information right over here and everything that I wrote up here, can we figure out the actual expressions for velocity as a function of time?"}, {"video_title": "Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3", "Sentence": "This is our unit for velocity. Let's say that we know that the position at time 2, at 2 seconds, is equal to negative 10 meters. If we're thinking in one dimension, if we're just moving along the number line, it's 10 to the left of the origin. Given this information right over here and everything that I wrote up here, can we figure out the actual expressions for velocity as a function of time? Not just velocity at time 3, but velocity generally as a function of time and position as a function of time. I encourage you to pause this video right now and try to figure it out on your own. Let's just work through this."}, {"video_title": "Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3", "Sentence": "Given this information right over here and everything that I wrote up here, can we figure out the actual expressions for velocity as a function of time? Not just velocity at time 3, but velocity generally as a function of time and position as a function of time. I encourage you to pause this video right now and try to figure it out on your own. Let's just work through this. What is, we know that velocity as a function of time is going to be the antiderivative of our acceleration as a function of time. Our acceleration is just 1. This is going to be, the antiderivative of this right over here is going to be T, and then we can't forget our constant, plus C. Now we can solve for C because we know V of 3 is negative 3."}, {"video_title": "Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3", "Sentence": "Let's just work through this. What is, we know that velocity as a function of time is going to be the antiderivative of our acceleration as a function of time. Our acceleration is just 1. This is going to be, the antiderivative of this right over here is going to be T, and then we can't forget our constant, plus C. Now we can solve for C because we know V of 3 is negative 3. Let's just write that down. V of 3 is going to be equal to 3, 3 plus C. Every place where I saw the T, or every place where I have the T, I replace it with this 3 right over here. Actually, let me make it a little bit clearer."}, {"video_title": "Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3", "Sentence": "This is going to be, the antiderivative of this right over here is going to be T, and then we can't forget our constant, plus C. Now we can solve for C because we know V of 3 is negative 3. Let's just write that down. V of 3 is going to be equal to 3, 3 plus C. Every place where I saw the T, or every place where I have the T, I replace it with this 3 right over here. Actually, let me make it a little bit clearer. V of 3 is equal to 3 plus C, and they tell us that that's equal to negative 3. That is equal to negative 3. What's C going to be?"}, {"video_title": "Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3", "Sentence": "Actually, let me make it a little bit clearer. V of 3 is equal to 3 plus C, and they tell us that that's equal to negative 3. That is equal to negative 3. What's C going to be? If we just look at this part of the equation, just this equation right over here, if you subtract 3 from both sides, you get C is equal to negative 6. Now we know the exact expression that defines velocity as a function of time. V of T is equal to T, T plus negative 6, or T minus 6."}, {"video_title": "Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3", "Sentence": "What's C going to be? If we just look at this part of the equation, just this equation right over here, if you subtract 3 from both sides, you get C is equal to negative 6. Now we know the exact expression that defines velocity as a function of time. V of T is equal to T, T plus negative 6, or T minus 6. We can verify that. The derivative of this with respect to time is just 1, and when time is equal to 3, 3 minus 6 is indeed negative 3. We've been able to figure out velocity as a function of time."}, {"video_title": "Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3", "Sentence": "V of T is equal to T, T plus negative 6, or T minus 6. We can verify that. The derivative of this with respect to time is just 1, and when time is equal to 3, 3 minus 6 is indeed negative 3. We've been able to figure out velocity as a function of time. Now let's do a similar thing to figure out position as a function of time. We know that position is going to be an antiderivative of the velocity function. Let's write that down."}, {"video_title": "Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3", "Sentence": "We've been able to figure out velocity as a function of time. Now let's do a similar thing to figure out position as a function of time. We know that position is going to be an antiderivative of the velocity function. Let's write that down. Position as a function of time is going to be equal to the antiderivative of V of T, dT, which is equal to the antiderivative of T minus 6, dT, which is equal to, well, the antiderivative of T is T squared over 2. T squared over 2, we've seen that before. The antiderivative of negative 6 is negative 6T."}, {"video_title": "Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3", "Sentence": "Let's write that down. Position as a function of time is going to be equal to the antiderivative of V of T, dT, which is equal to the antiderivative of T minus 6, dT, which is equal to, well, the antiderivative of T is T squared over 2. T squared over 2, we've seen that before. The antiderivative of negative 6 is negative 6T. Of course, we can't forget our constant. So plus C. This is what S of T is equal to. S of T is equal to all of this business right over here."}, {"video_title": "Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3", "Sentence": "The antiderivative of negative 6 is negative 6T. Of course, we can't forget our constant. So plus C. This is what S of T is equal to. S of T is equal to all of this business right over here. Now we can try to solve for our constant. We do that using this information right over here. At 2 seconds, our position is negative 10 meters."}, {"video_title": "Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3", "Sentence": "S of T is equal to all of this business right over here. Now we can try to solve for our constant. We do that using this information right over here. At 2 seconds, our position is negative 10 meters. S of 2, or I could just write it this way. Let me write it this way. S of 2 at 2 seconds is going to be equal to 2 squared over 2."}, {"video_title": "Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3", "Sentence": "At 2 seconds, our position is negative 10 meters. S of 2, or I could just write it this way. Let me write it this way. S of 2 at 2 seconds is going to be equal to 2 squared over 2. That is, let's see, that's 4 over 2. That's going to be 2 minus 6 times 2. So minus 12 plus C is equal to negative 10."}, {"video_title": "Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3", "Sentence": "S of 2 at 2 seconds is going to be equal to 2 squared over 2. That is, let's see, that's 4 over 2. That's going to be 2 minus 6 times 2. So minus 12 plus C is equal to negative 10. Is equal to negative 10. Let's see, we get 2 minus 12 is negative 10, plus C is equal to negative 10. You add 10 to both sides, you get C, in this case, is equal to 0."}, {"video_title": "Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3", "Sentence": "So minus 12 plus C is equal to negative 10. Is equal to negative 10. Let's see, we get 2 minus 12 is negative 10, plus C is equal to negative 10. You add 10 to both sides, you get C, in this case, is equal to 0. So we've figured out what our position function is as well. This C right over here is just going to be 0. So our position, as a function of time, is equal to T squared over 2 minus 6T."}, {"video_title": "Worked example motion problems (with definite integrals) AP Calculus AB Khan Academy.mp3", "Sentence": "You add 10 to both sides, you get C, in this case, is equal to 0. So we've figured out what our position function is as well. This C right over here is just going to be 0. So our position, as a function of time, is equal to T squared over 2 minus 6T. And you can verify. When T is equal to 2, 2 squared over 2 is 2, minus 12 is negative 10. You take the derivative here, you get T minus 6, and you can see, and we already verified, that V of 3 is negative 3."}, {"video_title": "Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "What we're going to do in this video is try to find the derivative with respect to x of x squared sine of x, all of that to the third power. And what's going to be interesting is that there's multiple ways to tackle it, and I encourage you to pause the video and see if you can work through it on your own. So there's actually multiple techniques. One path is to do the chain rule first. So I'll just say CR for chain rule first. And so I have, I'm taking the derivative with respect to x of something to the third power. So if I take the derivative, it would be the derivative with respect to that something, so that would be three times that something squared times the derivative with respect to x of that something, where the something in this case is x squared sine of x, x squared sine of x."}, {"video_title": "Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "One path is to do the chain rule first. So I'll just say CR for chain rule first. And so I have, I'm taking the derivative with respect to x of something to the third power. So if I take the derivative, it would be the derivative with respect to that something, so that would be three times that something squared times the derivative with respect to x of that something, where the something in this case is x squared sine of x, x squared sine of x. This is just an application of the chain rule. Now, the second part, what would that be? The second part here, this is another color, in orange."}, {"video_title": "Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So if I take the derivative, it would be the derivative with respect to that something, so that would be three times that something squared times the derivative with respect to x of that something, where the something in this case is x squared sine of x, x squared sine of x. This is just an application of the chain rule. Now, the second part, what would that be? The second part here, this is another color, in orange. Well, here I would apply the product rule. I have the product of two expressions, so I would take the derivative of, let me write this down. So this is gonna be the product rule, product rule."}, {"video_title": "Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "The second part here, this is another color, in orange. Well, here I would apply the product rule. I have the product of two expressions, so I would take the derivative of, let me write this down. So this is gonna be the product rule, product rule. I would take the derivative of the first expression, so x, derivative of x squared is two x, let me write a little bit to the right. This is gonna be two x times the second expression, sine of x, plus the first expression, x squared, times the derivative of the second one, cosine of x. That's just the product rule as applied to this part right over here."}, {"video_title": "Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So this is gonna be the product rule, product rule. I would take the derivative of the first expression, so x, derivative of x squared is two x, let me write a little bit to the right. This is gonna be two x times the second expression, sine of x, plus the first expression, x squared, times the derivative of the second one, cosine of x. That's just the product rule as applied to this part right over here. And all of that, of course, is being multiplied by this up front, which actually, let me just rewrite that. So all of this I could rewrite as, let's see. This would be three times, if I have the product of things raised to the second power, I could take each of them to the second power and then take their product."}, {"video_title": "Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "That's just the product rule as applied to this part right over here. And all of that, of course, is being multiplied by this up front, which actually, let me just rewrite that. So all of this I could rewrite as, let's see. This would be three times, if I have the product of things raised to the second power, I could take each of them to the second power and then take their product. So x squared squared is x to the fourth, and then sine of x squared is sine squared of x, and then all of that is being multiplied by that. And if we want, we can algebraically simplify, we can distribute everything out, in which case, what would we get? Well, let's see."}, {"video_title": "Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "This would be three times, if I have the product of things raised to the second power, I could take each of them to the second power and then take their product. So x squared squared is x to the fourth, and then sine of x squared is sine squared of x, and then all of that is being multiplied by that. And if we want, we can algebraically simplify, we can distribute everything out, in which case, what would we get? Well, let's see. Three times two is six. X to the fourth times x is x to the fifth. Sine squared of x times sine of x is sine of x to the third power."}, {"video_title": "Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, let's see. Three times two is six. X to the fourth times x is x to the fifth. Sine squared of x times sine of x is sine of x to the third power. And then, let's see, three, so plus three, x to the fourth times x squared is x to the sixth power, and then I have sine squared of x, sine squared of x, cosine of x. So there you have it. That's one strategy, chain rule first and then product rule."}, {"video_title": "Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Sine squared of x times sine of x is sine of x to the third power. And then, let's see, three, so plus three, x to the fourth times x squared is x to the sixth power, and then I have sine squared of x, sine squared of x, cosine of x. So there you have it. That's one strategy, chain rule first and then product rule. What would be another strategy? Pause the video and try to think of it. Well, we could just algebraically use our exponent properties first, in which case, this is going to be equal to the derivative with respect to x of, if I'm taking x squared times sine of x to the third power, instead I could say x to the third to the third power, which is going to be x to the sixth, and then sine of x to the third power, sine of x to the third power."}, {"video_title": "Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "That's one strategy, chain rule first and then product rule. What would be another strategy? Pause the video and try to think of it. Well, we could just algebraically use our exponent properties first, in which case, this is going to be equal to the derivative with respect to x of, if I'm taking x squared times sine of x to the third power, instead I could say x to the third to the third power, which is going to be x to the sixth, and then sine of x to the third power, sine of x to the third power. I'm using the same exponent property that we used right over here to simplify this. If I have, if I'm taking the product things to some exponent, well, that's the same thing of each of them raised to the exponent and then the product of the two. Now, how would we tackle this?"}, {"video_title": "Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Well, we could just algebraically use our exponent properties first, in which case, this is going to be equal to the derivative with respect to x of, if I'm taking x squared times sine of x to the third power, instead I could say x to the third to the third power, which is going to be x to the sixth, and then sine of x to the third power, sine of x to the third power. I'm using the same exponent property that we used right over here to simplify this. If I have, if I'm taking the product things to some exponent, well, that's the same thing of each of them raised to the exponent and then the product of the two. Now, how would we tackle this? Well, I, here, I would do the product rule first. So let's do that. So let's do the product rule."}, {"video_title": "Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Now, how would we tackle this? Well, I, here, I would do the product rule first. So let's do that. So let's do the product rule. So we're gonna take the derivative of the first expression. So derivative of x to the sixth is six x to the fifth times the second expression, sine to the third of x, or sine of x to the third power, plus the first x to the sixth times the derivative of the second, and I'm just gonna write that, d dx of sine of x to the third power. Now, to evaluate this right over here, it does definitely make sense to use the chain rule, use the chain rule."}, {"video_title": "Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So let's do the product rule. So we're gonna take the derivative of the first expression. So derivative of x to the sixth is six x to the fifth times the second expression, sine to the third of x, or sine of x to the third power, plus the first x to the sixth times the derivative of the second, and I'm just gonna write that, d dx of sine of x to the third power. Now, to evaluate this right over here, it does definitely make sense to use the chain rule, use the chain rule. And so what is this going to be? Well, I have the derivative of something to the third power, so that's going to be three times that something squared times the derivative of that something. So in this case, the something is sine of x, and the derivative of sine of x is cosine of x."}, {"video_title": "Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "Now, to evaluate this right over here, it does definitely make sense to use the chain rule, use the chain rule. And so what is this going to be? Well, I have the derivative of something to the third power, so that's going to be three times that something squared times the derivative of that something. So in this case, the something is sine of x, and the derivative of sine of x is cosine of x. And then I have all of this business over here. I have six x to the fifth, sine to the third, or sine of x to the third power, plus x to the sixth. And if I were to just simplify this a little bit, in fact, you see it very clearly, these two things are equivalent."}, {"video_title": "Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So in this case, the something is sine of x, and the derivative of sine of x is cosine of x. And then I have all of this business over here. I have six x to the fifth, sine to the third, or sine of x to the third power, plus x to the sixth. And if I were to just simplify this a little bit, in fact, you see it very clearly, these two things are equivalent. This term is exactly equivalent to this term, the way it's written. And then this is exactly, if you multiply three times x to the sixth, sine of x squared cosine of x. So the nice thing about math, if we're doing things that make logical sense, we should get to the same end point."}, {"video_title": "Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "And if I were to just simplify this a little bit, in fact, you see it very clearly, these two things are equivalent. This term is exactly equivalent to this term, the way it's written. And then this is exactly, if you multiply three times x to the sixth, sine of x squared cosine of x. So the nice thing about math, if we're doing things that make logical sense, we should get to the same end point. But the point here is that there's multiple strategies. You could use the chain rule first and then the product rule or you could use the product rule first and then the chain rule. In this case, you could debate which one is faster."}, {"video_title": "Applying the chain rule and product rule Advanced derivatives AP Calculus AB Khan Academy.mp3", "Sentence": "So the nice thing about math, if we're doing things that make logical sense, we should get to the same end point. But the point here is that there's multiple strategies. You could use the chain rule first and then the product rule or you could use the product rule first and then the chain rule. In this case, you could debate which one is faster. It looks like the one on the right might be a little bit faster. But sometimes, but these two are pretty close, but sometimes it'll be more clear than not which one is preferable. You really wanna minimize the amount of hairiness, the number of steps, the chances for careless mistakes you might have."}, {"video_title": "Worked example direct comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And as you can imagine, based on the context of where this video shows up on Khan Academy, that maybe we will do it using the comparison test. And at any point, if you feel like you can kind of take this to the finish line, feel free to pause the video and do so. So in order to kind of figure out or get a sense for this series right over here, it never hurts to kind of expand it out a little bit. So let's do that. So this would be equal to, when n equals one, this is going to be one over two to the one plus one. So it's gonna be one over two plus one, it's gonna be 1 3rd, plus, that's n equals one, when n equals two, it's gonna be one over two squared, which is four plus two, plus one over six, plus, let's see, when we go to three, n equals one, n equals two, n equals three, it's gonna be one over two to the third, which is eight plus three is 11. So one over 11, maybe I'll do one more term."}, {"video_title": "Worked example direct comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So let's do that. So this would be equal to, when n equals one, this is going to be one over two to the one plus one. So it's gonna be one over two plus one, it's gonna be 1 3rd, plus, that's n equals one, when n equals two, it's gonna be one over two squared, which is four plus two, plus one over six, plus, let's see, when we go to three, n equals one, n equals two, n equals three, it's gonna be one over two to the third, which is eight plus three is 11. So one over 11, maybe I'll do one more term. Two to the fourth power is going to be 16, plus four is 20, plus one over 20, and obviously we just keep going on and on and on. So in order to use, so it looks, it feels like this thing could converge. All of our terms are positive, but they are going, they're getting smaller and smaller quite fast, and if we really look at the, if we look at the behavior of this, the terms as n gets larger and larger, we see that the two to the n in the denominator will grow much, much faster than the n will."}, {"video_title": "Worked example direct comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So one over 11, maybe I'll do one more term. Two to the fourth power is going to be 16, plus four is 20, plus one over 20, and obviously we just keep going on and on and on. So in order to use, so it looks, it feels like this thing could converge. All of our terms are positive, but they are going, they're getting smaller and smaller quite fast, and if we really look at the, if we look at the behavior of this, the terms as n gets larger and larger, we see that the two to the n in the denominator will grow much, much faster than the n will. So this will start to behave, this kind of behaves like one over two to the n, which is a clue of something that we might be able to use for the comparison test. So let me just write that down. So we have one over, so the infinite series from n equals one to infinity of one over two to the n. And so when n equals one, this is going to be equal to, this is going to be equal to 1 1\u20442."}, {"video_title": "Worked example direct comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "All of our terms are positive, but they are going, they're getting smaller and smaller quite fast, and if we really look at the, if we look at the behavior of this, the terms as n gets larger and larger, we see that the two to the n in the denominator will grow much, much faster than the n will. So this will start to behave, this kind of behaves like one over two to the n, which is a clue of something that we might be able to use for the comparison test. So let me just write that down. So we have one over, so the infinite series from n equals one to infinity of one over two to the n. And so when n equals one, this is going to be equal to, this is going to be equal to 1 1\u20442. When n is equal to two, this is going to be equal to 1\u20444. When n is equal to three, this is equal to 1\u2078. When n is equal to four, this is equal to 1 16th."}, {"video_title": "Worked example direct comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So we have one over, so the infinite series from n equals one to infinity of one over two to the n. And so when n equals one, this is going to be equal to, this is going to be equal to 1 1\u20442. When n is equal to two, this is going to be equal to 1\u20444. When n is equal to three, this is equal to 1\u2078. When n is equal to four, this is equal to 1 16th. And you go on and on and on and on. And what's interesting about this is we recognize it. This is a geometric series, so let me be clear."}, {"video_title": "Worked example direct comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "When n is equal to four, this is equal to 1 16th. And you go on and on and on and on. And what's interesting about this is we recognize it. This is a geometric series, so let me be clear. This thing right over here, that is the same thing as the sum from n equals one to infinity of 1\u20442 to the n power. Just writing it in a different way. And since the absolute value of 1\u20442, which is just 1\u20442, so because the absolute value of 1\u20442 is less than one, we know that this geometric series converges."}, {"video_title": "Worked example direct comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "This is a geometric series, so let me be clear. This thing right over here, that is the same thing as the sum from n equals one to infinity of 1\u20442 to the n power. Just writing it in a different way. And since the absolute value of 1\u20442, which is just 1\u20442, so because the absolute value of 1\u20442 is less than one, we know that this geometric series converges. We know that it converges. And actually, we even have formulas for finding the exact sum of, or to find this, figure out what it converges to. And so we know this thing converges."}, {"video_title": "Worked example direct comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And since the absolute value of 1\u20442, which is just 1\u20442, so because the absolute value of 1\u20442 is less than one, we know that this geometric series converges. We know that it converges. And actually, we even have formulas for finding the exact sum of, or to find this, figure out what it converges to. And so we know this thing converges. And we see that actually these two series combined meet all of the constraints we need for the comparison test. So let's go back to what we wrote about the comparison test. So the comparison test, we have two series."}, {"video_title": "Worked example direct comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And so we know this thing converges. And we see that actually these two series combined meet all of the constraints we need for the comparison test. So let's go back to what we wrote about the comparison test. So the comparison test, we have two series. All of their terms are greater than or equal to zero. All of these terms are greater than or equal to zero. And then for the corresponding terms in one series, all of them are going to be less than or equal to the corresponding terms in the next one."}, {"video_title": "Worked example direct comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "So the comparison test, we have two series. All of their terms are greater than or equal to zero. All of these terms are greater than or equal to zero. And then for the corresponding terms in one series, all of them are going to be less than or equal to the corresponding terms in the next one. And so if we look over here, we could consider this one, the magenta series. This is kind of our infinite series of dealing with a sub n. And that this right over here is, well, I already did it in blue. This is kind of the blue series."}, {"video_title": "Worked example direct comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "And then for the corresponding terms in one series, all of them are going to be less than or equal to the corresponding terms in the next one. And so if we look over here, we could consider this one, the magenta series. This is kind of our infinite series of dealing with a sub n. And that this right over here is, well, I already did it in blue. This is kind of the blue series. And notice all their terms are non-negative. And the corresponding terms, 1\u20442 is greater than 1\u20443. 1\u20444 is greater than 1\u2076."}, {"video_title": "Worked example direct comparison test Series AP Calculus BC Khan Academy.mp3", "Sentence": "This is kind of the blue series. And notice all their terms are non-negative. And the corresponding terms, 1\u20442 is greater than 1\u20443. 1\u20444 is greater than 1\u2076. 1\u2078 is greater than 1\u2079. 1 over 2 to the n is always going to be greater than 1 over 2 to the n plus n for the n's that we care about here. And since we know that this converges, since we know that the larger one converges, it's a geometric series where the common ratio, the absolute value of the common ratio is less than one."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "What we want to do in this video is figure out what the limit as x approaches zero of one minus cosine of x over x is equal to. And we're going to assume we know one thing ahead of time. We're going to assume we know that the limit as x approaches zero of sine of x over x, that this is equal to one. And I'm not gonna reprove this in this video, but we have a whole other video dedicated to proving this famous limit. And we do it using the squeeze or the sandwich theorem. So let's see if we can work this out. So the first thing we're going to do is algebraically manipulate this expression a little bit."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And I'm not gonna reprove this in this video, but we have a whole other video dedicated to proving this famous limit. And we do it using the squeeze or the sandwich theorem. So let's see if we can work this out. So the first thing we're going to do is algebraically manipulate this expression a little bit. What I'm going to do is I'm gonna multiply both the numerator and the denominator by one plus cosine of x. So times the denominator, I have to do the same thing, one plus cosine of x. I'm not changing the value of the expression. This is just multiplying it by one."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So the first thing we're going to do is algebraically manipulate this expression a little bit. What I'm going to do is I'm gonna multiply both the numerator and the denominator by one plus cosine of x. So times the denominator, I have to do the same thing, one plus cosine of x. I'm not changing the value of the expression. This is just multiplying it by one. But what does that do for us? Well, I can rewrite the whole thing as the limit as x approaches zero. So one minus cosine of x times one plus cosine of x."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "This is just multiplying it by one. But what does that do for us? Well, I can rewrite the whole thing as the limit as x approaches zero. So one minus cosine of x times one plus cosine of x. Well, that is just going to be, let me do this in another color. That is going to be one squared, which is just one, minus cosine squared of x. Cosine squared of x, difference of squares. And then in the denominator, I am going to have these, which is just x times one plus cosine of x."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So one minus cosine of x times one plus cosine of x. Well, that is just going to be, let me do this in another color. That is going to be one squared, which is just one, minus cosine squared of x. Cosine squared of x, difference of squares. And then in the denominator, I am going to have these, which is just x times one plus cosine of x. Now what is one minus cosine squared of x? Well, this comes straight out of the Pythagorean identity, trig identity. This is the same thing as sine squared of x."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And then in the denominator, I am going to have these, which is just x times one plus cosine of x. Now what is one minus cosine squared of x? Well, this comes straight out of the Pythagorean identity, trig identity. This is the same thing as sine squared of x. So sine squared of x. And so I can rewrite all of this as being equal to the limit as x approaches zero. And let me rewrite this as, instead of sine squared of x, that's the same thing as sine of x times sine of x."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "This is the same thing as sine squared of x. So sine squared of x. And so I can rewrite all of this as being equal to the limit as x approaches zero. And let me rewrite this as, instead of sine squared of x, that's the same thing as sine of x times sine of x. Let me write it that way. Sine x times sine x. So I'll take the first sine of x."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "And let me rewrite this as, instead of sine squared of x, that's the same thing as sine of x times sine of x. Let me write it that way. Sine x times sine x. So I'll take the first sine of x. So I'll do, I'll take this one right over here and put it over this x. So sine of x over x times the second sine of x, let's say this one, over one plus cosine of x. Times sine of x over one plus cosine of x."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So I'll take the first sine of x. So I'll do, I'll take this one right over here and put it over this x. So sine of x over x times the second sine of x, let's say this one, over one plus cosine of x. Times sine of x over one plus cosine of x. All I've done is I've leveraged a trigonometric identity and I've done a little bit of algebraic manipulation. Well here, the limit of the product of these two expressions is going to be the same thing as the product of the limits. So I can rewrite this as being equal to the limit as x approaches zero of sine of x over x times the limit as x approaches zero of sine of x over one plus cosine of x."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Times sine of x over one plus cosine of x. All I've done is I've leveraged a trigonometric identity and I've done a little bit of algebraic manipulation. Well here, the limit of the product of these two expressions is going to be the same thing as the product of the limits. So I can rewrite this as being equal to the limit as x approaches zero of sine of x over x times the limit as x approaches zero of sine of x over one plus cosine of x. Now, we said going into this video that we're going to assume that we know what this is. We've proven it in other videos. What is the limit as x approaches zero of sine of x over x?"}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "So I can rewrite this as being equal to the limit as x approaches zero of sine of x over x times the limit as x approaches zero of sine of x over one plus cosine of x. Now, we said going into this video that we're going to assume that we know what this is. We've proven it in other videos. What is the limit as x approaches zero of sine of x over x? Well, that is equal to one. So this whole limit is just going to be dependent on whatever this is equal to. Well, this is pretty straightforward here."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "What is the limit as x approaches zero of sine of x over x? Well, that is equal to one. So this whole limit is just going to be dependent on whatever this is equal to. Well, this is pretty straightforward here. As x approaches zero, the numerator's approaching zero, sine of zero is zero. The denominator is approaching, cosine of zero is one, so the denominator is approaching two. So this is approaching zero over two or just zero."}, {"video_title": "Limit of (1-cos(x)) x as x approaches 0 Derivative rules AP Calculus AB Khan Academy.mp3", "Sentence": "Well, this is pretty straightforward here. As x approaches zero, the numerator's approaching zero, sine of zero is zero. The denominator is approaching, cosine of zero is one, so the denominator is approaching two. So this is approaching zero over two or just zero. So that's approaching zero. One times zero, well, this is just going to be equal to zero and we're done. Using that fact and a little bit of trig identities and a little bit of algebraic manipulation, we were able to show that our original limit, the limit as x approaches zero of one minus cosine of x over x is equal to zero."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "In the last video, we took the Maclaurin expansion of e to the x, and we saw that it looked like it was some type of a combination of the polynomial approximations of cosine of x and of sine of x. But it's not quite, because there were a couple of negatives in there, if we were to really add these two together, that we did not have when we took the representation of e to the x. But to reconcile these, I'll do a little bit of a, I don't know if you can even call it a trick. Let's see, if we take this polynomial expansion of e to the x, this approximation, what happens, and if we say e to the x is equal to this, especially as this becomes an infinite number of terms, then it becomes less of an approximation and more of an equality. What happens if I take e to the ix? And before, that might have been kind of a weird thing to do. Let me write it down, e to the ix."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let's see, if we take this polynomial expansion of e to the x, this approximation, what happens, and if we say e to the x is equal to this, especially as this becomes an infinite number of terms, then it becomes less of an approximation and more of an equality. What happens if I take e to the ix? And before, that might have been kind of a weird thing to do. Let me write it down, e to the ix. Because before, it's like, how do you define e to the ith power? That's a very bizarre thing to do, to take something to the xi power. How do you even comprehend some type of a function like that?"}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let me write it down, e to the ix. Because before, it's like, how do you define e to the ith power? That's a very bizarre thing to do, to take something to the xi power. How do you even comprehend some type of a function like that? But now that we can have a polynomial expansion of e to the x, we can maybe make some sense of it, because we can take i to different amounts, to different powers, and we know what that gives. i squared is negative 1, i to the third is negative i, so on and so forth. What happens if we take e to the ix?"}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "How do you even comprehend some type of a function like that? But now that we can have a polynomial expansion of e to the x, we can maybe make some sense of it, because we can take i to different amounts, to different powers, and we know what that gives. i squared is negative 1, i to the third is negative i, so on and so forth. What happens if we take e to the ix? Once again, it's just like taking the x up here and replacing it with an ix. Everywhere we see the x in its polynomial approximation, we would write an ix. Let's do that."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "What happens if we take e to the ix? Once again, it's just like taking the x up here and replacing it with an ix. Everywhere we see the x in its polynomial approximation, we would write an ix. Let's do that. e to the ix should be approximately equal to, and it will become more and more equal. This is more to give you an intuition. I'm not doing a rigorous proof here, but it's still profound."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let's do that. e to the ix should be approximately equal to, and it will become more and more equal. This is more to give you an intuition. I'm not doing a rigorous proof here, but it's still profound. Not to oversell it, but I don't think I can oversell what is about to be discovered or seen in this video. It would be equal to 1 plus, instead of an x, we'll have an ix, plus ix, plus, so what's ix squared? It's going to be, so let me write this down."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "I'm not doing a rigorous proof here, but it's still profound. Not to oversell it, but I don't think I can oversell what is about to be discovered or seen in this video. It would be equal to 1 plus, instead of an x, we'll have an ix, plus ix, plus, so what's ix squared? It's going to be, so let me write this down. What is ix squared over 2 factorial? Well, i squared is going to be negative 1, and then you have x squared over 2 factorial. It's going to be minus x squared over 2 factorial."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "It's going to be, so let me write this down. What is ix squared over 2 factorial? Well, i squared is going to be negative 1, and then you have x squared over 2 factorial. It's going to be minus x squared over 2 factorial. I think you might see where this is going to go. Then what is ix? Remember, everywhere we saw an x, we're going to replace it with an ix."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "It's going to be minus x squared over 2 factorial. I think you might see where this is going to go. Then what is ix? Remember, everywhere we saw an x, we're going to replace it with an ix. What is ix to the third power? Actually, let me write this out. Let me not skip some steps over here."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "Remember, everywhere we saw an x, we're going to replace it with an ix. What is ix to the third power? Actually, let me write this out. Let me not skip some steps over here. This is going to be ix squared over 2 factorial. Actually, let me, I want to do it just the way. Plus, plus ix squared over 2 factorial, plus ix to the third over 3 factorial, plus ix to the fourth over 4 factorial, and we can keep going, plus ix to the fifth over 5 factorial, and we could just keep going, so on and so forth."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let me not skip some steps over here. This is going to be ix squared over 2 factorial. Actually, let me, I want to do it just the way. Plus, plus ix squared over 2 factorial, plus ix to the third over 3 factorial, plus ix to the fourth over 4 factorial, and we can keep going, plus ix to the fifth over 5 factorial, and we could just keep going, so on and so forth. Let's evaluate these ix's raised to these different powers. This will be equal to 1 plus ix. ix squared, that's the same thing as i squared times x squared."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "Plus, plus ix squared over 2 factorial, plus ix to the third over 3 factorial, plus ix to the fourth over 4 factorial, and we can keep going, plus ix to the fifth over 5 factorial, and we could just keep going, so on and so forth. Let's evaluate these ix's raised to these different powers. This will be equal to 1 plus ix. ix squared, that's the same thing as i squared times x squared. i squared is negative 1. This is negative x squared over 2 factorial. Then this is going to be the same thing as i to the third times x to the third."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "ix squared, that's the same thing as i squared times x squared. i squared is negative 1. This is negative x squared over 2 factorial. Then this is going to be the same thing as i to the third times x to the third. i to the third is the same thing as i squared times i. It's going to be negative i. This is going to be minus i times x to the third over 3 factorial."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "Then this is going to be the same thing as i to the third times x to the third. i to the third is the same thing as i squared times i. It's going to be negative i. This is going to be minus i times x to the third over 3 factorial. Then plus, you're going to have, what's i to the fourth power? That's i squared squared. That's negative 1 squared."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "This is going to be minus i times x to the third over 3 factorial. Then plus, you're going to have, what's i to the fourth power? That's i squared squared. That's negative 1 squared. That's just going to be 1. i to the fourth is 1, and then you have x to the fourth. Plus x to the fourth over 4 factorial. Then you're going to have plus, i to the fifth."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "That's negative 1 squared. That's just going to be 1. i to the fourth is 1, and then you have x to the fourth. Plus x to the fourth over 4 factorial. Then you're going to have plus, i to the fifth. i to the fifth is going to be 1 times i. It's going to be i times x to the fifth over 5 factorial. Plus i times x to the fifth over 5 factorial."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "Then you're going to have plus, i to the fifth. i to the fifth is going to be 1 times i. It's going to be i times x to the fifth over 5 factorial. Plus i times x to the fifth over 5 factorial. I think you might see a pattern here. Coefficient is 1, then i, then negative 1, then negative i, then 1, then i, then negative 1. x to the sixth over 6 factorial. Then negative i, x to the seventh over 7 factorial."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "Plus i times x to the fifth over 5 factorial. I think you might see a pattern here. Coefficient is 1, then i, then negative 1, then negative i, then 1, then i, then negative 1. x to the sixth over 6 factorial. Then negative i, x to the seventh over 7 factorial. We have some terms. Some of them are imaginary. They have an i."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "Then negative i, x to the seventh over 7 factorial. We have some terms. Some of them are imaginary. They have an i. They're being multiplied by i. Some of them are real. Why don't we separate them out?"}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "They have an i. They're being multiplied by i. Some of them are real. Why don't we separate them out? Once again, e to the ix is going to be equal to this thing, especially as we add an infinite number of terms. Let's separate out the real and the non-real terms. Or the real and the imaginary terms, I should say."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "Why don't we separate them out? Once again, e to the ix is going to be equal to this thing, especially as we add an infinite number of terms. Let's separate out the real and the non-real terms. Or the real and the imaginary terms, I should say. This is real. This is real. This is real."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "Or the real and the imaginary terms, I should say. This is real. This is real. This is real. This right over here is real. Obviously, we could keep going on with that. The real terms here are 1 minus x squared over 2 factorial plus x to the fourth over 4 factorial."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "This is real. This right over here is real. Obviously, we could keep going on with that. The real terms here are 1 minus x squared over 2 factorial plus x to the fourth over 4 factorial. You might be getting excited now. Minus x to the sixth over 6 factorial. That's all I've done here."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "The real terms here are 1 minus x squared over 2 factorial plus x to the fourth over 4 factorial. You might be getting excited now. Minus x to the sixth over 6 factorial. That's all I've done here. They would keep going. Plus, so on and so forth. That's all of the real terms."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "That's all I've done here. They would keep going. Plus, so on and so forth. That's all of the real terms. What are the imaginary terms here? I'll just factor out the i over here. Actually, let me just factor out."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "That's all of the real terms. What are the imaginary terms here? I'll just factor out the i over here. Actually, let me just factor out. It's going to be plus i times, well this is ix, so this will be x. That's an imaginary term. This is an imaginary term."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "Actually, let me just factor out. It's going to be plus i times, well this is ix, so this will be x. That's an imaginary term. This is an imaginary term. We're factoring out the i. Minus x to the third over 3 factorial. The next imaginary term is right over there."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "This is an imaginary term. We're factoring out the i. Minus x to the third over 3 factorial. The next imaginary term is right over there. We factored out the i. Plus x to the fifth over 5 factorial. The next imaginary term is right there."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "The next imaginary term is right over there. We factored out the i. Plus x to the fifth over 5 factorial. The next imaginary term is right there. We factored out the i. It's minus x to the seventh over 7 factorial. Then we obviously would keep going."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "The next imaginary term is right there. We factored out the i. It's minus x to the seventh over 7 factorial. Then we obviously would keep going. Plus, minus, keep going, so on and so forth. Preferably to infinity so that we get as good of an approximation as possible. We have a situation where e to the ix is equal to all of this business here."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "Then we obviously would keep going. Plus, minus, keep going, so on and so forth. Preferably to infinity so that we get as good of an approximation as possible. We have a situation where e to the ix is equal to all of this business here. You probably remember from the last few videos, the real part, this was the polynomial, this was the Maclaurin approximation of cosine of x around 0. I should say the Taylor approximation around 0, or we could also call it the Maclaurin approximation. This and this are the same thing."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "We have a situation where e to the ix is equal to all of this business here. You probably remember from the last few videos, the real part, this was the polynomial, this was the Maclaurin approximation of cosine of x around 0. I should say the Taylor approximation around 0, or we could also call it the Maclaurin approximation. This and this are the same thing. This is cosine of x, especially when you add an infinite number of terms. Cosine of x. This over here is sine of x, the exact same thing."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "This and this are the same thing. This is cosine of x, especially when you add an infinite number of terms. Cosine of x. This over here is sine of x, the exact same thing. It looks like we're able to reconcile how you can add up cosine of x and sine of x to get something that's like e to the x. This right here is sine of x. If we take it for granted, I'm not rigorously proving it to you, that if you were to take an infinite number of terms here, that this will essentially become cosine of x."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "This over here is sine of x, the exact same thing. It looks like we're able to reconcile how you can add up cosine of x and sine of x to get something that's like e to the x. This right here is sine of x. If we take it for granted, I'm not rigorously proving it to you, that if you were to take an infinite number of terms here, that this will essentially become cosine of x. If you take an infinite number of terms here, this will become sine of x. It leads to a fascinating, fascinating formula. We could say that e to the ix is the same thing as cosine of x."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "If we take it for granted, I'm not rigorously proving it to you, that if you were to take an infinite number of terms here, that this will essentially become cosine of x. If you take an infinite number of terms here, this will become sine of x. It leads to a fascinating, fascinating formula. We could say that e to the ix is the same thing as cosine of x. It is cosine of x, and you should be getting goose pimples right around now, is equal to cosine of x plus i times sine of x. This is Euler's formula. I always pronounce them right over here."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "We could say that e to the ix is the same thing as cosine of x. It is cosine of x, and you should be getting goose pimples right around now, is equal to cosine of x plus i times sine of x. This is Euler's formula. I always pronounce them right over here. This right here is Euler's formula. If that by itself isn't exciting and crazy enough for you, because it really should be, because we've already done some pretty cool things. We're involving e, which we get from continuous compounding interest."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "I always pronounce them right over here. This right here is Euler's formula. If that by itself isn't exciting and crazy enough for you, because it really should be, because we've already done some pretty cool things. We're involving e, which we get from continuous compounding interest. We have cosine and sine of x, which are ratios of right triangles. It comes out of the unit circle. Somehow we've thrown in the square root of negative 1."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "We're involving e, which we get from continuous compounding interest. We have cosine and sine of x, which are ratios of right triangles. It comes out of the unit circle. Somehow we've thrown in the square root of negative 1. There seems to be this cool relationship here. It becomes extra cool, and we're going to assume we're operating in radians here, is if we assume Euler's formula, what happens when x is equal to pi? Just to throw in another wacky number in there, the ratio between the circumference and the diameter of a circle."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "Somehow we've thrown in the square root of negative 1. There seems to be this cool relationship here. It becomes extra cool, and we're going to assume we're operating in radians here, is if we assume Euler's formula, what happens when x is equal to pi? Just to throw in another wacky number in there, the ratio between the circumference and the diameter of a circle. What happens when we throw in pi? We get e to the i pi is equal to cosine of pi. Cosine of pi is what?"}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "Just to throw in another wacky number in there, the ratio between the circumference and the diameter of a circle. What happens when we throw in pi? We get e to the i pi is equal to cosine of pi. Cosine of pi is what? Cosine of pi is half way around the unit circle. Cosine of pi is negative 1. Then sine of pi is 0."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "Cosine of pi is what? Cosine of pi is half way around the unit circle. Cosine of pi is negative 1. Then sine of pi is 0. This term goes away. If you evaluate it at pi, you get something amazing. This is called Euler's identity."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "Then sine of pi is 0. This term goes away. If you evaluate it at pi, you get something amazing. This is called Euler's identity. I always have trouble pronouncing Euler. Euler's identity, which we could write like this, or we could add 1 to both sides, and we could write it like this. I'll write it in different colors for emphasis."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "This is called Euler's identity. I always have trouble pronouncing Euler. Euler's identity, which we could write like this, or we could add 1 to both sides, and we could write it like this. I'll write it in different colors for emphasis. E to the i times pi plus 1 is equal to, I'll do that in a neutral color, is equal to, I'm just adding 1 to both sides of this thing right over here, is equal to 0. This is thought provoking. Here we have, just so you see, this tells you that there's some connectedness to the universe that we don't fully understand, or at least I don't fully understand."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "I'll write it in different colors for emphasis. E to the i times pi plus 1 is equal to, I'll do that in a neutral color, is equal to, I'm just adding 1 to both sides of this thing right over here, is equal to 0. This is thought provoking. Here we have, just so you see, this tells you that there's some connectedness to the universe that we don't fully understand, or at least I don't fully understand. I is defined by engineers for simplicity so that they can find the roots of all sorts of polynomials as you could say the square root of negative 1. Pi is the ratio between the circumference of a circle and its diameter. Once again, another interesting number, but it seems like it comes from a different place as i. E comes from a bunch of different places."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "Here we have, just so you see, this tells you that there's some connectedness to the universe that we don't fully understand, or at least I don't fully understand. I is defined by engineers for simplicity so that they can find the roots of all sorts of polynomials as you could say the square root of negative 1. Pi is the ratio between the circumference of a circle and its diameter. Once again, another interesting number, but it seems like it comes from a different place as i. E comes from a bunch of different places. E, you can either think of it, it comes out of continuous compounding interest, super valuable for finance. It also comes from the notion that the derivative of e to the x is also e to the x, so another fascinating number, but once again, seemingly unrelated to how we came up with i and seemingly unrelated to how we came up with pi. Then of course, you have some of the most profound basic numbers right over here."}, {"video_title": "Euler's formula & Euler's identity Series AP Calculus BC Khan Academy.mp3", "Sentence": "Once again, another interesting number, but it seems like it comes from a different place as i. E comes from a bunch of different places. E, you can either think of it, it comes out of continuous compounding interest, super valuable for finance. It also comes from the notion that the derivative of e to the x is also e to the x, so another fascinating number, but once again, seemingly unrelated to how we came up with i and seemingly unrelated to how we came up with pi. Then of course, you have some of the most profound basic numbers right over here. You have 1. I don't have to explain why 1 is a cool number, and I shouldn't have to explain why 0 is a cool number. This right here connects all of these fundamental numbers in some mystical way that shows that there's some connectedness to the universe."}, {"video_title": "_-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3", "Sentence": "So let's say we have the indefinite integral of natural log of x to the 10th power, all of that over x, dx. Does u-substitution apply? And if so, how would we make that substitution? Well, the key for u-substitution is to see do I have some function and its derivative? And you might immediately recognize that the derivative of natural log of x is equal to one over x. To make it a little bit clearer, I could write this as the integral of natural log of x to the 10th power times one over x, dx. Now, it's clear."}, {"video_title": "_-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3", "Sentence": "Well, the key for u-substitution is to see do I have some function and its derivative? And you might immediately recognize that the derivative of natural log of x is equal to one over x. To make it a little bit clearer, I could write this as the integral of natural log of x to the 10th power times one over x, dx. Now, it's clear. We have some function, natural log of x, being raised to the 10th power, but we also have its derivative right over here, one over x. So we could make the substitution. We could say that u is equal to the natural log of x."}, {"video_title": "_-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3", "Sentence": "Now, it's clear. We have some function, natural log of x, being raised to the 10th power, but we also have its derivative right over here, one over x. So we could make the substitution. We could say that u is equal to the natural log of x. And the reason why I picked natural log of x is because I see something, I see its exact derivative here, something close to its derivative. In this case, it's its exact derivative. And so then I could say du, dx, du, dx is equal to one over x, which means that du is equal to one over x, dx."}, {"video_title": "_-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3", "Sentence": "We could say that u is equal to the natural log of x. And the reason why I picked natural log of x is because I see something, I see its exact derivative here, something close to its derivative. In this case, it's its exact derivative. And so then I could say du, dx, du, dx is equal to one over x, which means that du is equal to one over x, dx. And so here you have it. This right over here is du, and then this right over here is our u. And so this nicely simplifies to the integral of u to the 10th power, u to the 10th power, du."}, {"video_title": "_-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3", "Sentence": "And so then I could say du, dx, du, dx is equal to one over x, which means that du is equal to one over x, dx. And so here you have it. This right over here is du, and then this right over here is our u. And so this nicely simplifies to the integral of u to the 10th power, u to the 10th power, du. And so you would evaluate what this is, find the antiderivative here, and then you would back-substitute the natural log of x for u, and to actually evaluate this indefinite integral. Let's do another one. Let's say that we have the integral of, let's do something interesting here."}, {"video_title": "_-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3", "Sentence": "And so this nicely simplifies to the integral of u to the 10th power, u to the 10th power, du. And so you would evaluate what this is, find the antiderivative here, and then you would back-substitute the natural log of x for u, and to actually evaluate this indefinite integral. Let's do another one. Let's say that we have the integral of, let's do something interesting here. Let's say the integral of tangent x, dx. Does u-substitution apply here? And at first you say, well, I just have a tangent of x."}, {"video_title": "_-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3", "Sentence": "Let's say that we have the integral of, let's do something interesting here. Let's say the integral of tangent x, dx. Does u-substitution apply here? And at first you say, well, I just have a tangent of x. Where is its derivative? But one interesting thing to do is, well, we could rewrite tangent in terms of sine and cosine. So we could write this as the integral of sine of x over cosine of x, dx."}, {"video_title": "_-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3", "Sentence": "And at first you say, well, I just have a tangent of x. Where is its derivative? But one interesting thing to do is, well, we could rewrite tangent in terms of sine and cosine. So we could write this as the integral of sine of x over cosine of x, dx. And now you might say, well, where does u-substitution apply here? Well, there's a couple of ways to think about it. You could say the derivative of sine of x is cosine of x, but you're now dividing by the derivative as opposed to multiplying by it."}, {"video_title": "_-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3", "Sentence": "So we could write this as the integral of sine of x over cosine of x, dx. And now you might say, well, where does u-substitution apply here? Well, there's a couple of ways to think about it. You could say the derivative of sine of x is cosine of x, but you're now dividing by the derivative as opposed to multiplying by it. But more interesting, you could say the derivative of cosine of x is negative sine of x. We don't have a negative sine of x, but we can do a little bit of engineering. We can multiply by negative one twice."}, {"video_title": "_-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3", "Sentence": "You could say the derivative of sine of x is cosine of x, but you're now dividing by the derivative as opposed to multiplying by it. But more interesting, you could say the derivative of cosine of x is negative sine of x. We don't have a negative sine of x, but we can do a little bit of engineering. We can multiply by negative one twice. So we could say the negative of the negative sine of x. And I stuck one of the, you could say, negative ones outside of the integral, which comes straight from our integration properties. This is equivalent."}, {"video_title": "_-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3", "Sentence": "We can multiply by negative one twice. So we could say the negative of the negative sine of x. And I stuck one of the, you could say, negative ones outside of the integral, which comes straight from our integration properties. This is equivalent. I can put a negative on the outside and a negative on the inside so that this is the derivative of cosine of x. And so now this is interesting. In fact, let me rewrite this."}, {"video_title": "_-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3", "Sentence": "This is equivalent. I can put a negative on the outside and a negative on the inside so that this is the derivative of cosine of x. And so now this is interesting. In fact, let me rewrite this. This is going to be equal to negative, the negative integral of one over cosine of x times negative sine of x dx. Now, does it jump out at you what our u might be? Well, I have a cosine of x in a denominator and I have its derivative."}, {"video_title": "_-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3", "Sentence": "In fact, let me rewrite this. This is going to be equal to negative, the negative integral of one over cosine of x times negative sine of x dx. Now, does it jump out at you what our u might be? Well, I have a cosine of x in a denominator and I have its derivative. So what if I made u equal to cosine of x? u is equal to cosine of x and then du dx would be equal to negative sine of x. Or I could say that du is equal to negative sine of x dx."}, {"video_title": "_-substitution defining _ (more examples) AP Calculus AB Khan Academy.mp3", "Sentence": "Well, I have a cosine of x in a denominator and I have its derivative. So what if I made u equal to cosine of x? u is equal to cosine of x and then du dx would be equal to negative sine of x. Or I could say that du is equal to negative sine of x dx. And just like that, I have my du here. And this, of course, is my u. And so my whole thing is now simplified to, it's equal to the negative indefinite integral of one over u, one over u du, which is a much easier integral to evaluate."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "We approximated it using this polynomial. And we saw this pretty interesting pattern. Let's see if we can find a similar pattern if we try to approximate sine of x using a Maclaurin series. And once again, a Maclaurin series is really the same thing as a Taylor series, where we are centering our approximation around x is equal to 0. So it's just a special case of a Taylor series. So let's take f of x in this situation to be equal to sine of x. f of x is now equal to sine of x. And let's do the same thing that we did with cosine of x."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And once again, a Maclaurin series is really the same thing as a Taylor series, where we are centering our approximation around x is equal to 0. So it's just a special case of a Taylor series. So let's take f of x in this situation to be equal to sine of x. f of x is now equal to sine of x. And let's do the same thing that we did with cosine of x. Let's just take the different derivatives of sine of x really fast. So if you have the first derivative of sine of x is just cosine of x. The second derivative of sine of x is derivative of cosine of x, which is negative sine of x."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And let's do the same thing that we did with cosine of x. Let's just take the different derivatives of sine of x really fast. So if you have the first derivative of sine of x is just cosine of x. The second derivative of sine of x is derivative of cosine of x, which is negative sine of x. The third derivative is going to be the derivative of this. So I'll just write a 3 in parentheses there instead of doing all the prime, prime, prime. So the third derivative is the derivative of this, which is negative cosine of x."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "The second derivative of sine of x is derivative of cosine of x, which is negative sine of x. The third derivative is going to be the derivative of this. So I'll just write a 3 in parentheses there instead of doing all the prime, prime, prime. So the third derivative is the derivative of this, which is negative cosine of x. The fourth derivative is the derivative of this, which is positive sine of x again. So you see, just like cosine of x, it kind of cycles after you take the derivative enough time. And in order to do the Maclaurin series, we care about evaluating the function and each of these derivatives at x is equal to 0."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So the third derivative is the derivative of this, which is negative cosine of x. The fourth derivative is the derivative of this, which is positive sine of x again. So you see, just like cosine of x, it kind of cycles after you take the derivative enough time. And in order to do the Maclaurin series, we care about evaluating the function and each of these derivatives at x is equal to 0. So let's do that. So for this, let me do this in a different color, not that same blue. So I'll do it in this purple color."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And in order to do the Maclaurin series, we care about evaluating the function and each of these derivatives at x is equal to 0. So let's do that. So for this, let me do this in a different color, not that same blue. So I'll do it in this purple color. So f, that's hard to see, I think. So let's do this other blue color. So f of 0 in this situation is 0."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So I'll do it in this purple color. So f, that's hard to see, I think. So let's do this other blue color. So f of 0 in this situation is 0. And f, the first derivative evaluated at 0, is 1. Cosine of 0 is 1. Negative sine of 0 is going to be 0."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So f of 0 in this situation is 0. And f, the first derivative evaluated at 0, is 1. Cosine of 0 is 1. Negative sine of 0 is going to be 0. So f prime prime, the second derivative evaluated at 0 is 0. The third derivative evaluated at 0 is negative 1. Cosine of 0 is 1."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Negative sine of 0 is going to be 0. So f prime prime, the second derivative evaluated at 0 is 0. The third derivative evaluated at 0 is negative 1. Cosine of 0 is 1. You have a negative out there. It is negative 1. And then the fourth derivative evaluated at 0 is going to be 0 again."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Cosine of 0 is 1. You have a negative out there. It is negative 1. And then the fourth derivative evaluated at 0 is going to be 0 again. And we could keep going, but once again, it seems like there's a pattern. 0, 1, 0, negative 1, 0. Then you're going to go back to positive 1, so on and so forth."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And then the fourth derivative evaluated at 0 is going to be 0 again. And we could keep going, but once again, it seems like there's a pattern. 0, 1, 0, negative 1, 0. Then you're going to go back to positive 1, so on and so forth. So let's find its polynomial representation using the Maclaurin series. And just a reminder, this one up here, this was approximately cosine of x. And you'll get closer and closer to cosine of x. I'm not rigorously showing you how close and that it's definitely the exact same thing as cosine of x."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Then you're going to go back to positive 1, so on and so forth. So let's find its polynomial representation using the Maclaurin series. And just a reminder, this one up here, this was approximately cosine of x. And you'll get closer and closer to cosine of x. I'm not rigorously showing you how close and that it's definitely the exact same thing as cosine of x. But you get closer and closer and closer to cosine of x as you keep adding terms here. And if you go to infinity, you're going to be pretty much at cosine of x. Now, let's do the same thing for sine of x."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And you'll get closer and closer to cosine of x. I'm not rigorously showing you how close and that it's definitely the exact same thing as cosine of x. But you get closer and closer and closer to cosine of x as you keep adding terms here. And if you go to infinity, you're going to be pretty much at cosine of x. Now, let's do the same thing for sine of x. So I'll pick a new color. This green should be nice. So this is our new p of x."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Now, let's do the same thing for sine of x. So I'll pick a new color. This green should be nice. So this is our new p of x. So this is approximately going to be sine of x as we add more and more terms. And so the first term here, f of 0, that's just going to be 0. So we're not even going to need to include that."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So this is our new p of x. So this is approximately going to be sine of x as we add more and more terms. And so the first term here, f of 0, that's just going to be 0. So we're not even going to need to include that. The next term is going to be f prime of 0, which is 1 times x. So it's going to be x. Then the next term is f prime, the second derivative at 0, which we see here is 0."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So we're not even going to need to include that. The next term is going to be f prime of 0, which is 1 times x. So it's going to be x. Then the next term is f prime, the second derivative at 0, which we see here is 0. Let me scroll down a little bit. It is 0, so we won't have the second term. This third term right here, the third derivative of sine of x evaluated at 0 is negative 1."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Then the next term is f prime, the second derivative at 0, which we see here is 0. Let me scroll down a little bit. It is 0, so we won't have the second term. This third term right here, the third derivative of sine of x evaluated at 0 is negative 1. So we're now going to have a negative 1. Let me scroll down so you can see this. This is negative 1 in this case, times x to the third over 3 factorial."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "This third term right here, the third derivative of sine of x evaluated at 0 is negative 1. So we're now going to have a negative 1. Let me scroll down so you can see this. This is negative 1 in this case, times x to the third over 3 factorial. So negative x to the third over 3 factorial. And then the next term is going to be 0, because that's the fourth derivative. The fourth derivative evaluated at 0 is the next coefficient."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "This is negative 1 in this case, times x to the third over 3 factorial. So negative x to the third over 3 factorial. And then the next term is going to be 0, because that's the fourth derivative. The fourth derivative evaluated at 0 is the next coefficient. We see that that is going to be 0, so it's going to drop off. And what you're going to see here, and actually maybe I haven't done enough terms for you to feel good about this. Let me do one more term right over here, just so it becomes clear."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "The fourth derivative evaluated at 0 is the next coefficient. We see that that is going to be 0, so it's going to drop off. And what you're going to see here, and actually maybe I haven't done enough terms for you to feel good about this. Let me do one more term right over here, just so it becomes clear. f of the fifth derivative of x is going to be cosine of x again. The fifth derivative, let me do it in that same color, just so it's consistent. The fifth derivative evaluated at 0 is going to be 1."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Let me do one more term right over here, just so it becomes clear. f of the fifth derivative of x is going to be cosine of x again. The fifth derivative, let me do it in that same color, just so it's consistent. The fifth derivative evaluated at 0 is going to be 1. So the fourth derivative evaluated at 0 is 0. Then you go to the fifth derivative evaluated at 0 is going to be positive 1. And if I kept doing this, it would be positive 1 times, I don't have to write the 1 as a coefficient, times x to the fifth over 5 factorial."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "The fifth derivative evaluated at 0 is going to be 1. So the fourth derivative evaluated at 0 is 0. Then you go to the fifth derivative evaluated at 0 is going to be positive 1. And if I kept doing this, it would be positive 1 times, I don't have to write the 1 as a coefficient, times x to the fifth over 5 factorial. So there's something interesting going on here. And for cosine of x, I had 1, essentially 1 times x to the 0. Then I don't have x to the first power."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And if I kept doing this, it would be positive 1 times, I don't have to write the 1 as a coefficient, times x to the fifth over 5 factorial. So there's something interesting going on here. And for cosine of x, I had 1, essentially 1 times x to the 0. Then I don't have x to the first power. I don't have x to the odd powers, actually. And then I just essentially have x to all of the even powers. And whatever power it is, I'm dividing it by that factorial."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Then I don't have x to the first power. I don't have x to the odd powers, actually. And then I just essentially have x to all of the even powers. And whatever power it is, I'm dividing it by that factorial. And then the signs keep switching. And I shouldn't say this is an even power, because 0 really isn't. Well, I guess you can view it as an even number, because I won't go into all of that."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "And whatever power it is, I'm dividing it by that factorial. And then the signs keep switching. And I shouldn't say this is an even power, because 0 really isn't. Well, I guess you can view it as an even number, because I won't go into all of that. But it's essentially 0, 2, 4, 6, so on. And so forth. So this is interesting, especially when you compare it to this."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Well, I guess you can view it as an even number, because I won't go into all of that. But it's essentially 0, 2, 4, 6, so on. And so forth. So this is interesting, especially when you compare it to this. This is all of the odd powers. This is x to the first over 1 factorial. I didn't write it here."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So this is interesting, especially when you compare it to this. This is all of the odd powers. This is x to the first over 1 factorial. I didn't write it here. This is x to the third over 3 factorial, plus x to the fifth over 5 factorial. Yeah, 0 would be an even number. Anyway, I don't, that's almost, my brain is in a different place right now."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "I didn't write it here. This is x to the third over 3 factorial, plus x to the fifth over 5 factorial. Yeah, 0 would be an even number. Anyway, I don't, that's almost, my brain is in a different place right now. And you could keep going. If we kept this process up, you would then keep switching signs, x to the seventh over 7 factorial, plus x to the ninth over 9 factorial. So there's something interesting here."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "Anyway, I don't, that's almost, my brain is in a different place right now. And you could keep going. If we kept this process up, you would then keep switching signs, x to the seventh over 7 factorial, plus x to the ninth over 9 factorial. So there's something interesting here. You once again see this kind of complementary nature between sine and cosine here. You see almost this, they're filling each other's gaps over here. Cosine of x is all of the even powers of x, divided by that powers factorial."}, {"video_title": "Maclaurin series of sin(x) Series AP Calculus BC Khan Academy.mp3", "Sentence": "So there's something interesting here. You once again see this kind of complementary nature between sine and cosine here. You see almost this, they're filling each other's gaps over here. Cosine of x is all of the even powers of x, divided by that powers factorial. Sine of x, when you take its polynomial representation, is all of the odd powers of x, divided by its factorial, and you switch signs. In the next video, I'll do e to the x. And what's really fascinating is that e to the x starts to look like a little bit of a combination here, but not quite."}, {"video_title": "Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3", "Sentence": "So I have two different expressions here that I want to take the derivative of. And what I want you to do is pause the video and think about how you would first approach taking the derivative of this expression and how that might be the same or different as your approach in taking the derivative of this expression. The goal here isn't to compute the derivatives all the way, but really to just think about how we identify what strategies to use. Okay, so let's first tackle this one. And the key when looking at a complex expression like either of these is to look at the big picture structure of the expression. So one way to think about it is let's look at the outside rather than the inside details. So if we look at the outside here, we have the sign of something."}, {"video_title": "Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3", "Sentence": "Okay, so let's first tackle this one. And the key when looking at a complex expression like either of these is to look at the big picture structure of the expression. So one way to think about it is let's look at the outside rather than the inside details. So if we look at the outside here, we have the sign of something. So there's a sign of something going on here that I'm going to circle in red or in this pink color. So that's how my brain thinks about it. From the outside, I'm like, okay, big picture, I'm taking the sign of some stuff."}, {"video_title": "Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3", "Sentence": "So if we look at the outside here, we have the sign of something. So there's a sign of something going on here that I'm going to circle in red or in this pink color. So that's how my brain thinks about it. From the outside, I'm like, okay, big picture, I'm taking the sign of some stuff. I might be taking some stuff to some exponent. In this case, I'm inputting it into a trigonometric expression. But if you have a situation like that, it's a good sign that the chain rule is in order."}, {"video_title": "Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3", "Sentence": "From the outside, I'm like, okay, big picture, I'm taking the sign of some stuff. I might be taking some stuff to some exponent. In this case, I'm inputting it into a trigonometric expression. But if you have a situation like that, it's a good sign that the chain rule is in order. So let me write that down. So we would want to use, in this case, the chain rule, CR for chain rule. And how would we apply it?"}, {"video_title": "Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3", "Sentence": "But if you have a situation like that, it's a good sign that the chain rule is in order. So let me write that down. So we would want to use, in this case, the chain rule, CR for chain rule. And how would we apply it? Well, we would take the derivative of the outside with respect to this inside times the derivative of this inside with respect to x. And I'm gonna write it the way that my brain sometimes thinks about it. So we can write this as the derivative with respect to that something."}, {"video_title": "Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3", "Sentence": "And how would we apply it? Well, we would take the derivative of the outside with respect to this inside times the derivative of this inside with respect to x. And I'm gonna write it the way that my brain sometimes thinks about it. So we can write this as the derivative with respect to that something. I'm just gonna make that pink circle for the something rather than writing it all again, of sine of that something, sine of that something, not even thinking about what that something is just yet, times the derivative with respect to x of that something. This is just an application of the chain rule. No matter what was here in this pink colored circle, it might have been something with square roots and logarithms and whatever else, as long as it's being contained within the sine, I would move to this step."}, {"video_title": "Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3", "Sentence": "So we can write this as the derivative with respect to that something. I'm just gonna make that pink circle for the something rather than writing it all again, of sine of that something, sine of that something, not even thinking about what that something is just yet, times the derivative with respect to x of that something. This is just an application of the chain rule. No matter what was here in this pink colored circle, it might have been something with square roots and logarithms and whatever else, as long as it's being contained within the sine, I would move to this step. The derivative with respect to that something of sine of that something times the derivative with respect to x of the something. Now, what would that be tangibly in this case? Well, this first part, I will do it in orange, this first part would just be cosine of x squared plus five times cosine of x."}, {"video_title": "Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3", "Sentence": "No matter what was here in this pink colored circle, it might have been something with square roots and logarithms and whatever else, as long as it's being contained within the sine, I would move to this step. The derivative with respect to that something of sine of that something times the derivative with respect to x of the something. Now, what would that be tangibly in this case? Well, this first part, I will do it in orange, this first part would just be cosine of x squared plus five times cosine of x. So that's that circle right over there. Let me close the cosine right over there. And then times the derivative with respect to x, times the derivative with respect to x of all of this again, of x squared plus five times cosine of x."}, {"video_title": "Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3", "Sentence": "Well, this first part, I will do it in orange, this first part would just be cosine of x squared plus five times cosine of x. So that's that circle right over there. Let me close the cosine right over there. And then times the derivative with respect to x, times the derivative with respect to x of all of this again, of x squared plus five times cosine of x. And then I would close my brackets. And of course, I wouldn't be done yet. I have more derivative taking to do."}, {"video_title": "Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3", "Sentence": "And then times the derivative with respect to x, times the derivative with respect to x of all of this again, of x squared plus five times cosine of x. And then I would close my brackets. And of course, I wouldn't be done yet. I have more derivative taking to do. Here, now I would look at the big structure of what's going on. And I have two expressions being multiplied. I don't have just one big expression that's being an input into like a sine function or cosine function or one big expression that's taken to some exponent."}, {"video_title": "Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3", "Sentence": "I have more derivative taking to do. Here, now I would look at the big structure of what's going on. And I have two expressions being multiplied. I don't have just one big expression that's being an input into like a sine function or cosine function or one big expression that's taken to some exponent. I have two expressions being multiplied. I have this being multiplied by this. And so if I'm just multiplying two expressions, that's a pretty good clue that to compute this part, I would then use the product rule."}, {"video_title": "Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3", "Sentence": "I don't have just one big expression that's being an input into like a sine function or cosine function or one big expression that's taken to some exponent. I have two expressions being multiplied. I have this being multiplied by this. And so if I'm just multiplying two expressions, that's a pretty good clue that to compute this part, I would then use the product rule. And I could keep doing that and compute it, and I encourage you to do so, but this is more about the strategies and how do you recognize them. But now let's go to the other example. Well, this looks a lot more like this step of the first problem than the beginning of the original problem."}, {"video_title": "Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3", "Sentence": "And so if I'm just multiplying two expressions, that's a pretty good clue that to compute this part, I would then use the product rule. And I could keep doing that and compute it, and I encourage you to do so, but this is more about the strategies and how do you recognize them. But now let's go to the other example. Well, this looks a lot more like this step of the first problem than the beginning of the original problem. Here, I don't have a sine of a bunch of stuff or a bunch of stuff being raised to one exponent. Here, I have the product of two expressions, just like we saw over here. We have this expression being multiplied by this expression."}, {"video_title": "Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3", "Sentence": "Well, this looks a lot more like this step of the first problem than the beginning of the original problem. Here, I don't have a sine of a bunch of stuff or a bunch of stuff being raised to one exponent. Here, I have the product of two expressions, just like we saw over here. We have this expression being multiplied by this expression. So my brain just says, okay, I have two expressions. Then I'm going to use the product rule. Two expressions being multiplied, I'm going to use the product rule."}, {"video_title": "Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3", "Sentence": "We have this expression being multiplied by this expression. So my brain just says, okay, I have two expressions. Then I'm going to use the product rule. Two expressions being multiplied, I'm going to use the product rule. If it was one expression being divided by another expression, then I would use the quotient rule. But in this case, it's going to be the product rule. And so that tells me that this is going to be the derivative with respect to x of the first expression."}, {"video_title": "Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3", "Sentence": "Two expressions being multiplied, I'm going to use the product rule. If it was one expression being divided by another expression, then I would use the quotient rule. But in this case, it's going to be the product rule. And so that tells me that this is going to be the derivative with respect to x of the first expression. Just going to do that with the orange circle times the second expression. I'm going to do that with the blue circle plus the first expression, not taking its derivative, the first expression, times the derivative with respect to x of the second expression. Once again, here, this is just the product rule."}, {"video_title": "Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3", "Sentence": "And so that tells me that this is going to be the derivative with respect to x of the first expression. Just going to do that with the orange circle times the second expression. I'm going to do that with the blue circle plus the first expression, not taking its derivative, the first expression, times the derivative with respect to x of the second expression. Once again, here, this is just the product rule. You can substitute sine of x squared plus five where you see this orange circle. And you can substitute cosine of x where you see this blue circle. But the whole point here isn't to actually solve this or compute this, but really to just show how you identify the structures in these expressions to think about, well, do I use the chain rule first and then use the product rule here?"}, {"video_title": "Differentiating using multiple rules strategy AP Calculus AB Khan Academy.mp3", "Sentence": "Once again, here, this is just the product rule. You can substitute sine of x squared plus five where you see this orange circle. And you can substitute cosine of x where you see this blue circle. But the whole point here isn't to actually solve this or compute this, but really to just show how you identify the structures in these expressions to think about, well, do I use the chain rule first and then use the product rule here? Or in this case, do I use the product rule first? And even once you do this, you're not going to be done. Then to compute this derivative, you're going to have to use the chain rule."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "So our first reaction might just say, okay, well let's just use our limit properties a little bit. This is going to be the same thing as the limit as x approaches negative one of x plus one over, over the limit, the limit as x approaches negative one of square root of x plus five minus two. And then we could say, all right, this thing up here, x plus one, if we think about the graph y equals x plus one, it's continuous everywhere, especially at x equals negative one, and so to evaluate this limit, we just have to evaluate this expression at x equals negative one. So this numerator is just going to evaluate to negative one plus one. And then our denominator, square root of x plus five minus two isn't continuous everywhere, but it is continuous at x equals negative one, and so we can do the same thing. We can just substitute negative one for x. So this is going to be the square root of negative one plus five minus two."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "So this numerator is just going to evaluate to negative one plus one. And then our denominator, square root of x plus five minus two isn't continuous everywhere, but it is continuous at x equals negative one, and so we can do the same thing. We can just substitute negative one for x. So this is going to be the square root of negative one plus five minus two. Now what does this evaluate to? Well, in the numerator we get a zero, and in the denominator, negative one plus five is four, take the principal root is two minus two, we get zero again. So we get, we got zero over zero."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "So this is going to be the square root of negative one plus five minus two. Now what does this evaluate to? Well, in the numerator we get a zero, and in the denominator, negative one plus five is four, take the principal root is two minus two, we get zero again. So we get, we got zero over zero. Now when you see that, you might be tempted to give up. You say, oh look, there's a zero in the denominator, maybe this limit doesn't exist, maybe I'm done here, what do I do? And if this was non-zero up here in the numerator, if you're taking a non-zero value and dividing it by zero, that is undefined, and your limit would not exist."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "So we get, we got zero over zero. Now when you see that, you might be tempted to give up. You say, oh look, there's a zero in the denominator, maybe this limit doesn't exist, maybe I'm done here, what do I do? And if this was non-zero up here in the numerator, if you're taking a non-zero value and dividing it by zero, that is undefined, and your limit would not exist. But when you have zero over zero, this is indeterminate form, and it doesn't mean necessarily that your limit does not exist. And as we'll see in this video and many future ones, there are tools at our disposal to address this, and we will look at one of them. Now the tool that we're gonna look at is, is there another way of rewriting this expression so that we can evaluate its limit without getting the zero over zero?"}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "And if this was non-zero up here in the numerator, if you're taking a non-zero value and dividing it by zero, that is undefined, and your limit would not exist. But when you have zero over zero, this is indeterminate form, and it doesn't mean necessarily that your limit does not exist. And as we'll see in this video and many future ones, there are tools at our disposal to address this, and we will look at one of them. Now the tool that we're gonna look at is, is there another way of rewriting this expression so that we can evaluate its limit without getting the zero over zero? Well let's just rewrite, let's just take this, let me give it, so let's take this thing right over here, and let's say this is g of x. So essentially what we're trying to do is find the limit of g of x as x approaches negative one. So we could write g of x is equal to x plus one, and the whole reason why I'm defining it as g of x is just to be able to think of it more clearly as a function and manipulate the function, and then think about similar functions, over x plus five minus two, or x plus one over the square root of x plus five minus two."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "Now the tool that we're gonna look at is, is there another way of rewriting this expression so that we can evaluate its limit without getting the zero over zero? Well let's just rewrite, let's just take this, let me give it, so let's take this thing right over here, and let's say this is g of x. So essentially what we're trying to do is find the limit of g of x as x approaches negative one. So we could write g of x is equal to x plus one, and the whole reason why I'm defining it as g of x is just to be able to think of it more clearly as a function and manipulate the function, and then think about similar functions, over x plus five minus two, or x plus one over the square root of x plus five minus two. Now the technique we're gonna use is, when you get this indeterminate form, and if you have a square root in either the numerator or the denominator, it might help to get rid of that square root. And this is often called rationalizing expression. In this case you have a square root in the denominator, so it would be rationalizing the denominator."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "So we could write g of x is equal to x plus one, and the whole reason why I'm defining it as g of x is just to be able to think of it more clearly as a function and manipulate the function, and then think about similar functions, over x plus five minus two, or x plus one over the square root of x plus five minus two. Now the technique we're gonna use is, when you get this indeterminate form, and if you have a square root in either the numerator or the denominator, it might help to get rid of that square root. And this is often called rationalizing expression. In this case you have a square root in the denominator, so it would be rationalizing the denominator. And so this would be, the way we would do it, is we'd be leveraging our knowledge of difference of squares. We know, we know that a plus b times a minus b is equal to a squared minus b squared, you learned that in algebra a little while ago. Or, if we had the square root of a plus b, and we were to multiply that times the square root of a minus b, well that'd be the square root of a squared, which is just going to be a minus b squared."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "In this case you have a square root in the denominator, so it would be rationalizing the denominator. And so this would be, the way we would do it, is we'd be leveraging our knowledge of difference of squares. We know, we know that a plus b times a minus b is equal to a squared minus b squared, you learned that in algebra a little while ago. Or, if we had the square root of a plus b, and we were to multiply that times the square root of a minus b, well that'd be the square root of a squared, which is just going to be a minus b squared. So we can just leverage these ideas to get rid of this radical down here. The way we're going to do it, is we're gonna multiply the numerator and the denominator by the square root of x plus five plus two, right? We have the minus two, so we're gonna multiply it times the plus two."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "Or, if we had the square root of a plus b, and we were to multiply that times the square root of a minus b, well that'd be the square root of a squared, which is just going to be a minus b squared. So we can just leverage these ideas to get rid of this radical down here. The way we're going to do it, is we're gonna multiply the numerator and the denominator by the square root of x plus five plus two, right? We have the minus two, so we're gonna multiply it times the plus two. So let's do that. So we have square root of x plus five plus two, and we're gonna multiply the numerator times the same thing, because we don't want to change the value of the expression. This is one, so if we take the expression divided by the same expression, it's going to be one."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "We have the minus two, so we're gonna multiply it times the plus two. So let's do that. So we have square root of x plus five plus two, and we're gonna multiply the numerator times the same thing, because we don't want to change the value of the expression. This is one, so if we take the expression divided by the same expression, it's going to be one. So this is, so square root of x plus five plus two. And so this is going to be equal to, this is going to be equal to x plus one times the square root, times the square root of x plus five plus two. And then the denominator is going to be, well, it's going to be the square root of x plus five squared, which would be just x plus five, and then minus two squared, minus four."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "This is one, so if we take the expression divided by the same expression, it's going to be one. So this is, so square root of x plus five plus two. And so this is going to be equal to, this is going to be equal to x plus one times the square root, times the square root of x plus five plus two. And then the denominator is going to be, well, it's going to be the square root of x plus five squared, which would be just x plus five, and then minus two squared, minus four. And so this down here simplifies to x plus five minus four, it's just x plus one. So this is just, this is just x plus one. And it probably jumps out at you that both the numerator and the denominator have an x plus one in it, so maybe we can simplify."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "And then the denominator is going to be, well, it's going to be the square root of x plus five squared, which would be just x plus five, and then minus two squared, minus four. And so this down here simplifies to x plus five minus four, it's just x plus one. So this is just, this is just x plus one. And it probably jumps out at you that both the numerator and the denominator have an x plus one in it, so maybe we can simplify. So we could simplify and just say, well, g of x is equal to the square root of x plus five plus two. Now some of you might be feeling a little off here, and you would be correct. Your spider senses would be, say, is this, is this definitely the same thing as what we originally had before we canceled out the x plus ones?"}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "And it probably jumps out at you that both the numerator and the denominator have an x plus one in it, so maybe we can simplify. So we could simplify and just say, well, g of x is equal to the square root of x plus five plus two. Now some of you might be feeling a little off here, and you would be correct. Your spider senses would be, say, is this, is this definitely the same thing as what we originally had before we canceled out the x plus ones? And the answer is the way I just wrote it, it is not the exact same thing. It is the exact same thing everywhere, except at x equals negative one. This thing right over here is defined at x equals negative one."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "Your spider senses would be, say, is this, is this definitely the same thing as what we originally had before we canceled out the x plus ones? And the answer is the way I just wrote it, it is not the exact same thing. It is the exact same thing everywhere, except at x equals negative one. This thing right over here is defined at x equals negative one. This thing right over here is not defined at x equals negative one. And g of x was not, was not, so g of x right over here, you don't get a good result when you try x equals negative one. And so in order for this to truly be the same thing as g of x, the same function, we have to say for x not equal to negative one."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "This thing right over here is defined at x equals negative one. This thing right over here is not defined at x equals negative one. And g of x was not, was not, so g of x right over here, you don't get a good result when you try x equals negative one. And so in order for this to truly be the same thing as g of x, the same function, we have to say for x not equal to negative one. Now this is a simplified version of g of x. It is the same thing. For any input x that g of x is defined, this is going to give you the same output."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "And so in order for this to truly be the same thing as g of x, the same function, we have to say for x not equal to negative one. Now this is a simplified version of g of x. It is the same thing. For any input x that g of x is defined, this is going to give you the same output. And this has the exact same domain now, now that we've put this constraint in, as g of x. Now you might say, okay, well, how does this help us? Because we want to find the limit as x approaches negative one."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "For any input x that g of x is defined, this is going to give you the same output. And this has the exact same domain now, now that we've put this constraint in, as g of x. Now you might say, okay, well, how does this help us? Because we want to find the limit as x approaches negative one. And even here, I had to put this little constraint here that x cannot be equal to negative one. How do we think about this limit? Well, lucky for us, we know, lucky for us, we know that if we just take another function, f of x, if we say f of x is equal to the square root of x plus five plus two, well, then we know that f of x is equal to g of x for all x not equal to negative one, because f of x does not have that constraint."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "Because we want to find the limit as x approaches negative one. And even here, I had to put this little constraint here that x cannot be equal to negative one. How do we think about this limit? Well, lucky for us, we know, lucky for us, we know that if we just take another function, f of x, if we say f of x is equal to the square root of x plus five plus two, well, then we know that f of x is equal to g of x for all x not equal to negative one, because f of x does not have that constraint. And we know if this is true of two, if this is true of two functions, then the limit as x approaches, the limit, let me write this down, is since we know this, because of this, we know that the limit of f of x as x approaches negative one is going to be equal to the limit of g of x as x approaches negative one. And this, of course, is what we want to figure out, what was the beginning of the problem. And but we can now use f of x here, because only at x equals negative one that they are not the same."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "Well, lucky for us, we know, lucky for us, we know that if we just take another function, f of x, if we say f of x is equal to the square root of x plus five plus two, well, then we know that f of x is equal to g of x for all x not equal to negative one, because f of x does not have that constraint. And we know if this is true of two, if this is true of two functions, then the limit as x approaches, the limit, let me write this down, is since we know this, because of this, we know that the limit of f of x as x approaches negative one is going to be equal to the limit of g of x as x approaches negative one. And this, of course, is what we want to figure out, what was the beginning of the problem. And but we can now use f of x here, because only at x equals negative one that they are not the same. And if you were to graph g of x, it just has a point discontinuity, or a removable, or I should just say, yeah, a point discontinuity right over here at x equals negative one. And so what is the limit? And we are in the home stretch now."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "And but we can now use f of x here, because only at x equals negative one that they are not the same. And if you were to graph g of x, it just has a point discontinuity, or a removable, or I should just say, yeah, a point discontinuity right over here at x equals negative one. And so what is the limit? And we are in the home stretch now. What is the limit of f of x, or we could say the limit of the square root of x plus five plus two as x approaches negative one? Well, this expression is continuous, or this function is continuous at x equals negative one, so we can just evaluate it at x equals negative one. So this is going to be the square root of negative one plus five plus two."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "And we are in the home stretch now. What is the limit of f of x, or we could say the limit of the square root of x plus five plus two as x approaches negative one? Well, this expression is continuous, or this function is continuous at x equals negative one, so we can just evaluate it at x equals negative one. So this is going to be the square root of negative one plus five plus two. So this is four, square root, principle root of four is two. Two plus two is equal to four. So since the limit of f of x as x approaches negative one is four, the limit of g of x as x approaches negative one is also four."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "So this is going to be the square root of negative one plus five plus two. So this is four, square root, principle root of four is two. Two plus two is equal to four. So since the limit of f of x as x approaches negative one is four, the limit of g of x as x approaches negative one is also four. And if this little, this little, I guess you could say, leap that I just made right over here doesn't make sense to you, think about it visually. Think about it visually. So if this is my y-axis, and this is my x-axis, g of x looked something like this."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "So since the limit of f of x as x approaches negative one is four, the limit of g of x as x approaches negative one is also four. And if this little, this little, I guess you could say, leap that I just made right over here doesn't make sense to you, think about it visually. Think about it visually. So if this is my y-axis, and this is my x-axis, g of x looked something like this. The g of x, the g of x, let me draw it, g of x looked something, something like this. And it had a gap at negative one. So it had a gap right over there."}, {"video_title": "Limits by rationalizing Limits and continuity AP Calculus AB Khan Academy (2).mp3", "Sentence": "So if this is my y-axis, and this is my x-axis, g of x looked something like this. The g of x, the g of x, let me draw it, g of x looked something, something like this. And it had a gap at negative one. So it had a gap right over there. While f of x, f of x would have the same graph, except it wouldn't have, it wouldn't have the gap. And so if you're trying to find the limit, it seems completely reasonable. Well, let's just use f of x and evaluate what f of x would be to kind of fill that gap at x equals negative one."}, {"video_title": "Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3", "Sentence": "The graph of r in polar coordinates consists of two loops, as shown in the figure above. So let's think about why it has two loops. So as our theta, when theta is zero, r is zero, and then as our theta increases, we start tracing out this first loop all the way until when theta is equal to pi. So we traced out this first loop from theta is equal to zero to theta is equal to pi, and then the second loop has a larger r, so these are larger r's. This is when we're going from pi to two pi, and you might say, well, why doesn't it show up down here? Well, between sine of pi and sine of two pi, this part right over here is going to be negative, so it flips it over, the r, into this side, and the magnitude of the r is larger and larger because of this three theta, and so when we go from pi to two pi, we trace out the larger circle. Fair enough."}, {"video_title": "Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3", "Sentence": "So we traced out this first loop from theta is equal to zero to theta is equal to pi, and then the second loop has a larger r, so these are larger r's. This is when we're going from pi to two pi, and you might say, well, why doesn't it show up down here? Well, between sine of pi and sine of two pi, this part right over here is going to be negative, so it flips it over, the r, into this side, and the magnitude of the r is larger and larger because of this three theta, and so when we go from pi to two pi, we trace out the larger circle. Fair enough. That seems pretty straightforward. Point P is on the graph of r, right over there, and the y-axis. Find the rate of change of the x-coordinate with respect to theta at the point P. All right, so let's think about this a little bit."}, {"video_title": "Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3", "Sentence": "Fair enough. That seems pretty straightforward. Point P is on the graph of r, right over there, and the y-axis. Find the rate of change of the x-coordinate with respect to theta at the point P. All right, so let's think about this a little bit. They don't give us x as a function of theta. We have to figure out that from what they've given us. So just as a bit of a polar coordinates refresher, if this is our theta right over there, this is our r, and that would be a point on our curve for this theta."}, {"video_title": "Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3", "Sentence": "Find the rate of change of the x-coordinate with respect to theta at the point P. All right, so let's think about this a little bit. They don't give us x as a function of theta. We have to figure out that from what they've given us. So just as a bit of a polar coordinates refresher, if this is our theta right over there, this is our r, and that would be a point on our curve for this theta. Now, how do you convert that to x and y's? Well, you can construct a little bit of a right triangle right over here, and we know from our basic trigonometry that the length of this base right over here, this is going to be the hypotenuse. Let me just write that."}, {"video_title": "Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3", "Sentence": "So just as a bit of a polar coordinates refresher, if this is our theta right over there, this is our r, and that would be a point on our curve for this theta. Now, how do you convert that to x and y's? Well, you can construct a little bit of a right triangle right over here, and we know from our basic trigonometry that the length of this base right over here, this is going to be the hypotenuse. Let me just write that. That's going to be our x-coordinate. Our x-coordinate right over here is going to be equal to our hypotenuse, which is r times the cosine of theta. If we wanted the y-coordinate as a function of r and theta, it'd be y is equal to r sine of theta, but they don't want us to worry about y here, just the x-coordinate."}, {"video_title": "Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3", "Sentence": "Let me just write that. That's going to be our x-coordinate. Our x-coordinate right over here is going to be equal to our hypotenuse, which is r times the cosine of theta. If we wanted the y-coordinate as a function of r and theta, it'd be y is equal to r sine of theta, but they don't want us to worry about y here, just the x-coordinate. So we know this, but we want it purely in terms of theta. So how do we get there? Well, what we can do is take this expression for r. r itself is a function of theta, and replace it right over there."}, {"video_title": "Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3", "Sentence": "If we wanted the y-coordinate as a function of r and theta, it'd be y is equal to r sine of theta, but they don't want us to worry about y here, just the x-coordinate. So we know this, but we want it purely in terms of theta. So how do we get there? Well, what we can do is take this expression for r. r itself is a function of theta, and replace it right over there. And so what we can do is we can write, well, x of theta is going to be equal to r, which itself is three theta sine of theta times cosine of theta, times cosine of theta. And now we want to find the rate of change of the x-coordinate with respect to theta at a point. So let's just find the derivative of x with respect to theta."}, {"video_title": "Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3", "Sentence": "Well, what we can do is take this expression for r. r itself is a function of theta, and replace it right over there. And so what we can do is we can write, well, x of theta is going to be equal to r, which itself is three theta sine of theta times cosine of theta, times cosine of theta. And now we want to find the rate of change of the x-coordinate with respect to theta at a point. So let's just find the derivative of x with respect to theta. So x prime of theta is equal to, well, I have the product of three expressions over here. I have this first expression, three theta, then I have sine theta, and then I have cosine theta. So we can apply the product rule to find the derivative."}, {"video_title": "Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3", "Sentence": "So let's just find the derivative of x with respect to theta. So x prime of theta is equal to, well, I have the product of three expressions over here. I have this first expression, three theta, then I have sine theta, and then I have cosine theta. So we can apply the product rule to find the derivative. If you're using the product rule with the expression of three things, you essentially just follow the same pattern when you're taking the product of two things. The first term is going to be the derivative of the first of the expressions, three times the other two expressions. So we're gonna have three times sine of theta, cosine of theta, plus the second term is going to be the derivative of the middle term times the other two expressions."}, {"video_title": "Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3", "Sentence": "So we can apply the product rule to find the derivative. If you're using the product rule with the expression of three things, you essentially just follow the same pattern when you're taking the product of two things. The first term is going to be the derivative of the first of the expressions, three times the other two expressions. So we're gonna have three times sine of theta, cosine of theta, plus the second term is going to be the derivative of the middle term times the other two expressions. So we're gonna have three theta, and then derivative of sine theta is cosine theta, times another cosine theta, you're gonna have cosine squared of theta, or cosine of theta squared, just like that. And then you're gonna have the derivative of the last term is going to be the derivative of cosine theta times these other two expressions. Well, the derivative of cosine theta is negative sine theta."}, {"video_title": "Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3", "Sentence": "So we're gonna have three times sine of theta, cosine of theta, plus the second term is going to be the derivative of the middle term times the other two expressions. So we're gonna have three theta, and then derivative of sine theta is cosine theta, times another cosine theta, you're gonna have cosine squared of theta, or cosine of theta squared, just like that. And then you're gonna have the derivative of the last term is going to be the derivative of cosine theta times these other two expressions. Well, the derivative of cosine theta is negative sine theta. So if you multiply negative sine theta times three theta sine theta, you're going to have negative three theta sine squared theta. And so we want to evaluate this at point P. So what is theta at point P? Well, point P does happen on our first pass around, and so at point P, theta is equal to, theta right over here, is equal to pi over two."}, {"video_title": "Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3", "Sentence": "Well, the derivative of cosine theta is negative sine theta. So if you multiply negative sine theta times three theta sine theta, you're going to have negative three theta sine squared theta. And so we want to evaluate this at point P. So what is theta at point P? Well, point P does happen on our first pass around, and so at point P, theta is equal to, theta right over here, is equal to pi over two. So pi over two. So what we really just need to find is, well, what is x prime of pi over two? Well, that is going to be equal to three times sine of pi over two, sine of pi over two, which is one, times cosine of pi over two, which is zero, so this whole thing is zero, plus three times pi over two, this is three pi over two, times cosine squared of pi over two, or cosine of pi over two squared."}, {"video_title": "Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3", "Sentence": "Well, point P does happen on our first pass around, and so at point P, theta is equal to, theta right over here, is equal to pi over two. So pi over two. So what we really just need to find is, well, what is x prime of pi over two? Well, that is going to be equal to three times sine of pi over two, sine of pi over two, which is one, times cosine of pi over two, which is zero, so this whole thing is zero, plus three times pi over two, this is three pi over two, times cosine squared of pi over two, or cosine of pi over two squared. Well, that's just zero. So so far, everything is zero. Minus three times pi over two, three pi over two, times sine of pi over two squared."}, {"video_title": "Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3", "Sentence": "Well, that is going to be equal to three times sine of pi over two, sine of pi over two, which is one, times cosine of pi over two, which is zero, so this whole thing is zero, plus three times pi over two, this is three pi over two, times cosine squared of pi over two, or cosine of pi over two squared. Well, that's just zero. So so far, everything is zero. Minus three times pi over two, three pi over two, times sine of pi over two squared. Well, what's sine of pi over two? Well, that's one, you square it, you still get one. So all of this simplified to negative three pi over two."}, {"video_title": "Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3", "Sentence": "Minus three times pi over two, three pi over two, times sine of pi over two squared. Well, what's sine of pi over two? Well, that's one, you square it, you still get one. So all of this simplified to negative three pi over two. Now, it's always good to get a reality check. Does this make sense, that the rate of change of x with respect to theta is negative three pi over two? Well, think about what's happening."}, {"video_title": "Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3", "Sentence": "So all of this simplified to negative three pi over two. Now, it's always good to get a reality check. Does this make sense, that the rate of change of x with respect to theta is negative three pi over two? Well, think about what's happening. As theta increases a little bit, x is definitely going to decrease, so it makes sense that we have a negative out here. So right over here, rate of change of x with respect to theta, negative three pi over two. As theta increases, our x for sure is decreasing, but at least it does make intuitive sense."}, {"video_title": "Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3", "Sentence": "Well, think about what's happening. As theta increases a little bit, x is definitely going to decrease, so it makes sense that we have a negative out here. So right over here, rate of change of x with respect to theta, negative three pi over two. As theta increases, our x for sure is decreasing, but at least it does make intuitive sense."}, {"video_title": "Worked example differentiating polar functions AP Calculus BC Khan Academy.mp3", "Sentence": "As theta increases, our x for sure is decreasing, but at least it does make intuitive sense."}]